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Paul Monk 



~ Physical 




Physical Chemistry 

Understanding our Chemical World 

Physical Chemistry 

Understanding our Chemical World 

Paul Monk 

Manchester Metropolitan University, UK 

John Wiley & Sons, Ltd 

Copyright © 2004 John Wiley & Sons Ltd, The Atrium, Southern Gate, Chichester, 
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Library of Congress Cataloging-in-Publication Data 

Monk, Paul M. S. 

Physical chemistry : understanding our chemical world / Paul Monk, 
p. cm. 

Includes bibliographical references and index. 

ISBN 0-471-49180-2 (acid-free paper) - ISBN 0-471-49181-0 (pbk. : 
acid-free paper) 

1. Chemistry, Physical and theoretical. I. Title. 

QD453.3.M66 2004 

S41 _ dc22 


British Library Cataloguing in Publication Data 

A catalogue record for this book is available from the British Library 

ISBN 0-471-49180-2 hardback 
0-471-49181-0 paperback 

Typeset in 10.5/12.5pt Times by Laserwords Private Limited, Chennai, India 

Printed and bound in Great Britain by Antony Rowe Ltd, Chippenham, Wiltshire 

This book is printed on acid-free paper responsibly manufactured from sustainable forestry 

in which at least two trees are planted for each one used for paper production. 


Preface xv 

Etymological introduction xix 

List of symbols xxiii 
Powers of ten: standard prefixes xxviii 

1 Introduction to physical chemistry 1 

1.1 What is physical chemistry: variables, relationships and laws 1 

Why do we warm ourselves by a radiator? 1 

Why does water get hot in a kettle? 2 

Are these two colours complementary? 2 

Does my radio get louder if I vary the volume control? 3 
Why does the mercury in a barometer go up when the air pressure 

increases? 5 

Why does a radiator feel hot to the touch when 'on', and cold when 'off? 7 

1.2 The practice of thermodynamic measurement 9 

What is temperature? 9 

How long is a piece of string? 14 

How fast is 'greased lightning'? 15 

Why is the SI unit of mass the kilogram? 17 

Why is 'the material of action so variable'? 18 

1.3 Properties of gases and the gas laws 20 

Why do we see eddy patterns above a radiator? 20 

Why does a hot-air balloon float? 20 

How was the absolute zero of temperature determined? 21 

Why pressurize the contents of a gas canister? 23 

Why does thunder accompany lightning? 25 

How does a bubble-jet printer work? 26 


What causes pressure? 30 

Why is it unwise to incinerate an empty can of air freshener? 32 

1.4 Further thoughts on energy 33 

Why is the room warm? 33 

What do we mean by 'room temperature' ? 34 
Why do we get warmed-through in front of a fire, rather than just our 

skins? 35 

2 Introducing interactions and bonds 37 

2. 1 Physical and molecular interactions 37 

What is 'dry ice' ? 37 

How is ammonia liquefied? 38 

Why does steam condense in a cold bathroom? 39 

How does a liquid-crystal display work? 40 

Why does dew form on a cool morning? 42 
How is the three-dimensional structure maintained within the DNA double 

helix? 44 

How do we make liquid nitrogen? 47 

Why is petrol a liquid at room temperature but butane is a gas? 49 

2.2 Quantifying the interactions and their influence 50 

How does mist form? 50 

How do we liquefy petroleum gas? 52 

Why is the molar volume of a gas not zero at K? 54 

2.3 Creating formal chemical bonds 59 

Why is chlorine gas lethal yet sodium chloride is vital for life? 59 

Why does a bicycle tyre get hot when inflated? 59 

How does a fridge cooler work? 60 

Why does steam warm up a cappuccino coffee? 61 

Why does land become more fertile after a thunderstorm? 63 

Why does a satellite need an inert coating? 64 

Why does water have the formula H 2 0? 66 

Why is petroleum gel so soft? 67 

Why does salt form when sodium and chlorine react? 69 

Why heat a neon lamp before it will generate light? 69 

Why does lightning conduct through air? 72 

Why is argon gas inert? 74 

Why is silver iodide yellow? 75 

3 Energy and the first law of thermodynamics 77 

3.1 Introduction to thermodynamics: internal energy 77 

Why does the mouth get cold when eating ice cream? 77 

Why is skin scalded by steam? 79 

Why do we sweat? 81 

Why do we still feel hot while sweating on a humid beach? 83 


Why is the water at the top of a waterfall cooler than the water at its base? 85 

Why is it such hard work pumping up a bicycle tyre? 86 

Why does a sausage become warm when placed in an oven? 87 

Why, when letting down a bicycle tyre, is the expelled air so cold? 88 

Why does a tyre get hot during inflation? 89 

Can a tyre be inflated without a rise in temperature? 89 

How fast does the air in an oven warm up? 90 

Why does water boil more quickly in a kettle than in a pan on a stove? 91 

Why does a match emit heat when lit? 94 

Why does it always take 4 min to boil an egg properly? 95 

Why does a watched pot always take so long to boil? 98 

3.2 Enthalpy 99 

How does a whistling kettle work? 99 

How much energy do we require during a distillation? 102 

Why does the enthalpy of melting ice decrease as the temperature 

decreases? 104 

Why does water take longer to heat in a pressure cooker than in an open 

pan? 106 

Why does the temperature change during a reaction? 107 

Are diamonds forever? 109 

Why do we burn fuel when cold? Ill 

Why does butane burn with a hotter flame than methane? 114 

3.3 Indirect measurement of enthalpy 118 

How do we make 'industrial alcohol'? 118 

How does an 'anti-smoking pipe' work? 120 

Why does dissolving a salt in water liberate heat? 123 

Why does our mouth feel cold after eating peppermint? 125 

How does a camper's 'emergency heat stick' work? 127 

4 Reaction spontaneity and the direction of thermodynamic change 129 

4.1 The direction of physicochemical change: entropy 129 

Why does the colour spread when placing a drop of dye in a saucer of 

clean water? 129 

When we spill a bowl of sugar, why do the grains go everywhere and 

cause such a mess? 130 

Why, when one end of the bath is hot and the other cold, do the 

temperatures equalize? 131 

Why does a room containing oranges acquire their aroma? 133 

Why do damp clothes become dry when hung outside? 134 

Why does crystallization of a solute occur? 137 

4.2 The temperature dependence of entropy 139 

Why do dust particles move more quickly by Brownian motion in warm 

water? 139 

Why does the jam of a jam tart burn more than does the pastry? 139 


4.3 Introducing the Gibbs function 144 

Why is burning hydrogen gas in air (to form liquid water) a spontaneous 

reaction? 144 

How does a reflux condenser work? 144 

4.4 The effect of pressure on thermodynamic variables 148 

How much energy is needed? 148 

Why does a vacuum 'suck'? 151 

Why do we sneeze? 152 

How does a laboratory water pump work? 153 

4.5 Thermodynamics and the extent of reaction 156 

Why is a 'weak' acid weak? 156 

Why does the pH of the weak acid remain constant? 158 

Why does the voltage of a battery decrease to zero? 159 

Why does the concentration of product stop changing? 162 

Why do chicken eggs have thinner shells in the summer? 165 

4.6 The effect of temperature on thermodynamic variables 166 

Why does egg white denature when cooked but remain liquid at room 

temperature? 166 

At what temperature will the egg start to denature? 170 

Why does recrystallization work? 171 

5 Phase equilibria 177 

5.1 Energetic introduction to phase equilibria 177 

Why does an ice cube melt in the mouth? 177 

Why does water placed in a freezer become ice? 181 

Why was Napoleon's Russian campaign such a disaster? 182 

5.2 Pressure and temperature changes with a single-component system: 

qualitative discussion 184 

How is the 'Smoke' in horror films made? 184 

How does freeze-drying work? 185 

How does a rotary evaporator work? 188 

How is coffee decaffeinated? 189 

5.3 Quantitative effects of pressure and temperature change for a 
single-component system 192 

Why is ice so slippery? 192 

What is 'black ice'? 193 

Why does deflating the tyres on a car improve its road-holding on ice? 198 

Why does a pressure cooker work? 199 

5.4 Phase equilibria involving two-component systems: partition 205 

Why does a fizzy drink lose its fizz and go flat? 205 

How does a separating funnel work? 207 

Why is an ice cube only misty at its centre? 208 

How does recrystallization work? 209 

Why are some eggshells brown and some white? 211 


5.5 Phase equilibria and colligative properties 212 

Why does a mixed-melting-point determination work? 212 

How did the Victorians make ice cream? 216 

Why boil vegetables in salted water? 217 

Why does the ice on a path melt when sprinkled with salt? 218 

5.6 Phase equilibria involving vapour pressure 221 

Why does petrol sometimes have a strong smell and sometimes not? 221 

How do anaesthetics work? 222 

How do carbon monoxide sensors work? 224 

Why does green petrol smell different from leaded petrol? 224 

Why do some brands of 'green' petrol smell different from others? 225 
Why does a cup of hot coffee yield more steam than above a cup of 

boiling water at the same temperature? 229 

How are essential oils for aromatherapy extracted from plants? 229 

6 Acids and Bases 233 

6.1 Properties of Lowry-Br0nsted acids and bases 233 

Why does vinegar taste sour? 233 

Why is it dangerous to allow water near an electrical appliance, if water is 

an insulator? 235 

Why is bottled water 'neutral' ? 236 

What is 'acid rain' ? 237 

Why does cutting an onion make us cry? 239 

Why does splashing the hands with sodium hydroxide solution make them 

feel 'soapy' ? 239 

Why is aqueous ammonia alkaline? 240 

Why is there no vinegar in crisps of salt and vinegar flavour? 241 

How did soldiers avoid chlorine gas poisoning at the Second Battle of 

Ypres? 242 

How is sherbet made? 244 

Why do steps made of limestone sometimes feel slippery? 244 

Why is the acid in a car battery more corrosive than vinegar? 245 

Why do equimolar solutions of sulphuric acid and nitric acid have 

different pHs? 250 

What is the pH of a 'neutral' solution? 25 1 

What do we mean when we say blood plasma has a 'pH of 7.4' ? 251 

6.2 'Strong' and 'weak' acids and bases 253 

Why is a nettle sting more painful than a burn from ethanoic acid? 253 

Why is 'carbolic acid' not in fact an acid? 254 

Why does carbonic acid behave as a mono-protic acid? 259 

Why is an organic acid such as trichloroethanoic acid so strong? 260 

6.3 Titration analyses 261 

Why does a dock leaf bring relief after a nettle sting? 261 

How do indigestion tablets work? 262 


6.4 pH buffers 267 

Why does the pH of blood not alter after eating pickle? 267 

Why are some lakes more acidic than others? 267 

How do we make a 'constant-pH solution' ? 270 

6.5 Acid-base indicators 273 

What is 'the litmus test' ? 273 

Why do some hydrangea bushes look red and others blue? 274 

Why does phenolphthalein indicator not turn red until pH 8.2? 276 

7 Electrochemistry 279 

7.1 Introduction to cells: terminology and background 279 

Why does putting aluminium foil in the mouth cause pain? 279 

Why does an electric cattle prod cause pain? 281 

What is the simplest way to clean a tarnished silver spoon? 282 

How does 'electrolysis' stop hair growth? 283 
Why power a car with a heavy-duty battery yet use a small battery in a 

watch? 283 

How is coloured ('anodized') aluminium produced? 285 

How do we prevent the corrosion of an oil rig? 286 

What is a battery? 288 

Why do hydrogen fuel cells sometimes 'dry up' ? 289 

Why bother to draw cells? 291 

Why do digital watches lose time in the winter? 293 

Why is a battery's potential not constant? 294 

What is a 'standard cell' ? 295 

Why aren't electrodes made from wood? 300 

Why is electricity more dangerous in wet weather? 302 

7.2 Introducing half-cells and electrode potentials 303 

Why are the voltages of watch and car batteries different? 303 

How do 'electrochromic' car mirrors work? 305 

Why does a potential form at an electrode? 306 

7.3 Activity 308 

Why does the smell of brandy decrease after dissolving table salt in it? 308 

Why does the smell of gravy become less intense after adding salt to it? 308 

Why add alcohol to eau de Cologne? 309 

Why does the cell em/ alter after adding LiCl? 312 

Why does adding NaCl to a cell alter the emf, but adding tonic water 

doesn't? 314 

Why does MgCl 2 cause a greater decrease in perceived concentration than 

KC1? 315 

Why is calcium better than table salt at stopping soap lathering? 316 

Why does the solubility of AgCl change after adding MgS0 4 ? 318 

7.4 Half-cells and the Nernst equation 321 

Why does sodium react with water yet copper doesn't? 321 


Why does a torch battery eventually 'go flat' ? 325 

Why does E Ag ci,Ag change after immersing an SSCE in a solution of salt? 326 

Why 'earth' a plug? 328 

7.5 Concentration cells 333 

Why does steel rust fast while iron is more passive? 333 

How do pH electrodes work? 336 

7.6 Transport phenomena 339 

How do nerve cells work? 339 

What is a 'salt bridge' ? 342 

7.7 Batteries 343 

How does an electric eel produce a current? 343 

What is the earliest known battery? 345 

8 Chemical kinetics 349 

8.1 Kinetic definitions 349 

Why does a 'strong' bleach clean faster than a weaker one does? 349 

Why does the bleaching reaction eventually stop? 351 

Why does bleach work faster on some greases than on others? 354 

Why do copper ions amminate so slowly? 356 

How fast is the reaction that depletes the ozone layer? 358 
Why is it more difficult to breathe when up a mountain than at ground 

level? 359 

8.2 Qualitative discussion of concentration changes 361 

Why does a full tank of petrol allow a car to travel over a constant 

distance? 361 

Why do we add a drop of bromine water to a solution of an alkene? 362 

When magnesium dissolves in aqueous acid, why does the amount of 

fizzing decrease with time? 364 

8.3 Quantitative concentration changes: integrated rate equations 368 

Why do some photographs develop so slowly? 368 

Why do we often refer to a 'half-life' when speaking about radioactivity? 378 

How was the Turin Shroud 'carbon dated' ? 382 

How old is Otzi the iceman? 385 
Why does the metabolism of a hormone not cause a large chemical change 

in the body? 387 

Why do we not see radicals forming in the skin while sunbathing? 388 

8.4 Kinetic treatment of complicated reactions 393 

Why is arsenic poisonous? 393 
Why is the extent of Walden inversion smaller when a secondary alkyl 

halide reacts than with a primary halide? 394 

Why does 'standing' a bottle of wine cause it to smell and taste better? 397 

Why fit a catalytic converter to a car exhaust? 399 

Why do some people not burn when sunbathing? 400 

How do Reactolite sunglasses work? 403 


8.5 Thermodynamic considerations: activation energy, absolute reaction rates 

and catalysis 408 

Why prepare a cup of tea with boiling water? 408 

Why store food in a fridge? 408 

Why do the chemical reactions involved in cooking require heating? 409 

Why does a reaction speed up at higher temperature? 411 

Why does the body become hotter when ill, and get 'a temperature' ? 415 

Why are the rates of some reactions insensitive to temperature? 416 

What are catalytic converters? 420 

9 Physical chemistry involving light: spectroscopy and 

photochemistry 423 

9.1 Introduction to photochemistry 423 

Why is ink coloured? 423 

Why do neon streetlights glow? 424 

Why do we get hot when lying in the sun? 425 

Why is red wine so red? 426 

Why are some paints red, some blue and others black? 427 

Why can't we see infrared light with our eyes? 429 

How does a dimmer switch work? 433 

Why does UV-b cause sunburn yet UV-a does not? 434 

How does a suntan protect against sunlight? 436 

How does sun cream block sunlight? 439 

Why does tea have a darker colour if brewed for longer? 442 
Why does a glass of apple juice appear darker when viewed against a 

white card? 442 

Why are some paints darker than others? 444 

What is ink? 445 

9.2 Photon absorptions and the effect of wavelength 446 

Why do radical reactions usually require UV light? 446 

Why does photolysis require a powerful lamp? 452 

Why are spectroscopic bands not sharp? 453 

Why does hydrogen look pink in a glow discharge? 455 

Why do surfaces exposed to the sun get so dusty? 457 

Why is microwave radiation invisible to the eye? 458 

9.3 Photochemical and spectroscopic selection rules 459 

Why is the permanganate ion so intensely coloured? 459 

Why is chlorophyll green? 461 

Why does adding salt remove a blood stain? 462 

What is gold-free gold paint made of? 462 

What causes the blue colour of sapphire? 463 

Why do we get hot while lying in the sun? 464 

What is an infrared spectrum? 467 

Why does food get hot in a microwave oven? 469 

Are mobile phones a risk to health? 471 


9.4 Photophysics: emission and loss processes 472 

How are X-rays made? 472 

Why does metal glow when hot? 473 

How does a light bulb work? 474 

Why is a quartz-halogen bulb so bright? 474 

What is 'limelight' ? 476 

Why do TV screens emit light? 476 

Why do some rotting fish glow in the dark? 478 

How do 'see in the dark' watch hands work? 479 

How do neon lights work? 480 

How does a sodium lamp work? 481 

How do 'fluorescent strip lights' work? 482 

9.5 Other optical effects 483 

Why is the mediterranean sea blue? 483 

Do old-master paintings have a 'fingerprint' ? 485 

10 Adsorption and surfaces, colloids and micelles 487 

10.1 Adsorption and definitions 487 

Why is steam formed when ironing a line-dried shirt? 487 

Why does the intensity of a curry stain vary so much? 489 

Why is it difficult to remove a curry stain? 492 

Why is iron the catalyst in the Haber process? 494 
Why is it easier to remove a layer of curry sauce than to remove a curry 

stain! 496 

How does water condense onto glass? 497 

How does bleach remove a dye stain? 498 

How much beetroot juice does the stain on the plate contain? 499 

Why do we see a 'cloud' of steam when ironing a shirt? 503 

10.2 Colloids and interfacial science 504 

Why is milk cloudy? 504 

What is an 'aerosol' spray? 505 

What is 'emulsion paint' ? 506 

Why does oil not mix with water? 508 

10.3 Colloid stability 509 

How are cream and butter made? 509 

How is chicken soup 'clarified' by adding eggshells? 510 

How is 'clarified butter' made? 510 

Why does hand cream lose its milky appearance during hand rubbing? 511 

Why does orange juice cause milk to curdle? 512 

How are colloidal particles removed from waste water? 513 

10.4 Association colloids: micelles 514 

Why does soapy water sometimes look milky? 514 

What is soap? 517 

Why do soaps dissolve grease? 518 


Why is old washing-up water oily when cold but not when hot? 519 

Why does soap generate bubbles? 521 

Why does detergent form bubbles? 522 

Answers to SAQs 525 

Bibliography 533 

Index 565 


This book 

Some people make physical chemistry sound more confusing than it really is. One of 
their best tricks is to define it inaccurately, saying it is 'the physics of chemicals'. This 
definition is sometimes quite good, since it suggests we look at a chemical system and 
ascertain how it follows the laws of nature. This is true, but it suggests that chemistry 
is merely a sub-branch of physics; and the notoriously mathematical nature of physics 
impels us to avoid this otherwise useful way of looking at physical chemistry. 

An alternative and more user-friendly definition tells us that physical chemistry 
supplies 'the laws of chemistry', and is an addition to the making of chemicals. This 
is a superior lens through which to view our topic because we avoid the bitter aftertaste 
of pure physics, and start to look more closely at physical chemistry as an applied 
science: we do not look at the topic merely for the sake of looking, but because 
there are real-life situations requiring a scientific explanation. Nevertheless, most 
practitioners adopting this approach are still overly mathematical in their treatments, 
and can make it sound as though the science is fascinating in its own right, but will 
sometimes condescend to suggest an application of the theory they so clearly relish. 

But the definition we will employ here is altogether simpler, 
and also broader: we merely ask 'why does it happen?' as we 
focus on the behaviour of each chemical system. Every example 
we encounter in our everyday walk can be whittled down into 
small segments of thought, each so simple that a small child can 
understand. As a famous mystic of the 14th century once said, T 
saw a small hazelnut and I marvelled that everything that exists could be contained 
within it'. And in a sense she was right: a hazelnut looks brown because of the way 
light interacts with its outer shell - the topic of spectroscopy (Chapter 9); the hazelnut 
is hard and solid - the topic of bonding theory (Chapter 2) and phase equilibria 
(Chapter 5); and the nut is good to eat - we say it is readily metabolized, so we think 

Now published as Rev- 
elations of Divine Love, 
by Mother Julian of 


of kinetics (Chapter 8); and the energetics of chemical reactions (Chapters 2-4). The 
sensations of taste and sight are ultimately detected within the brain as electrical 
impulses, which we explain from within the rapidly growing field of electrochemistry 
(Chapter 7). Even the way a nut sticks to our teeth is readily explained by adsorption 
science (Chapter 10). Truly, the whole of physical chemistry can be encompassed 
within a few everyday examples. 

So the approach taken here is the opposite to that in most other books of physical 
chemistry: each small section starts with an example from everyday life, i.e. both the 
world around us and also those elementary observations that a chemist can be certain 
to have pondered upon while attending a laboratory class. We then work backwards 
from the experiences of our hands and eyes toward the cause of why our world is 
the way it is. 

Nevertheless, we need to be aware that physical chemistry is not a closed book in 
the same way of perhaps classical Latin or Greek. Physical chemistry is a growing 
discipline, and new experimental techniques and ideas are continually improving the 
data and theories with which our understanding must ultimately derive. 

Inevitably, some of the explanations here have been over-simplified because phys- 
ical chemistry is growing at an alarming rate, and additional sophistications in theory 
and experiment have yet to be devised. But a more profound reason for caution is 
in ourselves: it is all too easy, as scientists, to say 'Now I understand!' when in fact 
we mean that all the facts before us can be answered by the theory. Nevertheless, if 
the facts were to alter slightly - perhaps we look at another kind of nut - the theory, 
as far as we presently understand it, would need to change ever so slightly. Our 
understanding can never be complete. 

So, we need a word about humility. It is said, probably too often, that science is 
not an emotional discipline, nor is there a place for any kind of reflection on the 
human side of its application. This view is deeply mistaken, because scientists limit 
themselves if they blind themselves to any contradictory evidence when sure they 
are right. The laws of physical chemistry can only grow when we have the humility 
to acknowledge how incomplete is our knowledge, and that our explanation might 
need to change. For this reason, a simple argument is not necessary the right one; but 
neither is a complicated one. The examples in this book were chosen to show how 
the world around us manifests Physical Chemistry. The explanation of a seemingly 
simple observation may be fiendishly complicated, but it may be beautifully simple. It 
must be admitted that the chemical examples are occasionally artificial. The concept 
of activity, for example, is widely misunderstood, probably because it presupposes 
knowledge from so many overlapping branches of physical chemistry. The examples 
chosen to explain it may be quite absurd to many experienced teachers, but, as 
an aid to simplification, they can be made to work. Occasionally the science has 
been simplified to the point where some experienced teachers will maintain that it is 
technically wrong. But we must start from the beginning if we are to be wise, and 
only then can we progress via the middle . . . and physical chemistry is still a rapidly 
growing subject, so we don't yet know where it will end. 

While this book could be read as an almanac of explanations, it provides students 
in further and higher education with a unified approach to physical chemistry. As a 


teacher of physical chemistry, I have found the approaches and examples here to be 
effective with students of HND and the early years of BSc and MChem courses. It has 
been written for students having the basic chemical and mathematical skills generally 
expected of university entrants, such as rearrangement of elementary algebra and a 
little calculus. It will augment the skills of other, more advanced, students. 

To reiterate, this book supplies no more than an introduction to physical chemistry, 
and is not an attempt to cover the whole topic. Those students who have learned 
some physical chemistry are invited to expand their vision by reading more special- 
ized works. The inconsistencies and simplifications wrought by lack of space and 
style in this text will be readily overcome by copious background reading. A com- 
prehensive bibliography is therefore included at the end of the book. Copies of the 
figures and bibliography, as well as live links can be found on the book's website at 

A ckn o w ledge m en ts 

One of the more pleasing aspects of writing a text such as this is the opportunity to 
thank so many people for their help. It is a genuine pleasure to thank Professor Seamus 
Higson of Cranfield University, Dr Roger Mortimer of Loughborough University, 
and Dr Michele Edge, Dr David Johnson, Dr Chris Rego and Dr Brian Wardle from 
my own department, each of whom read all or part of the manuscript, and whose 
comments have been so helpful. 

A particular 'thank you' to Mrs Eleanor Riches, formerly a high-school teacher, 
who read the entire manuscript and made many perceptive and helpful comments. 

I would like to thank the many students from my department who not only saw 
much of this material, originally in the form of handouts, but whose comments helped 
shape the material into its present form. 

Please allow me to thank Michael Kaufman of The Campaign for a Hydrogen Econ- 
omy (formerly the Hydrogen Association of UK and Ireland) for helpful discussions 
to clarify the arguments in Chapter 7, and the Tin Research Council for their help in 
constructing some of the arguments early in Chapter 5. 

Concerning permission to reproduce figures, I am indebted to The Royal Society of 
Chemistry for Figures 1.8 and 8.26, the Open University Press for Figure 7.10, Else- 
vier Science for Figures 4.7 and 10.3, and John Wiley & Sons for Figures 7.19, 10.11 
and 10.14. Professor Robin Clarke frs of University College London has graciously 
allowed the reproduction of Figure 9.28. 

Finally, please allow me to thank Dr Andy Slade, Commissioning Editor of Wiley, 
and the copy and production editors Rachael Ballard and Robert Hambrook. A special 
thank you, too, to Pete Lewis. 

Paul Monk 

Department of Chemistry & Materials 
Manchester Metropolitan University 



The hero in The Name of the Rose is a medieval English monk. He acts as sleuth, 
and is heard to note at one point in the story how, 'The study of words is the 
whole of knowledge'. While we might wish he had gone a little 

"Etymology" means 
the derivation of a 
word's meaning. 

further to mention chemicals, we would have to agree that many 
of our technical words can be traced back to Latin or Greek roots. 
The remainder of them originate from the principal scientists who 
pioneered a particular field of study, known as etymology. 

Etymology is our name for the science of words, and describes the sometimes- 
tortuous route by which we inherit them from our ancestors. In fact, most words 
change and shift their meaning with the years. A classic example describes how King 
George III, when first he saw the rebuilt St Paul's Cathedral in London, described it 
as 'amusing, artificial and awful', by which he meant, respectively, it 'pleased him', 
was 'an artifice' (i.e. grand) and was 'awesome' (i.e. breathtaking). 

Any reader will soon discover the way this text has an unusual etymological empha- 
sis: the etymologies are included in the belief that taking a word apart actually helps us 
to understand it and related concepts. For example, as soon as we know the Greek for 
'green' is Mows, we understand better the meanings of the proper nouns c/z/orophyll 
and chlorine, both of which are green. Incidentally, phyll comes from the Greek for 
'leaf, and ine indicates a substance. 

Again, the etymology of the word oxygen incorporates much historical informa- 
tion: oxys is the Greek for 'sharp', in the sense of an extreme sensory experience, 
such as the taste of acidic vinegar, and the ending gen comes from gignesthaw (pro- 
nounced 'gin-es-thaw'), meaning 'to be produced'. The classical roots of 'oxygen' 
reveal how the French scientists who first made the gas thought they had isolated the 
distinguishing chemical component of acids. 

The following tables are meant to indicate the power of this approach. There are 
several dozen further examples in the text. The bibliography on p. 533 will enable 
the interested reader to find more examples. 






















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List of Symbols 

Symbols for variables 

a Activity 

a van der Waals constant 

A optical absorbance 

A Area 

b van der Waals constant 

B' virial coefficient 

c concentration 

c the intercept on the v-axis of a graph 

c constant of proportionality 

C virial coefficient 

C heat capacity 

C p heat capacity determined at constant 

Cy heat capacity determined at constant 

E energy 

E potential 

E n activation energy 

£( ea ) electron affinity 

£j junction potential 

£(ioad) potential of a battery or cell when 

passing a current 
emf potential of a cell, determined at zero 



electrode potential for the couple 

+ Me" = R 

c O,R 

standard electrode potential for the 

couple + ne~ = R 


electrode potential of the negative 

electrode in a cell 


electrode potential of the positive 

electrode in a cell 








Gibbs function 


change in Gibbs function 


standard Gibbs function 


Gibbs function of activation 






change in enthalpy 


enthalpy of adsorption 


standard enthalpy 

A // BE 

bond enthalpy 

A// c 

enthalpy of combustion 

AH f 

enthalpy of formation 


enthalpy of activation 


electrical current 


intensity of light following absorption 



intensity of incident light beam 



ionic strength 



ionization energy 



rotational quantum number; rotational 
quantum number of an excited 



n m 


rotational quantum number of ground 




force constant of a bond 



proportionality constant 



rate constant 



pseudo rate constant 



rate constant of an rath-order reaction 



rate constant for the back reaction of 


an nth-order reaction 



rate constant of the nth process in a 


multi-step reaction 



rate constant of adsorption 



rate constant of desorption 



Henry's law constant 


equilibrium constant 



correction constant of an ion-selective 




K a 

acidity constant ('acid dissociation' 





acidity constant for the nth 


dissociation reaction 

5 e 

K b 

basicity constant 


K c 

equilibrium constant formulated in 


terms of concentration 


K P 

equilibrium constant formulated in 



terms of pressure 



equilibrium constant of solubility 


(sometimes called 'solubility 


product' or 'solubility constant') 


K K 

autoprotolysis constant of water 



equilibrium constant of forming a 

transition state 'complex' 






gradient of a graph 


relative molar mass 

number of moles 

number of electrons in a redox 

amount of material in an adsorbed 


partial pressure of component i 
vapour pressure of pure i 
standard pressure of 10 Pa 
heat energy 

reaction quotient 
separation between ions 
radius of a circle or sphere 
bond length 
bond length in an optically excited 

equilibrium bond length 
electrical resistance 

stoichiometric ratio 

change in entropy 
standard entropy 
entropy of activation 
half life 


optical transmittance 

optical transmittance without a sample 

Krafft temperature 

internal energy 

change in internal energy, e.g. during 

quantum-number of vibration 
quantum-number of vibration in an 

excited-state species 


v" quantum-number of vibration in 

a ground-state species 

V volume 

V voltage, e.g. of a power pack 

V Coulomb potential energy 
V m molar volume 

w work 

x controlled variable on the horizontal 

axis of a graph 
x deviation of a bond from its 

equilibrium length 
Xi mole fraction of i 

y observed variable on the vertical 

of a graph 
z charge on ion (so z + for a cation 

and z~ for an anion) 
Z compressibility 






frequency (the reciprocal of the period 

of an event) 
frequency following transmission 

(in Raman spectroscopy) 
extent of reaction 

electrical conductivity 
standard deviation 
electric field strength (electrostatic 

work function of a metal 
primary quantum yield 
quantum yield of a reaction 
wavenumber of a vibration 

(determined as co — k 4- c) 









^ (max) 



activity coefficient 

mean ionic activity coefficient 

fugacity coefficient 

surface tension 

small increment 

partial differential 

change in a variable (so 

Symbols for constants 

AX = X 

(final form) 



form) ) 

extinction coefficient ('molar decadic 

relative permittivity 
permittivity of free space 
adsorption isotherm 

ionic conductivity 
the wavelength of a peak in a 

reduced mass 
chemical potential of i 
standard chemical potential of i 
stoichiometric constant 


Debye-Hiickel 'A' factor 


the speed of light in vacuo 


standard concentration 


charge on an electron, of value 

1.6 x 10" 19 C 


mathematical operator ('function of) 


Faraday constant, of value 

96485 CmoP 1 


Boltzmann constant, of value 

1.38 x 1(T 23 


Avogadro constant, of value 

6.022 x 10 23 mor 1 

N A 

Avogadro number, of value 

6.022 x 10 23 mor 1 


acceleration due to gravity, of value 

9.81 m s~ 2 


Planck constant, of value 

6.626 x 10" 34 J s 


gas constant, of value 

8.314 J K" 1 mol -1 


Symbols for units 















angstrom, length of value 10~ 10 m 

standard pressure of 10 5 Pa 

(non-SI unit) 

centigrade (non-SI) 
millimetre of mercury (non-SI unit 

of pressure) 

second (SI unit) 

Acronyms and abbreviations 


charge transfer 


differential operator (which never 

appears on its own) 


3 highest occupied molecular orbital 


intelligence quotient 




1 International Union of Pure and 

Applied Chemistry 

IVF in vitro fertilization 

LCD liquid crystal display 

LHS left-hand side 

LUMO lowest unoccupied molecular 

MLCT metal-to-ligand charge transfer 
MRI magnetic resonance imaging 

NIR near-infra red 

NMR nuclear magnetic resonance 
O general oxidized form of a redox 

p mathematical operator, 

— log 10 [variable], so 

P H=-log 10 [H+] 
PEM proton exchange membrane 
R general reduced form of a redox 

RHS right-hand side 

s.t.p. standard temperature and 

SAQ self-assessment question 

SCE saturated calomel electrode 

SCUBA self-contained underwater breathing 

SHE standard hydrogen electrode 

SHM simple harmonic motion 
SI Systeme Internationale 

Sn 1 unimolecular nucleophilic substitution 

Sn2 bimolecular nucleophilic substitution 

SSCE silver-silver chloride electrode 
TS transition state 

TV television 

UPS UV-photoelectron spectroscopy 

UV ultraviolet 

UV-vis ultraviolet and visible 
XPS X-ray photoelectron spectroscopy 


Standard subscripts (other 



than those where a word or 

+ • 

radical cation 

phrase is spelt in full) 


standard state 



adsorption; adsorbed 



Chemicals and materials 



at equilibrium 


general anion 












N, 7V-dimethylformamide 


left-hand side of a cell 






deoxyribonucleic acid 


at constant pressure 




platinum (usually, as an electrode) 


ethylenediamine tetra-acetic acid 




general Lowry-Br0nsted acid 


right-hand side of cell 


liquid petroleum gas 




general cation 




methylene blue 


at time t (i.e. after a reaction or 


methyl viologen 

process has commenced) 

(1,1 ' -dime thy 1-4 , 4' -bipyridilium) 


at constant volume 


general oxidized form of a redox 

initially (i.e. at time t — 0) 



measurement taken after an infinite 


propylene carbonate 

length of time 


phenyl substituent 


general alkyl substituent 


general reduced form of a redox 

Standard superscripts (other 


than those where a word or 


sodium dodecyl sulphate 

phrase is spelt in full) 


tetrafluoroacetic acid 


particle emitted during radioactive 
disintegration of nucleus 


activated quantity 




particle emitted during radioactive 



disintegration of nucleus 


p.Yritp.H state 


high-energy photon (gamma ray) 


















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Introduction to physical 


In this, our introductory chapter, we start by looking at the terminology of phys- 
ical chemistry. Having decided what physical chemistry actually is, we discuss the 
nature of variables and relationships. This discussion introduces the way relationships 
underlying physical chemistry are formulated. 

We also introduce the fundamental (base) units of the Systeme Internationale (SI), 
and discuss the way these units are employed in practice. 

We look at the simple gas laws to explore the behaviour of systems with no inter- 
actions, to understand the way macroscopic variables relate to microscopic, molecular 
properties. Finally, we introduce the statistical nature underlying much of the physical 
chemistry in this book when we look at the Maxwell-Boltzmann relationship. 

1.1 What is physical chemistry: variables, 
relationships and laws 

Why do we warm ourselves by a radiator? 

Cause and effect 

We turn on the radiator if we feel cold and warm ourselves in front of it. We become 
warm because heat travels from the radiator to us, and we absorb its heat energy, 
causing our own energy content to rise. At root, this explains why we feel more 

While this example is elementary in the extreme, its importance lies in the way it 
illustrates the concept of cause and effect. We would not feel warmer if the radiator 


was at the same temperature as we were. We feel warmer firstly because the radiator 
is warmer than us, and secondly because some of the heat energy leaves the radiator 
and we absorb it. A transfer of energy occurs and, therefore, a change. Without the 
cause, the effect of feeling warmer could not have followed. 

We are always at the mercy of events as they occur around 
us. The physical chemist could do nothing if nothing happened; 
chemists look at changes. We say a physical chemist alters vari- 
ables, such as pressure or temperature. Typically, a chemist causes 
one variable to change and looks at the resultant response, if any. 
Even a lack of a response is a form of result, for it shows us what is and what is not 
a variable. 

A variable is an exper- 
imental parameter we 
can change or 'tweak'. 

The fearsome-looking 
word 'physicochemi- 
cal' means 'relating to 
physical chemistry'. 

Why does water get hot in a kettle? 

Physicochemical relationships 

Putting water into an electric kettle does not cause the water to get 
hot. The water stays cold until we turn on the power to the kettle 
element, which converts electrical energy from the mains into heat energy. The heat 
energy from the kettle element is then absorbed by the water, which gets hot as a 
direct consequence. 

The temperature of the water does not increase much if a small amount of elec- 
trical energy is consumed; conversely, the water gets hotter if a greater amount of 
energy is consumed and thereafter passed to the water. A physi- 
cal chemist says a 'relationship' exists (in this case) between heat 
input and temperature, i.e. the temperature of the water depends on 
the amount of energy consumed. 

Mathematically, we demonstrate the existence of a relationship 

by writing T = /(energy), where T is temperature and the small 

/ means 'is a function of. 

So the concept of variables is more powerful than just changing 

parameters; nor do physical chemists merely vary one parameter and see what happens 

to others. They search for 'physicochemical' relationships between the variables. 

In words, the symbols 
T = /"(energy) means T 
is a function of energy'. 
Note how variables are 
usually printed in italic 

lese two colours complementary? 

Qualitative and quantitative measurements 

We often hear this question, either at the clothes shop or at a paint merchant. Either 
someone wants to know if the pink colour of a sweatshirt matches the mauve of a 
skirt, or perhaps a decorator needs to know if two shades of green will match when 
painted on opposing bedroom walls. 


Complementary means 
'to make complete'. 

But while asking questions concerning whether a series of 
colours are complementary, we are in fact asking two questions 
at once: we ask about the colour in relation to how dark or light 
it is ('What is the brightness of the colour?'); but we also ask a 
more subjective question, saying 'Is the pink more red or more white: what kind of 
pink is it?' We are looking for two types of relationship. 

In any investigation, we first look for a qualitative relationship. In effect, we ask 
questions like, 'If I change the variable x, is there is a response in a different variable 
yV We look at what kind of response we can cause - a scientist wants to know about 
the qualities of the response, hence QUAL-itative. An obvious question relating to 
qualitative relationships is, 'If I mix solutions of A and B, does a reaction occurV 

Only after we know whether or not there is a response (and of what general kind) 
does a physical chemist ask the next question, seeking a quantitative assessment. He 
asks, 'How much of the response is caused?' In effect, physical chemists want to 
know if the magnitude (or quantity) of a response is big, small or intermediate. We 
say we look for a QUANT-itative aspect of the relationship. An obvious question 
relating to quantitative relationships is, 'I now know that a reaction occurs when I 
mix solutions of A and B, but to what extent does the reaction occur; what is the 
chemical yield!' 

Does my radio get louder if I vary the volume control? 

Observed and controlled variables 

We want to turn up the radio because it's noisy outside, and we want to hear what is 
broadcast. We therefore turn the volume knob toward 'loud'. At its most basic, the 
volume control is a variable resistor, across which we pass a current from the battery, 
acting much like a kettle element. If we turn up the volume control then a larger 
current is allowed to flow, causing more energy to be produced by the resistor. As 
a listener, we hear a response because the sound from the speakers becomes louder. 
The speakers work harder. 

But we must be careful about the way we state these relationships. We do not 'turn 
up the volume' (although in practice we might say these exact words and think in these 
terms). Rather, we vary the volume control and, as a response, our ears experience 
an increase in the decibels coming through the radio's speakers. The listener controls 
the magnitude of the noise by deciding how far the volume-control knob needs to be 
turned. Only then will the volume change. The process does not occur in reverse: we 
do not change the magnitude of the noise and see how it changes 
the position of the volume-control knob. 

While the magnitude of the noise and the position of the volume 
knob are both variables, they represent different types, with one 
depending on the other. The volume control is a controlled variable 
because the listener dictates its position. The amount of noise is the 
observed variable because it only changes in response to variations 
in the controlled variable, and not before. 

We consciously, care- 
fully, vary the magni- 
tude of the controlled 
variable and look at 
the response of the 
observed variable. 


Relationships and graphs 

Physical chemists often depict relationships between variables by 
drawing graphs. The controlled variable is always drawn along the 
x-axis, and the observed variable is drawn up the y-axis. 

Figure 1.1 shows several graphs, each demonstrating a different 
kind of relationship. Graph (a) is straight line passing through the 
origin. This graph says: when we vary the controlled variable x, 
the observed variable y changes in direct proportion. An obvious 
example in such a case is the colour intensity in a glass of black- 
currant cordial: the intensity increases in linear proportion to the 
concentration of the cordial, according to the Beer-Lambert law 
(see Chapter 9). Graph (a) in Figure 1.1 goes through the origin 
because there is no purple colour when there is no cordial (its 
concentration is zero). 
Graph (b) in Figure 1.1 also demonstrates the existence of a relationship between 
the variables x and y, although in this case not a linear relationship. In effect, the graph 
tells us that the observed variable y increases at a faster rate than does the controlled 
variable x. A simple example is the distance travelled by a ball as a function of time 
t as it accelerates while rolling down a hill. Although the graph is not straight, we 
still say there is a relationship, and still draw the controlled variable along the x-axis. 

The x-axis (horizontal) 
is sometimes called 
the abscissa and the 
/-axis (vertical) is the 
ordinate. A simple way 
to remember which 
axis is which is to say, 
'an expanse of road 
goes horizontally along 
the x-axis', and 'a Yo- 
Yo goes up and down 
the /-axis'. 





















Controlled var 



Controlled variable x 










o o 











o o o 









O o o o 

Controlled variable 


Controlled variable x 



Figure 1.1 Graphs of observed variable (along the y-axis) against controlled variable (along the 
x-axis). (a) A simple linear proportionality, so y = constant x x; (b) a graph showing how y is not 
a simple function of x, although there is a clear relationship; (c) a graph of the case where variable 
y is independent of variable x ; (d) a graph of the situation in which there is no relationship between 
y and x, although y does vary 


Graph (c) in Figure 1.1 is a straight-line graph, but is horizontal. In other words, 
whatever we do to the controlled variable x, the observed variable y will not change. 
In this case, the variable y is not a function of x because changing x will not change 
y. A simple example would be the position of a book on a shelf as a function of time. 
In the absence of other forces and variables, the book will not move just because it 
becomes evening. 

Graph (d) in Figure 1.1 shows another situation, this time the 
data do not demonstrate a straightforward relationship; it might 
demonstrate there is no relationship at all. The magnitude of the 
controlled variable x does not have any bearing on the observed 

Data is plural; the sin- 
gular is datum. 

When two variables are 
multiplied together, we 
call them a compound 

variable y. We say the observed variable y is independent of the 
controlled variable x. Nevertheless, there is a range of results for 
y as x varies. Perhaps x is a compound variable, and we are being 
simplistic in our analysis: an everyday example might be a stu- 
dent's IQ as x and his exam performance as y, suggesting that, 
while IQ is important, there must be another variable controlling the magnitude of 
the exam result, such as effort and commitment. Conversely, the value of y might be 
completely random (so repeating the graphs with the same values of x would gen- 
erate a different value of y - we say it is irreproducible). An example of this latter 
situation would be the number of people walking along a main road as a function 
of time. 

Why does the mercury in a barometer go up when the 
air pressure increases? 

Relationships between variables 

The pressure p of the air above any point on the Earth's surface relates ultimately 
to the amount of air above it. If we are standing high up, for example on the top 
of a tall mountain, there is less air between us and space for gravity to act upon. 
Conversely, if we stand at the bottom of the Grand Canyon (one of the lowest places 
on Earth) then more air separates us from space, causing the air pressure p to be 
much greater. 

A barometer is an instrument designed to measure air pressure p. It consists of 
a pool of liquid mercury in a trough. A long, thin glass tube (sealed at one end) 
is placed in the centre of the trough with its open-side beneath the surface of the 
liquid; see Figure 1.2. The pressure of the air acts as a force on the surface of the 
mercury, forcing it up and into the capillary within the tube. If the air pressure is 
great, then the force of the air on the mercury is also great, causing much mer- 
cury up the tube. A lower pressure is seen as a shorter length h of mercury in 
the tube. 

By performing experiments at different pressures, it is easy to prove the existence 
of a relationship between the air pressure p and the height h of the mercury column 




glass tube * 

Trough of mercury 

Figure 1.2 A barometer is a device for measuring pressures. A vacuum-filled glass tube (sealed 
at one end) is placed in a trough of mercury with its open end beneath the surface of the liquid 
metal. When the tube is erected, the pressure of the external air presses on the surface and forces 
mercury up the tube. The height of the mercury column h is directly proportional to the external 
pressure p 

In fact, the value 
of the constant c in 
Equation (1.1) com- 
prises several natural 
constants, including 
the acceleration due 
gravity g and the den- 
sity p of the mercury. 

in the tube. This relationship follows Equation (1.1): 

h = c x p 


where c is merely a proportionality constant. 

In practice, a barometer is merely an instrument on which 
we look at the length of the column of mercury h and, via 
Equation (1.1), calculate the air pressure p. The magnitude of h is 
in direct relation to the pressure p. We ascertain the magnitude of 
h if we need to know the air pressure p. 
While physical chemistry can appear to be horribly mathematical, in fact the mathe- 
matics we employ are simply one way (of many) to describe the relationships between 
variables. Often, we do not know the exact nature of the function until a later stage 
of our investigation, so the complete form of the relationship has to be discerned 
in several stages. For example, perhaps we first determine the existence of a linear 
equation, like Equation (1.1), and only then do we seek to measure 
an accurate value of the constant c. 

But we do know a relationship holds, because there is a response. 
We would say there was no relationship if there was no response. 
For example, imagine we had constructed a poor-quality barometer 
(meaning it does not follow Equation (1.1)) and gave it a test run. If 
we could independently verify that the pressure p had been varied 
over a wide range of values yet the length of the mercury h in 
the barometer did not change, then we would say no relationship 
existed between p and h. 

We might see this 
situation written math- 
ematically as, h =£ f(p), 
where the V means 
'is not equal to'. In 
other words, h is not a 
function of p in a poor 


Why does a radiator feel hot to the touch when x on', 
and cold when *off r ? 


Laws and the minus-oneth law of thermodynamics 

A 'law' in physical 
chemistry relates to 
a wide range of situa- 

Feeling the temperature of a radiator is one of the simplest of 

experiments. No one has ever sat in front of a hot radiator and 

felt colder. As a qualitative statement, we begin with the excellent 

generalization, 'heat always travels from the hotter to the colder 

environment' . We call this observation a law because it is universal. 

Note how such a law is not concerned with magnitudes of change 

but simply relays information about a universal phenomenon: energy in the form of 

heat will travel from a hotter location or system to a place which is colder. Heat 

energy never travels in the opposite direction. 

We can also notice how, by saying 'hotter' and 'colder' rather than just 'hot' and 
'cold', we can make the law wider in scope. The temperature of a radiator in a living 
room or lecture theatre is typically about 60 °C, whereas a human body has an ideal 
temperature of about 37 °C. The radiator is hotter than we are, so heat travels to us 
from the radiator. It is this heat emitted by the radiator which we absorb in order to 
feel warmer. 

Conversely, now consider placing your hands on a colder radiator having a tem- 
perature of 20 °C (perhaps it is broken or has not been switched on). In this second 
example, although our hands still have the same temperature of 37 °C, this time the 
heat energy travels to the radiator from our hands as soon as we touch it. The direction 
of heat flow has been reversed in response to the reversal of the relative difference 
between the two temperatures. The direction in which the heat energy is transferred 
is one aspect of why the radiator feels cold. We see how the movement of energy 
not only has a magnitude but also a direction. 

Such statements concerning the direction of heat transfer are 
sometimes called the minus-oneth law of thermodynamics, which 
sounds rather daunting. In fact, the word 'thermodynamics' here 
may be taken apart piecemeal to translate it into everyday English. 
First the simple bit: 'dynamic' comes from the Greek word 
dunamikos, which means movement. We obtain the conventional 
English word 'dynamic' from the same root; and a cyclist's 
'dynamo' generates electrical energy from the spinning of a bicycle wheel, i.e. from a 
moving object. Secondly, thermo is another commonly encountered Greek root, and 
means energy or temperature. We encounter the root thermo incorporated into such 
everyday words as 'thermometer', 'thermal' and 'thermos flask'. A 'thermodynamic' 
property, therefore, relates to events or processes in which there are 'changes in heat 
or energy'. 

The 'minus-oneth law 
of thermodynamics' 
says, 'heat always 
travels from hot to 



We need to explain the bizarre name of this law, which is really an accident of history. 
Soon after the first law of thermodynamics was postulated in the mid nineteenth century, 
it was realized how the law presupposed a more elementary law, which we now call 
the zeroth law (see below). We call it the 'zeroth' because zero comes before one. But 
scientists soon realized how even the zeroth law was too advanced, since it presupposed a 
yet more elementary law, which explains why the minus-oneth law had to be formulated. 

How does a thermometer work? 

Thermal equilibrium and the zeroth law of thermodynamics 

The word 'thermometer' 
has two roots: meter 
denotes a device to 
measure something, 
and thermo means 
'energy' or 'temper- 
ature'. Thus, a 'ther- 
mometer' is a device for 
measuring energy as 
a function of tempera- 

A fever is often the first visible sign of someone developing an 
illness. The body's temperature rises - sometimes dramatically - 
above its preferred value of 37 °C. As a good generalization, the 
temperature is hotter when the fever is worse, so it is wise to 
monitor the temperature of the sick person and thereby check the 
progress of the illness. A thermometer is the ideal instrument for 
this purpose. 

When measuring a temperature with a thermometer, we place 
the mercury-containing end into the patient's mouth or armpit and 
allow the reading to settle. The mercury is encased within a thin- 
walled glass tube, which itself is placed in contact with the patient. 
A 'reading' is possible because the mercury expands with increas- 
ing temperature: we take the length / of the mercury in the tube to be an accurate 
function of its temperature T. We read the patient's temperature from the thermometer 
scale only when the length of the mercury has stopped changing. 

But how does the thermometer work in a thermodynamic sense, since at no time 
can the toxic mercury be allowed to touch the patient? 

Consider the flow of heat: heat energy first flows from the patient 
to the glass, and thence flows through the glass into the mercury. 
Only when all three - mercury, glass and patient - are at the same 
temperature can the thermometer reading become steady. We say 
we have thermal equilibrium when these three have the same tem- 
perature; see Figure 1.3. 
Although in some respects a trivial example, a thermometer helps us see a profound 
truth: only when both (i) the mercury and the glass, and (ii) the glass and the patient 
are at thermal equilibrium can the patient and the mercury truly be said to be at the 
same temperature. By this means, we have measured the temperature of the patient by 

Bodies together at the 
same temperature are 
said to be in 'thermal 


Patient's tongue 

Mercury-in-glass thermometer 

Figure 1.3 The zeroth law states, 'Imagine three bodies, A, B and C. If A and B are in thermal 
equilibrium, and B and C are in thermal equilibrium, then A and C are also in thermal equilibrium' 
(see inset). A medic would rephrase the law, 'If mercury is in thermal equilibrium with the glass 
of a thermometer, and the glass of a thermometer is in thermal equilibrium with a patient, then the 
mercury and the patient are also in thermal equilibrium' 

utilizing a temperature-dependent property of the liquid metal inside the thermometer, 
yet at no time do we need to expose the patient to the toxic mercury. 

We begin to understand the power of thermodynamics when we 
realize how often this situation arises: in effect, we have made 
an indirect measurement - a frequent occurrence - so we need to 
formulate another law of thermodynamics, which we call the zeroth 
law. Imagine three bodies, A, B and C. If A and B are in thermal 
equilibrium, and B and C are in thermal equilibrium, then A and 
C are also in thermal equilibrium. 

While sounding overly technical, we have in fact employed the 
zeroth law with the example of a thermometer. Let us rephrase 
the definition of the zeroth law and say, 'If mercury is in thermal 
equilibrium with the glass of a thermometer, and the glass of a 
thermometer is in thermal equilibrium with a patient, then the mercury and the patient 
are also in thermal equilibrium'. A medic could not easily determine the temperature 
of a patient without this, the zeroth law. 

From now on we will assume the zeroth law is obeyed each time we use the phrase 
'thermal equilibrium'. 

The zeroth law of ther- 
modynamics says: 
imagine three bodies, 
A, BandC. If Aand Bare 
in thermal equilibrium, 
and B and C are also 
in thermal equilibrium, 
then A and C will be in 
thermal equilibrium. 

1.2 The practice of thermodynamic 

What is temperature? 

Scientific measurement 

Although the answer to the simple question 'what is temperature?' seems obvious at 
first sight, it is surprisingly difficult to answer to everyone's satisfaction. In fact, it is 



'Corollary' means a 
deduction following on 
from another, related, 
fact or series of facts. 

The word 'thermo- 
chemistry' has two 
roots: thermo, mean- 
ing 'temperature or 
energy', and chem- 
istry, the science of 
the combination of 
chemicals. We see 
how 'thermochemistry' 
studies the energy and 
temperature changes 
accompanying chemi- 
cal changes. 

generally easier to state the corollary, 'a body has a higher temper- 
ature if it has more energy, and a lower temperature if it has less 
energy' . 

We have been rather glib so far when using words such as 
'heat' and 'temperature', and will be more careful in future. Heat 
is merely one way by which we experience energy. Everything contains energy in 
various amounts, although the exact quantity of the energy is not only unknown 
but unknowable. 

Much of the time, we, as physical chemists, will be thinking 
about energy and the way energetic changes accompany chemical 
changes (i.e. atoms, ions or whole groups of atoms combine, or 
add, or are being lost, from molecules). While the total energy 
cannot be known, we can readily determine the changes that occur 
in tandem with chemical changes. We sometimes give the name 
thermochemistry to this aspect of physical chemistry. 

In practice, the concept of temperature is most useful when deter- 
mining whether two bodies are in thermal equilibrium. Firstly, we 
need to appreciate how these equilibrium processes are always 
dynamic, which, stated another way, indicates that a body simulta- 
neously emits and absorbs energy, with these respective amounts 
of energy being equal and opposite. Furthermore, if two bodies 
participate in a thermal equilibrium then we say that the energy 
emitted by the first body is absorbed by the second; and the first 
body also absorbs a similar amount of energy to that emitted by 
the second body. 

Temperature is most conveniently visualized in terms of the 
senses: we say something is hotter or is colder. The first ther- 
mometer for studying changes in temperature was devised in 1631 
by the Frenchman Jean Rey, and comprised a length of water in a 
glass tube, much like our current-day mercury-in-glass thermometers but on a much 
bigger scale. The controlled variable in this thermometer was temperature T, and the 
observed variable was the length I of the water in the glass tube. 

Rey's thermometer was not particularly effective because the density of water is so 
low, meaning that the volume of the tube had to be large. And the tube size caused 
an additional problem. While the water expanded with temperature (as required for 
the thermometer to be effective), so did the glass encapsulating it. In consequence 
of both water and glass expanding, although the water expanded in a straightforward 
way with increasing temperature, the visible magnitude of the expansion was not in 
direct proportion to the temperature rise. 

Although we could suggest that a relationship existed between 
the length / and the temperature T (saying one is a function of the 
other), we could not straightforwardly ascertain the exact nature 
of the function. In an ideal thermometer, we write the mathemat- 
ical relationship, / = f(T). Because Rey's thermometer contained 

A body in 'dynamic equi- 
librium' with another 
exchanges energy with 
it, yet without any net 

Scientists use the word 
'ideal' to mean obeying 
the laws of science. 



water, Rey was not able to observe a linear dependence of I on T for 
so he could not write I = aT + b (where a and b are constants). 

For a more dense liquid, such as mercury, the relationship bet- 
ween I and T is linear - at least over a relatively narrow range 
of temperatures - so a viable mercury-in-glass thermometer may 
be constructed. But, because the temperature response is only lin- 
ear over a narrow range of temperatures, we need to exercise 

If we assume the existence of a linear response for such a ther- 
mometer, then the thermometer is 'calibrated' by correlating the 
readings of length / using the known properties of the standard, as 
follows. First, the thermometer is placed in a trough of pure ice at 
its melting temperature, and the end of the mercury bead marked 
as °C. The same thermometer is then placed in water at its boil- 
ing point and the end of the mercury bead marked as 100 °C. The 
physical distance between these two extremes is subdivided into 
100 equal portions, each representing a temperature increment of 
1 °C. This centigrade scale is satisfactory for most purposes. (The 
same scale is sometimes called Celsius after a Swedish physicist 
who championed its use.) 

This formulation of the centigrade scale presupposed a linear 
relationship between length / and temperature T (i.e. the straight 
line (a) on the graph in Figure 1.4), but we must be aware the 
relationship might only be approximately linear (e.g. the curved 
line (b) on the graph in Figure 1.4). The straight and the curved 
lines only agree at the two temperatures 0°C and 100 °C merely 
because they were defined that way. 

his thermometer, 


' in 








To 'calibrate' an instru- 
ment such as a ther- 
mometer, we correlate 
a physico-chemical 
property (such as the 
length / of the mercury) 
using the temperature- 
dependent properties 
of a known standard. 

The 'centigrade' scale 
was first proposed in 
1694 by Renaldi. Centi 
is a Latin prefix mean- 
ing 'hundred'. 

2 E 

— o 

2 E 

-§ $ 

> >- 


(a) Ideal response 

(b) More realistic response 

0°C 100°C 

Controlled variable (temperature, T) 

Figure 1.4 In using a thermometer, we assume the existence of a linear response between the 
length / of the mercury and the controlled variable temperature T. Trace (a) shows such a rela- 
tionship, and trace (b) shows a more likely situation, in which there is a close approximation to a 
linear relationship between length / and temperature T 


This last paragraph inevitably leads to the questions, 'So how do we know what 
the exact temperature is?' and 'How do I know if my thermometer follows profile 
(a) or profile (b) in Figure 1.4?' Usually, we do not know the answer. If we had a 
single thermometer whose temperature was always accurate then we could use it as a 
primary standard, and would simply prepare a calibrated thermometer against which 
all others are calibrated. 

But there are no ideal (perfect) thermometers in the real world. In practice, we 
generally experiment a bit until we find a thermometer for which a property X is as 
close to being a linear function of temperature as possible, and call it a standard ther- 
mometer (or 'ideal thermometer'). We then calibrate other thermometers in relation to 
this, the standard. There are several good approximations to a standard thermometers 
available today: the temperature-dependent (observed) variable in a gas thermometer 
is the volume of a gas V. Provided the pressure of the gas is quite low (say, one- 
hundredth of atmospheric pressure, i.e. 100 Pa) then the volume V and temperature 
T do indeed follow a fairly good linear relationship. 

A second, popular, standard is the platinum-resistance thermometer. Here, the elec- 
trical resistance R of a long wire of platinum increases with increased temperature, 
again with an essentially linear relationship. 

Worked Example 1.1 A platinum resistance thermometer has a resistance R of 3.0 x 
10" 4 12 at 0°C and 9.0 x 10" 4 12 at 100 °C. What is the temperature if the resistance R 
is measured and found to be 4.3 x 10~ 4 12? 

We first work out the exact relationship between resistance R and temperature T. We 
must assume a linear relationship between the two to do so. 

The change per degree centigrade is obtained as 'net change in 

These discussions are 
expressed in terms of 
centigrade, although 
absolute temperatures 
are often employed - 
see next section. 

resistance 4- net change in temperature'. The resistance R increases 
by 6.0 x 10~ 4 12 while the temperature is increased over the 100 °C 
range; therefore, the increase in resistance per degree centigrade is 
given by the expression 

6.0 x 10~ 4 12 , , 

R per°C = = 6x 10" 6 12 "C" 1 

100 °C 

Next, we determine by how much the resistance has increased in going to the new 
(as yet unknown) temperature. We see how the resistance increases by an amount (4.3 — 
3.0) x 10~ 4 12 = 1.3 x 10- 4 12. 

The increase in temperature is then the rise in resistance divided by the change in 

resistance increase per degree centigrade. 

We obtain 

1.3 x 10" 4 12 

6 x lO" 6 ^^- 1 
so the new temperature is 21.7 °C. 


We should note, before proceeding, firstly how the units of £2 on both top and 
bottom of this fraction cancel; and secondly, how °C _1 is in the denominator of the 
fraction. As a consequence of it being on the bottom of the fraction, it is inverted 
and so becomes °C. In summary, we see how a simple analysis of the units in this 
sum automatically allows the eventual answer to be expressed in terms of °C. We are 
therefore delighted, because the answer we want is a temperature, and the units tell 
us it is indeed a temperature. 


This manipulation of units is sometimes called dimensional analysis. Strictly speaking, 
though, dimensional analysis is independent of the units used. For example, the units 
of speed may be in metres per second, miles per hour, etc., but the dimensions of speed 
are always a length [L] divided by a time [T]: 

[speed] = [L] - [T] 

Dimensional analysis is useful in two respects. (1) It can be used to determine the 
units of a variable in an equation. (2) Using the normal rules of algebra, it can be used 
to determine whether an equation is dimensionally correct, i.e. the units should balance 
on either side of the equation. All equations in any science discipline are dimensionally 
correct or they are wrong! 

A related concept to dimensional analysis is quantity calculus, a method we find 
particularly useful when it comes to setting out table header rows and graph axes. 
Quantity calculus is the handling of physical quantities and their units using the normal 
rules of algebra. A physical quantity is defined by a numerical value and a unit: 

physical quantity = number x unit 


AH = 40.7 U mol" 1 

which rearranges to 

Atf/kJ mol -1 = 40.7 

SAQ 1.1 A temperature is measured with the same platinum-resistance 
thermometer used in Worked Example 1.1, and a resistance R = 11.4 x 
10~ 4 Q, determined. What is the temperature? 



The word 'philosoph- 
ical' comes from the 
Greek words philos 
meaning 'love' and 
sophia meaning 'wis- 
dom'. Philosophy is 
therefore the love of 
wisdom. This same 
usage of 'wisdom' is 
seen with the initials 
PhD, which means a 
'philosophy doctorate'. 

Some people might argue that none of the discussion above actu- 
ally answers the philosophical question, 'What is temperature?' 
We will never come to a completely satisfactory answer; but we 
can suppose a body has a higher temperature if it contains more 
energy, and that it has a lower temperature if it has less energy. 
More importantly, a body will show a rise in temperature if its 
energy content rises, and it will show a lower temperature if its 
energy content drops. This is why we sit in front of a fire: we want 
to absorb energy, which we experience as a higher temperature. 

How long is a p/e< 


This definition of 'more 
energy means hotter' 
needs to be handled 
with care: consider 
two identical weights 
at the same tempera- 
ture. The higher weight 
has a greater potential 

The SI unit of length 

A common problem in Anglo Saxon England, as well as much 
of contemporary Europe, was the way cloth merchants could so 
easily cheat the common people. At a market, it was all too easy 
to ask for a yard of cloth, to see it measured against the merchant's 
yardstick, and pay for the cloth only to get home to learn just how 
short the merchant's stick was. Paying for 10 yards and coming 
home with only 9 yards was common, it seems; and the problem 
was not restricted to just cloth, but also to leather and timber. 
According to legend, the far-sighted English King Edgar (ad 959-975) solved the 
problem of how to stop such cheating by standardizing the length. He took 100 foot 
soldiers and measured the length of the right foot of each, one after the other, as 
they stood in line along the floor of his threshing hall. This overall length was then 
subdivided into 100 equal parts to yield the standard length, the foot. The foot is still 
commonly employed as a unit of length in Britain to this day. Three of these feet 
made up 1 yard. The king was said to keep in his treasury a rod of gold measuring 
exactly 1 yard in length. This is one theory of how the phrase 'yardstick' originated. 
Any merchant accused of cheating was required to bring his yardstick and to compare 
its length against that of the king. Therefore, a merchant whose stick was shorter was 
a cheat and paid the consequences. A merchant whose stick was longer was an idiot. 
While feet and yards are still used in Britain and other countries, 
the usual length is now the metre. At the time of the French Rev- 
olution in the 18th century and soon after, the French Academy 
of Sciences sought to systemize the measurement of all scientific 
quantities. This work led eventually to the concept of the Systeme 
Internationale, or SI for short. Within this system, all units and 
definitions are self-consistent. The SI unit of length is the metre. 

The original metre rule was kept in the International Bureau 
of Weights and Measures in Sevres, near Paris, and was a rod of 

SI units are self- 
consistent, with all 
units being defined in 
terms a basis of seven 
fundamental units. The 
SI unit of length / is the 
metre (m). 



platinum -iridium alloy on which two deep marks were scratched 
1 m apart. It was used in exactly the same way as King Edgar's 
yardstick 10 centuries earlier. 

Unfortunately, platinum -iridium alloy was a poor choice, for it 
has the unusual property of shrinking (albeit microscopically) with 
time. This SI metre rule is now about 0.3 per cent too short. King 
Edgar's yardstick, being made of gold, would still be the same 
length today as when it was made, but gold is too ductile, and 
could have been stretched, bent or re-scored. 

In 1960, the SI unit of length was redefined. While keeping 
the metre as the unit of length, it is now defined as 1 650763.73 
wavelengths of the light emitted in vacuo by krypton-86. This is a 
sensible standard, because it can be reproduced in any laboratory 
in the world. 

'Ductile' means the 
ability of a metal to 
be drawn to form a 
wire, or to be worked. 
Ductile is the opposite 
of 'brittle'. 

In vacuo is Latin for 
'in a vacuum'. Many 
properties are mea- 
sured in a vacuum to 
avoid the complication 
of interference effects. 

How fast is "greased lightning'? 

Other SI standards 

In comic books of the 1950s, one of the favourite phrases of super-heroes such as 
Superman was 'greased lightning!' The idea is one of extreme speed. The lightning we 
see, greased or otherwise, is a form of light and travels very, very fast. For example, 
it travels through a vacuum at 3 x 10 8 m s _1 , which we denote with the symbol c. 
But while the speed c is constant, the actual speed of light may not be: in fact, it 
alters very slightly depending on the medium through which it travels. We see how 
a definition of time involving the speed of light is inherently risky, explaining why 
we now choose to define time in terms of the duration (or fractions and multiples 
thereof) between static events. And by 'static' we mean unchanging. 

SI 'base units' 

Time is one of the so-called 'base units' within the SI system, and so is length. 
Whereas volume can be expressed in terms of a length (for example, a cube has a 
volume P and side of area I 2 ), we cannot define length in terms of 
something simpler. Similarly, whereas a velocity is a length per unit 
time, we cannot express time in terms of something simpler. In fact, 
just as compounds are made up of elements, so all scientific units 
are made up from seven base units: length, time, mass, temperature, 
current, amount of material and luminous intensity. 

Table 1.1 summarizes the seven base (or 'fundamental') SI phys- 
ical quantities and their units. The last unit, luminous intensity, will 
not require our attention any further. 

The SI unit of 'time' t is the second. The second was originally 
defined as 1/86 400th part of a mean solar day. This definition is 

There are seven base 
SI units: length, time, 
mass, temperature, 
current, luminous 
intensity and amount 
of material. 

The SI unit of 'time' r is 
the second (s). 


Table 1.1 The seven fundamental SI physical quantities and their units 

Physical quantity 

Symbol 3 

SI unit 














Electrical current 




Thermodynamic temperature 




Amount of substance 




Luminous intensity 




a Notice how the abbreviation for each quantity, being a variable, is always italicized, 
whereas the abbreviation for the unit, which is not a variable, is printed with an upright 
typeface. None of these unit names starts with a capital. 

again quite sensible because it can be reproduced in any laboratory in the world. 
While slight changes in the length of a solar year do occur, the word 'mean' in our 
definition obviates any need to consider them. Nevertheless, it was felt necessary 
to redefine the second; so, in the 1960s, the second was redefined as 9 192631 770 
periods of the radiation corresponding to the transition between two of the hyperfine 
levels in the ground state of the caesium- 133 atom. Without discussion, we note how 
the heart of a so-called 'atomic clock' contains some caesium-133. 

In a similar way, the Systeme Internationale has 'defined' other 

The SI unit of 'tempera 
ture' T is the kelvin (K) 

common physicochemical variables. The SI unit of 'temperature' 
T is the kelvin. We define the kelvin as 1/273. 16th part of the 
thermodynamic temperature difference between absolute zero (see 
Section 1.4) and the triple point of water, i.e. the temperature at which liquid water 
is at equilibrium with solid water (ice) and gaseous water (steam) provided that the 
pressure is 610 Pa. 

The SI unit of 'current' / is the ampere (A). An ampere was 

The SI unit of 'current' 
I is the ampere (A). 

first defined as the current flowing when a charge of 1 C (coulomb) 
passed per second through a perfect (i.e. resistance-free) conductor. 
The SI definition is more rigorous: 'the ampere is that constant 
current which, if maintained in two parallel conductors (each of negligible resistance) 
and placed in vacuo 1 m apart, produces a force between of exactly 2 x 10~ 7 N per 
metre of length'. We will not employ this latter definition. 

The SI unit of the 'amount of substance' n is the mole. Curi- 

The SI unit of 'amount 
of substance' n is the 
mole (mol). 

ously, the SI General Conference on Weights and Measures only 
decided in 1971 to incorporate the mole into its basic set of funda- 
mental parameters, thereby filling an embarrassing loophole. The 
mole is the amount of substance in a system that contains as many 
elementary entities as does 0.012 kg (12 g) of carbon- 12. The amount of substance 
must be stated in terms of the elementary entities chosen, be they photons, electrons, 
protons, atoms, ions or molecules. 

The number of elementary entities in 1 mol is an experimentally determined quan- 
tity, and is called the 'Avogadro constant' L, which has the value 6.022 x 10 23 mol -1 . 
The Avogadro constant is also (incorrectly) called the 'Avogadro number'. It is 


Table 1.2 Several of the more common units that are not members of the Systeme 


Non-SI unit 


Conversion from non-SI to SI 




1 cal = 4.184 J 




1 A= 10 -10 m 




1 atm = 101 325 Pa 




1 bar = 10 5 Pa 



dm 3 

1 dm 3 = 1(T 3 m 3 

increasingly common to see the Avogadro constant given a different symbol than 
L. The most popular alternative symbol at present seems to be N&. 

Non-SI units 

It is important to be consistent with units when we start a calculation. An enormously 
expensive spacecraft crashed on the surface of the planet Mars in 1999 because a 
distance was calculated by a NASA scientist in terms of inches rather than centimetres. 
Several non-SI units persist in modern usage, the most common being listed in 
Table 1.2. A calculation performed wholly in terms of SI units will be self-consistent. 
Provided we know a suitable way to interchange between the SI and non-SI units, 
we can still employ our old non-SI favourites. 


In addition to the thermodynamic temperature T there is also the Celsius temperature 
f, defined as 

t = T-T 

where T = 273.15 K. 

Sometimes, to avoid confusion with the use of t as the symbol for time, the Greek 
symbol 9 (theta) is substituted for the Celsius temperature t instead. 

Throughout this book we adopt T to mean temperature. The context will make clear 
whether T is required to be in degrees Celsius or kelvin. Beware, though, that most 
formulae require the use of temperature in kelvin. 

Why is the SI unit of mass the kilogram? 

Multiples and the SI unit of mass m 

The definition of mass in the Systeme Internationale scheme departs from the stated 
aim of formulating a rigorous, self-consistent set of standards. The SI unit of 'mass' 



Table 1.3 Selection of a few physicochemical 
parameters that comprise combinations of the 
seven SI fundamental quantities 



SI units 







m 2 



kgm~ 3 



kg m s~ 2 



kg ITT 1 s~ 2 






m 3 

The SI unit of 'mass' m 
is the kilogram (kg). 

In the SI system, 1 g 
is defined as the mass 
of 5.02 x 10 22 atoms of 
carbon-12. This num- 
ber comes from L/12, 
where L is the Avo- 
gadro number. 

m is the 'kilogram'. Similar to the metre, the original SI standard 
of mass was a block of platinum metal in Sevres, near Paris, which 
weighted exactly 1 kg. The current SI definition is more compli- 
cated: because 12.000 g in the SI system represents exactly 1 mol 
of carbon-12, then 1 g is one-twelfth of a mole of carbon-12. 

The problem with the SI base unit being a kilogram is the 'kilo' 
part. The philosophical idea behind the SI system says any param- 
eter (physical, chemical, mechanical, etc.) can be derived from 
a suitable combination of the others. For example, the SI unit of 
velocity is metres per second (m s _1 ), which is made up of the two 
SI fundamental units of length (the metre) and time (the second). 
A few of these combinations are cited in Table 1.3. 

Why is 'the material of action so variable'? 

Writing variables and phrases 

The classical author Epictetus (ca 50-ca 138 ad) once said, 'The materials of action 
are variable, but the use we make of them should be constant'. How wise. 

When we build a house, we only require a certain number of 
building materials: say, bricks, tubes and window panes. The quan- 
tity surveyor in charge of the building project decides which materi- 
als are needed, and writes a quantity beside each on his order form: 
10000 bricks, 20 window panes, etc. Similarly, when we have a 
velocity, we have the units of 'm' and 's -1 ', and then quantify 
it, saying something like, 'The man ran fast, covering a distance 
of 10 metres per second'. By this means, any parameter is defined 
both qualitatively (in terms of its units) and quantitatively (in terms 
of a number). With symbols, we would write v = 10 ms -1 . 

We give the name 
'compound unit' to 
several units written 
together. We leave a 
space between each 
constituent unit when 
we write such a com- 
pound unit. 



A variable (mass, length, velocity, etc.) is written in a standard format, according 
to Equation (1.2): 

Variable or physicochemical quantity = number x units 


We sometimes call it a 'phrase'. Because some numbers are huge 
and others tiny, the SI system allows us a simple and convenient 
shorthand. We do not need to write out all the zeros, saying the 
velocity of light c is 300000000 ms" 1 : we can write it as c = 3 x 
10 8 ms -1 or as 0.3 Gms -1 , where the capital 'G' is a shorthand 
for 'giga', or 1 000000000. The symbol G (for giga) in this context 
is called a 'factor'. In effect, we are saying 300000000 ms -1 = 
0.3 Gms -1 . The standard factors are listed on pp. xxviii-xxxi. 

Most people find that writing 300000000 ms -1 is a bit long 
winded. Some people do not like writing simple factors such as G 
for giga, and prefer so-called scientific notation. In this style, we 
write a number followed by a factor expressed as ten raised to an 
appropriate power. The number above would be 3.0 x 10 8 ms -1 . 

Worked Example 1.2 Identify the variable, number, factor and unit 
in the phrase, 'energy = 12 kJmol -1 '. 

'Giga' comes from the 
Latin gigas, meaning 
'giant' or 'huge'. We 
also get the every- 
day words 'giant' and 
'gigantic' from this 

In physical chemistry, 
a 'factor' is a number 
by which we multiply 
the numerical value of 
a variable. Factors are 
usually employed with 
a shorthand notation. 



J mol 

Variable Number Factor Compound 



Variable - in simple mathematical 'phrases' such as this, we almost always write 

the variable on the left. A variable is a quantity whose value can be altered. 
Number - the easy part! It will be made up of numbers 1, 2, 3, . . . , 0. 
Factor - if we need a factor, it will always be written between the number and 

the units (compound or single). A comprehensive list of the simple factors is 

given on pp. xxviii-xxxi. 
Units - the units are always written on the right of a phrase such as this. There 

are two units here, joules (J) and moles (as 'mol~ ', in this case). We should 

leave a space between them. 

A factor is simply shorthand, and is dispensable. We could have dispensed with the 
factor and written the number differently, saying energy = 12000 Jmol~ . This same 
energy in scientific notation would be 12 x 10 3 Jmol~ . But units are not dispensable. 


SAQ 1.2 Identify the variable, number, factor and unit in the phrase, 
'length = 3.2 km'. 

1.3 Properties of gases and the gas laws 

Why do we see eddy patterns above a radiator? 

The effects of temperature on density 

The air around a hot radiator soon acquires heat. We explain this observation from 
the 'minus oneth law of thermodynamics' (see Section 1.1), since heat travels from 
hot to cold. 

The density of a gas depends quite strongly on its temperature, so hot air has 
a smaller density than does cold air; colder air is more dense than hot air. From 
everyday experience, we know that something is dense if it tries to drop, which is 
why a stone drops to the bottom of a pond and a coin sinks to the bottom of a pan of 
water. This relative motion occurs because both the stone and the coin have higher 
densities than does water, so they drop. Similarly, we are more dense than air and 
will drop if we fall off a roof. 

Just like the coin in water, cold air sinks because it is denser than warmer air. 
We sometimes see this situation stated as warm air 'displaces' the cold air, which 
subsequently takes its place. Alternatively, we say 'warm air rises', which explains 
why we place our clothes above a radiator to dry them, rather than below it. 

Light entering the room above the radiator passes through these pockets of warm 
air as they rise through colder air, and therefore passes through regions of different 
density. The rays of light bend in transit as they pass from region to region, much in 
the same way as light twists when it passes through a glass of water. We say the light 
is refracted. The eye responds to light, and interprets these refractions and twists as 
different intensities. 

So we see swirling eddy (or 'convective') patterns above a radiator because the 
density of air is a function of temperature. If all the air had the same temperature, then 
no such difference in density would exist, and hence we would see no refraction and 
no eddy currents - which is the case in the summer when the radiator is switched off. 
Then again, we can sometimes see a 'heat haze' above a hot road, which is caused 
by exactly the same phenomenon. 

Why does a hot-air balloon float? 

The effect of temperature on gas volume 

A hot-air balloon is one of the more graceful sights of summer. A vast floating 
ball, powered only by a small propane burner, seems to defy gravity as it floats 
effortlessly above the ground. But what is it causing the balloon to fly, despite its 
considerable weight? 



The small burner at the heart of the balloon heats the air within 
the canvas hood of the balloon. The densities of all materials - 
solid, liquid or gas - alter with temperature. Almost universally, 
we find the density p increases with cooling. Density p is defined 
as the ratio of mass m to volume V, according to 

'Density' p is defined as 
mass per unit volume. 

density p 

mass, m 

volume, V 


It is not reasonable to suppose the mass m of a gas changes by heating or cooling it 
(in the absence of chemical reactions, that is), so the changes in p caused by heating 
must have been caused by changes in volume V. On the other hand, if the volume 
were to decrease on heating, then the density would increase. 

So the reason why the balloon floats is because the air inside its voluminous hood 
has a lower density than the air outside. The exterior air, therefore, sinks lower 
than the less-dense air inside. And the sinking of the cold air and the rising of the 
warm air is effectively the same thing: it is movement of the one relative to the 
other, so the balloon floats above the ground. Conversely, the balloon descends back 
to earth when the air it contains cools to the same temperature as the air outside 
the hood. 

How was the absolute zero of temperature 

Charles's law 

J. A. C. Charles (1746-1823) was an aristocratic amateur scientist of the 18th century 
He already knew that the volume V of a gas increased with increasing temperature T 
and was determined to find a relationship between these variables. 
The law that now bears his name can be stated as, 'The ratio of 
volume and temperature for a fixed mass of gas remains constant', 
provided the external pressure is not altered. 
Stated mathematically, Charles demonstrated 




According to 'Charles's 
law', a linear relation- 
ship exists between 
V and 7" (at constant 
pressure p). 

where the value of the constant depends on both the amount and 
the identity of the gas. It also depends on the pressure, so the data 
are obtained at constant pressure p. 

This is one form of 'Charles's law'. (Charles's law is also called 
'Gay-Lussac's law'.) Alternatively, we could have multiplied both 
sides of Equation (1.4) by T, and rewritten it as 

constant x T 


A 'straight line' will 
always have an equa- 
tion of the type y = 
mx + c, where m is the 
gradient and c is the 
intercept on the /-axis 
(i.e. when the value of 
x = 0). 



Lord Kelvin (1824-1907) was a great thermodynamicist whom we shall meet quite 
often in these pages. He noticed how the relationship in Equation (1.5) resembles the 
equation of a straight line, i.e. takes the form 

y = mx + c 
observed gradient controlled constant 




except without an intercept, i.e. c = 0. Kelvin obtained good-quality data for the 
volume of a variety of gases as a function of temperature, and plotted graphs of 
volume V (as y) against temperature T (as x) for each; curiously, however, he was 
unable to draw a graph with a zero intercept for any of them. 

Kelvin then replotted his data, this time extrapolating each graph 
till the volume of the gas was zero, which he found to occur at a 
temperature of —273.15 °C; see Figure 1.5. He then devised a new 
temperature scale in which this, the coldest of temperatures, was 
the zero. He called it absolute zero, and each subsequent degree 
was equal to 1 °C. This new scale of temperature is now called 
the thermodynamic (or absolute) scale of temperature, and is also 
sometimes called the Kelvin scale. 
The relationship between temperatures T on the centigrade and 
the absolute temperature scales is given by 

Note: degrees in the 
Kelvin scale do not 
have the degree sym- 
bol. The units have a 
capital K, but the noun 
'kelvin' has a small 

T in °C= T in K- 273.15 


Equation (1.7) demonstrates how 1 °C = 1 K. 






200 300 

Temperature 77K 

Figure 1.5 A graph of the volume V of a gas (as y) against temperature T (as x) is linear. 
Extrapolating the gas's volume suggests its volume will be zero if the temperature is — 273.15°C 
(which we call K, or absolute zero) 



SAQ 1.3 What is the temperature T expressed in kelvin 
if the temperature is 30 °C? 

SAQ 1.4 What is the centigrade temperature correspon- 
ding to 287.2 K? 

SAQ 1.5 The data in the table below relate to gaseous 
helium. Demonstrate the linear relationship between the 
volume V and the temperature T. 

We divide each tem- 
perature, both kelvin 
and centigrade, by 
its respective unit 
to obtain a number, 
rather than the tem- 

Temperature 77K 
Volume V/m 3 










Charles's law is often expressed in a slightly different form than 
Equation (1.4), as 

Tx T 2 


which is generally regarded as superior to Equation (1.4) because 
we do not need to know the value of the constant. 

Equation (1.8) is also preferred in situations where the volume of 
a fixed amount of gas changes in response to temperature changes 
(but at constant pressure). The subscripts refer to the two situations; 
so, for example, the volume at temperature T\ is V\ and the volume 
at temperature T 2 is V 2 . 

SAQ 1.6 The gas inside a balloon has a volume l/i of 
1 dm 3 at 298 K. It is warmed to 350 K. What is the vol- 
ume following warming? Assume the pressure remained 

Note how we write 
the controlled vari- 
able along the top row 
of a table, with the 
observed following. (If 
the table is vertical, 
we write the controlled 
variable on the far left.) 

The subscripts written 
to the right of a variable 
are called 'descriptors'. 
They are always written 
as a subscript, because 
a superscripted number 
means a power, i.e. V 2 
means V x V. 

Why pressurize the contents of a gas canister? 

The effect of pressure on gas volume: Boyle's law 

It is easy to buy canisters of gas of many sizes, e.g. as fuel when we wish to camp 
in the country, or for a portable welding kit. The gas will be w-butane if the gas is 
for heating purposes, but might be oxygen or acetylene if the gas is to achieve the 
higher temperatures needed for welding. 

Typically, the components within the can are gaseous at most temperatures. The 
typical volume of an aerosol can is about 0.3 dm 3 (3 x 10~ 4 m 3 ), so it could contain 
very little gas if stored at normal pressure. But if we purchase a canister of gas and 
release its entire contents at once, the gas would occupy a volume similar that of 



Care: a small p indi- 
cates pressure, yet a 
big P is the symbol for 
the element phospho- 
rus. Similarly, a big V 
indicates volume and a 
small v is the symbol 
for velocity. 

an entire living room. To ensure the (small) can contains this (large) 
amount of gas, we pressurize it to increase its capacity. We see 
how volume and pressure are interrelated in a reciprocal way: the 
volume decreases as the pressure increases. 

Robert Boyle was the first to formulate a relationship between 
p and V. Boyle was a contemporary of the greatest scientist the 
world has ever seen, the 17th-century physicist Sir Isaac Newton. 
Boyle's law was discovered in 1660, and states 

pV = constant 


where the numerical value of the constant on the right-hand side of the equation 
depends on both the identity and amount of the gas, as well as its temperature T. 

Figure 1.6 shows a graph of pressure p (as y) against volume V 
(as x) for 1 mol of neon gas. There are several curves, each repre- 
senting data obtained at a different temperature. The temperature 
per curve was constant, so we call each curve an isotherm. The 
word isotherm has two Greek roots: iso means 'same' and thermo 
means temperature or energy. An isotherm therefore means at the 
same energy. 

The actual shape of the curves in Figure 1.6 are those of recipro- 
cals. We can prove this mathematical form if we divide both sides 
of Equation (1.9) by V, which yields 

An 'isotherm' is a line 
on a graph represent- 
ing values of a variable 
obtained at constant 

'Reciprocal' means to 
turn a fraction upside 
down. X can be thought 
of as 'Xh- 1', so its 
reciprocal is 1/X (i.e. 
1- X). 

p = — x constant 


Figure 1.7 shows a graph of volume p (as y) against 1/volume 
V (as x), and has been constructed with the same data as used for 

100 000 

80 000 

"S. 60 000 


g 40 000 


20 000 

Volume V/m 3 

Figure 1.6 Graph of pressure p (as y) against volume V (as x) for 1 mol of an ideal gas as a 
function of temperature: ( ) 200 K; ( ) 400 K; ( ) 600 K; ( ) 800 K 




1 .00E+05 


40 60 

1/(l/7m 3 ) 

Figure 1.7 Graph of pressure p (as y) against reciprocal volume 1 -f- V (as x) for 1 mol of 
an ideal gas as a function of temperature. The data are the same as those from Figure 1.6. The 
temperatures are indicated. We need to appreciate how plotting the same data on a different set of 
axes yields a linear graph, thereby allowing us to formulate a relationship between p and 1 -f- V 

Figure 1.6. Each of the lines on the graph is now linear. Again, we find these data are 
temperature dependent, so each has the gradient of the respective value of 'constant'. 

At constant temperature T, an increase in pressure (so pi > 
Pi) causes a decrease in volume (so V2 < ^i)- This observation 
explains why the graph for the gas at the higher temperatures has 
a smaller value for the constant. 

An alternative way of writing Equation (1.9) is 

P\V\ = P2V2 


At constant temper- 
ature T, an increase 
in pressure (p 2 > Pi) 
causes a decrease in 
volume (V 2 < l/i). 

SAQ 1.7 The usual choice of propellant within an aerosol of air freshener 
is propane gas. What is the volume of propane following compression, if 
1 dm 3 of gaseous propane is compressed from a pressure of 1 atm to a 
pressure of 2.5 atm? Assume the temperature is kept constant during the 

Why does thunder accompany lightning? 

Effect of changing both temperature and pressure on gas 

Lightning is one of the most impressive and yet frightening manifestations of nature. 
It reminds us just how powerful nature can be. 


'Experiential' means 
the way we notice 
something exists fol- 
lowing an experience 
or sensation. 

Lightning is quite a simple phenomenon. Just before a storm 
breaks, perhaps following a period of hot, fine weather, we often 
note how the air feels 'tense'. In fact, we are expressing an expe- 
riential truth: the air contains a great number of ions - charged 
particles. The existence of a large charge on the Earth is mirrored 
by a large charge in the upper atmosphere. The only difference 
between these two charges is that the Earth bears a positive charge and the atmosphere 
bears a negative charge. 

Accumulation of a charge difference between the Earth and the upper atmosphere 
cannot proceed indefinitely. The charges must eventually equalize somehow: in prac- 
tice, negative charge in the upper atmosphere passes through the air to neutralize 
the positive charge on the Earth. The way we see this charge conducted between 
the Earth and the sky is lightning: in effect, air is ionized to make it a conductor, 
allowing electrons in the clouds and upper atmosphere to conduct through the air 
to the Earth's surface. This movement of electrical charge is a current, which we 
see as lightning. Incidentally, ionized air emits light, which explains why we see 
lightning (see Chapter 9). Lightning comprises a massive amount of energy, so the 
local air through which it conducts tends to heat up to as much as a few thousand 
degrees centigrade. 

And we have already seen how air expands when warmed, e.g. as described math- 
ematically by Charles's law (Equation (1.6)). In fact, the air through which the 
lightning passes increases in volume to an almost unbelievable extent because of 
its rise in temperature. And the expansion is very rapid. 

SAQ 1.8 Show, using the version of Charles's law in Equation (1.8), how 
a rise in temperature from 330 K to 3300 K is accompanied by a tenfold 
increase in volume. 

We hear the sensation of sound when the ear drum is moved by compression waves 
travelling through the air; we hear people because their speech is propagated by subtle 
pressure changes in the surrounding air. In a similar way, the huge increase in air 
volume is caused by huge changes in air pressure, itself manifested as sound: we hear 
the thunder caused by the air expanding, itself in response to lightning. 

And the reason why we see the lightning first and hear the thunder later is because 
light travels faster than sound. The reason why thunder accompanies lightning, then, 
is because pressure p, volume V and temperature T are interrelated. 

How does a bubble-jet printer work? 

The ideal-gas equation 

A bubble-jet printer is one of the more useful and versatile inventions of the last 
decade. The active component of the printer is the 'head' through which liquid ink 
passes before striking the page. The head moves from side to side over the page. When 



the 'head' is positioned above a part of the page to which an image is required, the 
computer tells the head to eject a tiny bubble of ink. This jet of ink strikes the page 
to leave an indelible image. We have printing. 

The head is commonly about an inch wide, and consists of a row of hundreds of 
tiny pores (or 'capillaries'), each connecting the ink reservoir (the cartridge) and the 
page. The signals from the computer are different for each pore, allowing different 
parts of the page to receive ink at different times. By this method, images or letters 
are formed by the printer. 

The pores are the really clever bit of the head. Half-way along each pore is a 
minute heater surrounded by a small pocket of air. In front of the heater is a small 
bubble of ink, and behind it is the circuitry of the printer, ultimately connecting the 
heater to the computer. One such capillary is shown schematically in Figure 1.8. 

Just before the computer instructs the printer to eject a bubble of ink, the heater 
is activated, causing the air pocket to increase in temperature T at quite a rapid 
rate. The temperature increase causes the air to expand to a greater volume V. This 
greater volume increases the pressure p within the air pocket. The enhanced air 
pressure p is sufficient to eject the ink bubble from the pore and onto the page. This 
pressure-activated ejection is similar to spitting. 

This ejection of ink from a bubble-jet printer ingeniously utilizes the interconnect- 
edness of pressure p, volume V and temperature T. Experiments with simple gases 
show how p, T and V are related by the relation 

P V 

= constant 


which should remind us of both Boyle's law and Charles's law. 

Inkjet nozzle 

t> 5 us 

t~ 10 us 


Figure 1.8 Schematic diagram of a capillary (one of hundreds) within the printing 'head' of a 
bubble-jet printer. The resistor heats a small portion of solution, which boils thereby increasing the 
pressure. Bubbles form within 5 (is of resistance heating; after 10 (jls the micro-bubbles coalesce 
to force liquid from the aperture; and a bubble is ejected a further 10 p,s later. The ejected bubble 
impinges on the paper moments afterwards to form a written image. Reproduced by permission of 


If there is exactly 1 mol of gas, the pressure is expressed in pascals (Pa), the 
temperature is in kelvin and the volume is in cubic metres (both SI units), then the 
value of the constant is 8.314 JK _1 mol -1 . We call it the gas constant and give it 
the symbol R. (Some old books may call R the 'universal gas constant', 'molar gas 
constant' or just 'the gas constant'. You will find a discussion about R on p. 54) 
More generally, Equation (1.12) is rewritten as 

Equation (1.13) tells us 
the constant in Boyle's 
law is y nRT' and the 
(different) constant in 
Charles's law is 

P V = nRT (1.13) 

where n is the number of moles of gas. Equation (1.13) is called 
the ideal-gas equation (or, sometimes, in older books the 'universal 
gas equation'). The word 'ideal' here usually suggests that the gas 
in question obeys Equation (1.13). 

Worked Example 1.3 What is the volume of 1 mol of gas at a room temperature of 
25 °C at an atmospheric pressure of 10 5 Pa? 

First, we convert the data into the correct SI units. In this example, only the temperature 
needs to be converted. From Equation (1.7), the temperature is 298 K. 

Secondly, we rearrange Equation (1.13) to make V the subject, by dividing both sides 
by p: 


V = 

and then insert values: 


1 mol x 8.314 JKT 1 mok 1 x 298 K 
10 5 Pa 

So the volume V = 0.0248 m 3 . 

If we remember how there are 1000 dm in 1 m 3 , we see how 1 mol of gas at room 
temperature and standard pressure has a volume of 24.8 dm . 

SAQ 1.9 2 mol of gas occupy a volume V = 0.4 m 3 at a temperature 
T — 330 K. What is the pressure p of the gas? 

An alternative form of Equation (1.13) is given as 

piVi p 2 V 2 

T 2 


and is used when we have to start with a constant number of moles of gas n housed in a 
volume V\. Its initial pressure is p\ when the temperature is T\. Changing one variable 
causes at least one of the two to change. We say the new temperature is T 2 , the new 



pressure is p2 and the new volume is V2. Equation (1.14) then holds provided the number 
of moles does not vary. 

Worked Example 1.4 Nitrogen gas is housed in a sealed, hollow cylinder at a pressure 
of 10 5 Pa. Its temperature is 300 K and its volume is 30 dm . The volume within the 
cylinder is increased to 45 dm 3 , and the temperature is increased at the same time to 
310 K. What is the new pressure, />2? 

We first rearrange Equation (1.14) to make the unknown volume p 2 the subject, writing 

PiViT 2 


TiV 2 

We insert values into the rearranged equation: 

10 5 Pax 30 dm 3 x 310 K 

Pi = 

300 K x 45 dm 3 

Note how the units of 
volume cancel, mean- 
ing we can employ any 
unit of volume provided 
the units of \A and V 2 
are the same. 

so P2 — 0.69 x 10 5 Pa. The answer demonstrates how the pressure drops by about a third 
on expansion. 

SAQ 1.10 The pressure of some oxygen gas is doubled from 1.2 x 10 5 Pa 
to 2.4 x 10 5 Pa. At the same time, the volume of the gas is decreased 
from 34 dm 3 to 29 dm 3 . What is the new temperature T2 if the initial 
temperature 7~i was 298 K? 

Justification Box 1.1 

We start with n moles of gas at a temperature T\ , housed in a volume V\ at a pressure 
of p\ . Without changing the amount of material, we change the volume and temperature 
to V2 and T 2 respectively, therefore causing the pressure to change to p 2 . 

The number of moles remains unaltered, so we rearrange Equation (1.13) to make n 
the subject: 

n = p l V l ~RT l 

Similarly, the same number of moles n under the second set of conditions is 

n = p 2 V 2 + RT 2 

Again, although we changed the physical conditions, the number of moles n remains 
constant, so these two equations must be the same. We say 


RT 2 


As the value of R (the gas 
multiplying both sides by R 

constant) does 
to obtain 



not vary, 

p 2 V 2 
T 2 







which is Equation (1.14). 

What causes pressure? 

Motion of particles in the gas phase 

The question, 'What is pressure?' is another odd question, but is not too difficult 
to answer. 

The constituent particles of a substance each have energy. In practice, the energy 
is manifested as kinetic energy - the energy of movement - and explains why all 
molecules and atoms move continually as an expression of that kinetic energy. This 
energy decreases as the temperature decreases. The particles only stop moving when 
cooled to a temperature of absolute zero: K or —273.15 °C. 

The particles are not free to move throughout a solid substance, but can vibrate 
about their mean position. The frequency and amplitude of such vibration increases as 
the temperature rises. In a liquid, lateral motion of the particles is possible, with the 
motion becoming faster as the temperature increases. We call this energy translational 
energy. Furthermore, as the particles acquire energy with increased temperature, so 
the interactions (see Chapter 2) between the particles become comparatively smaller, 
thereby decreasing the viscosity of the liquid and further facilitating rapid motion of 
the particles. When the interactions become negligible (comparatively), the particles 
can break free and become gaseous. 

And each particle in the gaseous state can move at amazingly high speeds; indeed, 
they are often supersonic. For example, an average atom of helium travels at a mean 
speed of 1204 ms" 1 at 273.15 K. Table 1.4 lists the mean speeds of a few other gas 
molecules at 273.15 K. Notice how heavier molecules travel more slowly, so carbon 
dioxide has a mean speed of 363 ms -1 at the same temperature. This high speed 
of atomic and molecular gases as they move is a manifestation of 
their enormous kinetic energy. It would not be possible to travel 
so fast in a liquid or solid because they are so much denser - we 
call them condensed phases. 

The separation between each particle in gas is immense, and 
usually thousands of times greater than the diameter of a single 
gas particle. In fact, more than 99 per cent of a gas's volume 
is empty space. The simple calculation in Worked Example 1.5 
demonstrates this truth. 

The gas particles are 
widely separated. 

Particles of gas travel 
fast and in straight 
lines, unless they col- 


Table 1.4 The average speeds of gas 
molecules at 273.15 K, given in order of 
increasing molecular mass. The speeds 
c are in fact root-mean-square speeds, 
obtained by squaring each velocity, tak- 
ing their mean and then taking the square 
root of the sum 


Speed c/m s ' 

Monatomic gases 







Diatomic gases 









Carbon monoxide 




Polyatomic gases 







Carbon dioxide 




Worked Example 1.5 What is the molar volume of neon, assuming it to be a straight- 
forward solid? 

We must first note how the neon must be extremely cold if it is to be 
a solid - probably no colder than about 20 K. 

We know that the radius of a neon atom from tables of X-ray crystal- 
lographic data is about 10 -10 m, so the volume of one atom (from the 

The 'molar volume' is 
the name we give to 
the volume 'per mole'. 

equation of a sphere, V — ^rcr 3 ) is 4.2 x 10 -30 m 3 . If we assume the 
neon to be a simple solid, then 1 mol of neon would occupy a volume of 4.2 x 10 -30 m 3 
per atom x 6.022 x 10 23 atoms per mole = 2.5 x 10 -6 m 3 mol -1 . This volume represents 
2.5 cm 3 mol -1 . 

A volume of 2.5 cm 3 mol -1 is clearly much smaller than the value we calculated 
earlier in Worked Example 1.3 with the ideal-gas equation, Equation (1.13). It is 
also smaller than the volume of solid neon made in a cryostat, 
suggesting the atoms in a solid are also separated by much empty 
space, albeit not so widely separated as in a gas. 

In summary, we realize how each particle of gas has enor- 
mous kinetic energy and are separated widely. Yet, like popcorn 
in a popcorn maker, these particles cannot be classed as wholly 
independent, one from another, because they collide. They collide 

By corollary, if the gas 
particles move fast and 
the gas is ideal, the gas 
particles must travel in 
straight lines between 



Newton's first law 
states that every action 
has an equal but oppo- 
site reaction. His sec- 
ond law relates the 
force acting on an 
object to the product of 
its mass multiplied by 
its acceleration. 

The pressure of a 
gas is a 'macroscopic' 
manifestation of the 
'microscopic' gas parti- 
cles colliding with the 
internal walls of the 

The surface area inside 
a cylinder of radius r 
and height h is 2nrh. 
Don't forget to include 
the areas of the two 
ends, each of which is 
Ttr 2 . 

firstly with each other, and secondly with the internal walls of the 
container they occupy. 

Just like the walls in a squash court, against which squash balls 
continually bounce, the walls of the gas container experience a 
force each time a gas particle collides with them. From Newton's 
laws of motion, the force acting on the wall due to this incessant 
collision of gas particles is equal and opposite to the force applied 
to it. If it were not so, then the gas particles would not bounce 
following a collision, but instead would go through the wall. 

We see how each collision between a gas particle and the internal 
walls of the container causes the same result as if we had applied a 
force to it. If we call the area of the container wall A and give the 
symbol F to the sum of the forces of all the particles in the gas, then 
the pressure p exerted by the gas-particle collisions is given by 

pressure, p = 

force, F 
area, A 


In summary, the pressure caused by a container housing a gas is 
simply a manifestation of the particles moving fast and colliding 
with the container walls. 

SAQ 1.11 A cylindrical can contains gas. Its height is 
30 cm and its internal diameter is 3 cm. It contains gas 
at a pressure of 5 x 10 5 Pa. First calculate the area of the 
cylinder walls (you will need to know that 1 m = 100 cm, 
so 1 m 2 = 10 4 cm 2 ), and then calculate the force neces- 
sary to generate this pressure. 


A popular misconception says a molecule in the gas phase travels faster than when in 
a liquid. In fact, the molecular velocities will be the same in the gas and liquid phases 
if the temperatures are the same. Molecules only appear to travel slower in a liquid 
because of the large number of collisions between its particles, causing the overall 
distance travelled per unit time to be quite short. 

Why is it unwise to incinerate an empty can of air 

The molecular basis of the gas laws 

The writing printed on the side of a can of air freshener contains much information. 
Firstly, it cites the usual sort of advertising prose, probably saying it's a better product 



CFC stands for chlo- 
rofluorocarbon. Most 
CFCs have now been 
banned because of 
their ability to dam- 
age the ozone layer in 
the upper atmosphere. 

than anyone else's, and smells nicer. Few people seem to bother reading these bits. 
But in most countries, the law says the label on the can should also gives details of 
the can's contents, both in terms of the net mass of air freshener it contains and also 
perhaps a few details concerning its chemical composition. Finally, a few words of 
instruction say how to dispose safely of the can. In this context, the 
usual phrase printed on the can is, 'Do not incinerate, even when 
empty'. But why? 

It is common for the can to contain a propellant in addition to the 
actual components of the air freshener mixture. Commonly, butane 
or propane are chosen for this purpose, although CFCs were the 
favoured choice in the recent past. 

Such a can is thrown away when it contains no more air fresh- 
ener, although it certainly still contains much propellant. Inciner- 
ation of the can leads to an increase in the kinetic energy of the 
remaining propellant molecules, causing them to move faster and 
faster. And as their kinetic energy increases, so the frequency with 
which they strike the internal walls of the can increases. The force 
of each collision also increases. In fact, we rediscover the ideal gas 
equation, Equation (1.13), and say that the pressure of the gas (in 
a constant-volume system) increases in proportion to any increase 
in its temperature. In consequence, we should not incinerate an old 
can of air freshener because the internal pressure of any residual 
propellant increases hugely and the can explodes. Also note the 
additional scope for injury afforded by propane' s flammability. 

Pressure increases with 
increasing temperature 
because the collisions 
between the gas parti- 
cles and the container 
wall are more ener- 
getic and occur more 

1.4 Further thoughts on energy 

Why is the room warm? 

The energy of room temperature 

Imagine coming into a nice, warm room after walking outside in the snow. We 
instantly feel warmer, because the room is warmer. But what exactly is the energy 
content of the room? Stated another way, how much energy do we get from the air 
in the room by virtue of it being at its own particular temperature? 

For simplicity, we will consider only the molecules of gas. Each molecule of gas 
will have kinetic energy (the energy of movement) unless the temperature is absolute 
zero. This energy may be transferred through inelastic molecules collisions. But how 
much kinetic energy does the gas have? 

At a temperature T, 1 mol of gas has a kinetic energy of jRT, where T is the ther- 
modynamic temperature and R is the gas constant. This energy is directly proportional 
to the thermodynamic temperature, explaining why we occasionally call the kinetic 
energy 'thermal motion energy'. This simple relationship says that temperature is 
merely a measure of the average kinetic energy of gas molecules moving chaotically. 



It is important to appreciate that this energy relates to the average energy of 1 mol 
of gas molecules. The concept of temperature has no meaning when considering a 
single molecule or atom. For example, the velocity (and hence the kinetic energy) 
of a single particle changes with time, so in principle its temperature also changes. 
Temperature only acquires any thermodynamic meaning when we consider average 
velocities for a large number of particles. 

Provided we know the temperature of the gas, we know its energy - the energy it 
has simply by existing at the temperature T. 

Worked Example 1.6 What is the energy of 1 mol of gas in a warm room at 310 K? 
The energy per mole is | x R x T; so, inserting values, energy = 

The Voom energy' §R7 
derives from the kinetic 
(movement) energy of 
a gas or material. 

| x 8.314 JK" 1 mol" 1 x 310 K. 

Energy = 3866 JmoP 1 % 3.9 kJmok 

The molar energy of these molecules is about 4 kJ moP ' , which is 
extremely slight compared with the energy of the bonds connecting 

the respective atoms within a molecule (see Chapters 2 and 3). There is little chance of 

this room energy causing bonds to break or form. 

SAQ 1.12 What is the room energy per mole on a cold winter's day, at 
-8°C (265 K)? 

What do we mean by 'room temperature'? 

Standard temperature and pressure 

Suppose two scientists work on the same research project, but one resides in the far 
north of the Arctic Circle and the other lives near the equator. Even if everything 
else is the same - such as the air pressure, the source of the chemicals and the 
manufacturers of the equipment - the difference between the temperatures in the two 
laboratories will cause their results to differ widely. For example, the 'room energy' 
RT will differ. One scientist will not be able to repeat the experiments of the other, 
which is always bad science. 

An experiment should always be performed at known tempera- 
ture. Furthermore, the temperature should be constant throughout 
the course of the experiment, and should be noted in the labora- 
tory notebook. 

But to enable complete consistency, we devise what is called a set 
of standard conditions. 'Standard pressure' is given the symbol p & , 
and has a value of 10 5 Pa. We sometimes call it '1 bar'. Atmospheric pressure has a 
value of 101 325 Pa, so it is larger than p e . We often give atmospheric pressure the 
symbol 'atm'. 

An experiment should 
always be performed at 
a known, fixed temper- 



'Standard temperature' has the value of 298 K exactly, which 
equates to just below 25 °C. If both the pressure and the temper- 
ature are maintained at these standard conditions, then we say the 
measurement was performed at 'standard temperature and pres- 
sure', which is universally abbreviated to 's.t.p.' If the scientists 
at the equator and the Arctic Circle perform their work in thermo- 
statically controlled rooms, both at s.t.p., then the results of their 
experiments will be identical. 

A 'thermostat' is a 
device for maintain- 
ing a temperature. 
Thermo is Greek for 
'energy' or 'tempera- 
ture', and 'stat' derives 
from the Greek root 
statikos, meaning 'to 
stand', i.e. not move or 

Why do we get w 'armed-through in front of a fire, 
rather than just our skins? 

The Maxwell- Boltzmann distribution of energies 


If no heat was distributed, then our faces and those parts closest to the fire 
would quickly become unbearably hot, while the remainder of our flesh would 
continue to feel cold. Heat conducts through the body principally by the fire 
warming the blood on the surface of the skin, which is then 
pumped to other parts of the body through the circulatory 
system. The energy in the warmed blood is distributed within 
cooler, internal tissues. 

It is important to note how the heat energy is distributed around 
the body, i.e. shared and equalized. Nature does not like diversity 
in terms of energetic content, and provides many mechanisms by 
which the energy can be 'shared'. We shall discuss this aspect of 
thermochemistry in depth within Chapter 4. 

We can be certain that molecules do not each have the same 
energy, but a distribution of energies. The graph in Figure 1.9 
concerns the energies in a body. The x-axis gives the range of 
energies possible, and the y-axis represents the number of particles 
in the body (molecules, atoms, etc.) having that energy. The graph 
clearly shows how few particles possess a large energy and how 
a few particles have a tiny energy, but the majority have lesser 
energies. We call this spread of energies the 'Maxwell-Boltzmann 

All speeds are found at all temperatures, but more molecules 
travel at faster speeds at the higher temperatures. 

The distribution law depicted in Figure 1.9 may be modelled 
mathematically, to describe the proportions of molecules of molar 
mass M with energies E in the range E to E + AE that exist in 

We often see this rela- 
tionship called merely 
the 'Boltzmann dis- 
tribution', after the 
Austrian Physicist Lud- 
wig Boltzmann (1844- 
1906), who played a 
pivotal role in marrying 
thermodynamics with 
statistical and molecu- 
lar physics. 

The thermodynamic 
temperature is the sole 
variable required to 
define the Maxwell- 
Boltzmann distribution: 
raising the temperature 
increases the spread of 






Velocity Wm s 

Figure 1.9 Molecular energies follow the Maxwell-Boltzmann distribution: energy distribution 
of nitrogen molecules (as y) as a function of the kinetic energy, expressed as a molecular velocity 
(as x). Note the effect of raising the temperature, with the curve becoming flatter and the maximum 
shifting to a higher energy 

thermal equilibrium with each other at a temperature T: 

f(E) = An 




E exp 

Ms l \ 
2RT ) 


where / on the far left indicates the 'function' that is to be applied to the variable E : 
the mathematical nature of this function is given by the right-hand side of the equation. 
So, in summary, we feel warmer in front of a fire because energy is distributed 
between those parts facing the flames and the more hidden tissues within. 


Introducing interactions 
and bonds 


We look first at deviations from the ideal-gas equation, caused by inter-particle 
interactions. Having described induced dipoles (and hydrogen bonds) the interaction 
strengths are quantified in terms of the van der Waals and virial equations of state. 

Next, formal bonds are described, both covalent (with electrons shared between 
participating atoms) and ionic (in which electrons are swapped to form charged ions; 
these ions subsequently associate in response to electrostatic forces). Several under- 
lying factors are expounded, such as ionization energy / and electron affinity E^. 
The energy changes occurring while forming these interactions are alluded to, but are 
treated properly in Chapter 3. 

2.1 Physical and molecular interactions 

What is 'dry ice'? 

Deviations from the ideal-gas equation 

We call solid carbon dioxide (CO2) 'dry ice'. To the eye, it looks 

just like normal ice, although it sometimes appears to 'smoke'; 

see below. Carbon dioxide is a gas at room temperature and only 

solidifies (at atmospheric pressure) if the temperature drops to about 

—78 °C or less, so we make dry ice by cooling gaseous CO2 below 

its freezing temperature. We call it dry ice because, unlike normal 

ice made with water, warming it above its melting temperature 

leaves no puddle of liquid, because the CO2 converts directly to a gas. We say 

it sublimes. 

Gases become denser as we lower their temperature. If CO2 was still a gas at 
— 90 °C, then its molar volume would be 15 200 cm 3 . In fact, the molar volume of 

Substances sublime if 
they pass directly from 
a solid to form a gas 
without being a liquid 
as an intermediate 
phase; see Chapter 5. 



solid CO2 at this temperature is about 30 cm 3 . We deduce that CO2 does not obey 
the ideal-gas equation (Equation (1.13)) below its freezing temperature, for the very 
obvious reason that it is no longer a gas. 

SAQ 2.1 Show that the volume of 1 mol of C0 2 would be 15 200 cm 3 at 
p^ and -90 °C (183 K). [Hint: use the ideal-gas equation. To express this 
answer in cubic metres, you will need to remember that 1 m 3 = 10 3 dm 3 
and 10 6 cm 3 .] 

Although solidifying CO2 is an extreme example, it does show how deviations 
from the ideal-gas equation occur. 

How is ammonia liquefied? 

We sometimes call 
a solid or a liquid a 
'condensed phase'. 

'Intermolecular' means 
'between molecules'. 

Intermolecular forces 

Compressing ammonia gas under high pressure forces the molecules into close 
proximity. In a normal gas, the separation between each molecule is generally 
large - approximately 1000 molecular diameters is a good 
generalization. By contrast, the separation between the molecules in 
a condensed phase (solid or liquid) is more likely to be one to two 
molecular diameters, thereby explaining why the molar volume of 
a solid or liquid is so much smaller than the molar volume of a gas. 
As a direct consequence of the large intermolecular separations, 
we can safely say no interactions form between the molecules in 
ammonia gas. The molecules are simply too far apart. We saw 
in the previous chapter how the property known as pressure is 
a macroscopic manifestation of the microscopic collisions occurring between gas 
particles and, say, a solid object such as a container's walls. But the gas particles can 
also strike each other on the same microscopic scale: we say the resultant interactions 
between molecules are intermolecular. 

Intermolecular interactions only operate over relatively short dis- 
tances, so we assume that, under normal conditions, each molecule 
in a gas is wholly unaffected by all the others. By contrast, when 
the gas is compressed and the particles come to within two or three 
molecular diameters of each other, they start to 'notice' each other. 
We say the outer-shell electrons on an atom are perturbed by the 
charges of the electrons on adjacent atoms, causing an interaction. 
We call these interactions bonds, even though they may be too 
weak to be formal bonds such as those permanently connecting 
the atoms or ions in a molecule. 
The intermolecular interactions between molecules of gas are generally attractive; 
so, by way of response, we find that, once atoms are close enough to interact, they 
prefer to remain close - indeed, once a tentative interaction forms, the atoms or 

A 'formal bond' involves 
the permanent involve- 
ment of electrons in 
covalent or ionic bonds; 
see p. 64. Interactions 
between molecules 
in a compressed gas 
are temporary. 



Translational motion 
is movement through 
space, rather than a 
vibration about a mean 
point or a rotation 
about an axis. 

molecules generally draw themselves closer, which itself makes 
the interaction stronger. We see a simple analogy with everyday 
magnets: once two magnets are brought close enough to induce an 
interaction, we feel the attractive force dragging them closer still. 

As soon as the particles of a gas attract, the inertia of the 
aggregate species increases, thereby slowing down all translational 
motion. And slower particles, such as these aggregates, are an eas- 
ier target for further collisions than fast-moving gas atoms and 
molecules. The same principle explains why it is impossible to 
catch someone who is running very fast during a playground game 
of 'tig'. Only when the runners tire and slow down can they 
be caught. In practice, as soon as an aggregate forms, we find 
that other gas particles soon adhere to it, causing eventual coa- 
lescence and the formation of a droplet of liquid. We say that 
nucleation occurs. 

With the same reasoning as that above, we can force the molecules of ammonia 
still closer together by applying a yet larger pressure, to form a denser state such as 
a solid. 

Formation of an ag- 
gregate facilitates fur- 
ther coalescence (even- 
tually forming a con- 
densed phase). We say 
'nucleation' occurs. 

Why does steam condense in a cold bathroom? 

Elastic and inelastic collisions 

In the previous example, we looked at the interactions induced when changing the 
external pressure, forcing the molecules into close proximity. We look here at the 
effects of changing the temperature. 

A bathroom mirror is usually colder than the temperature of the steam rising from 
a hot bath. Each molecule of steam (gaseous water) has an enormous energy, which 
comes ultimately from the boiler that heats the water. The particles of steam would 
remain as liquid if they had less energy. In practice, particles evaporate from the 
bath to form energetic molecules of steam. We see this energy as kinetic energy, so 
the particles move fast (see p. 30). The typical speeds at which gas particles move 
make it inevitable that steam molecules will collide with the mirror. 
We say such a collision is elastic if no energy transfers during the 
collision between the gas particle and the mirror; but if energy does 
transfer - and it usually does - we say the collision is inelastic. 

The energy transferred during an inelastic collision passes from 
the hot molecule of steam to the cooler mirror. This energy flows 
in this direction because the steam initially possessed more energy 
per molecule than the mirror as a consequence of its higher temperature. It is merely a 
manifestation of the minus-oneth law of thermodynamics, as discussed in Chapter 1. 

But there are consequences to the collisions being inelastic: the molecules of steam 
have less energy following the collision because some of their energy has transferred. 

No energy is exchang- 
ed during an 'elastic' 
collision, but energy is 
exchanged during an 
'inelastic' collision. 



Energy is never lost 
or gained, only trans- 
ferred or converted; 
see Chapter 3. 

We generally assume 
that all particles in an 
ideal gas do not inter- 
act, meaning that the 
gas obeys the ideal-gas 
equation. This assump- 
tion is sometimes poor. 

We perceive this lower energy as a cooler temperature, meaning 
that the water vapour in a steam-filled bathroom will cool down; 
conversely, the mirror (and walls) become warmer as they receive 
the energy that was previously possessed by the steam. These 
changes in the temperatures of gas and mirror occur in a com- 
plementary sense, so no energy is gained or lost. 

These changes in temperature represent a macroscopic proof that 
microscopic processes do occur. Indeed, it is difficult to envisage 
a transfer of energy between the gas particles with the cold mirror 
without these microscopic interactions. 

We spent quite a lot of time looking at the concept of an ideal gas 
in Chapter 1 . The simplest definition of an ideal gas is that it obeys 
the ideal-gas equation (Equation (1.13)). Most gases can be con- 
sidered as ideal most of the time. The most common cause of a gas 
disobeying the ideal-gas equation is the formation of interactions, 
and the results of intermolecular collisions. 

How does a liquid-crystal display work? 

Electronegativity and electropositivity 

Liquid crystals are organic compounds that exhibit properties somewhere between 
those of a solid crystal and a liquid. Compounds I and II in Figure 2.1 both form 
liquid crystals at room temperature. 

We observe that liquid crystals can flow like any other viscous liquid, but they also 
possess some of the properties of crystalline solids, such as physical order, rather 
than random chaos. Unlike most other liquids, liquid crystals have some properties 


CiqH 2 


Figure 2.1 Compounds that form room-temperature liquid crystals 



Light transmitted 

Lower polarizer blocks 

the transmission 

of light 

No voltage applied Voltage applied 

Figure 2.2 The transparent electrodes in an LCD are coated with crossed polarizers. The liquid 
crystals (depicted as slender lozenges) form helices, thereby 'guiding' polarized light from the upper 
electrode through the LCD, enabling transmission through to the lower polarizer. This is why the 
display has no colour. The helical structure is destroyed when a voltage is applied, because the polar 
liquid crystals align with the electrodes' field. No light can transmit, so the display looks black 

A physicist would say 
the liquid crystal 
adopted a twisted 
nematic structure. 

that depend on the direction of measurement, because of the alignment of their long, 
rod-like structures. 

In a liquid-crystal display (LCD) device, the two electrodes 
are parallel and separated by a thin layer of liquid crystal (see 
Figure 2.2). The liquid crystals in this layer naturally adopt a heli- 
cal structure. 

Light can be represented as a transverse electromagnetic wave 
made up of fluctuating electric and magnetic fields, moving in 
mutually perpendicular directions (see Chapter 9). Ordinary light is made up of waves 
that fluctuate at all possible angles, which normally cannot be separated. A polarizer 
is a material that allows only light with a specific angle of vibration to transmit. 
We place a light polarizer on one side of either transparent electrode in the LCD, 
each similar to one lens in a pair of polaroid sunglasses. The helix of the liquid 
crystal twists the polarized light as it transmits through the LCD, guiding it from the 
upper polarizer and allowing it unhindered passage through the 'sandwich' and lower 
polarizer. The transmitting state of an LCD (at zero voltage) is thus 'clear'. 

Applying a voltage to a pixel within the cell causes the mole- 
cules to move, aligning themselves parallel with the electric field 
imparted by the electrodes. This realignment destroys the helical 
structure, precluding the unhindered transmission of light, and the 
display appears black. 

Molecules of this type are influenced by an external electric 
field because they possess a dipole: one end of the molecule is 
electron withdrawing while the other is electron attracting, with the result that one 
end possesses a higher electron density than the other. As a result, the molecule 
behaves much like a miniature bar magnet. Applying a voltage between the two 

'Pixel' is short for 'pic- 
ture element'. An LCD 
image comprises many 
thousands of pixels. 



electrodes of the LCD causes the 'magnet' to reorientate in just the same way as a 
magnet moves when another magnet is brought close to it. 

These dipoles form because of the way parts of the molecule 
attract electrons to differing extents. The power of an element 
(when part of a compound) to attract electrons is termed its 'elec- 
tronegativity' x- Highly electron-attracting atoms tend to exert 
control over the outer, valence electrons of adjacent atoms. The 
most electronegative elements are those placed near the right-hand 
side of the periodic table, such as oxygen and sulphur in Group 
VI(b) or the halogens in Group VII(b). 

There have been a large number of attempts to quantify elec- 
tronegativities x » either theoretically or semi-empirically, but none 
has been wholly successful. All the better methods rely on bond 
strengths or the physical dimensions of atoms. 

Similar to the concept of electronegativity is the electropositivity of an element, which 
is the power of its atoms (when part of a compound) to lose an electron. The most 
electropositive elements are the metals on the far-left of the periodic table, particularly 
Groups 1(a) and 11(a), which prefer to exist as cations. Being the opposite concept to 
electronegativity, electropositivity is not employed often. Rather, we tend to say that an 
atom such as sodium has a tiny electronegativity instead of being very electropositive. 

Atoms or groups are 
'electronegative' if they 
tend to acquire neg- 
ative charge at the 
expense of juxtaposed 
atoms or groups. Groups 
acquiring a positive 
charge are 'electropos- 

Why does dew form on a cool morning? 

Van der Waals forces 

Many people love cool autumn mornings, with the scent of the cool air and a rich 
dew underfoot on the grass and paths. The dew forms when molecules of water 
from the air coalesce, because of the cool temperature, to form minute aggregates 
that subsequently nucleate to form visible drops of water. These water drops form a 
stable colloid (see Chapter 10). 

Real gases are never wholly ideal: there will always be some extent of non-ideality. 
At one extreme are the monatomic rare gases such as argon and neon, which are non- 
polar. Hydrocarbons, like propane, are also relatively non-polar, thereby precluding 
stronger molecular interactions. Water, at the opposite extreme, is very polar because 
some parts of the molecule are more electron withdrawing than 
others. The central oxygen is relatively electronegative and the two 
hydrogen atoms are electropositive, with the result that the oxygen 
is more negative than either of the hydrogen atoms. We say it has 
a slight excess charge, which we write as S~. Similar reasoning 
shows how the hydrogen atoms are more positive than the oxygen, 
with excess charges of <5 + . 
These excess charges form in consequence of the molecule incorporating a variety 
of atoms. For example, the magnitude of S~ on the chlorine of H-Cl is larger than 
the excess charges in the F-Cl molecule, because the difference in electronegativity 

The symbol 8 means 'a 
small amount of . . .', so 
\5 - ' is a small amount 
of negative charge. 



Table 2.1 Values of electronegativity x for some main-group elements 



















































X between H and CI is greater than the difference between F and CI. There will be no 
excess charge in the two molecules H-H or Cl-Cl because the atoms in both are the 
same - we say they are homonuclear. Table 2.1 contains a few electronegativities. 

SAQ 2.2 By looking at the electronegativities in Table 2.1, suggest wheth- 
er the bonds in the following molecules will be polar or non-polar: (a) hydro- 
gen bromide, HBr; (b) silicon carbide, SiC; (c) sulphur dioxide, 0=S=0; and 
(d) sodium iodide, Nal. 

The actual magnitude of the excess charge is generally unknown, 
although we do know they are small. Whereas some calculations 
suggest that S is perhaps as much as 0.1 of a full, formal charge, 
others suggest about 0.01 or even less. 

While debate persists concerning the magnitudes of each excess 
charge within a molecule, it is certain that the overall charge on the 
molecule is zero, meaning that the two positive charges in water 
cancel out the central negative charge on the oxygen. We reason 
this by saying that water is a neutral molecule. 

Figure 2.3 shows the 'V shape of the water molecule. The top 
of the molecule (as drawn) has a negative excess charge and the 
bottom is positive. The S + and S~ charges are separated spatially, 
which we call a dipole. Such dipoles are crucial when explain- 
ing why water vapour so readily forms a liquid: those parts of 
the molecule bearing a slight positive charge (<5 + ) attract those 
parts of adjacent molecules that bear a slight negative charge (<5~). 
The interaction is electrostatic, and forms in much a similar man- 
ner to the north pole of a magnet attracting the south pole of 
another magnet. 

Electrostatic interactions of this type are called 'dipole-dipole 
interactions', or 'van der Waals forces' after the Dutch physicist 
Johannes Diderik van der Waals (1837-1923) who first postulated 
their existence. A van der Waals force operates over a relatively 

Water is a neutral 
molecule, so the cen- 
tral negative charge 
in the water molecule 
counteracts the two 
positive charges. 

A 'dipole' forms when 
equal and opposite 
charges are separated 
by a short distance. 'Di' 
means two, and 'pole' 
indicates the two ends 
of a magnet. 

'Van der Waals forces' 
are electrostatic inter- 
actions between di- 
poles. (Note how we 
pronounce 'Waals' as 



Figure 2.3 The water molecule has a 'V shape. Experiments show 
that gaseous water has an O-H length of 0.957 18 A; the H-O-H angle 
is 104.474°. Water is polar because the central oxygen is electronega- 
tive and the two hydrogen atoms are electropositive. The vertical arrow 
indicates the resultant dipole, with its head pointing toward the more 
negative end of the molecule 

-O— - H O- 



\ I /O 

\ H' K ,- H 

/ H 

H / H >— H 

0' / 

/ \ H H 

H " 

Figure 2.4 Water would be a gas rather than a liquid at room temperature if no van der Waals 
forces were present to 'glue' them together, as indicated with dotted lines in this two-dimensional 
representation. In fact, water coalesces as a direct consequence of this f/jree-dimensional network 
of dipole-dipole interactions. Note how all the O-H • • • O bonds are linear 

short distance because the influence of a dipole is not large. In practice, we find that 
the oxygen atoms can interact with hydrogen atoms on an adjacent molecule of water, 
but no further. 

The interactions between the two molecules helps to 'glue' them together. It is a 
sobering thought that water would be a gas rather than a liquid if hydrogen bonds 
(which are merely a particularly strong form of van der Waals forces) did not promote 
the coalescence of water. The Earth would be uninhabitable without them. Figure 2.4 
shows the way that liquid water possesses a three-dimensional network, held together 
with van der Waals interactions. 

Each H2O molecule in liquid water undergoes at least one interaction with another 
molecule of H2O (sometimes two). Nevertheless, the interactions are not particularly 
strong - perhaps as much as 20 kJ mol~ . 

Whereas the dipoles themselves are permanent, van der Waals interactions are 
not. They are sufficiently weak that they continually break and re-form as part of a 
dynamic process. 

How is the three-dimensional structure maintained 
within the DNA double helix? 

Hydrogen bonds 

DNA is a natural polymer. It was first isolated in 1869 by Meischer, but its role 
in determining heredity remained unrecognized until 1944, by which time it was 



appreciated that it is the chromosomes within a cell nucleus that 
dictate hereditary traits. And such chromosomes consist of DNA 
and protein. In 1944, the American bacteriologist Oswald Avery 
showed how it was the DNA that carried genetic information, not 
the protein. 

The next breakthrough came in 1952, when Francis Crick and 
Donald Watson applied X-ray diffraction techniques to DNA and 
elucidated its structure, as shown schematically in Figure 2.5. They 
showed how its now famous 'double helix' is held together via 
a series of unusually strong dipole-dipole interactions between 
precisely positioned organic bases situated along the DNA poly- 
mer's backbone. 

There are four bases in DNA: guanine, thymine, cytosine and 
adenine. Each has a ketone C=0 group in which the oxygen is quite 
electronegative and bears an excess negative charge S~, and an 
amine in which the electropositive hydrogen atoms bear an excess 

The word 'theory' comes 
from the Greek theoreo, 
meaning 'I look at'. A 
theory is something we 
look at, pending accep- 
tance or rejection. 

The rules of 'base pair- 
ing' (or nucleotide pair- 
ing) in DNA are: ade- 
nine (A) always pairs 
with thymine (T); cyto- 
sine (C) always pairs 
with guanine (G). 

3.4 nm 



0--H-N N ^H 

\ // \ / *C 

c-c c-c , 

// \ // \\ 

H-C N — H --N C 

\ / \ / 

N-C C=N 

** W / 

DNA backbone 0--H — N 





DNA backbone 


H 3 C 


-N N H 

\ // 

\ / *c 


C-C 7 

// \ 

" * N 
N C-\ 
\ / ^ 

H-C N- 


\ / 


C = N DNA backbone 

S* * 


DNA backbone 





Figure 2.5 (a) The structure of the 'double helix' at the heart of DNA. The slender 'rods' represent 
the hydrogen bonds that form between the organic bases situated on opposing strands of the helix, 
(b) Hydrogen bonds (the dotted lines) link adenine with thymine, and guanine with cytosine 



Table 2.2 The energies of hydrogen bonds 

Atoms in H-bond 

Typical energy/kj mol 




positive charge S + . Since the hydrogen atom is so small and so electropositive, its 
excess charge leads to the formation of an unusually strong dipole, itself leading 
to a strong van der Waals bond. The bond is usually permanent (unlike a typical 
dipole-dipole interaction), thereby 'locking' the structure of DNA into its pair of 
parallel helices, much like the interleaving teeth of a zip binding together two pieces 
of cloth. 

We call these extra-strong dipole-dipole bonds 'hydrogen bonds', 
and these are defined by the IUPAC as 'a form of association between 
an electronegative atom and a hydrogen atom attached to a second, 
relatively electronegative atom'. All hydrogen bonds involve two 
dipoles: one always comprises a bond ending with hydrogen; the 
other terminates with an unusually electronegative atom. It is best 
considered as an electrostatic interaction, heightened by the small 
size of hydrogen, which permits proximity of the interacting dipoles 
or charges. Table 2.2 contains typical energies for a few hydrogen 
bonds. Both electronegative atoms are usually (but not necessarily) 
from the first row of the periodic table, i.e. N, O or F. Hydrogen 
bonds may be intermolecular or intramolecular. 

Finally, as a simple illustration of how weak these forces are, 
note how the energy required to break the hydrogen bonds in 
liquid hydrogen chloride (i.e. the energy required to vaporize it) 
is 16kJmol _1 , yet the energy needed break the chemical bond 
between atoms of hydrogen and chlorine in H-Cl is almost 30 
times stronger, at 431 kJmol -1 . 

IUPAC is the Interna- 
tional Union of Pure 
and Applied Chemistry. 
It defines terms, quan- 
tities and concepts in 

Strictly, the dipole- 
dipole interactions dis- 
cussed on p. 42 are 
also hydrogen bonds, 
since we discussed 
the interactions arising 
between H-O-bonded 


The ancient Greeks recognized that organisms often pass on traits to their offspring, but 
it was the experimental work of the Austrian monk Gregor Mendel (1822-1884) that 
led to a modern hereditary 'theory'. He entered the Augustinian monastery at Briinn 
(now Brno in the Czech Republic) and taught in its technical school. 

He cultivated and tested the plants at the monastery garden for 7 years. Starting in 
1856, painstakingly analysing seven pairs of seed and plant characteristics for at least 
28 000 pea plants. These tedious experiments resulted in two generalizations, which 
were later called Mendel's laws of heredity. Mendel published his work in 1866, but 
it remained almost unnoticed until 1900 when the Dutch botanist Hugo Marie de Vries 
referred to it. The full significance of Mendel's work was only realized in the 1930s. 



His observations also led him to coin two terms that still persist in present-day 
genetics: dominance for a trait that shows up in an offspring, and recessiveness for a 
trait masked by a dominant gene. 

How do we make liquid nitrogen? 

London dispersion forces 

A 'homonuclear' mole- 
cule comprises atoms 
from only one element; 
homo is Greek for 
'same'. Most molecules 
are 'heteronuclear' and 
comprise atoms from 
several elements; het- 
ero is Greek for 'other' 
or 'different'. 

Liquid nitrogen is widely employed when freezing sperm or eggs 
when preparing for the in vitro fertilization (IVF). It is also essential 
for maintaining the cold temperature of the superconducting magnet 
at the heart of the NMR spectrometers used for structure elucidation 
or magnetic resonance imaging (MRI). 

Nitrogen condenses to form a liquid at — 196 °C (77 K), which 
is so much lower than the temperature of 373.15 K at which water 
condenses that we suspect a different physicochemical process is in 
evidence. Below — 196 °C, molecules of nitrogen interact, causing 
condensation. That there is any interaction at all should surprise us, 
because the dipoles above were a feature of heteronuclear bonds, 
but the di-nitrogen molecule (N=N) is homonuclear, meaning both 
atoms are the same. 

We need to invoke a new type of interaction. The triple bond 
between the two nitrogen atoms in the di-nitrogen molecule incor- 
porates a huge amount of electron density. These electrons are 
never still, but move continually; so, at any instant in time, one 
end of a molecule might be slightly more negative than the other. 
A fraction of a second later and the imbalance departs. But a tiny 
dipole forms during the instant while the charges are unbalanced: 
we call this an induced dipole; see Figure 2.6. 

The electron density changes continually, so induced dipoles 
never last more than about 10~ n s. Nevertheless, they last suf- 
ficiently long for an interaction to form with the induced dipole of another nitrogen 
molecule nearby. We call this new interaction the London dispersion force after Fritz 
London, who first postulated their existence in 1930. 

London dispersion forces form between all molecules, whether polar or non-polar. 
In a large atom or molecule, the separation between the nucleus and valence elec- 
trons is quite large; conversely, the nucleus -electron separation in a lighter atom or 
molecule is smaller, implying that the electrons are more tightly held. The tighter 
binding precludes the ready formation of an induced dipole. For this reason, larger 
(and therefore heavier) atoms and molecules generally exhibit stronger dispersion 
forces than those that are smaller and lighter. 

Dipoles are usually a 
feature of heteronu- 
clear bonds, although a 
fuller treatment needs 
to consider the elec- 
tronic environment 
of atoms and groups 
beyond the bond of 



Figure 2.6 Schematic diagram to show how an induced dipole forms when polarizable electrons 
move within their orbitals and cause a localized imbalance of charge (an 'induced dipole' in which 
the negative electrons on one atom attract the positive nucleus on another). The dotted line represents 
the electrostatic dipole interaction 


The existence of an attractive force between non-polar molecules was first recognized 
by van der Waals, who published his classic work in 1873. The origin of these forces 
was not understood until 1930 when Fritz London (1900-1954) published his quantum- 
mechanical discussion of the interaction between fluctuating dipoles. He showed how 
these temporary dipoles arose from the motions of the outer electrons on the two 

We often use the term 'dispersion force' to describe these attractions. Some texts 
prefer the term 'London-van der Waals' forces. 

The ease with which 
the electron distribu- 
tion around an atom 
or molecule can be 
distorted is called its 

The weakest of all the 
intermolecular forces in 
nature are always Lon- 
don dispersion forces. 


The electrons in a molecule's outer orbitals are relatively free to 
move. If we could compare 'snapshots' of the molecule at two 
different instants in time then we would see slight differences in 
the charge distributions, reflecting the changing positions of the 
electrons in their orbitals. The ease with which the electrons can 
move with time depends on the molecule's polarizability, which 
itself measures how easily the electrons can move within their 

In general, polarizability increases as the orbital increases in size: 
negative electrons orbit the positive nucleus at a greater distance 
in such atoms, and consequently experience a weaker electrostatic 
interaction. For this reason, London dispersion forces tend to be 
stronger between molecules that are easily polarized, and weaker 
between molecules that are not easily polarized. 



Why is petrol a liquid at room temperature but butane 
a gas? ^^|^^^^^^^^^^^H 

The magnitude of London dispersion forces 

The major component of the petrol that fuels a car is octane, present as a mixture of 
various isomers. It is liquid at room temperature because its boiling point temperature 
7(boii) is about 125.7 °C. The methane gas powering an oven has a T^oii) = —162.5 °C 
and the butane propellant in a can of air freshener has r^oii) = —0.5 °C. Octadecane 
is a gel and paraffin wax is a solid. Figure 2.7 shows the trend in T(boii) for a series of 
straight-chain alkanes Each hydrocarbon experiences exactly the same intermolecular 
forces, so what causes the difference in T^oU)? 

Interactions always form between molecules because London forces cannot be erad- 
icated. Bigger molecules experience greater intermolecular forces. These dispersion 
forces are weak, having a magnitude of between 0.001 and 0.1 per cent of the strength 
of a typical covalent bond binding the two atoms in diatomic molecule, H2. These 
forces, therefore, are so small that they may be ignored within molecules held together 
by stronger forces, such as network covalent bonds or large permanent dipoles or ions. 

The strength of the London dispersion forces becomes stronger with increased 
polarizability, so larger molecule (or atoms) form stronger bonds. This observation 
helps explain the trends in physical state of the Group VII(b) halogens: L is a solid, 
Br2 is a liquid, and CI2 and F2 are gases. 

But the overall dispersion force strength also depends on the total number of elec- 
trons in the atom or molecule. It is a cumulative effect. Butane contains 14 atoms and 
58 electrons, whereas octane has 26 atoms and 114 electrons. The greater number of 
electrons increases the total number of interactions possible and, since both melting 




'S -50 





O CsH, 
O C 7 H 16 
O C 6 H 14 
O C 5 H 12 

—I "I 1 - 

40 60C 4 H 10 80 

O C 3 H 8 


OC,H 6 

O CH 4 

Molecular mass/g mol 1 



Figure 2.7 The boiling temperature of simple linear hydrocarbons increases as a function of 
molecular mass, as a consequence of a greater number of induced dipoles 



and boiling points depend directly on the strength of intermolecular bonds, the overall 
strength of the London forces varies as the molecule becomes larger. 


When is a dispersion force sufficiently strong that we can safely call it a hydrogen bond? 
Hydrogen bonds are much stronger than London dispersion forces for two princi- 
pal reasons: 

(1) The induced dipole is permanent, so the bond is permanent. 

(2) The molecule incorporates a formal H-X covalent bond in which X is a 
relatively electronegative element (see p. 42). 

We call an interaction 'a hydrogen bond' when it fulfils both criteria. 

2.2 Quantifying the interactions 
and their influence 

How does mist form? 

Condensation and the critical state 

Why is it that no dew forms if the air pressure is low, however cool the air tempe- 

To understand this question, we must first appreciate how molecules come closer 
together when applying a pressure. The Irish physical chemist Thomas Andrews 
(1813-1885) was one of the first to study the behaviour of gases as they liquefy: 
most of his data refer to CO2. In his most famous experiments, he observed liquid CO2 
at constant pressure, while gradually raising its temperature. He readily discerned a 
clear meniscus between condensed and gaseous phases in his tube at low temperatures, 
but the boundary between the phases vanished at temperatures of about 31 °C. Above 
this temperature, no amount of pressure could bring about liquefaction of the gas. 

Andrews suggested that each gas has a certain 'critical' tem- 
perature, above which condensation is impossible, implying that 
no liquid will form by changes in pressure alone. He called this 
temperature the 'critical temperature' ^critical)- 

Figure 2.8 shows a Boyle' s-law plot of pressure p (as y) against 
volume V (as x) for carbon dioxide. The figure is drawn as a 
function of temperature. Each line on the graph represents data 
obtained at a single, constant temperature, and helps explain why we call each line 
an isotherm. The uppermost isotherm represents data collected at 31.5 °C. Its shape is 
essentially straightforward, although it clearly shows distortion. The middle trace (at 

The 'critical temper- 
ature' 7" (C riticai) is that 
temperature above 
which it is impossible 
to liquefy a gas. 



32 36 40 44 48 52 

Molar volume/cm 3 moh 1 

56 60 64 

Figure 2.8 Isotherms of carbon dioxide near the critical point of 3 1.013 °C. The shaded parabolic 
region indicates those pressures and volumes at which it is possible to condense carbon dioxide 

the cooler temperature of 30.4 °C) is more distorted - it even shows a small region 
where the pressure experienced is independent of temperature, which explains why 
the graph is linear and horizontal. And in the bottom trace the isotherm at 29.9 °C is 
even more distorted still. 

The area enclosed by the parabola at the bottom of Figure 2.8 represents those 
values of pressure and temperature at which CO2 will condense. If the temperature 
is higher than that at the apex of the parabola (i.e. warmer than 
31.013 °C) then, whatever the pressure, CO2 will not liquefy. For 
this reason, we call 31.013 °C the critical temperature T( cr iticai) of 
CO2. Similarly, the gas will not liquefy unless the pressure is above 
a minimum that we call the 'critical pressure' /? (critical) • 

Critical fluids are dis- 
cussed in Chapter 5, 
where values of 7" (crit icai) 
are listed. 



As the temperature rises above the critical temperature and the pressure drops 
below the critical pressure, so the gas approximates increasingly to an ideal gas, 
i.e. one in which there are no interactions and which obeys the ideal-gas equation 
(Equation (1.13)). 

How do we liquefy petroleum gas? 

We often call a gas that 
is non-ideal, a real gas. 

Quantifying the non-ideality 

Liquefied petroleum gas (LPG) is increasingly employed as a fuel. We produce it by 
applying a huge pressure (10-20 x p & ) to the petroleum gas obtained from oil fields. 
Above a certain, critical, pressure the hydrocarbon gas condenses; we say it reaches 
the dew point, when droplets of liquid first form. The proportion of the gas liquefying 
increases with increased pressure until, eventually, all of it has liquefied. 

Increasing the pressure forces the molecules closer together, and the intermolecular 
interactions become more pronounced. Such interactions are not particularly strong 
because petroleum gas is a non-polar hydrocarbon, explaining why it is a gas at room 
temperature and pressure. We discuss other ramifications later. 

The particles of an ideal gas (whether atoms or molecules) do 
not interact, so the gas obeys the ideal-gas equation all the time. 
As soon as interactions form, the gas is said to be non-ideal, 
with the result that we lose ideality, and the ideal-gas equation 
(Equation (1.13)) breaks down. We find that pV ^ nRT. 

When steam (gaseous water) is cooled below a certain tem- 
perature, the molecules have insufficient energy to maintain their 
high-speed motion and they slow down. At these slower speeds, 
they attract one another, thereby decreasing the molar volume. 

Figure 2.9 shows a graph of the quotient pV 4- nRT (as y) 
against pressure p (as x). We sometime call such a graph an 
Andrews plot. It is clear from the ideal-gas equation (Equa 
tion (1.13)) that if pV = nRT then pV 4 nRT should always 
equal to one: the horizontal line drawn through y = 1, therefore, 
indicates the behaviour of an ideal gas. 

But the plots in Figure 2.9 are not horizontal. The deviation of a 
trace from the y = 1 line quantifies the extent to which a gas devi- 
ates from the ideal-gas equation; the magnitude of the deviation 
depends on pressure. The deviations for ammonia and ethene are 
clearly greater than for nitrogen or methane: we say that ammo- 
nia deviates from ideality more than does nitrogen. Notice how 
the deviations are worse at high pressure, leading to the empiri- 
cal observation that a real gas behaves more like an ideal gas at 
lower pressures. 

Figure 2.10 shows a similar graph, and displays Andrews plots 
for methane as a function of temperature. The graph clearly 

The physical chem- 
istry underlying the 
liquefaction of a gas 
is surprisingly compli- 
cated, so we shall not 
return to the question 
until Chapter 5. 

We form a 'quotient' 
when dividing one thing 
by another. We meet 
the word frequently 
when discussing a 
person's IQ, their 'intel- 
ligence quotient', which 
we define as: (a per- 
son's score in an intel- 
ligence test h- the aver- 
age score) x 100. 











H 2 

V N 2 

CH 4 

/ C2H4 

NH 3 

200 400 600 800 

Applied pressure/p 6 " 



Figure 2.9 An Andrews plot of P V -h nRT (as y) against pressure p (as x) for a series of real 
gases, showing ideal behaviour only at low pressures. The function on the y-axis is sometimes 
called the compressibility Z 



o 20 

>, 1.5 





200 K 

300 K 
400 K 

— 600 K 



200 400 600 

Applied pressure/p® 



Figure 2.10 An Andrews plot of PV -H nRT (as y) against pressure p (as x) for methane gas as 
a function of temperature. Methane behaves more like an ideal gas at elevated temperatures 

demonstrates how deviations from ideality become less severe with increasing 
temperature. In fact, we should expect the deviations to decrease as the temperature 
increases, because a higher temperature tells us how the particles have more energy, 
decreasing the likelihood of interparticle interactions being permanent. 

Drawing graphs such as Figure 2.10 for other gases suggests a 
second empirical law, that gases behave more like ideal gases as 
the temperature rises. The ideal-gas equation (Equation (1.13)) is 
so useful that we do not want to lose it. Accordingly, we adapt it 

Gases behave more like 
ideal gases at higher 



somewhat, writing it as 

We sometimes call the 
function "pV + nRT' 
the 'compressibility' 
or 'compressibility fac- 
tor' Z. 

pV = Z x (nRT) 


where Z is the compressibility or compressibility factor. The value 
of Z will always be one for an ideal gas, but Z rarely has a value 
of one for a real gas, except at very low pressures. As soon as p 
increases, the gas molecules approach close enough to interact and 
pV 7^ nRT. The value of Z tells us a lot about the interactions 
between gas particles. 


The gas constant R is generally given the value 8.314 JK _1 mol _1 , but in fact this 
numerical value only holds if each unit is the SI standard, i.e. pressure expressed in 
pascals, temperature in kelvin and volume in cubic metres. 

The value of R changes if we express the ideal-gas equation (Equation (1.13)) with 
different units. Table 2.3 gives values of R in various other units. We must note an 
important philosophical truth here: the value of the gas constant is truly constant, but 
the actual numerical value we cite will depend on the units with which we express it. 
We met a similar argument before on p. 19, when we saw how a standard prefix (such as 
deca, milli or mega) will change the appearance of a number, so V = 1 dm = 10 3 cm 3 . 
In reality, the number remains unaltered. 

We extend this concept here by showing how the units themselves alter the numerical 
value of a constant. 

Table 2.3 Values of the gas constant R 
expressed with various units 3 

8.3145 jr'mor 1 

2 calKT'mor 1 

0.083 145 dm 3 barmor 1 Kr 1 

l v-i 

83.145 cm J barmor'K 
0.082058 dm 3 atmmor 1 K 

i v-i 

82.058 cm J atmmor 1 K 


a l bar = p 6 = 10 5 Pa. 1 atm = 1.013 25 x 10 5 
Pa. The 'calorie' is a wholly non-SI unit of energy; 
1 cal = 4.157 J. 

Why is the molar volume of a gas not zero at K? 

The van der Waals equation 

In Chapter 1 we recalled how Lord Kelvin devised his temperature scale after cooling 
gases and observing their volumes. If the simplistic graph in Figure 1.5 was obeyed, 



then a gas would have a zero volume at — 273.15°C. In fact, the molar volume of 
a gas is always significant, even at temperatures close to absolute zero. Why the 
deviation from Kelvin's concept? 

Every gas consists of particles, whether as atoms (such as neon) or as molecules 
(such as methane). To a relatively good first approximation, any atom can be regarded 
as a small, incompressible sphere. The reason why we can compress a gas relates to 
the large separation between the gas particles. The first effect of compressing a gas 
is to decrease these interparticle distances. 

Particles attract whenever they approach to within a minimum distance. Whatever 
the magnitude of the interparticle attraction, energetic molecules will separate and 
continue moving after their encounter; but, conversely, molecules of lower energy do 
not separate after the collision because the attraction force is enough to overwhelm the 
momentum that would cause the particles to bounce apart. The process of coalescence 
has begun. 

Compressing a gas brings the particles into close proximity, thereby increasing the 
probability of interparticle collisions, and magnifying the number of interactions. At 
this point, we need to consider two physicochemical effects that operate in opposing 
directions. Firstly, interparticle interactions are usually attractive, encouraging the 
particles to get closer, with the result that the gas has a smaller molar volume than 
expected. Secondly, since the particles have their own intrinsic volume, the molar 
volume of a gas is described not only by the separations between particles but also 
by the particles themselves. We need to account for these two factors when we 
describe the physical properties of a real gas. 

The Dutch scientist van der Waals was well aware that the ideal- 
gas equation was simplistic, and suggested an adaptation, which 
we now call the van der Waals equation of state: 


n a 

P + ^2 

(V - nb) = nRT 


The a term reflects 
the strength of the 
interaction between 
gas particles, and the 
b term reflects the 
particle's size. 

where the constants a and b are called the 'van der Waals constants', the values 
of which depend on the gas and which are best obtained experimentally. Table 2.4 
contains a few sample values. The constant a reflects the strength of the interaction 
between gas molecules; so, a value of 18.9 for benzene suggests a strong interaction 
whereas 0.03 for helium represents a negligible interaction. Incidentally, this latter 
value reinforces the idea that inert gases are truly inert. The magnitude of the constant 
b reflects the physical size of the gas particles, and are again seen 
to follow a predictable trend. The magnitudes of a and b dictate 
the extent to which the gases deviate from ideality. 

Note how Equation (2.2) simplifies to become the ideal-gas equa- 
tion (Equation (1.13)) if the volume V is large. We expect this 
result, because a large volume not only implies a low pressure, 
but also yields the best conditions for minimizing all instances of 
interparticle collisions. 

Equation (2.2) simpli- 
fies to become the 
ideal-gas equation 
(Equation (1.13)) 
whenever the volume 
V is large. 



Table 2.4 Van der Waals constants for various gases 


a/(mol dm 3 ) 2 bar 

fc/(mol dm" 3 )- 1 




0.034 589 

0.023 733 



0.017 383 







Diatomic gases 










Carbon monoxide 







0.037 847 



0.043 067 



0.065 144 










The value of p calcu- 
lated with the ideal-gas 
equation is 3.63 x 
10 5 Pa, or 3.63 bar. 

Worked Example 2.1 0.04 mol of methane gas is enclosed within 
a flask of volume 0.25 dm 3 . The temperature is 0°C. From Table 2.4, 
a = 2.3026 dm 6 bar mol" 2 and b = 0.043 067 dm 3 mol - ' . What is the 
pressure p exerted? 

We first rearrange Equation (2.2), starting by dividing both sides by 
the term (V — nb), to yield 

Calculations with the 
van der Waals equation 
are complicated be- 
cause of the need to 
convert the units to 
accommodate the SI 
system. The value of/? 
comes from Table 2.3. 



V 2 (V-nb) 
We then subtract (n 2 a) H- V 2 from both sides: 

P = 



Next, we insert values and convert to SI units, i.e. 0°C is expressed 
as 273.15 K. 

P = 

0.04 molx(8.314xl0" 2 dm 3 bar K" 1 mol _1 )x273.15 K 
(0.25 dm 3 - 0.04 mol x 0.043 067 dm 3 mol" 1 ) 

0.04 molY 
0.25 dm 3 / 


x 2.3026 (dm J mol" 1 ) 2 bar 



p = 3.65887 bar - 0.05895 bar 
p = 3.59992 bar 

The pressure calculated with the ideal-gas equation (Equation (1.13)) is 3.63 bar, so 
the value we calculate with the van der Waals equation (Equation (2.2)) is 1 per cent 
smaller. The experimental value is 3.11 bar, so the result with the van der Waals equation 
is superior. 

The lower pressure causes coalescence of gas particles, which decreases their kinetic 
energy. Accordingly, the impact between the aggregate particle and the container's 
walls is less violent, which lowers the observed pressure. 

The virial equation 

An alternative approach to quantifying the interactions and deviations from the ideal- 
gas equation is to write Equation (1.13) in terms of 'virial coefficients': 


RT (1 + B'p + C'p 2 + ...) 


The word 'virial' comes 
from the Latin for force 
or powerful. 

where the V 4- n term is often rewritten as V m and called the molar volume. 

Equation (2.3) is clearly similar to the ideal-gas equation, 
Equation (1.13), except that we introduce additional terms, each 
expressed as powers of pressure. We call the constants, B' ', C etc., 
'virial coefficients', and we determine them experimentally. We 
call B' the second virial coefficient, C" the third, and so on. 

Equation (2.3) becomes the ideal-gas equation if both B' and C are tiny. In fact, 
these successive terms are often regarded as effectively 'fine-tuning' the values of p or 
V m . The C coefficient is often so small that we can ignore it; and D' is so minuscule 
that it is extremely unlikely that we will ever include a fourth virial coefficient in any 
calculation. Unfortunately, we must exercise care, because B' constants are themselves 
a function of temperature. 

Worked Example 2.2 What is the molar volume V m of oxygen gas 
at 273 K and p^l Ignore the third and subsequent virial terms, and 
take B' = -4.626 x 10" 2 bar" 1 . 

From Equation (2.3) 


x (1 + B'p) 

Care: the odd-looking 
units of B' require us 
to cite the gas con- 
stant R in SI units with 



The value of V m cal- 
culated with the ideal- 
gas equation (Equation 
(1.13)) is 4.4 per cent 

Inserting numbers (and taking care how we cite the value of R) yields 
83.145 cm 3 barmor' KT 1 x 273 K 

1 bar 

x (1 - 4.626 x 1(T 2 bar" 1 x 1 bar) 

V m = 22 697 x 0.954 cm 3 

V m = 21647 cm 3 

1 cm 3 represents a volume of 1 x 10~ 6 m 3 , so expressing this value 
of V m in SI units yields 21.6 x 10~ 3 m 3 . 

SAQ 2.3 Calculate the temperature at which the molar volume of oxygen 
is 24 dm 3 . [Hint: you will need some of the data from Worked Example 2.2. 
Assume that B' has not changed, and be careful with the units, i.e. 
V m = 24 000 cm 3 .] 

The relationship bet- 
ween 6 and 6' is B' = 
(6- RT). 

An alternative form of the virial equation is expressed in terms 
of molar volume V m rather than pressure: 

A positive virial coef- 
ficient indicates repul- 
sive interactions 
between the particles. 
The magnitude of B 
indicates the strength 
of these interaction. 

B C 

PV m = RT\\ + — + — + 


Note that the constants in Equation (2.4) are distinguishable from 
those in Equation (2.3) because they lack the prime symbol. For 
both Equations (2.3) and (2.4), the terms in brackets represents 
the molar compressibility Z. Table 2.5 lists a few virial coeffi- 

SAQ 2.4 Calculate the pressure of 1 mol of gaseous argon housed within 
2.3 dm 3 at 600 K. Take B= 11.9 cm 3 mol _1 , and ignore the third virial 

term, C. [Hint: 
volume to m 3 .] 

take care with all units; e.g. remember to convert the 

Table 2.5 Virial coefficients B for real gases as a function 
of temperature, and expressed in units of cm 3 mol -1 


100 K 

273 K 

373 K 



























The word 'chlorine' 
derives from the Greek 
chloros, meaning 

2.3 Creating formal chemical bonds 

Why is chlorine gas lethal yet sodium chloride is vital 
for life? 

The interaction requires electrons 

Chlorine gas is very reactive, and causes horrific burns to the eyes 
and throat; see p. 243. The two atoms are held together by means of 
a single, non-polar covalent bond. Q2 has a yellow -green colour 
and, for a gas, is relatively dense at s.t.p. Conversely, table salt 
(sodium chloride) is an ionic solid comprising Na + and Cl~ ions, 
held together in a three-dimensional array. What is the reason for 
their differences in behaviour? 

The outer shell of each 'atom' in CI2 possesses a full octet of electrons: seven 
electrons of its own (which explains why it belongs to Group VII(b) of the periodic 
table) and an extra electron from covalent 'sharing' with the other atom in the CI2 
molecule. The only other simple interactions in molecular chlorine are the inevitable 
induced dipolar forces, which are too weak at room temperature to allow for the 
liquefying of Cl2( g ) . 

Each chloride ion in NaCl also has eight electrons: again, seven electrons come 
from the element prior to formation of a chloride ion, but the extra eighth electron 
comes from ionizing the sodium counter ion. This extra electron resides entirely on 
the chloride ion, so no electrons are shared. The interactions in solid NaCl are wholly 
ionic in nature. Induced dipoles will also exist within each ion, but their magnitude 
is utterly negligible when compared with the strength of the formal charges on the 
Na + and Cl~ ions. We are wise to treat them as absent. 

So, in summary, the principal differences between Cl2( g ) and NaCl( s ) lie in the 
location and the interactions of electrons in the atoms' outer shells. We say these 
electrons reside in an atom's frontier orbitals, meaning that we can ignore the inner 
electrons, which are tightly bound to the nucleus. 

Why does a bicycle tyre get hot when inflated? 

Bonds and interactions involve energy changes 

A bicycle tyre gets quite hot during its inflation. The work of 
inflating the tyre explains in part why the temperature increases, but 
careful calculations (e.g. see pp. 86 and 89) show that additional 
factors are responsible for the rise in temperature. 

On a macroscopic level, we say we compress the gas into the 
confined space within the tyre; on a microscopic level, interparti- 
cle interactions form as soon as the gas particles come into close 

We look on p. 86 at the 
effect of performing 
'work' while inflating a 
bicycle tyre, and the 
way work impinges on 
the internal energy of 
the gas. 



All matter seeks to minimize its energy and entropy; see Chapter 4. This concept 
explains, for example, why a ball rolls down a hill, and only stops when it reaches 
its position of lowest potential energy. These interparticle interactions form for a 
similar reason. 

When we say that two atoms interact, we mean that the outer electrons on the two 
atoms 'respond' to each other. The electrons within the inner orbitals are buried too 
deeply within the atom to be available for interactions or bonding. We indicate this 
situation by saying the electrons that interact reside within the 'frontier' orbitals. 

And this interaction always occurs in such as way as to minimize the energy. We 
could describe the interaction schematically by 

A + B 

product + energy 


where A and B are particles of gas which interact when their frontier orbitals are 
sufficiently close to form a 'product' of some kind; the product is generally a molecule 
or association complex. (A less naive view should also accommodate changes in 
entropy; see Chapter 4.) 

We saw earlier (on p. 33) that measuring the temperature is the 
simplest macroscopic test for an increased energy content. There- 
fore, we understand that the tyre becomes warmer during inflation 
because interactions form between the particles with the concurrent 
release of energy (Equation (2.5)). 

Energy is liberated 
when bonds and inter 
actions form. 

How does a fridge cooler work? 

Introduction to the energetics of bond formation 

At the heart of a fridge's cooling mechanism is a large flask containing volatile 
organic liquids, such as alkanes that have been partially fluorinated and/or chlorinated, 
which are often known as halons or chlorofluorocarbons (CFCs). We place this flask 
behind the fridge cabinet, and connect it to the fridge interior with a thin-walled pipe. 
The CFCs circulate continually between the fridge interior and the rear, through a 
heat exchanger. 

Now imagine placing a chunk of cheese in the refrigerator. We need to cool the 
cheese from its original temperature to, say, 5 °C. Because the cheese is warmer 
than the fridge interior, energy in the form of heat transfers from 
the cheese to the fridge, as a consequence of the zeroth law of 
thermodynamics (see p. 8). This energy passes ultimately to the 
volatile CFCs in the cooling system. 

The CFC is initially a liquid because of intermolecular interac- 
tions (of the London dispersion type). Imagine that the interactions 
involves 4 kJ of energy but cooling the cheese to 5 °C we liberate 
about 6 kJ of energy: it should be clear that more energy is liber- 
ated than is needed to overcome the induced dipoles. We say that 

Converting the liquid 
CFC to a gas (i.e. boil 
ing) is analogous to 
putting energy into a 
kettle, and watching 
the water boil off as 



absorption of the energy from the cheese 'overcomes' the interactions - i.e. breaks 
them - and enables the CFC to convert from its liquid form to form a gas: 

CFC(i) + energy 




The fridge pump circulates the CFC, so the hotter (gaseous) CFC is removed from 
the fridge interior and replaced with cooler CFC (liquid). We increase the pressure of 
the gaseous CFC with a pump. The higher pressure causes the CFC to condense back 
to a liquid. The heat is removed from the fridge through a so-called heat exchanger. 
Incidentally, this emission of heat also explains why the rear of a 
fridge is generally warm - the heat emitted is the energy liberated 
when the cheese cooled. 

In summary, interactions form with the liberation of energy, but 
adding an equal or greater amount of energy to the system can 
break the interactions. Stated another way, forming bonds and inter- 
actions liberates energy; breaking bonds and interactions requires 
the addition of energy. 

A similar mechanism operates at the heart of an air-conditioning mechanism in a 
car or office. 

Forming bonds and 
interactions liberates 
energy; breaking bonds 
and interactions re- 
quires the addition 
of energy. 

Why does steam warm up a cappuccino coffee? 

Forming a bond releases energy: introducing calorimetry 

To make a cappuccino coffee, pass high-pressure steam through a cup of cold milk 
to make it hot, then pour coffee through the milk froth. The necessary steam comes 
from a kettle or boiler. 

A kettle or boiler heats water to its boiling point to effect the process: 

H2OQ + heat energy 

H 2 

2 u (g) 


Water is a liquid at room temperature because cohesive forces bind together the 
molecules; the bonds in this case are hydrogen bonds - see p. 44. To effect the phase 
transition, liquid — > gas, we overcome the hydrogen bonds, which explains why we 
must put energy into liquid water to generate gaseous steam. Stated another way, 
steam is a high-energy form of water. 

Much of the steam condenses as it passes through the cool milk. This condensation 
occurs in tandem with forming the hydrogen bonds responsible for 
the water being a liquid. These bonds form concurrently with the 
liberation of energy. This energy transfers to the milk, explaining 
why its temperature increases. 

Calorimetry is the measurement of energy changes accompany- 
ing chemical or physical changes. We usually want to know how 
much energy is liberated or consumed per unit mass or mole of 

The word 'calorimetry' 
comes from the Latin 
calor, which means 
heat. We get the word 
'calorie' from the same 



substance undergoing the process. Most chemists prefer data to be 
presented in the form of energy per mole. In practice, we measure 
accurately the amount of heat energy liberated or consumed by a 
known amount of steam while it condenses. 

A physical chemist reading from a data book learns that 40.7 kJ 
mol -1 of energy are liberated when 1 mol of water condenses and 
will 'translate' this information to say that when 1 mol (18 g) of 
steam condenses to form liquid water, bonds form concurrently 
with the liberation of 40 700 J of energy. 

As 40.7 kJmol - is the molar energy (the energy per mole), we 

can readily calculate the energy necessary, whatever the amount of water involved. 

In fact, every time the experiment is performed, the same amount of energy will be 

liberated when 18 g condense. 

Worked Example 2.3 How much energy is liberated when 128 g of water con- 

Strictly, this amount 
of energy is liber- 
ated only when the 
temperature remains 
at 100°C during the 
condensation process. 
Any changes in tem- 
perature need to be 
considered separately. 

Note the way the units 
of 'g' cancel, to leave n 
expressed in the units 
of moles. 

Firstly, we calculate the amount of material n involved using 

mass in grams 

amount of material n — (2.8) 

molar mass in grams per mole 

so, as 1 mol has a mass of 18 gmol -1 

128 g 

n — r 

18 gmol" 1 
n — 7.11 mol 

Secondly, the energy liberated per mole is 40.7 kJ mol ' , so the overall amount of energy 
given out is 40.7 kJmor 1 x 7.11 mol = 289 kJ. 

SAQ 2.5 How much energy will be liberated when 21 g 
of water condense? 

Cappuccino coffee is 
named after Marco 
d'Aviano, a 'Capuchin' 
monk who was recently 
made a saint. He 
entered a looted Turk- 
ish army camp, and 
found sacks of roasted 
coffee beans. He mixed 
it with milk and honey 
to moderate its bit- 
ter flavour. 

A physical chemist will go one stage further, and say that this 
energy of 40.7 kJmol -1 relates directly to processes occurring dur- 
ing the condensation process. In this case, the energy relates to the 
formation of hydrogen bonds. 

As each water molecule forms two hydrogen bonds, so 1 mol of 
water generates 2 mol of hydrogen bonds. The energy per hydrogen 
bond is therefore (40.7 kJmol - 4-2); so the energy of forming a 
hydrogen bond is 20.35 kJmol -1 . 

In summary, the macroscopic changes in energy measured in an 
experiment such as this are a direct reflection of microscopic energy 
changes occurring on the molecular level. The milk of a cappuccino 
coffee is warmed when steam passes through it because the steam 



condenses to form liquid water; and the water is a liquid because of the formation of 
intermolecular forces in the form of hydrogen bonds. 

The reaction of elemen- 
tal nitrogen to form 
compounds that can 
be readily metabolized 
by a plant is termed 
'fixing'. All the principal 
means of fixing nitro- 
gen involve bacteria. 

Why does land become more fertile after a 
thunderstorm ? 

Breaking bonds requires an input of energy 

A plant accumulates nutrients from the soil as it grows. Such accumulation depletes 
the amount of nutrient remaining in the soil; so, harvesting an arable crop, such as 
maize, barley or corn, removes nutrients from the field. A farmer 
needs to replenish the nutrients continually if the land is not to 
become 'exhausted' after a few seasons. 

In the context here, 'nutrients' principally comprise compounds 
of nitrogen, most of which come from bacteria that employ natu- 
rally occurring catalysts (enzymes) which feed on elemental nitro- 
gen - a process known as fixing. An example is the bacterium 
Rhizobium which lives on beans and peas. The bacteria convert 
atmospheric nitrogen into ammonia, which is subsequently avail- 
able for important biological molecules such as amino acids, pro- 
teins, vitamins and nucleic acids. 

Other than natural fixing, the principal sources of nutrients are the man-made fertil- 
izers applied artificially by the farmer, the most common being inorganic ammonium 
nitrate (NH4NO3), which is unusually rich in nitrogen. 

But lightning is also an efficient fertilizer. The mixture of gases 
we breathe comprises nitrogen (78 per cent), oxygen (21 per cent) 
and argon (1 per cent) as its principal components. The nitro- 
gen atoms in the N2 molecule are bound together tightly via a 
triple bond, which is so strong that most reactions occurring dur- 
ing plant growth (photosynthesis) cannot cleave it: N2 is inert. But 
the incredible energies unleashed by atmospheric lightning are able 
to overcome the N=N bond. 

The actual mechanism by which the N=N molecule cleaves is 
very complicated, and is not fully understood yet. It is nevertheless 
clear that much nitrogen is oxidized to form nitrous oxide, NO. This NO dissolves in 
the water that inevitably accompanies lightning and forms water-soluble nitrous acid 
HNO2, which further oxidizes during the storm to form nitric acid, HNO3. Nitric acid 
functions as a high-quality fertilizer. It has been estimated that a thunderstorm can 
yield many tonnes of fertilizer per acre of land. 

To summarize, the NsN bond in the nitrogen molecule is very 
strong and cannot be cleaved unless a large amount of energy is 
available to overcome it. Whereas bacteria can fix nitrogen, the 
biological processes within crops, such as corn and maize, can- 
not provide sufficient energy. But the energy unleashed during a 

Notice the difference 
between the two words 
'princiPAL' (meaning 
'best', 'top' or 'most 
important') and 'prin- 
ciPLE' (meaning 'idea', 
'thought' or 'concept'). 

We require energy to 
cleave bonds: bond 
energies are discussed 
on p. 114. 



thunderstorm easily overcomes the N=N bond energy, fixing the nitrogen without 
recourse to a catalyst. 


In the high mountains of Pashawa in Pakistan, near the border with Afghanistan, thun- 
derstorms are so common that the soil is saturated with nitrates deriving from the 
nitric acid formed by lightning. The soil is naturally rich in potassium compounds. 
Ion-exchange processes occur between the nitric acid and potassium ions to form large 
amounts of potassium nitrate, KNO3, which forms a thick crust of white crystals on the 
ground, sometimes lending the appearance of fresh snow. 

High concentrations of KNO3 are relatively toxic to plant growth because the ratio 
of K + to Na + is too high, and so the soil is not fertile. 

The hydrogen atoms in 
space form a 'hydride' 
with the materials on 
the surface of the satel- 

Why does a satellite need an inert coating? 

Covalency and bond formation 

A satellite, e.g. for radio or TV communication, needs to be robust to withstand its 
environment in space. In particular, it needs to be protected from the tremendous 
gravitational forces exerted during take off, from the deep vacuum of space, and from 
atoms in space. 

Being a deep vacuum, there is a negligible 'atmosphere' sur- 
rounding a satellite as it orbits in space. All matter will exist solely 
as unattached atoms (most of them are hydrogen). These atoms 
impinge on the satellite's outer surface as it orbits. On Earth, hydro- 
gen atoms always seek to form a single bond. The hydrogen atoms 
in space interact similarly, but with the satellite's tough outer skin. 
Such interactions are much stronger than the permanent hydrogen 
bonds or the weaker, temporary induced dipoles we met in Section 2.1. They form a 
stronger interaction, which we call a covalent bond. 

The great American scientist G. N. Lewis coined the word covalent, early in the 
20th century. He wanted to express the way that a bond formed by means of electron 
sharing. Each covalent bond comprises a pair of electrons. This pairing is permanent, 
so we sometimes say a covalent bond is a formal bond, to distinguish it from weak 
and temporary interactions such as induced dipoles. 

The extreme strength of the covalent bond derives from the way electrons accumu- 
late between the two atoms. The space occupied by the electrons as they accumulate 
is not random; rather, the two electrons occupy a molecular orbital that is orientated 
spatially in such a way that the highest probability of finding the electronic charges 
is directly between the two atomic nuclei. 

As we learn about the distribution of electrons within a covalent bond, we start 
with a popular representation known as a Lewis structure. Figure 2.11 depicts the 



Atomic nucleus 

• X Electrons 

Figure 2.11 Lewis structure of the covalent hydrogen molecule in 
which electrons are shared 

Lewis structure of the hydrogen molecule, in which each atom of 
hydrogen (atomic number 1) provides a single electron. The resul- 
tant molecule may be defined as two atoms held together by means 
of sharing electrons. Incidentally, we note that the glue holding the 
two atoms together (the 'bond') involves two electrons. This result 
is common: each covalent bond requires two electrons. 

We often call these 
Lewis structures 'dot- 
cross diagrams'. 


Why call it a 'molecule'? 

Colloids are discussed 
in Chapter 10. 

The word 'molecule' has a long history. The word itself comes from the old French 
molecule, itself derived from the Latin molecula, the diminutive of moles, meaning 

One of the earliest cited uses of the word dates from 1794, when Adams wrote, 
'Fermentation disengages a great quantity of air, that is disseminated among the fluid 
molecules'; and in 1799, Kirwen said, 'The molecules of solid abraded and carried 
from some spots are often annually recruited by vegetation'. In modern parlance, both 
Kirwen and Adams meant 'very small particle'. 

Later, by 1840, Kirwen' s small particle meant "micro- 
scopic particle' . For example, the great Sir Michael Fara- 
day described a colloidal suspension of gold, known then 
as Purple of Cassius, as comprising molecules which 
were 'small particles'. The surgeon William Wilkinson said in 1851, 'Molecules are 
merely indistinct granules; but under a higher magnifying power, molecules become 
[distinct] granules'. 

Only in 1859 did the modern definition come into being, when the Italian scientist 
Stanislao Cannizarro (1826-1910) defined a molecule as 'the smallest fundamental unit 
comprising a group of atoms of a chemical compound'. This statement arose while 
Cannizarro publicized the earlier work of his compatriot, the chemist and physicist 
Amedeo Avogadro (1776-1856). 

This definition of a molecule soon gained popularity. Before modern theories of 
bonding were developed, Tyndall had clearly assimilated Cannizarro' s definition of a 
molecule when he described the way atoms assemble, when he said, 'A molecule is a 
group of atoms drawn and held together by what chemists term affinity'. 


Why does water have the formula H 2 OP 

Covalent bonds and valence 

The water molecule always has a composition in which two hydrogen atoms combine 
with one oxygen. Why? 

The Lewis structure in Figure 2.1 1 represents water, H2O. Oxygen is element num- 
ber eight in the periodic table, and each oxygen atom possesses six electrons in its 
outer shell. Being a member of the second row of the periodic table, each oxygen 
atom seeks to have an outer shell of eight electrons - we call this trend the 'octet 
rule'. Each oxygen atom, therefore, has a deficiency of two electrons. As we saw 
immediately above, an atom of hydrogen has a single electron and, being a row 1 
element, requires just one more to complete its outer shell. 

The Lewis structure in Figure 2.12 shows the simplest way in 

The concept of the full 
outer shell is crucial if 
we wish to understand 
covalent bonds. 

which nature satisfies the valence requirements of each element: 
each hydrogen shares its single electron with the oxygen, meaning 
that the oxygen atom now has eight electrons (six of its own - the 
crosses in the figure) and two from the hydrogen atoms. Looking 
now at each hydrogen atom (the two are identical), we see how 

each now has two electrons: its own original electron (the dot in the diagram) together 

with one extra electron each from the oxygen (depicted as crosses). 
We have not increased the number of electrons at all. All we have done is shared 

them between the two elements, thereby enabling each atom to have a full outer shell. 

This approach is known as the electron-pair theory. 

Valence bond theory 

The valence bond theory was developed by Linus Pauling (1901-1992) and others in 
the 1930s to amalgamate the existing electron-pair bonding theory of G. N. Lewis and 
new data concerning molecular geometry. Pauling wanted a single, unifying theory. 
He produced a conceptual framework to explain molecular bonding, but in practice 
it could not explain the shapes of many molecules. 

^ Atomic nucleus 
• X Electrons 

Figure 2.12 Lewis structure of the covalent water molecule. The inner shell of the oxygen atom 
has been omitted for clarity 



Nevertheless, even today, we often discuss the bonding of organic compounds in 
terms of Lewis structures and valence bond theory. 

Why is petroleum gel so soft? 

Properties of covalent compounds 

Clear petroleum gel is a common product, comprising a mixture of simple hydro- 
carbons, principally n-octadecane (III). It is not quite a solid at room temperature; 
neither is it really a liquid, because it is very viscous. We call it a gel. Its principal 
applications are to lubricate (in a car) or to act as a water-impermeable barrier (e.g. 
between a baby and its nappy, or on chapped hands). 


We saw on p. 52 how methane is a gas unless condensed by 
compression at high pressure or frozen to low temperatures. But 
octadecane is neither a solid nor a gas. Why? 

There are several, separate types of interaction in III: both cova- 
lent bonds and dipoles. Induced dipoles involve a partial charge, 
which we called <5 + or <5~, but, by contrast, covalent bonds involve 
whole numbers of electrons. A normal covalent bond, such as that 
between a hydrogen atom and one of the carbon atoms in the back- 
bone of III, requires two electrons. A 'double bond' consists simply 
of two covalent bonds, so four electrons are shared. Six electrons 
are incorporated in each of the rare instances of a covalent 'triple 
bond'. A few quadruple bonds occur in organometallic chemistry, 
but we will ignore them here. 

Most covalent bonds are relatively non-polar. Some are com- 
pletely non-polar: the diatomic hydrogen molecule is held together 
with two electrons located equidistantly from the two hydrogen 
nuclei. Each of the two atoms has an equal 'claim' on the elec- 
trons, with the consequence that there is no partial charge on the 
atoms: each is wholly neutral. Only homonuclear molecules such 
as H2, F2, O2 or N2 are wholly non-polar, implying that the major- 
ity of covalent bonds do possess a slight polarity, arising from an 
unequal sharing of the electrons bound up within the bond. 

We see the possibility of a substance having several types of 
bond. Consider water for example. Formal covalent bonds hold 
together the hydrogen and oxygen atoms, but the individual water 
molecules cohere by means of hydrogen bonds. Conversely, paraf- 
fin wax (n-C^f^) is a solid. Each carbon is bonded covalently 

Molecules made of only 
one element are called 
'homonuclear', since 
homo is Greek for 
'same'. Examples of 
homonuclear molecules 
are H 2 , N 2 , S 8 and ful- 
lerene C6o- 

Even a covalent bond 
can possess a perma- 
nent induced dipole. 

Covalent compounds 
tend to be gases or liq- 
uids. Even when solid, 
they tend to be soft. 
But many covalent 
compounds are only 
solid at lower temper- 
atures and/or higher 
pressures, i.e. by max- 
imizing the incidence of 
induced dipoles. 



Table 2.6 Typical properties covalent compounds 



Low melting point 
Low boiling point 

Physically soft 
Malleable, not brittle 
Low electrical conductivity 
Dissolve in non-polar solvents 
Insoluble in polar solvents 

Ice melts in the mouth 

Molecular nitrogen is a gas at room 

We use petroleum jelly as a lubricant 
Butter is easily spread on a piece of bread 
We insulate electrical cables with plastic" 
We remove grease with methylated spirit b 
Polyurethane paint protects the window frame 

from rain 

"The polythene coating on an electrical wire comprises a long-chain alkane. 

b 'Methylated spirit' is the industrial name for a mixture of ethanol and methanol. 


Figure 2.13 Diamond has a giant macroscopic structure in which 
each atom is held in a rigid three-dimensional array. Other covalent 
solids include silica and other p-block oxides such as AI2O3 

to one or two others to form a linear chain; the hydrogen atoms are bound to this 
backbone, again with covalent bonds. But the wax is a solid because dispersion 
forces 'glue' together the molecules. Table 2.6 lists some of the common properties 
of covalent compounds. 

Finally, macromolecular covalent solids are unusual in comprising atoms held 
together in a gigantic three-dimensional array of bonds. Diamond and silica are 
the simplest examples; see Figure 2.13. Giant macroscopic structures are always 


The word covalent was coined in 1919 when the great American Chemist Irving Lang- 
muir said, 'it is proposed to define valence as the number of pairs of electrons which 
a given atom shares with others. In view of the fact . . . that 'valence' is very often 
used to express something quite different, it is recommended that the word covalence 
be used to denote valence defined as above.' He added, 'In [ionic] sodium chloride, the 
covalence of both sodium and chlorine is zero'. 

The modern definition from IUPAC says, 'A covalent bond is a region of rela- 
tively high electron density between nuclei which arises (at least partially) from shar- 
ing of electrons, and gives rise to an attractive force and characteristic inter-nuclear 



Why does salt form when sodium and chlorine react? 

Bond formation with ions 

Ionic interactions are electrostatic by nature, and occur between ions of opposite charge. 
The overwhelming majority of ionic compounds are solids, although a few biological 
exceptions do occur. Table 2.7 lists a few typical properties of ionic compounds. 

It is generally unwise to think of ionic compounds as holding together with physical 
bonds; it is better to think of an array of point charges, held together by the balance of 
their mutual electrostatic interactions. (By 'mutual' here, we imply equal numbers of 
positive and negative ions, which therefore impart an overall charge of zero to the solid.) 

Ionic compounds generally form following the reaction of metal- 
lic elements; non-metals rarely have sufficient energy to provide 
the necessary energy needed to form ions (see p. 123). 

The structure in Figure 2.14 shows the result of an ionic reac- 
tion: sodium metal has reacted with chlorine gas to yield white 
crystalline sodium chloride, NaCl. Each Na atom has lost an elec- 
tron to form an Na + cation and each chlorine atom has gained an 
electron and is hence a Cl~ anion. In practice, the new electron 
possessed by the chloride came from the sodium atom. 

The electron has transferred and in no way is it shared. Sodium 
chloride is a compound held together with an ionic bond, the 
strength of the bond coming from an electrostatic interaction bet- 
ween the positive and negative charges on the ions. 

Care: chlorlNE is an 
elemental gas; chlo- 
rlDE is a negatively 
charged anion. 

The chloride ion has 
a negative charge 
because, following ion- 
ization, it possesses 
more electrons than 

Why heat a neon lamp before it will generate light? 

Ionization energy 

Neon lamps generate a pleasant pink-red glow. Gaseous neon within the tube (at 
low pressure) is subjected to a strong electric discharge. One electron per neon atom 

Table 2.7 Typical properties of ionic compounds 



High melting point 

High boiling point 

Physically hard 

Often physically brittle 

High electrical conductivity in 

Dissolve in polar solvents 
Insoluble in non-polar solvents 

We need a blast furnace to melt metals 

A lightning strike is needed to volatilize some substances 

Ceramics (e.g. plates) can bear heavy weights 

Table salt can be crushed to form a powder 

Using a hair dryer in the bath risks electrocution 

Table salt dissolves in water 

We dry an organic solvent by adding solid CaCl2 



Sodium cation, Na + 

Chloride anion, CI 

^ Atomic nucleus 
• X Electrons 

Figure 2.14 Lewis structure of ionic sodium chloride. Note how the outer shell of the sodium 
ion is empty, so the next (inner) shell is full 

is lost, forming positively charged Ne + ions: 



-> Ne + (g) + e" 



Generally, the flask 
holding the neon gas 
contains a small amount 
of sodium to catalyse 
('kick start') the ion- 
ization process - see 
p. 481. 

'Monatomic' is an ab- 
breviation for 'mono- 
atomic', meaning the 
'molecule' contains 
only one atom. The 
word generally applies 
to the Group VIII(a) 
rare gases. 

We 'ionized' the neon atoms to form Ne + cations, i.e. each bears a 
positive charge. (On p. 480 we discuss in detail the photochemical 
processes occurring at the heart of the neon lamp.) 

We generally need quite a lot of energy to ionize an atom or 
molecule. For example, 2080 kJ of energy are required to ionize 
1 mol of monatomic neon gas. This energy is large and explains 
the need to heat the neon strongly via a strong electric discharge. 
We call this energy the ionization energy, and give it the symbol 
/ (some people symbolize it as I e ). Ionization energy is defined 
formally as the minimum energy required to ionize 1 mol of an 
element, generating 1 mol of electrons and 1 mol of positively 
charged cations. 

The energy required will vary slightly depending on the condi- 
tions employed, so we need to systematize our terminology. While 
the definition of / is simple enough for neon gas, we need to 
be more careful for elements that are not normally gaseous. For 
example, consider the process of ionizing the sodium catalyst at 
the heart of the neon lamp. In fact, there are two energetically 
distinct processes: 

(1) Vaporization of the sodium, to form a gas of sodium atoms: Na( S ) 

(2) Ionization of gaseous atoms to form ions: Na( g ) — »■ Na + ( g ) + e~(g). 



To remove any possible confusion, we further refine the definition of ionization 
energy, and say that / is the minimum energy required to ionize 1 mol of a gaseous 
element. The ionization energy / relates to process (2); process (1) is additional. 



Table 2.8 Ionization energies / ( „). For convenience, the figures in the table are given in MJ mol 
rather than the more usual kj moP 1 to emphasize their magnitudes 
















































































115.389 131.442 


40 60 

Atomic number 

Figure 2.15 

(as x) 

The first ionization energies / of the first 105 elements (as y) against atomic number 

Table 2.8 lists several ionization energies: notice that all of them are positive. 
Figure 2.15 depicts the first ionization energies 1^ (as y) for the elements hydrogen 
to nobelium (elements 1-102) drawn as a function of atomic number (as x). 

It is clear from Figure 2.15 that the rare gases in Group VIII(b) have the highest 
values of /, which is best accounted for by noting that they each have a full outer shell 
of electrons and, therefore, are unlikely to benefit energetically from being ionized. 
Similarly, the halogens in Group VII(b) have high values of / because their natural 
tendency is to accept electrons and become anions X~, rather than to lose electrons. 

The alkali metals in Group 1(a) have the lowest ionization energies, which is again 
expected since they always form cations with a +1 valence. There is little variation 
in / across the d-block and f-block elements, with a slight increase in / as the atomic 
number increases. 



Why does lightning conduct through air? 

Electron affinity 

Lightning is one of the more impressive manifestations of the power in nature: the 
sky lights up with a brilliant flash of light, as huge amounts of electrical energy pass 
through the air. 

As an excellent generalization, gases may be thought of as electrical insulators, so 
why do we see the lightning travel through the air? How does it conduct? Applying a 
huge voltage across a sample of gas generates an electric discharge, which is apparent 
by the appearance of light. In fact, the colour of the light depends on the nature of the 
gas, so neon gives a red colour, krypton gives a green colour and helium is invisible 
to the eye, but emits ultraviolet light. 

The source of the light seen with an electric discharge is the plasma formed by 
the electricity, which is a mixture of ions and electrons, and unionized atoms. If, 
for example, we look solely at nitrogen, which represents 78 per cent of the air, an 
electric discharge would form a plasma comprising N2" 1 ", N + , electrons e~, nitrogen 
radicals N*, as well as unreacted N2. Incidentally, the formation of these ions explains 
how air may conduct electricity. 

Very soon after the electric discharge, most of the electrons and nitrogen cations 
reassemble to form uncharged nitrogen, N2. The recombination produces so much 
energy that we see it as visible light - lightning. Some of the electrons combine with 
nitrogen atoms to form nitrogen anions N~, via the reaction 

N (g) + e (g) 




Care: do not con- 
fuse the symbols for 
electron affinity E (ea) 
and activation energy 
E a from kinetics (see 
Chapter 8). 

and, finally, some, N2" 1 " cations react with water or oxygen in the 
air to form ammonium or hydroxy lamine species. 

The energy exchanged during the reaction (in Equation (2.10)) 
is called the 'electron affinity' £( ea )- This energy (also called the 
electron attachment energy) is defined as the change in the inter- 
nal energy that occurs when 1 mol of atoms in the gas phase are 
converted by electron attachment to form 1 mol of gaseous ions. 

The negative ions formed in Equation (2.10) are called anions. 
Most elements are sufficiently electronegative that their electron 
affinities are negative, implying that energy is given out during 
the electron attachment. For example, the first electron affinity 
of nitrogen is only 7 kJmol -1 , but for chlorine £ (ea ) = 
—364 kJmol - . Table 2.9 lists the electron affinities of gaseous 
halogens, and Figure 2.16 depicts the electron affinities for the first 
20 elements (hydrogen-calcium). 

The electron affinity measures the attractive force between the incoming electron 
and the nucleus: the stronger the attraction, the more energy there is released. The 
factors that affect this attraction are exactly the same as those relating to ionization 

The negative value of 
£ ( ea) illustrates how the 
energy of the species 
X~ (g) is lower than that 
of its precursor, X (g) . 












Table 2.9 The first electron affini- 
ties of the Group VII(a) elements 


^(eaj/kJ mol 

F 2 

Cl 2 

Br 2 



Note: there is much disagreement in the 
literature about the exact values of elec- 
tron affinity. These values are taken from 
the Chemistry Data Book by Stark and 
Wallace. If we use a different data source, 
we may find slightly different numbers. 
The trends will be the same, whichever 
source we consult. 

Atomic number 

Figure 2.16 Graph of the electron affinities E (ea) of the first 30 elements (as y) against atomic 
number (as x) 

energies - nuclear charge, the distance between the nucleus and the electron, as 
well as the number of electrons residing between the nucleus and the outer, 
valence electrons. 


We must be careful with the definition above: in many older textbooks, the electron 
affinity is defined as the energy released when an electron attaches to a neutral atom. 
This different definition causes £ (ea ) to change its sign. 



Krypton, xenon and 
radon will form a 
very limited number 
of compounds, e.g. 
with fluorine, but only 
under quite excep- 
tional conditions. 

Why is argon 

First electron affinity and reactivity 

Gases such as helium, neon and argon are so unreactive that we 
call them the inert gases. They form no chemical compounds, and 
their only interactions are of the London dispersion force type. 
They cannot form hydrogen bonds, since they are not able to bond 
with hydrogen and are not electronegative. 

The outer shell of the helium atom is full and complete: the 
shell can only accept two electrons and, indeed, is occupied by 
two electrons. Similarly, argon has a complete octet of electrons 
in its outer shell. Further reaction would increase the number of 
electrons if argon were to undergo a covalent bond or become an anion, or would 
decrease the number of electrons below the 'perfect' eight if a cation were to form. 
There is no impetus for reaction because the monatomic argon is already at its position 
of lowest energy, and we recall that bonds form in order to decrease the energy. 

Sodium atoms always seek to lose a single electron to form the Na + monocation, 
because the outer valence shell contains only one electron - that is why we assign 
sodium to Group 1(a) of the periodic table. This single electron helps us explain why 
it is so favourable, energetically, to form the Na + cation: loss of the electron empties 
the outer shell, to reveal a complete inner shell, much like removing the partial skin 
of an onion to expose a perfectly formed inner layer. So, again, removal of sodium's 
single outer electron occurs in order to generate a full shell of electrons. 

But if we look at an element like magnesium, there are several ionization pro- 
cesses possible: 

(1) Formation of a monocation: Mg ( •. —> Mg + ( g) + e ( g ). 

(2) Formation of a dication: Mg" 1 




(g) + e (g ). 

The energy change in reaction (1) is called the first ionization 
energy and the energy associated with reaction (2) is the second 
ionization energy. We symbolize the two processes as I^ and Iq) 

The second ionization energy is always larger than the first, 
because we are removing a negative electron from a positively 
charged cation, so we need to overcome the attractive force bet- 
ween them. The value of 7(i) for a magnesium atom is 734 kJ 
mol -1 , but 1(2) for removing an electron from the Mg + monoca- 
tion is 1451 kJmol -1 . Both ionization energies are huge, but 7(2) is clearly much the 
larger. Table 2.8 contains many other ionization energies for elements 1-10. 

It is clear from Table 2.8 that each ionization energy is larger than the one before. 
Also note that the last two ionization energies of an element are always larger than the 
others. The sudden rise follows because the last two energies represent the removal 
of the two 1 s electrons: removal of electrons from the 2s and 2p orbitals is easier. 

Care: do not con- 
fuse the symbols for 
molecular iodine I 2 and 
the second ionization 
energy 7 (2) . Hint: note 
carefully the use of 
italic type. 



Why is silver iodide yellow? 

Mixed bonding 

Silver chloride is white; silver bromide is pale yellow; and silver iodide has a rich 
yellow colour. We might first think that the change in colour was due to Agl incor- 
porating the iodide anion, yet Nal or HI are both colourless, so the colour does not 
come from the iodide ions on their own. We need to find a different explanation. 

Silver iodide also has other anomalous properties: it is physically soft - it can 
even be beaten into a sheet, unlike the overwhelming majority of ionic compounds. 
More unusual still, it is slightly soluble in ethanol. Clearly, silver iodide is not a 
straightforward ionic compound. In fact, its properties appear to overlap between 
covalent (see Table 2.6) and ionic (see Table 2.7). 

Silver iodide is neither wholly covalent nor wholly ionic; its bonding shows con- 
tributions from both. In fact, most formal chemical bonds comprise a contribution 
from both covalent and ionic forces. The only exceptions to this general rule are 
homonuclear molecules such as hydrogen or chlorine, in which the bonding is 100 
per cent covalent. The extent of covalency in compounds we prefer 
to think of as ionic will usually be quite small: less than 0.1 per 
cent in NaCl. For example, each C-H bond in methane is about 4% 
ionic, but the bonding can be quite unusual in compounds compris- 
ing elements from the p- and d-blocks of the periodic table. For 
example, aluminium chloride, AICI3, has a high vapour pressure 
(see p. 221); tungsten trioxide will sublime under reduced pres- 
sures to form covalent W3O9 trimers; sulphur trioxide is a gas but 
will dissolve in water. Each, therefore, demonstrates a mixture of 
ionic and covalent bonding. 

In other words, the valence bonds approach is suitable for com- 
pounds showing purely ionic or purely covalent behaviour; we 
require molecular orbitals for a more mature description of the 
bonding in such materials. So the yellow colour of silver iodide 
reflects the way the bonding is neither ionic nor covalent. We find, 
in fact, that the charge clouds of the silver and iodide ions overlap 
to some extent, allowing change to transfer between them. We will 
look at charge transfer in more detail on p. 459. 

A 'trimer' is a species 
comprising three com- 
ponents (the Latin 
tri means 'three'). 
The W3O9 trimer has 
a triangle structure, 
with a WO3 unit at 
each vertex. 

We require 'molec- 
ular orbitals' for a 
more mature descrip- 
tion of the bonding in 
such materials. 

Oxidation numbers 

Valency is the number of electrons lost, borrowed or shared in a 
chemical bond. Formal charges are indicated with Arabic numerals, 
so the formal charge on a copper cation is expressed as Cu 2+ , 
meaning each copper cation has a deficiency of two electrons. In 
this system of thought, the charge on the central carbon of methane 
is zero. 

Numbers written as 
1,2,3,..., etc. are called 


Table 2.10 Rule for assigning oxidation numbers 

1 . In a binary compound, the metal has a positive oxidation number and, if a 
non-metal, it has a negative oxidation number. 

2. The oxidation number of a free ion equals the charge on the ion, e.g. in 
Na + the sodium has a +1 oxidation number and chlorine in the Cl~ ion 
has an oxidation number of —I. The oxidation number of the MnOzt~ ion 
is —I, oxide 2 ~ is —II and the sulphate SOzt 2 ~ ion is —II. 

3. The sum of the oxidation numbers in a polyatomic ion equals the 
oxidation number of the ions incorporated: e.g. consider MnOzt~ ion. 
Overall, its oxidation number is —I (because the ion's charge is —1). Each 
oxide contributes —II to this sum, so the oxidation number of the central 
manganese must be +VII. 

4. The oxidation number of a neutral compound is zero. The oxidation 
number of an uncombined element is zero. 

5. Variable oxidation numbers: 

H = +1 (except in the case of hydrides) 

CI = —I (except in compounds and ions containing oxygen) 

O = —II (except in peroxides and superoxides) 

Unfortunately, many compounds contain bonds that are a mixture of ionic and 

covalent. In such a case, a formal charge as written is unlikely to represent the 

actual number of charges gained or lost. For example, the complex ferrocyanide anion 

[Fe(CN) 6 ] 4 ~ is prepared from aqueous Fe 2+ , but the central iron atom in the complex 

definitely does not bear a +2 charge (in fact, the charge is likely 

to be nearer +1.5). Therefore, we employ the concept of oxidation 

Numbers written as I, 
II, III, . .. etc. are called 
'Roman numerals'. 

number. Oxidation numbers are cited with Roman numbers, so the 
oxidation number of the iron atom in the ferrocyanide complex 
is +11. The IUPAC name for the complex requires the oxidation 
number: we call it hexacyanoferrate (II). 
Considering the changes in oxidation number during a reaction can dramatically 
simplify the concept of oxidation and reduction: oxidation is an increase in oxidation 
number and reduction is a decrease in oxidation number (see Chapter 7). Be aware, 
though, oxidation numbers rarely correlate with the charge on an ion. For example, 
consider the sulphate anion S0 4 ~ (IV). 


= S = 


The central sulphur has eight bonds. The ion has an overall charge of —2. The 
oxidation number of the sulphur is therefore 8 — 2 = +6. We generally indicate oxi- 
dation numbers with roman numerals, though, so we write S(VI). Table 2.10 lists the 
rules required to assign an oxidation number. 


Energy and the first law 
of thermodynamics 


In this chapter we look at the way energy may be converted from one form to another, 
by breaking and forming bonds and interactions. We also look at ways of measuring 
these energy changes. 

While the change in internal energy AU is relatively easy to visualize, chemists 
generally concentrate on the net energy AH, where H is the enthalpy. AH relates to 
changes in AU after adjusting for pressure -volume expansion work, e.g. against the 
atmosphere and after transfer of energy q into and out from the reaction environment. 

Finally, we look at indirect ways of measuring these energies. Both internal energy 
and enthalpy are state functions, so energy cycles may be constructed according to 
Hess's law; we look also at Born-Haber cycles for systems in which ionization 
processes occur. 

3.1 Introduction to thermodynamics: 
internal energy 

Why does the mouth get cold when eating ice cream? 


Eating ice cream soon causes the mouth to get cold, possibly to 
the extent of making it feel quite uncomfortable. The mouth of a 
normal, healthy adult has a temperature of about 37 °C, and the ice 
cream has a maximum temperature of °C, although it is likely to 
be in the range —5 to —10 °C if it recently came from the freezer. 
A large difference in temperature exists, so energy transfers from 
the mouth to the ice cream, causing it to melt. 

Ice cream melts as it 
warms in the mouth 
and surpasses its nor- 
mal melting tempera- 
ture; see Chapter 5. 



The evidence for such a transfer of energy between the mouth and the ice cream is 
the change in temperature, itself a response to the minus-oneth law of thermodynamics 
(p. 7), which says heat travels from hot to cold. Furthermore, the zeroth law (p. 8) 
tells us energy will continue to transfer from the mouth (the hotter object) to the 
ice cream (the colder) until they are at the same temperature, i.e. when they are in 
thermal equilibrium. 

We cannot know how 
much energy a body 
or system has Mocked' 
within it. Experimen- 
tally, we can only study 
changes in the internal 
energy, AU. 

Internal energy U 

Absolutely everything possesses energy. We cannot 'see' this energy directly, nor do 
we experience it except under certain conditions. It appears to be invisible because it 
is effectively 'locked' within a species. We call the energy possessed by the object 
the 'internal energy', and give it the symbol U. 

The internal energy U is defined as the total energy of a body's 
components. Unfortunately, there is no way of telling how much 
energy is locked away. In consequence, the experimentalist can 
only look at changes in U. 

The energy is 'locked up' within a body or species in three prin- 
cipal ways (or 'modes'). First, energy is locked within the atomic 
nuclei. The only way to release it is to split the nucleus, as hap- 
pens in atomic weapons and nuclear power stations to yield nuclear 
energy. The changes in energy caused by splitting nuclei are mas- 
sive. We will briefly mention nuclear energy in Chapter 8, but the 
topic will not be discussed otherwise. It is too rare for most physical 
chemists to consider further. 

This second way in which energy is locked away is within chem- 
ical bonds. We call this form of energy the chemical energy, which 
is the subject of this chapter. Chemical energies are smaller than 
nuclear energies. 

And third, energy is possessed by virtue of the potential energy, 
and the translational, vibrational, rotational energy states of the 
atoms and bonds within the substance, be it atomic, molecular or ionic. The energy 
within each of these states is quantized, and will be discussed in greater detail in 
Chapter 9 within the subject of spectroscopy. These energies are normally much 
smaller than the energies of chemical bonds. 

As thermodynamicists, we generally study the second of these 
modes of energy change, following the breaking and formation of 
bonds (which are held together with electrons), although we occa- 
sionally consider potential energy. The magnitude of the chemical 
energy will change during a reaction, i.e. while altering the number 
and/or nature of the bonds in a chemical. We give the name calorimetry to the study 
of energy changes occurring during bond changes. 

The energy E locked 
into the atomic nucleus 
is related to its mass 
m and the speed of 
light c, according to 
the Einstein equation, 
E = mc 2 . 

Strictly, the bonds 
are held together with 
'outer-shell' electrons. 



Chemists need to understand the physical chemistry underlying 
these changes in chemical energy. We generally prefer to write in 
shorthand, so we don't say 'changes in internal energy' nor the 
shorter phrase 'changes in [/', but say instead 'At/'. But we need 
to be careful: the symbol A does not just mean 'change in'. We 
define it more precisely with Equation (3.1): 

AU t 



(final state) 


(initial state) 


where the phrases 'initial state' and 'final state' can refer to a single 
chemical or to a mixture of chemicals as they react. This way, AC/ 
has both a magnitude and a sign. 

Placing the Greek letter 
A (Delta) before the 
symbol for a parameter 
such as U indicates 
the change in U while 
passing from an initial 
to a final state. We 
define the change in a 
parameter X as AX = 

-^(flnal state) — -^(initial state)- 


In some texts, Equation (3.1) is assumed rather than defined, so we have to work out 
which are the final and initial states each time, and remember which comes first in 
expressions like Equation (3.1). In other texts, the final state is written as a subscript 
and the initial state as a superscript. The value of AU for melting ice cream would 

be written as AU 

(final state) 
(initial state) ' 

• Le - A ^("e 1 crea^ e befor ) melting)- Zt ma Y even be abbreviated to 

AI/ S , where s = solid and 1 = liquid. 

To further complicate matters, other books employ yet another notation. They retain 
the sub- and super- scripts, but place them before the variable, so the last expression in 
the previous paragraph would be written as A\U. 

We will not use any of these notation styles in this book. 

Why is skin scalded by steam? 

Exothermic reactions 

Water in the form of a gas is called 'steam'. Two things hap- 
pen concurrently when human skin comes into close contact with 
steam - it could happen, for example, when we get too close to 
a boiling kettle. Firstly, the flesh in contact with the steam gets 
burnt and hurts. Secondly, steam converts from its gaseous form 
to become liquid water. We say it condenses. We summarize the 
condensation reaction thus: 




The 'condensation' 
reaction is one of the 
simplest forms of a 
'phase change', which 
we discuss in greater 
depth in Chapter 5. 





We will use the word 
'process' here to mean 
any physical chemistry 
requiring a change in 

From Equation (3.1), which defines the changes to internal energy, 
AC/ for the process in Equation (3.2) is A C/ (condensation) = £/ (water , i) 

C/ (water, g) • 

As we saw in Chapter 2, the simplest way of telling whether some- 
thing gains energy is to ascertain whether its temperature goes up. 
The temperature of the skin does increase greatly (so it feels hot); its 
energy increases following the condensation reaction. Conversely, 
the temperature of the water decreases - indeed, its temperature decreases to below its 
boiling temperature, so it condenses. The water has lost energy. In summary, we see 
how energy is transferred, with energy passing/rom the steam to the skin. 

When energy passes from one body to another, we say the 
process is thermodynamic. The condensation of water is a ther- 
modynamic process, with the energy of the water being lower fol- 
lowing condensation. Stated another way, the precursor steam had 
more energy than the liquid water product, so C/(fi na i) is lower than 
^(initial) - Figure 3.1 represents this situation visually, and clearly 
shows how the change in internal energy AC/ during steam con- 
densation is negative. We say the change in U is exothermic. 

The energy lost by the steam passes to the skin, which therefore 

gains energy. We experience this excess energy as burning: with the 

skin being an insulator, the energy from the steam remains within 

the skin and causes damaging thermal processes. Nerve endings in the skin report the 

damage to the brain, which leads to the experience of pain. 

But none of the energy is lost during condensation, so exactly the same amount of 
energy is given out by the steam as is given to the skin. (In saying this, we assume 
no other thermodynamic processes occur, such as warming of the surrounding air. 
Even if other thermodynamic processes do occur, we can still say confidently that no 
energy is lost. It's just more difficult to act as an 'energy auditor', and thereby follow 
where it goes.) 

The word 'exother- 
mic' comes from two 
Greek roots: thermo, 
meaning 'energy' or 
'temperature', and exo 
meaning 'outside' or 
'beyond'. An exother- 
mic process therefore 
gives out energy. 






(before reaction) 

(after reaction) 

Figure 3.1 In an exothermic process, the final product has less energy than the initial starting 
materials. Energy has been given out 



Worked Example 3.1 Use Equation (3.1) to demonstrate that At/ 
is negative for the condensation of steam if, say, t/(fi na i) = 12 J and 

^(initial) = 25 J. 

Inserting values into Equation (3.1): 

At/ = (/(fi na l) — (/(initial) 

At/ = (12-25) J 
At/ = -13 J 

Important: although 
we have assigned 
numerical values to 

(/(final) and (/(initial)/ it 

is, in fact, impossible 
to know their values. 
In reality, we only 
know the difference 
between them. 

So we calculate the value of A U as — 13 J. The change in U is negative 
and, therefore, exothermic, as expected. 

We see that A(/ is negative. We could have reasoned this result 
by saying t/( nn ai) < (/(initial), and subtracting a larger energy from a 
smaller one generates a deficit. 

The symbol 'J' here 
means joule, which is 
the SI unit of energy. 

Why do we sweat? 

Endothermic reactions 

We all sweat at some time or other, e.g. after running hard, living 
in a hot climate or perhaps during an illness when our temperature 
is raised due to an infection (which is why we sometimes say, we 
have 'got a temperature'). 

Producing sweat is one of the body's natural ways of cooling 
itself, and it operates as follows. Sweat is an aqueous solution 
of salt and natural oils, and is secreted by glands just below the 
surface of the skin. The glands generate this mixture whenever the 
body feels too hot. Every time air moves over a sweaty limb, from 
a mechanical fan or natural breeze, the skin feels cooler following 
evaporation of water from the sweat. 

When we say the water evaporates when a breeze blows, we 
mean it undergoes a phase transition from liquid to vapour, i.e. a 
phase transition proceeding in the opposite direction to that in the 
previous example, so Equation (3.2) occurs backwards. When we 
consider the internal energy changes, we see (/(final) = (/(water, g) 
and (/(initial) = t^(water, i) . so the final state of the water here is more 
energetic than was its initial state. Figure 3.2 shows a schematic 
representation of the energy change involved. 

We need the salt in 
sweat to decrease the 
water's surface tension 
in order to speed up the 
evaporation process 
(we feel cooler more 
quickly). The oils in 
sweat prevents the 
skin from drying out, 
which would make it 
susceptible to sunburn. 

Evaporation is also 
called 'vaporization'. 
It is a thermodynamic 
process, because en- 
ergy is transferred. 












(before reaction) 

(after reaction) 

Figure 3.2 In an endothermic process, the final product has more energy than the initial starting 
materials. Energy has been taken in 

A process is endo- 
thermic if the final 
state has more energy 
than does the ini- 
tial state. The word 
derives from the Greek 
roots thermo (mean- 
ing 'energy' or 'tem- 
perature') and endo 
(meaning 'inside' or 
'within'). An endother- 
mic process takes in 

Worked Example 3.2 What is the change in internal energy dur- 
ing sweating? 

The definition of At/ in Equation (3.1) is AU — U t 


the value of AU, 


is obtained as U, 

(water, g) 


U(innml), SO 

. We 

(water, 1) 

already know that the final state of the water is more energetic than 
its initial state, so the value of AU is positive. We say such a process 
is endothermic. 

We feel cooler when sweating because the skin loses energy 
by transferring it to the water on its surface, which then evapo- 
rates. This process of water evaporation (sweating) is endothermic 
because energy passes from the skin to the water, and a body 
containing less energy has a lower temperature, which is why we 
feel cooler. 


Heat is absorbed from the surroundings while a liquid evaporates. This heat does not 
change the temperature of the liquid because the energy absorbed equates exactly to 
the energy needed to break intermolecular forces in the liquid (see Chapter 2). Without 
these forces the liquid would, in fact, be a gas. 

At constant temperature, the heat absorbed during evaporation is often called the 
latent heat of evaporation. This choice of words arises from the way evaporation occurs 
without heating of the liquid; 'latent' is Latin for 'hidden', since the energy added to is 
not 'seen' as a temperature rise. 


Why do we still feel hot while sweating 
on a humid beach? 

State functions 

Sometimes we feel hot even when sweating, particularly in a humid environment 
like a beach by the sea on a hot day. Two processes occur in tandem on the skin: 
evaporation (liquid water — »■ gaseous water) and condensation (gaseous water — »■ 
liquid water). It is quite possible that the same water condenses on our face as 
evaporated earlier. In effect, then, a cycle of 'liquid — »■ gas — >■ liquid' occurs. The 
two halves of this cycle operate in opposite senses, since both exo- and endo-thermic 
processes occur simultaneously. The net change in energy is, therefore, negligible, 
and we feel no cooler. 

These two examples of energy change involve water. The only difference between 
them is the direction of change, and hence the sign of At/. But these two factors are 
related. If we were to condense exactly 1 mol of steam then the amount of energy 
released into the skin would be 40700 J. The change in internal energy AU (ignoring 
volume changes) is negative because energy is given out during the condensation 
process, so AU = -40700 J. 

Conversely, if we were to vaporize exactly 1 mol of water from the skin of a 
sweaty body, the change in internal energy would be +40 700 J. In other words, the 
magnitude of the change is identical, but the sign is different. 

While the chemical substance involved dictates the magnitude of AU (i.e. the 
amount of it), its sign derives from the direction of the thermodynamic process. We 
can go further: if the same mass of substance is converted from state A to state B, 
then the change in internal energy is equal and opposite to the same process occurring 
in the reverse direction, from B to A. This essential truth is depicted schematically 
in Figure 3.3. 

The value of AU when condensing exactly 1 mol of water is termed the molar 
change in internal energy. We will call it AU m (condensation) » where the small 'm' 
indicates that a mole is involved in the thermodynamic process. Similarly, the molar 


Figure 3.3 The change in internal energy when converting a material from state A to state B is 
equal and opposite to the change in U obtained when performing the same process in reverse, from 
B to A 




often omit 



II 'm'; so, 



on, we assume 


changes in 



energy are 



change in energy during vaporization can be symbolized as 
AU m (vaporization)- If we compare the molar energies for these two 
similar processes, we see the following relation: 


m (condensation) 


in (vaporization) 


The two energies are equal and opposite because one process occurs 
in the opposite direction to the other, yet the same amount of 
material (and hence the same amount of energy) is involved in both. 
Following from Equation (3.3), we say that internal energy is a 
state function. A more formal definition of state function is, 'A ther- 
modynamic property (such as internal energy) that depends only on 
the present state of the system, and is independent of its previous 
history'. In other words, a 'state function' depends only on those 
variables that define the current state of the system, such as how 
much material is present, whether it is a solid, liquid or gas, etc. 

The concept of a state function can be quite difficult, so let us 
consider a simple example from outside chemistry. Geographical 
position has analogies to a thermodynamic state function, insofar as it does not matter 
whether we have travelled from London to New York via Athens or flew direct. 
The net difference in position is identical in either case. Figure 3.4 shows this truth 
diagrammatically. In a similar way, the value of AU for the process A — > C is the 
same as the overall change for the process A — > B — > C. We shall look further at the 
consequences of U being a state function on p. 98. 

Internal energy U is 
a 'state function' be- 
cause: (1) it is a ther- 
modynamic property; 
and (2) its value 
depends only on the 
present state of the 
system, i.e. is indepen- 
dent of the previous 




Figure 3.4 If geographical position were a thermodynamic variable, it would be a state function 
because it would not matter if we travelled from London to New York via Athens or simply 
flew direct. The net difference in position would be identical. Similarly, internal energy, enthalpy, 
entropy and the Gibbs function (see Chapter 4) are all state functions 



Furthermore, because internal energy is a state function, the over- 
all change in U is zero following a series of changes described by 
a closed loop. As an example, imagine three processes: a change 
from A — > B, then B — > C and finally from C — > A. The only rea- 
son why the net value of AU for this cycle is zero is because we 
have neither lost nor picked up any energy over the cycle. We can 
summarize this aspect of physical chemistry by saying, 'energy can- 
not be created or destroyed, only converted' - a vital truth called 
the first law of thermodynamics . 

If we measure AU over a thermodynamic cycle and obtain a non-zero value, 
straightaway we know the cycle is either incomplete (with one or more processes not 
accounted for) or we employed a sloppy technique while measuring AU. 

The 'first law of ther- 
modynamics' says 
energy can neither be 
created nor destroyed, 
only converted from 
one form to another. 


William Rankine was the first to propose the first law of thermodynamics explicitly, in 
1853 (he was famous for his work on steam engines). The law was already implicit in the 
work of other, earlier, thermodynamicists, such as Kelvin, Helmholtz and Clausius. None 
of these scientists sought to prove their theories experimentally; only Joule published 
experimental proof of the first law. 

op of a waterfall cooler 

Why is the water at the 
the water at its base? 

The mechanical equivalence of work and energy 

Two of the architects of modern thermodynamics were William Thompson (better 
known as Lord Kelvin) and his friend James Prescott Joule - a scientist of great 
vision, and a master of accurate thermodynamic measurement, as well as being some- 
thing of an English eccentric. For example, while on a holiday in Switzerland in 1847, 
Thompson met Joule. Let Thompson describe what he saw: 

I was walking down from Chamonix to commence a tour of Mont Blanc, and whom 
should I meet walking up but Joule, with a long thermometer in his hand and a carriage 
with a lady in, not far off. He told me that he had been married since we parted in 
Oxford [two weeks earlier] and that he was going to try for the elevation of temperature 
in waterfalls. 

Despite it being his honeymoon, Joule possessed a gigantic thermometer fully 4 to 
5 feet in length (the reports vary). He spent much of his spare time during his honey- 
moon in making painstaking measurements of the temperature at the top and bottom 
of elongated Swiss waterfalls. He determined the temperature difference between the 


water at the bottom and top of the waterfall, finding it to be about 1 °F warmer at 
the bottom. In Joule's own words, 'A [water]fall of 817 feet [249 m] will generate 
one degree [Fahrenheit] of temperature' . This result is not attributable to colder air at 
the top of the waterfall, nor due to friction or viscous drag, or other effects occurring 
during the water's descent, but is wholly due to a change in internal energy. The 
water was simply changing its altitude. 

The potential energy of a raised object is given by the expression 

potential energy = mgh (3.4) 

where m is the mass, g is the acceleration due to gravity and h the height by which it 
is raised. The potential energy of the water decreases during descent because its height 
decreases. This energy is liberated; and, as we have noted several times already, the 
simplest way to tell if the internal energy has increased is to determine its temperature. 
Joule showed the temperature of the water of the waterfalls had indeed increased. 

We could summarize by saying that thermodynamic work w is energetically equiv- 
alent to the lowering or raising of a weight (like the water of the waterfall, above), 
as discussed below. 

Why is it such hard work pumping up a bicycle tyre? 

Thermodynamic work 

No one who has pumped up a bicycle tyre says it's easy. Pumping a car tyre is harder 
still. It requires a lot of energy, and we really have to work at it. 

We saw in Chapter 1 how increasing the amount of a gas causes 
its volume to increase. This increase in volume is needed to oppose 
any increases in pressure. It also explains why blowing into a party 
balloon causes it to get bigger. By contrast, a car tyre cannot expand 
greatly during pumping, so increasing the amount of gas it contains 
will increase its internal pressure. In a fully inflated car tyre, the 
internal pressure is about 10 times greater than 'standard pressure' 
p & , where p & has a value of 10 5 Pa. 

The first law of thermodynamics states that energy may be con- 
verted between forms, but cannot be created or destroyed. Joule 
was a superb experimentalist, and performed various types of work, 
each time generating energy in the form of heat. In one set of exper- 
iments, for example, he rotated small paddles immersed in a water 
trough and noted the rise in temperature. This experiment was 
apparently performed publicly in St Anne's Square, Manchester. 
Joule discerned a relationship between energy and work (symbol w). We have to 
perform thermodynamic work to increase the pressure within the tyre. Such work 
is performed every time a system alters its volume against an opposing pressure or 
force, or alters the pressure of a system housed within a constant volume. 

The pressure inside a 
party balloon is higher 
than the external, 
atmospheric pressure, 
as evidenced by the 
way it whizzes around a 
room when punctured. 

Work is a form of ener- 
gy. The word 'energy' 
comes from the Greek 
en ergon, meaning 
'from work'. 



Work and energy can be considered as interchangeable: we per- 
form work whenever energy powers a physical process, e.g. to 
propel a car or raise a spoon to the mouth. The work done on 
a system increases its energy, so the value of U increases, itself 
causing AU to be positive). Work done by a system corresponds 
to a negative value of AU. 

Work done on a system 
increases its energy, 
so AU is positive. Work 
done by a system cor- 
responds to a negative 
value of AU. 

Why does a sausage become warm when placed 
in an oven? 

Isothermal changes in heat and work 

At first sight, the answer to our title question is obvious: from the minus-oneth law 
of thermodynamics, heat travels from the hot oven to the cold(er) item of food we 
place in it. Also, from the zeroth law, thermal equilibrium is attained only when the 
sausage and the oven are at the same temperature. So the simplest answer to why 
a sausage gets hot is to say the energy content of the sausage (in the form of heat) 
increases, causing its internal energy to rise. And, yet again, we see how the simplest 
test of an increasing internal energy is an increased temperature. 

We can express this truth by saying the sausage gets warmer as 
the magnitude of its internal energy increases; so, from Equation 

(3.1), AC/ = C/(fi na i) — C/(i n i t i a i), hence AU = [/(after heating) _ 

[^(before heating)- We see how the value of AU is positive since 

U (after heating) ^ U (before heating) • 

But we can now be more specific. The internal energy U changes 
in response to two variables, work w and heat energy q, as defi- 
ned by 

AU = q + w (3.5) 

We have already met the first law of thermodynamics. Equa- 
tion (3.5) here is the definitive statement of this law, and is ex- 
pressed in terms of the transfer of energy between a system and its 
environment. In other words, the magnitude of AU is the sum of 
the changes in the heat q added (or extracted) from a system, and 
the work w performed by (or done to) it. 

The internal energy can increase or decrease even if one or other 
of the two variables, q and w, remains fixed. Although the sausage 
does no work w in the oven, the magnitude of AU increases 
because the food receives heat energy q from the oven. 

Worked Example 3.3 What is the energy of the sausage after heat- 
ing, if its original energy is 4000 J, and 20 000 J is added to it? 

Care: in the past, 
Equation (3.5) was 
often written as AU = 
q- w, where the minus 
sign is intended to 
show how the internal 
energy decreases fol- 
lowing work done by 
a system. We will use 
Equation (3.5), which 
is the more usual form. 

Care: The symbol of 
the joule is J. A small 'j' 
does not mean joules; 
it represents another 
variable from a com- 
pletely different branch 
of physical chemistry. 

No work is done, so w — 0. 



The definition of AC/ is given by Equation (3.1): 

Care: this is a highly 
artificial calculation and 
is intended for illus- 
trative purposes only. 
In practice, we never 
know values of U, only 
changes in U, i.e. AU. 

The word 'isothermal' 
can be understood by 
looking at its Greek 
roots. Iso means 'same' 
and thermo means 
'energy' or 'tempera- 
ture', so a measure- 
ment is isothermal 
when performed at a 
constant temperature. 

AU — (/(fi na i) — (/(initial) 

Rearranging to make (/( nna i) the subject, we obtain 

^(final) = ^(initial) + AC/ 

Equation (3.5) is another expression for AC/. Substituting for AC/ in 
Equation (3.1) allows us to say 

t/(final) = ^/(initial) + 1 + W 

Inserting values into this equation, we obtain 

C/(tinai) = 4000 J + 20000 J = 24000 J 

The example above illustrates how energy flows in response 
to the minus-oneth law of thermodynamics, to achieve thermal 
equilibrium. The impetus for energy flow is the equalization of 
temperature (via the zeroth law), so we say that the measurement 
is isothermal. 

We often want to perform thermodynamic studies isothermally 
because, that way, we need no subsequent corrections for inequal- 
ities in temperature; isothermal measurements generally simplify 
our calculations. 

Why, when letting down a bicycle tyre, is the expelled 
air so cold? 

Thermodynamic work 

When a fully inflated car tyre is allowed to deflate, the air streaming through the 
nozzle is cold to the touch. The pressure of the air within the tyre is fairly high, so 
opening the tyre valve allows it to leave the tyre rapidly - the air movement may even 
cause a breeze. We could feel a jet of cold air on our face if we were close enough. 
As it leaves the tyre, this jet of air pushes away atmospheric air, which requires 
an effort. We say that work is performed. (It is a form of pres- 
sure-volume work, and will be discussed in more depth later, in 
Section 3.2.) 

The internal energy of the gas must change if work is performed, 
because AC/ = q + w. It is unlikely that any energy is exchanged 
so, in this simplistic example, we assume that q = 0. 

Energy is consumed because work w is performed by the gas, 
causing the energy of the gas to decrease, and the change in internal 

Energy added to, or 
work done on, a sys- 
tem is positive. Energy 
removed from, or work 
done by, a system is 



energy is negative. If AU is negative and q = 0, then w is also negative. By corollary, 
the value of w in Equation (3.5) is negative whenever the gas performs work. 

From Chapter 2, we remember again that the simplest way to tell whether the 
internal energy decreases is to check whether the temperature also decreases. We see 
that the gas coming from the tyre is cold because it performs work, which decreases 
its internal energy. 

Why does a tyre get hot during inflation? 

The temperature of 
a tyre also increases 
when inflated, and is 
caused by interparticle 
interactions forming; 
see p. 59. 

Adiabatic changes 

Anyone inflating a tyre with a hand pump will agree that much 
hard work is needed. A car or bicycle tyre usually gets hot during 
inflation. In the previous example, the released gas did thermody- 
namic work and the value of w was negative. In this example, work 
is done to the gas in the tyre, so the value of w is positive. Again, 
we assume that no energy is transferred, which again allows us to 
take q as zero. 

Looking again at Equation (3.5), AU = q + w, we see that if 
q = and w increases (w is positive), then AU increases. This increase in AU 
explains why the temperature of the gas in the tyre increases. 

Let us return to the assertion that q is zero, which implies that the system is ener- 
getically closed, i.e. that no energy can enter or leave the tyre. This statement is 
not wholly true because the temperature of the gas within the tyre will equilibrate 
eventually with the rubber of the tyre, and hence with the outside air, so the tyre 
becomes cooler in accordance with the minus-oneth and zeroth laws of thermody- 
namics. But the rubber with which tyre is made is a fairly good thermal insulator, 
and equilibration is slow. We then make the good approximation that the system is 
closed, energetically. We say the change in energy is adiabatic. 

Energetic changes are adiabatic if they can be envisaged to occur 
while contained within a boundary across which no energy can 
pass. In other words, the energy content within the system stays 
fixed. For this reason, there may be a steep temperature jump in 
going from inside the sealed system to its surroundings - the gas 
in the tyre is hot, but the surrounding air is cooler. 

In fact, a truly adiabatic system cannot be attained, since even 
the most insulatory materials will slowly conduct heat. The best 
approximations are devices such as a Dewar flask (sometimes called a 'vacuum 

A thermodynamic 


cess is adiabatic 


it occurs within 


(conceptual) bound- 

ary across which 


energy can flow. 

Can a tyre be inflated without a rise in temperature? 

Thermodynamic reversibility 

A tyre can indeed be inflated without a rise in temperature, most simply by filling 
it with a pre-cooled gas, although some might regard this 'adaptation' as cheating! 



'Infinitesimal' is the 
reciprocal of infinite, 
i.e. incredibly small. 

Alternatively, we could consider inflating the tyre with a series of, say, 100 short 
steps - each separated by a short pause. The difference in pressure before and after 
each of these small steps would be so slight that the gas within the tyre would be 
allowed to reach equilibrium with its surroundings after adding each increment, and 
before the next. Stated a different way, the difference between the pressure of the gas 
in the hand pump and in the tyre will always be slight. 

We can take this idea further. We need to realize that if there is 
no real difference in pressure between the hand pump and the gas 
within the tyre, then no work would be needed to inflate because 
there would never be the need to pump against a pressure. Alter- 
natively, if the inflation were accomplished at a rate so slow that it 
was infinitesimally slow, then there would never be a difference in pressure, ensuring 
w was always zero. And if w was zero, then U would stay constant per increment. (We 
need to be aware that this argument requires us to perform the process isothermally.) 
It should be clear that inflating a tyre under such conditions is never going to occur 
in practice, because we would not have the time, and the inflation would never be 
complete. But as a conceptual experiment, we see that working at an infinitesimally 
slow rate does not constitute work in the thermodynamic sense. 

It is often useful to perform thought experiments of this type, 
changing a thermodynamic variable at an infinitely slow rate: we 
say we perform the change reversibly. (If we perform a process in a 
non-reversible manner then we say it is 'irreversible'.) As a simple 
definition, a process is said to be reversible if the change occurs 
at an infinitesimal rate, and if an infinitesimal change in an exter- 
nal variable (such as pressure) could change the direction of the 
thermodynamic process. It is seen that a change is only reversible 
if it occurs with the system and surroundings in equilibrium at 
all times. In practice this condition is never attained, but we can 
sometimes come quite close. 

Reversibility can be a fairly difficult concept to grasp, but it 
is invaluable. In fact, the amount of work that can be performed 
during a thermodynamic process is maximized when performing 
it reversibly. 

The discussion here has focused on work done by changes in 
pressure, but we could equally have discussed it in terms of volume 
changes, electrical work (see Chapter 7) or chemical changes (see 
Chapter 4). 

A thermodynamic pro- 
cess is reversible if an 
infinitesimal change in 
an external variable 
(e.g. pressure) can 
change the direction 
in which the process 

The amount of work 
that can be performed 
during a thermody- 
namic process is maxi- 
mized by performing it 

How fast does the air in an oven warm up? 

Absorbing energy 

The air inside an oven begins to get warm as soon as we switch it on. We can 
regard the interior of the oven as a fixed system, so the internal energy U of the 



gases increases as soon as energy is added, since AU = q + w (Equation (3.5)). 
For simplicity, in this argument we ignore the expansion of the atmosphere inside 
the oven. 

But what is the temperature inside the oven? And by how much does the temperature 
increase? To understand the relationship between the temperature and the amount of 
heat entering the system, we must first appreciate that all energies are quantized. The 
macroscopic phenomenon of temperature rise reflects the microscopic absorption of 
energy. During absorption, quanta of energy enter a substance at the lowest energy 
level possible, and only enter higher quantal states when the lower energy states are 
filled. We see the same principle at work when we fill a jar with marbles: the first 
marbles fall to the jar bottom (the position of lowest potential energy); and we only 
see marbles at the top of the jar when all the lower energy levels are filled. Continuing 
the analogy, a wide jar fills more slowly than does a narrower jar, even when we add 
marbles at a constant rate. 

On a macroscopic level, the rate at which the quantal states are filled as a body 
absorbs energy is reflected by its heat capacity C. We can tell how quickly the 
quantum states are occupied because the temperature of a body is in direct proportion 
to the proportion of states filled. A body having a large number of quantum states 
requires a large number of energy quanta for the temperature to increase, whereas a 
body having fewer quantum states fills more quickly, and becomes hot faster. 

Why does water boil more quickly in a kettle 
than in a pan on a stove? 

Heat capacity 

The SI unit of power is 
the watt (W). A heater 
rated at a power of 1 W 
emits 1 Js _1 . 

Most modern kettles contain a powerful element (the salesman's word for 'heater') 
operating at a power of 1000 W or more. A heater emits 1 W if it gives out 1 Js _1 
so, a heater rated at 1000 W emits 1000 Js _1 . We may see this 
power expressed as 1 kW (remember that a small 'k' is shorthand 
for kilo, meaning 1000). By contrast, an electrical ring on the stove 
will probably operate between 600 and 800 W, so it emits a smaller 
amount of heat per second. Because the water absorbs less heat 
energy per unit time on a stove, its temperature rises more slowly. 

The amount of energy a material or body must absorb for its 
temperature to increase is termed its 'heat capacity' C. A fixed 
amount of water will, therefore, get warmer at a slower rate if the 
amount of heat energy absorbed is smaller per unit time. 

Equation (3.6) expresses the heat capacity C in a mathematical 

c *= df )„ a6) 

The heat capacity C 
of a material or body 
relates the amount of 
energy absorbed when 
raising its temperature. 



The expression in Equation (3.6) is really a partial differential: the value of U depends 
on both T and V, the values of which are connected via Equation (1.13). Accordingly, 
we need to keep one variable constant if we are unambiguously to attribute changes in 
Cv to the other. The two subscript ' V terms tell us C is measured while maintaining 
the volume constant. When the derivative is a partial derivative, it is usual to write 
the 'd' as '3'. 

We call Cv 'the heat capacity at constant volume'. With the 
volume constant, we measure Cv without performing any work 
(so w = 0), so we can write Equation (3.6) differently with Aq 
rather than dU. 

Unfortunately, the value of Cv changes slightly with tempera- 
ture; so, in reality, a value of Cy is obtained as the tangent to 
the graph of internal energy (as v) against temperature (as x); see 
Figure 3.5. 

If the change in temperature is small, then we can usually assume 
that Cv has no temperature dependence, and write an approximate 
form of Equation (3.6), saying 

We also call C v the 
isochoric heat capacity. 

A tangent is a straight 
line that meets a curve 
at a point, but not 
does cross it. If the 
heat capacity changes 
slightly with temper- 
ature, then we obtain 
the value of Cv as the 
gradient of the tangent 
to a curve of AU (as y) 
against T (as x). 

C v = 



Analysing Equations (3.6) and (3.7) helps us remember how the 
SI unit of heat capacity Cv is J K _1 . Chemists usually cite a heat 
capacity after dividing it by the amount of material, calling it the specific heat capacity, 
either in terms of J K _1 mol~ orJ K _1 g _1 . As an example, the heat capacity of water 
is 4.18 JK _1 g _1 , which means that the temperature of 1 g of water increases by 1 K 
for every 4.18 J of energy absorbed. 

SAQ 3.1 Show that the molar heat capacity of water is 75.24 J K 1 mol 1 
if Cv — 4.18 J K _1 g _1 . [Hint: first calculate the molar mass of H2O.] 

Tangential gradient = heat capacity, C v 

Figure 3.5 The value of the heat capacity at constant volume Cv changes slightly with temper- 
ature, so its value is best obtained as the gradient of a graph of internal energy (as y) against 
temperature (as x) 



Worked Example 3.4 An electrical heater warms 12 g of water. Its initial temperature 
is 35.0 °C. The heater emits 15 W for 1 min. What is the new temperature of the water? 

Answer strategy. Firstly, we will calculate the energy produced by the 
heater, in joules. Secondly, knowing the heat capacity of the water C, 
we divide this energy by C to obtain the temperature rise. 

(I) To calculate the energy produced by the heater. Remember that 
1 W = Us -1 , so a wattage of 15 W means 15 Js _1 . The heater oper- 
ates for 1 min (i.e. 60 s), so the energy produced is 15 Js x 60s = 
900 J. 

This amount of energy is absorbed by 12 g of water, so the energy 
absorbed per gram is 

The word 'strategy' 
comes from the Greek 
stratos meaning 'army'. 
Strategy originally con- 
cerned military 

900 J 

75 J g" 

(2) To calculate the temperature rise. The change in temperature AT is sufficiently small 

that we are justified in assuming that the value of Cy is independent of temperature. This 

assumption allows us to employ the approximate equation, Equation (3.7). We rearrange 

it to make AT the subject: 


AT = 

C v 

Inserting values: 



75 J g- 1 
4.18 JK-'g" 1 

A! = 17.9 K 

As 1 °C = 1 K, the final temperature of the water is (25.0 + 17.9) °C = 42.9 °C. 

SAQ 3.2 How much energy must be added to 1.35 kg of water in a pan if 
it is to be warmed from 20 °C to its boiling temperature of 100 °C? Assume 

Cv does not vary from 4.18 J K mol . 

The heat capacity Cy is an extensive quantity, so its value depends on how much 
of a material we want to warm up. As chemists, we usually want a value of Cy 
expressed per mole of material. A molar heat capacity is an intensive quantity. 


Another heat capacity is C p , the heat capacity measured at constant pressure (which is 
also called the isobaric heat capacity). The values of C p and Cy will differ, by perhaps 
as much as 5-10 per cent. We will look at C p in more depth in the next section. 



Why does a match emit heat when lit? 

Reintroducing calorimetry 

'Lighting' a match means initiating a simple combustion reaction. Carbohydrates 
in the wood combine chemically with oxygen in the air to form water and carbon 
dioxide. The amount of heat liberated is so great that it catches fire (causing the water 
to form as steam rather than liquid water). 

Heat is evolved because the internal energy of the system changes during the 
combustion reaction. Previously, the oxygen was a gaseous element characterized by 
0=0 bonds, and the wood was a solid characterized by C-C, C-H and C-0 bonds. 
The burning reaction completely changes the number and type of bonds, so the internal 
energies of the oxygen and the wood alter. This explains the change in At/. 

We know from Equation (3.5) that AU = q + w. Because AU changes, one or 
both of q and w must change. It is certain that much energy is liberated because we 
feel the heat, so the value of q is negative. Perhaps work w is also performed because 
gases are produced by the combustion reaction, causing movement of the atmosphere 
around the match (i.e. w is positive). 

The simplest way to measure the change in internal energy AU is to perform a 
reaction in a vessel of constant volume and to look at the amount of heat evolved. 
We perform a reaction in a sealed vessel of constant volume called a calorimeter. In 
practice, we perform the reaction and look at the rise in temperature. The calorimeter is 
completely immersed in a large reservoir of water (see Figure 3.6) and its temperature 
is monitored closely before, during, and after the reaction. If we know the heat 

Water stirrer 





Bomb Sample 

Figure 3.6 Schematic representation of the bomb calorimeter for measuring the changes in internal 
energy that occur during combustion. The whole apparatus approximates to an adiabatic chamber, 
so we enclose it within a vacuum jacket (like a Dewar flask) 



capacity C of both the calorimeter itself and the surrounding water, then we can 
readily calculate the change in energy AC/ accompanying the reaction. 

Why does it always take 4 min to boil an egg properly? 


Most people prefer their eggs to be lightly boiled, with the yellow yolk still liquid 
and the albumen solid and white. We say the egg white has been 'denatured'. The 
variation in egg size is not great. An average egg contains essentially a constant 
amount of yolk and albumen, so the energy necessary to heat both the yolk and 
albumen (and to denature the albumen) is, more or less, the same for any egg. 

If the energy required to cook an egg is the same per egg, then the simplest way 
to cook the egg perfectly every time is to ensure carefully that the same amount of 
energy is absorbed. Most people find that the simplest way to do this is to immerse 
an egg in boiling water (so the amount of energy entering the egg per unit time is 
constant), and then to say, 'total energy = energy per second x number of seconds'. 
In practice, it seems that most people prefer an egg immersed in boiling water for 
about 240 s, or 4 min. 

This simple example introduces the topic of thermochemistry. 
In a physical chemist's laboratory, we generally perform a similar 
type of experiment but in reverse, placing a sample in the calorime- 
ter and measuring the energy released rather than absorbed. The 
most commonly performed calorimetry experiment is combustion 
inside a bomb calorimeter (Figure 3.6). We place the sample in the 
calorimeter and surround it with oxygen gas at high pressure, then 
seal the calorimeter securely to prevent its internal contents leaking 
away, i.e. we maintain a constant volume. An electrical spark then 
ignites the sample, burning it completely. A fearsome amount of energy is liberated 
in consequence of the ignition, which is why we call this calorimeter a 'bomb'. 

The overall heat capacity of the calorimeter is a simple function of the amount of 
steel the bomb comprises and the amount of water surrounding it. If the mass is m 

Thermochemistry is 
the branch of thermo- 
dynamics concerned 
with the way energy 
is transferred, released 
or consumed during a 
chemical reaction. 

and the heat capacity is C, then the overall heat capacity is expressed by 

C (overall) = \ m (steel) x C m (steel)) + \ m (water) x C m (water)) 


If the amount of compound burnt in the calorimeter is n, and 
remembering that no work is done, then a combination of Equa- 
tions (3.7) and (3.8) suggests that the change in internal energy 
occurring during combustion is given by 

Qoveraii) is the heat 
capacity of the reac- 
tion mixture and the 



m (combustion) 







where the 'm' means 'molar'. The negative sign arises from the conventions above, 
since heat is given out if the temperature goes up, as shown by A T being positive. 
It is wise first to calibrate the calorimeter by determining an 
accurate value of C( ove raii)- This is achieved by burning a com- 
pound for which the change in internal energy during combustion 
is known, and then accurately warming the bomb and its reservoir 
with an electrical heater. Benzoic acid (I) is the usual standard of 
choice when calibrating a bomb calorimeter. 

Al/ (C ombustion) for ben- 
zoic acid is -3.2231 MJ 
mor 1 at 298 K. 


The electrical energy passed is q, defined by 

q = V x I x t 


where V is the voltage and / the current of the heater, which operates for a time of 
t seconds. 

SAQ 3.3 A voltage of 10 V produces a current of 1.2 A when applied 
across a heater coil. The heater is operated for 2 min and 40 s. Show that 
the energy produced by the heater is 1920 J. 

We can assume C p is 
constant only if AT is 
small. For this reason, 
we immerse the 'bomb' 
in a large volume of 
water. This explains 
why we need to oper- 
ate the heater for a 
long time. 


Worked Example 3.5 A sample of glucose (10.58 g) is burnt com- 
pletely in a bomb calorimeter. What is the change in internal energy 
At/ if the temperature rises by 1.224 K? The same heater as that in 
SAQ 3.3 is operated for 11 240 s to achieve a rise in temperature of 
1.00 K. 

Firstly, we calculate the energy evolved by the reaction. From 
Equation (3.10), the energy given out by the heater is q — 10 V x 
1.2Ax 11240s = 134880J. 

Secondly, we determine the value of C( 0ve raii) for the calorimeter, 
saying from Equation (3.9) 


energy released 

134 880 J 

change in temperature 1.00 K 

C= 134 880 J K" 



Thirdly, we determine the amount of glucose consumed n. We obtain the value of n 

as 'amount = mass 4- molar mass'. The molar mass of glucose is 180 gmor 
number of moles is 5.88 x 10~ 2 mol. 

Finally, we calculate the value of At/ from Equation (3.9). Insert- 
ing values: 

so the 






134 880 JKT 1 x 1.224K 
0.0588 mol 

= -2.808 MJmol" 

Notice how this value of A U is negative. As a good generalization, the 
change in internal energy AU liberated during combustion is negative, 
which helps explain why so many fires are self-sustaining (although 
see Chapter 4). 

The value of A f/( Com b us tion) for glucose is huge, but most values of 
AU are smaller, and are expressed in kilo joules per mole. 

SAQ 3.4 A sample of anthracene (C14H10, II) was burnt 
in a bomb calorimeter. A voltage of 10 V and a current 
of 1.2 A were passed for exactly 15 min to achieve the 
same rise in temperature as that caused by the burning 
of 0.40 g. Calculate the molar energy liberated by the 

The minus sign is a 
consequence of the 
way Equation (3.9) is 

An energy change of MJ 
mor 1 is exceptional. 
Most changes in AU are 
smaller, of the order of 
kJ mor 1 . 

We often calculate a 
volume of AU but 
cite the answer after 
adjusting for pres- 
sure-volume work; 
see p. 102. 



The large value of At/ in Worked Example 3.5 helps explain why sweets, meals and 
drinks containing sugar are so fattening. If we say a single spoonful of sugar comprises 
5 g of glucose, then the energy released by metabolizing it is the same as that needed 
to raise a 3.5 kg weight from the ground to waist level 7000 times. 

(We calculate the energy per lift with Equation (3.4), saying E = m x g x h, where 
m is the mass, g is the acceleration due to gravity and h is the height through which 
the weight is lifted.) 



Why does a watched pot always take so long to boil? 

Introduction to Hess's law 

We sometimes say, 'A watched pot never boils'. This empirical observation - that 
we get bored waiting a long time for the pot to boil - follows because we need to put 
a lot of energy (heat) in order for the water to boil. The amount 
of energy we can put into the water per unit time was always low 
in the days of coal and wood fires. Accordingly, a long time was 
required to boil the water, hence the long wait. 

Imagine we want to convert 1 mol of water starting at a room 
temperature of, say, 25 °C to steam. In fact we must consider 
two separate thermodynamic processes: we first consider the heat 
needed to warm the water from 25 °C to its boiling temperature 
of 100 °C. The water remains liquid during this heating process. 
Next, we convert 1 mol of the liquid water at 100 °C to gaseous 
water (i.e. we boil it), but without altering the temperature. 

We will at the moment ignore once more the problems caused 
by volume changes. The change in internal energy AC/( 0V eraii) f° r 

The popular saying 
A watched pot never 
boils' arose when most 
fires were wood or 
coal, neither of which 
generates heat as fast 
as, say, a modern 1 kW 

This argument relies 
only on words. In real- 
ity, the situation is 
somewhat more com- 
plicated because water 
expands slightly on 
heating, and greatly on 

the overall process H20(i) at 25 C 
separated into two components: 

H 2 (g) at 100 C can be 

Energy AC/i relates to the process 
H 2 (1) at 25 °C ► H 2 0(i) at 100 °C 

Energy A C/ 2 relates to the process 

H 2 ( i) at 100 °C ► H 2 (g) at 100 °C 

so AUi relates to warming the water until it reaches the boiling temperature, and 
AC/ 2 relates to the actual boiling process itself. 
We can obtain A[/( overa ii) algebraically, according to 

We can obtain the 
answer in several dif- 
ferent ways because 
internal energy is a 
'state function'. 

A [/ (0V e r all) = AC/] + AC/ 2 


Hess's law states that 
the value of an energy 
obtained is indepen- 
dent of the number of 
intermediate reaction 
steps taken. 

In practice, we could have measured A [/(overall) directly in the 
laboratory. Alternatively, we could have measured AU\ or AC/ 2 
in the laboratory and found the AC/ values we did not know in a 
book of tables. Either way, we will get the same answer from these 
two calculation routes. 

Equation (3.11) follows directly from AC/ being a state function, 
and is an expression of Hess's law. The great German fhermody- 
namicist Hess observed in 1840 that, 'If a reaction is performed in 
more than one stage, the overall enthalpy change is a sum of the 
enthalpy changes involved in the separate stages'. 

We shall see shortly how the addition of energies in this way 
provides the physical chemist with an extremely powerful tool. 


Hess's law is a restatement if the first law of thermodynamics. We do not need to 
measure an energy change directly but can, in practice, divide the reaction into several 
constituent parts. These parts need not be realizable, so we can actually calculate the 
energy change for a reaction that is impossible to perform in the laboratory. The only 
stipulation is for all chemical reactions to balance. 

The importance of Hess's law lies in its ability to access information about a 
reaction that may be difficult (or impossible) to obtain experimentally, by looking at 
a series of other, related reactions. 

3.2 Enthalpy 

How does a whistling kettle work? 

Pressure-volume work 

The word 'work' in the question above could confuse. In common parlance, we say 
a kettle works or does not work, meaning it either functions as a kettle or is useless. 
But following the example in the previous section, we now realize how the word 
'work' has a carefully defined thermodynamic meaning. 'Operate' would be a better 
choice in this context. In fact, a kettle does not perform any work at all, since it has 
no moving parts and does not itself move. 

In a modern, automatic kettle, an electric heater warms the water inside the ket- 
tle - we call it the 'element'. The electric circuit stops when the water reaches 100 °C 
because a temperature-sensing bimetallic strip is triggered. But the energy for a more 
old-fashioned, whistling kettle comes from a gas or a coal hob. The water boils on 
heating and converts to form copious amounts of gas (steam), which passes through 
a small valve in the kettle lid to form a shrill note, much like in a football ref- 
eree's whistle. 

The whistle functions because boiling is accompanied by a change in volume, so 
the steam has to leave the kettle. And the volume change is large: the volume per 
mole of liquid water is 18 cm 3 (about the size of a small plum) but the volume of a 
mole of gaseous water (steam) is huge. 

SAQ 3.5 Assuming steam to be an ideal gas, use the ideal-gas equation 
(Equation (1.13)) to prove that 1 mol of steam at 100°C (373 K) and 
standard pressure (p s = 10 5 Pa) has a volume of is 0.031 m 3 . 

A volume of 0.031 m 3 corresponds to 31 dm 3 , so the water increases its volume by 
a factor of almost 2000 when boiled to generate steam. This staggering result helps 
us realize just how great the increase in pressure is inside the kettle when water boils. 

The volume inside a typical kettle is no more than 2 dm 3 . To avoid a rapid build 
up of pressure within the kettle (which could cause an explosion), the steam seeks 
to leave the kettle, exiting through the small aperture in the whistle. All the vapour 
passes through this valve just like a referee blowing 'time' after a game. And the 


large volume ensures a rapid exit of steam, so the kettle produces an intense, shrill 
whistle sound. 

The steam expelled from the kettle must exert a pressure against the air as it leaves 
the kettle, pushing it aside. Unless stated otherwise, the pressure of the air surrounding 
the kettle will be 10 5 Pa, which we call 'standard pressure' p^. The value of p^ is 
10 5 Pa. The steam must push against this pressure when leaving the kettle. If it 
does not do so, then it will not move, and will remain trapped within the kettle. This 
pushing against the air represents work. Specifically, we call it pressure-volume work, 
because the volume can only increase by exerting work against an external pressure. 

The magnitude of this pressure- volume work is w, and is expressed by 

w = -A( P V) (3.12) 

where A(pV) means a change in the product of p x V. Work is done to the gas 
when it is compressed at constant pressure, i.e. the minus sign is needed to make 
w positive. 

Equation (3.12) could have been written as Ap x A V if both the pressure and the 
volume changed at the same time (an example would be the pushing of a piston 
in a car engine, to cause the volume to decrease at the same time as the pressure 
increases). In most of the physicochemical processes we will consider here, either 
p or V will be constant so, in practice, there is only one variable. And with one 
variable, Equation (3.12) becomes either w = p x AV or w = AV x p, depending 
on whether we hold p or V constant. 

Most chemists perform experiments in which the contents of our 

For most purposes, a 
chemist can say w = 
p x AV. 

beaker, flask or apparatus are open to the air - obvious examples 
include titrations and refluxes, as well as the kinetic and electro- 
chemical systems we consider in later chapters. The pressure is 
the air pressure (usually p & ), which does not change, so any pres- 
sure-volume work is the work necessary to push back the atmosphere. For most 
purposes, we can say w = pAV. 

It should be obvious that the variable held constant - whether p or V - cannot be 
negative, so the sign of w depends on which of the variables we change, so the sign 
of w in Equation (3.12) depends on the sign(s) of Ap or AV. The sign of w will be 
negative if we decrease the volume or pressure while performing work. 

Worked Example 3.6 We generate 1 mol of water vapour in a kettle by boiling liquid 
water. What is the work w performed by expansion of the resultant steam? 

We have already seen how 1 mol of water vapour occupies a volume of 0.031 m 3 (see 
SAQ 3.5). This volume of air must be pushed back if the steam is to leave the kettle. The 
external pressure is p e , i.e. 10 5 Pa. 
The change in volume 

AV — V(final) — ^(initial) 


AV — V(„ a ter, g) — V( water , 1) 



Inserting numbers yields 

AV = 




volume per mole 
of steam 

18 x 10~ 6 

J m 3 = 0. 

031m 3 


volume per mole 
of liquid water 

Each aliquot of 1 cm 3 
represents a volume of 
1 x 10 6 m 3 . 

Inserting values into Equation (3.12), w — pAV: 

w = 10 5 Pax 0.031m 3 

io = 3.1x 10 3 J 

so the pressure -volume work is 3100 J. 

We often encounter energies of the order of thousands of joules. 
As a shorthand, we often want to abbreviate, so we rewrite the 
answer to Worked Example 3.6, and say w = 3.1 x 1000 J. Next, 
we substitute for the factor of 1000 with an abbreviation, generally 
choosing a small letter 'k'. We rewrite, saying w = 3.1 kJ. 

SAQ 3.6 What is the work done when the gas from a 
party balloon is released? Assume the inflated balloon 
has a volume of 2 dm 3 , and a volume of 10 cm 3 when 
deflated. Assume there is no pressure change, so p = p & . 
[Hint: 1 1 = lx KHm 3 ]. 

An energy expressed 
with a letter 'k' in the 
answer means "thou- 
sands of joules'. We 
say kilojoules. The 
choice of 'k' comes 
from the Greek word 
for thousand, which is 

Worked Example 3.7 What is the work performed when inflating a car tyre from p f 
to 6 x p e . Assume the volume inside the tyre stays constant at 0.3 m 3 . 

Firstly, we calculate the change in pressure, from an equation like Equation (3.1), Ap = 

P(flnal) - P(initial), SO Ap = (6 - 1) X p e , i.e. 

Ap — 5 x p e 
A P = 5x 10 5 Pa 

Then, inserting values into Equation (3.5), w — Ap x V: 

w = 5 x 10 5 Pax 0.3m 3 


w = 15000 J = 15kJ 


How much energy do we require during a distillation? 

The effect of work on AU: introducing enthalpy H 

Performing a simple distillation experiment is every chemist's delight. We gently 
warm a mixture of liquids, allowing each component to boil off at its own charac- 
teristic temperature (the 'boiling temperature' 7(b ii))- Each gaseous component cools 
and condenses to allow collection. Purification and separation are thereby effected. 

Although we have looked already at boiling and condensation, until now we have 
always assumed that no work was done. We now see how invalid this assumption 
was. A heater located within the distillation apparatus, such as an isomantle, supplies 
heat energy q to molecules of the liquid. Heating the flask increases the internal 
energy U of the liquids sufficiently for it to vaporize and thence become a gas. 

But not all of the heater's energy q goes into raising U. We need some of it to 
perform pressure -volume work, since the vapour formed on boiling works to push 
back the external atmosphere. The difference between the internal energy U and the 
available energy (the enthalpy) is given by 

AH = AU + pAV (3.13) 

H is a state function since p, V and U are each state functions. As a state function, 
the enthalpy is convenient for dealing with systems in which the pressure is constant 
but the volume is free to change. This way, an enthalpy can be equated with the 
energy supplied as heat, so q = AH. 

Worked Example 3.8 A mole of water vaporizes. What is the change in enthalpy, A HI 
Take pressure as p^ . 

We have already seen in the previous section that AU — +40.7 kJ per mole of water, 
and from SAQ 3.5 the volume of 1 mol of water vapour is 0.031 m 3 per mole of water. 
Inserting values into Equation (3.13): 


40700 J mol" 1 \+( 10 5 Pa x 0.031m 3 mol" 1 J 

/ / / 

AU p AV 


AH = 40700 Jmol" 1 + (SlOOJmor 1 ) 


AH = 45.8 kJmol" 1 

In this example, the difference between AU and AH is about 11 per cent. 
The magnitude of the difference will increase as the values of AH and AU get 



Justification Box 3.1 

We saw above how the work w performed by a gas is pV. Because performing work 
will decrease the internal energy, we say 

w = —pV 


Substitution of this simple relationship into the definition of internal energy in Equation 
(3.5) yields 

U = q-pV (3.15) 

and rearranging Equation (3.15) yields q = U + pV . 
This combination of variables occurs so often in phys- 
ical chemistry, that we give it a name: we call it the 
enthalpy, and give it the symbol H. Accordingly, we 
rewrite Equation (3.15) as: 

We often call a col- 
lection of variables a 
'compound variable'. 

H = U + pV 


The change in enthalpy AH during a thermodynamic process is defined in terms of 
internal energy and pressure-volume work by 

AH = AU + A(pV) 


Because it is usual to perform a chemical experiment with the top of the beaker 
open to the open air, the pressure p during most chemical reactions and thermodynamic 
processes is the atmospheric pressure p e ', Furthermore, this pressure will not vary. In 
other words, we usually simplify A(pV) saying pAV because only the volume changes. 

Accordingly, Equation (3.17) becomes 

AH = AU + pAV 


The equation in the form of Equation (3.18) is the usual form we use. Changes in U 
are not equal to the energy supplied as heat (at constant pressure p) because the system 
employs some of its energy to push back the surroundings as they expand. The pV term 
for work is, therefore, a correction for the loss of energy as work. Because many, if 
not most, physicochemical measurements occur under conditions of constant pressure, 
changes in enthalpy are vitally important because it automatically corrects for the loss 
of energy to the surroundings. 


One of the most common mistakes we make during calculations of this kind is forgetting 
the way 'k' stands for '1000'. Think of it this way: a job advertisement offers a salary 
of £14 k. We would be very upset if, at the end of the first year, we were given just 
£14 and the employer said he 'forgot' the 'k' in his advert! 



Why does the enthalpy of melting ice decrease 
as the temperature decreases? 

Temperature dependence of enthalpy 

The enthalpy of freezing water is —6.00 kJmol -1 at its normal freezing temperature 
of 273.15 K. The value is negative because energy is liberated during freezing. But 
the freezing temperature of water changes if the external pressure 
is altered; so, for example, water freezes at the lower temperature 
of 253 K when a pressure of about 100 x p & is applied. This high 
pressure is the same as that along the leading edge of an aeroplane 
wing. At this lower temperature, AH( me \ t ) =5.2 kJmol - . 
The principal cause of A//( me it) changing is the decreased temperature. The magni- 
tude of an enthalpy depends on the temperature. For this reason, we need to cite the 
temperature at which an enthalpy is determined. If the conditions are not cited, we 
assume a temperature of 298 K and a pressure of p* . We recognize these conditions 

The word 'normal' in 
this context means 'at 
a pressure of p*'. 

as s.t.p. Values of enthalpy are often written as A//, 

r 298 K 

for this reason. 

The temperature dependence of the standard enthalpy is related by Kirchhoff's law: 

AH* = AH* + 


AC p (T)dT 


Reminder. The 'curly d' 
symbols 3 tells us the 
bracketed term in the 
equation is a 'partial 

C P (T) means C p as a 
function of thermody- 
namic temperature. 

where C p is the molar heat capacity at constant pressure of the 
substance in its standard state at a temperature of T . We define C p 
according to 



C p = 


The value of C p is itself a function of temperature (see p. 140), 
which explains why we integrate C P {T) rather than C p alone. 

The Kirchhoff law is a direct consequence of the heat capacity at 
constant pressure being the derivative of enthalpy with respect to 
temperature. It is usually sufficient to assume that the heat capacity 
C p is itself independent of temperature over the range of temperatures required, in 
which case Equation (3.19) simplifies to 

AH t \ = AH; Ti + AC P (T 2 - TO 


The experimental scientist should ensure the range of temperatures is slight if calcu- 
lating with Equation (3.21). 

Worked Example 3.9 The standard enthalpy of combustion AH* for benzoic acid 
(I) is —3223.1 kJmol -1 at 20°C. What is AH C 298 K ? The change in C p during the 


reaction is 118.5 JKr'moE 1 . Assume this value is temperature independent over this 
small temperature interval. 

Inserting data into Equation (3.21): 

A// r*298 k = -3223.7 kJmor 1 + 118.5 JKT 1 mol _1 (298 - 293) K 
A// r ° 29g K = -3223.7 kJmor 1 + 592.5 JmoP 1 


A/f r ° 298 K = -3223.1 kJmoP'or- 3.2231 MJmol" 1 

SAQ 3.7 Ethane burns completely in oxygen to form carbon dioxide and 
water with an enthalpy of AH & = -1558.8 kJ mol" 1 at 25 °C. What is AH* 
at 80 °C? First calculate the change in heat capacity C p from the data in 
the following table and Equation (3.22). 


C p at 80°C/J K^mol -1 

C2H6(g) 02(g) C02(g) H 2 ( |) 

52.6 29.4 37.1 75.3 

ac p = J2 vC p- J2 vC p (3 - 22 ^ 

products reactants 

where the upper-case Greek letter Sigma £ means 'sum of, and the 
lower-case Greek letters v (nu) represent the stoichiometric number of 
each species, which are the numbers of each reagent in a fully balanced 
equation. In the convention we adopt here, the values of v are positive for 
products and negative for reactants. 

Justification Box 3.2 

Starting with the definition of heat capacity in Equation 


\w) p = Cp 

This equation represents C p for a single, pure substance. 


; the variables 


dH = C p dT 

Then we integrate between limits, saying the enthalpy is H\ at T\ 

and #2 

at T 2 : 

/ dH = / C p dT 



Integrating yields 

(H 2 - HO = CJT 2 - TO 


where the term on the left-hand side is AH. Equation (3.23) relates to a single, pure sub- 

If we consider a chemical reaction in which several chemicals combine, we can write 
an expression like this for each chemical. Each chemical has a unique value of AH and 
C p , but the temperature change (T 2 — 7i) remains the same for each. 

We combine each of the AH terms to yield AH r T (i.e. AH r e at T 2 ) and AH T T . 
Combining the C p terms according to Equation (3.22) yields AC p . Accordingly, Equa- 
tion (3.23) then becomes Equation (3.21), i.e.: 


r T 2 

Aff* + AC„(72 - 7i) 

Why does water take longer to heat in a pressure 
cooker than in an open pan? 

The differences between C v and C p 


A pressure cooker is a sealed cooking pan. Being sealed, as soon as boiling occurs, 
the pressure of steam within the pan increases dramatically, reaching a maximum 
pressure of about 6 x p & , causing the final boiling temperature to 
increase (see Fig. 5.12 on p. 200). Unlike other pans, the internal 
volume is fixed and the pressure can vary; the pressure in most 
pans is atmospheric pressure (~ p & ), but the volume of the steam 
increases continually. 

The heat capacity of the contents in a pressure cooker is Cy 
because the internal volume is constant. By contrast, the heat capacity of the food 
or whatever inside a conventional pan is C p . The water is a pressure cooker warms 
slower because the value of C p is always smaller than Cy. And being smaller, the 
temperature increases faster per unit input of energy. 

In fact, the relationship between Cy and C p is given by 

See p. 199 to see why 
a pressure cooker can 
cook faster than a con- 
ventional, open pan. 

It is relatively rare 
that we need C v val- 
ues; most reactions are 
performed at constant 
pressure, e.g. refluxing 
a flask at atmospheric 

Cy — C r 



where we have met all terms previously. 

Worked Example 3.10 What is the heat capacity Cy of 1 mol of 
water? Take the value of C p from SAQ 3.7. 

Rearranging Equation (3.24) slightly yields 

C v — nR + C B 


Inserting values: 

C v = (1 x 8.3 + 75.3) J KT'mor 1 

so C v = 83.6 JK-'mor 1 . 

The value of Cv is 11 per cent higher than C p , so the water in the pressure cooker 
will require 1 1 per cent more energy than if heated in an open pan. 

Justification Box 3.3 

Starting with the definition of enthalpy in Equation (3.16): 

H = U + pV 

The pV term can be replaced with 'nRT' via the ideal-gas equation (Equation (1.13)), 

H = U + nRT 

The differential for a small change in temperature is 

The values of n and R 
dH = dU + nRdT are constants and do 

not change. 

dividing throughout by dr yields 

™) = ( 9 JL) +nR 

dT J \dT J 
The first bracket equals CV and the second bracket equals C p , so 

Cy = C p + nR 

which is just Equation (3.24) rearranged. Dividing throughout by n yields the molar 
heat capacities: 

C v = C P + R (3.25) 

Why does the temperature change during a reaction? 

Enthalpies and standard enthalpies of reaction: AH r and AH* 

One of the simplest definitions of a chemical reaction is 'changes in the bonds'. All 
reactions proceed with some bonds cleaving concurrently with others forming. Each 


bond requires energy to form, and each bond liberates energy when breaking (see 
p. 63 ff). Typically, the amount of energy consumed or liberated is characteristic of 
the bond involved, so each C-H bond in methane releases about 220 kJmol -1 of 
energy. And, as we have consistently reported, the best macroscopic indicator of a 
microscopic energy change is a change in temperature. 

Like internal energy, we can never know the enthalpy of a reagent; only the change 
in enthalpy during a reaction or process is knowable. Nevertheless, we can think of 
changes in H. Consider the preparation of ammonia: 

N 2(g ) + 3H 2(g) ► 2NH 3(g) (3.26) 

We obtain the standard enthalpy change on reaction AH^ as a sum of the molar 
enthalpies of each chemical participating in the reaction: 

products reactants 

The values of v for the reaction in Equation (3.26) are V(nh 3 ) = +2, V(h 2 ) = —3 and 
v (N2 ) = — 1. We obtain the standard molar enthalpy of forming ammonia after inserting 
values into Equation (3.27), as 

AH; = 2/ONH 3 ) " t#m (N 2 ) + 3/OH 2 )] 

SAQ 3.8 Write out an expression for AHf for the reaction 2NO + 2 -> 
2N0 2 in the style of Equation (3.27). 

Unfortunately, we do not know the enthalpies of any reagent. All we can know is a 
change in enthalpy for a reaction or process. But what is the magnitude of this energy 
change? As a consequence of Hess's law (see p. 98), the overall change in enthalpy 
accompanying a reaction follows from the number and nature of the bonds involved. 
We call the overall enthalpy change during a reaction the 'reaction enthalpy' AH V , 
and define it as 'the change in energy occurring when 1 mol of reaction occurs'. In 
consequence, its units are J mol -1 , although chemists will usually want to express 
AH in kJ mol" 1 . 

In practice, we generally prefer to tighten the definition of AH V above, and look 
at reagents in their standard states. Furthermore, we maintain the temperature T at 
298 K, and the pressure p at p e . We call these conditions standard temperature and 
pressure, or s.t.p. for short. We need to specify the conditions because temperature and 
pressure can so readily change the physical conditions of the reactants and products. 
As a simple example, elemental bromine is a liquid at s.t.p., so we say the standard 
state of bromine at s.t.p. is Br 2 (i). If a reaction required gaseous bromine Br 2 ( g ) then 
we would need to consider an additional energy - the energy of vaporization to effect 
the process Br 2 (i) — »■ Br 2 ( g ). Because we restricted ourselves to s.t.p. conditions, we 
no longer talk of the reaction enthalpy, but the 'standard reaction enthalpies' A// r e , 
where we indicate the standard state with the plimsoll sign ' e '. 



In summary, the temperature of a reaction mixture changes because energy is 
released or liberated. The temperature of the reaction mixture is only ever constant in 
the unlikely event of AH^ being zero. (This argument requires an adiabatic reaction 
vessel; see p. 89.) 

Some standard enthalpies have special names. We consider below some of the more 
important cases. 

Are diamonds forever? 

Some elements exist in 
several different crys- 
tallographic forms. The 
differing crystal forms 
are called allotropes. 

Enthalpies of formation 

We often hear it said that 'diamonds are forever'. There was even a James Bond novel 
and film with this title. Under most conditions, a diamond will indeed last forever, 
or as near 'for ever' as makes no difference. But is it an absolute statement of fact? 

Diamond is one of the naturally occurring allotropes of carbon, 
the other common allotrope being graphite. (Other, less common, 
allotropes include buckminster fullerine.) If we could observe a 
diamond over an extremely long time scale - in this case, several 
billions of years - we would observe a slow conversion from bril- 
liant, clear diamond into grey, opaque graphite. The conversion 
occurs because diamond is slightly less stable, thermodynamically, 
than graphite. 

Heating graphite at the same time as compressing it under enormous pressure 
will yield diamond. The energy needed to convert 1 mol of graphite to diamond is 
2.4 kJmol - . We say the 'enthalpy of formation' AHf for the diamond is +2.4 kJ 
mol -1 because graphite is the standard state of carbon. 

We define the 'standard enthalpy of formation' AH^ as the 
enthalpy change involved in forming 1 mol of a compound from 
its elements, each element existing in its standard form. Both T 
and p need to be specified, because both variables influence the 
magnitude of AH. Most books and tables cite AH^ at standard 
pressure p e and at a temperature of 298 K. Table 3.1 cites a few 
representative values of AH f . 

It will be immediately clear from Table 3.1 that several val- 
ues of AHf are zero. This value arises from the definition we 
chose, above: as AHf relates to forming a compound from its con- 
stituent elements, it follows that the enthalpy of forming an element 
can only be zero, provided it exists in its standard state. Inci- 
dentally, it also explains why A//f(Br2, 1) = but A//f(Br2, g) = 
29.5 kJmol -1 , because the stable form of bromine is liquid at s.t.p. 

For completeness, we stipulate that the elements must exist in 
their standard states. This sub-clause is necessary, because whereas 
most elements exist in a single form at s.t.p. (in which case their 
enthalpy of formation is zero), some elements, such as carbon 

The 'standard enthalpy 
of formation' A/-/ f & is 
the enthalpy change 
involved in forming 
1 mol of a compound 
or non-stable allotrope 
from its elements, each 
element being in its 
standard form, at s.t.p. 

We define the enthalpy 
of formation of an ele- 
ment (in its normal 
state) as zero. 


Table 3.1 Standard enthalpies of formation AH f at 298 K 

Compound AH f /kj moP 1 


methane (CH4, g) 

ethane (CH3CH3, g) 

propane (CH 3 CH 2 CH 3 , g) 

w-butane (C4H10, g) 

ethane (CH 2 =CH 2 , g) 

ethyne (CH=CH, g) 

c/i-2-butene (C4H8, g) 

?ra«.s-2-butene (C4H8, g) 

w-hexane (Cf,Hi4, 1) 

cyclohexane (C6H12, 1) 


methanol (CH 3 OH, 1) 
ethanol (C 2 H 5 OH, 1) 

Aromatic s 

benzene (C6H 6 , 1) 
benzene (CeHj,, g) 
toluene (CH 3 C 6 H 5 ,1) 


a-D-glucose (C6H 12 06, s) 
jS-D-glucose^gH^Oe, s) 
sucrose (Ci 2 H 22 0n, s) 


bromine (Br 2 , 1) 
bromine (Br 2 , g) 
chlorine (Cl 2 , g) 
chlorine (CI, g) 
copper (Cu, s) 
copper (Cu, g) 
fluorine (F 2 , g) 
fluorine (F, g) 
iodine (I 2 , s) 
iodine (I 2 , g) 
iodine (I, g) 
nitrogen (N, g) 
phosphorus (P, white, s) 
phosphorus (P, red, s) 
sodium (Na, g) 
sulphur (S, rhombic, s) 
sulphur (S, monoclinic, s) 


carbon (diamond, s) 
carbon monoxide (CO, g) 

— ;t.o 





































Table 3.1 (continued) 


AH*/kJ moP 1 

carbon dioxide (CO2, g) 


copper oxide (CuO, s) 


hydrogen oxide (H 2 0, 1) 


hydrogen oxide (H2O, g) 


hydrogen fluoride (HF, g) 


hydrogen chloride (HC1, g) 


nitrogen hydride (NH 3 , g) 


nitrogen hydride (NH3, aq) 


nitrogen monoxide (NO, g) 


nitrogen dioxide (NO2, g) 


phosphine (PH 3 , g) 


silicon dioxide (SiC>2, s) 


sodium hydroxide (NaOH, s) 


sulphur dioxide (SO2, g) 


sulphur trioxide (SO3, g) 


sulphuric acid (H2SO4, 1) 


(above), sulphur or phosphorus, have allotropes. The enthalpy of formation for the 
stable allotrope is always zero, but the value of A//f for the non-stable allotropes will 
not be. In fact, the value of A//f for the non-stable allotrope is cited with respect to 
the stable allotrope. As an example, A//f for white phosphorus is zero by definition 
(it is the stable allotrope at s.t.p.), but the value of A//f for forming red phosphorus 
from white phosphorus is 15.9 kJmol -1 . 

If the value of A//f is determined within these three constraints of standard T, 
standard p and standard allotropic form, we call the enthalpy a standard enthalpy, 
which we indicate using the plimsoll symbol '*' as A// f & . 

To conclude: are diamonds forever? No. They convert slowly into graphite, which 
is the stablest form of carbon. Graphite has the lowest energy for any of the allotropes 
of carbon, and will not convert to diamond without the addition of energy. 

Why do we burn fuel when cold? 

Enthalpies of combustion 

A common picture in any book describing our Stone Age forebears shows short, hairy 
people crouched, warming themselves round a flickering fire. In fact, fire was one of 
the first chemical reactions discovered by our prehistoric ancestors. Primeval fire was 
needed for warmth. Cooking and warding off dangerous animals with fire was a later 
'discovery' . 

But why do they burn wood, say, when cold? The principal reactions occurring 
when natural materials burn involve chemical oxidation, with carbohydrates 
combining with elemental oxygen to yield water and carbon dioxide. Nitrogen 



compounds yield nitrogen oxide, and sulphur compounds yield sulphur dioxide, which 
itself oxidizes to form SO3. 

Let us simplify and look at the combustion of the simplest hydrocarbon, methane. 
CH4 reacts with oxygen according to 

CH4(g) + 20 


> C0 2 ( £) + 2H 2 



The reaction is very exothermic, which explains why much of the developed world 
employs methane as a heating fuel. We can measure the enthalpy change accompany- 
ing the reaction inside a calorimeter, or we can calculate a value with thermochemi- 
cal data. 

This enthalpy has a special name: we call it the enthalpy of 
combustion, and define it as the change in enthalpy accompanying 
the burning of methane, and symbolize it as A//( com b ust ion) or just 
AH C . In fact, we rarely perform calculations with AH C but with the 
standard enthalpy of combustion AH^, where the plimsoll symbol 
' e ' implies s.t.p. conditions. 

Table 3.2 contains values of A// c & for a few selected organic 
compounds. The table shows how all value of AH^ are nega- 
tive, reminding us that energy is given out during a combustion reaction. We say 
combustion is exothermic, meaning energy is emitted. All exothermic reactions are 
characterized by a negative value of A// c °. 

But we do not have to measure each value of A// c & : we can 
calculate them if we know the enthalpies of formation of each 
chemical, product and reactant, we can adapt the expression in 
eq. (3.27), saying: 

Most authors abbre- 
viate 'combustion' to 
just V, and symbolize 
the enthalpy change 
as AHf . Others write 

A "(comb)- 

We can use equations 
like Equation (3.27) for 
any form of enthalpy, 
not just combustion. 


= J2 vA// f e - Yl vAH ? 



We could not perform 
cycles of this type 
unless enthalpy was 
a stare function. 

The word 'calorific' 
means heat contain- 
ing, and comes from 
the Latin calor, mean- 
ing 'heat'. 

where each AH term on the right-hand side of the equation is a 
molar enthalpy of formation, which can be obtained from tables. 

Worked Example 3.11 The wood mentioned in our title question 
is a complicated mixture of organic chemicals; so, for simplicity, we 
update the scene. Rather than prehistoric men sitting around a fire, we 
consider the calorific value of methane in a modern central-heating 
system. Calculate the value of AH C for methane at 25 °C using molar 
enthalpies of formation A H f . 
The necessary values of AH f are: 

Species (all as gases) 
AHf/kJmoT 1 

CH 4 


o 2 

co 2 


H 2 



Table 3.2 Standard enthalpies of combustion AH C 
for a few organic compounds (all values are at 298 K) 


AH*/kJ moP 1 


methane (CH 4 , g) 


ethane (CH 3 CH 3 , g) 


propane (CH 3 CH 2 CH 3 , g) 


n-butane (C4H10, g) 


cyclopropane (C 3 H(,, g) 


propene (C 3 H 6 , g) 


1-butene (C 4 H 8 , g) 


«'.?-2-butene (C4H8, g) 


fra«.?-2-butene (C4H8, g) 



methanol (CH 3 OH, 1) 


ethanol (C 2 H 5 OH, 1) 


Aromatic s 

benzene (C^Hs, 1) 


toluene (CH 3 C 6 H 5 ,1) 


naphthalene (CioHs, s) 



methanoic (HC0 2 H, 1) 


ethanoic (CH 3 C0 2 H,1) 


oxalic (HC0 2 • C0 2 H, s) 


benzoic (C 6 H 5 ■ C0 2 H, s) 



a-D-glucose (CgH 12 06, s) 


yS-D-glucose (C 6 H 12 6 , s) 


sucrose (Ci 2 H 22 0n, s) 


O2 is an element, so its value of AH f is zero. The other values of A// f are exothermic. 
Inserting values into Equation (3.29): 

A// c & = [(-393.51) + (2 x -285.83)] - [(1 x -74.81) + (2 x 0)] kJmor 1 
A// c & = -965.17 - (-74.81) kJmor 1 
A// C e = -886.36 kJmor 1 

which is very close to the experimental value of —890 kJmol -1 in 
Table 3.2. 

SAQ 3.9 Calculate the standard enthalpy of combustion 
AHc for burning ^S-D-glucose, C 6 Hi 2 06. The required val- 
ues of AHf may be found in Table 3.1. 

The massive value 
of AHf for glucose 
explains why ath- 
letes consume glucose 
tablets to provide them 
with energy. 



We defined the value of AH* during combustion as ff (final) — H (initial)) so a neg- 
ative sign for AH* suggests the final enthalpy is more negative after combustion. 
In other words, energy is given out during the reaction. Our Stone Age forebears 
absorbed this energy by their fires in the night, which is another way of saying 'they 
warmed themselves'. 

Why does butane burn with a hotter flame 
than methane? 

Bond enthalpies 

Methane is easily bottled for transportation because it is a gas. It burns with a 
clean flame, unlike coal or oil. It is a good fuel. The value of AH* for methane is 
—886 kJmol -1 , but AH* for ra-butane is —2878 kJmol -1 . Burning butane is clearly 
far more exothermic, explaining why it burns with a hotter flame. In other words, 
butane is a better fuel. 

The overall enthalpy change during combustion is AH*. An alternative way of 
calculating an enthalpy change during reaction dispenses with enthalpies of forma- 
tion AH* and looks at the individual numbers of bonds formed and broken. We 
saw in Chapter 2 how we always need energy to break a bond, and release energy 
each time a bond forms. Its magnitude depends entirely on the enthalpy change 
for breaking or making the bonds, and on the respective numbers of each. For 
example, Equation (3.28) proceeds with six bonds cleaving (four C-H bonds and 
two 0=0 bonds) at the same time as six bonds form (two C=0 bonds and four H-0 

A quick glance at Worked Example 3.11 shows how the energy released during 
combustion is associated with forming the CO2 and H2O. If we 
could generate more CO2 and H2O, then the overall change in 
AH would be greater, and hence the fuel would be superior. In 
fact, many companies prefer butane to methane because it releases 
more energy per mole. 

We can calculate an enthalpy of reaction with bond enthalpies by 
assuming the reaction consists of two steps: first, bonds break, and 
then different bonds form. This approach can be simplified further 
if we consider the reaction consists only of reactive fragments, 
and the products form from these fragments. The majority of the 
molecule can remain completely unchanged, e.g. we only need to 
consider the hydroxyl of the alcohol and the carboxyl of the acid 
during a simple esterification reaction. 

To simplify the cal- 
culation, we pretend 
the reaction proceeds 
with all bonds break- 
ing at once; then, an 
instant later, different 
bonds form, again all at 
once. Such an idea is 
mechanistic nonsense 
but it simplifies the 

Worked Example 3.12 What fragments do we need to consider during the esterification 
of 1-butanol with ethanoic acid? 



We first draw out the reaction in full: 


The butyl ethanoate 
produced by Equation 
(3.30) is an ester, and 
smells of pear drops. 


,0, , .0. 

^CH 3 +H H 


Second, we look for those parts that change and those that remain unchanged. In this 
example, the bonds that cleave are the O— H bond on the acid and the C-0 bond on the 
alcohol. Such cleavage will require energy. The bonds that form are an O-H bond (to 
yield water) and a C-0 bond in the product ester. All bonds release energy as they form. 
In this example, the bonds outside the box do not change and hence do not change their 
energy content, and can be ignored. 

The value of A// r relates to bond changes. In this example, equal numbers of O-H 
and C-O bonds break as form, so we expect an equal amount of energy to be released as 
is consumed, leading to an enthalpy change of zero. In fact, the value of AH r is tiny at 
-12kJmol _1 . 

In some texts, A/-/g E is 
written simply as 'BE'. 

Worked Example 3.12 is somewhat artificial, because most reac- 
tions proceed with differing numbers of bonds breaking and form- 
ing. A more rigorous approach quantifies the energy per bond - the 
'bond enthalpy' AH BE (also called the 'bond dissociation energy'). 
AH BE is the energy needed to cleave 1 mol of bonds. For this reason, values of AH BE 
are always positive, because energy is consumed. 

The chemical environment of a given atom in a molecule will influence the mag- 
nitude of the bond enthalpy, so tabulated data such as that in Table 3.3 represent 
average values. 

We can calculate a value of A// r & with an adapted form of Equation (3.29): 



E yA// i 




where the subscripted i means those bonds that cleave or form within each reactant 
or product species during the reaction. We need the minus sign because of the way 
we defined the bond dissociation enthalpy. All the values in Table 3.3 are positive 

because AH BE relates to bond dissociation. 

The stoichiometric numbers v here can be quite large unless the molecules are small. 
The combustion of butane, for example, proceeds with the loss of 10 C-H bonds. A 


Table 3.3 Table of mean bond enthalpies AH — as a function of bond order and atoms. All 

r BE 

values cited in kj moP 1 and relate to data obtained (or corrected) to 298 K 
























































803 a 

590 b 




























590 c 



a 728 if -C=0. 
b 406if-NO 2 ; 368 if-N0 3 . 
c 506 if alternating - and =. 

moment's thought suggests an alternative way of writing Equation (3.31), i.e.: 

Values of &H? can 
vary markedly from 
experimental values if 
calculated in terms of 


/bonds formed bonds broken 

A// r° = " E VA <E- E VA// B*E (3-32) 

Note again the minus sign, which we retain for the same reason as 
for Equation (3.31). 

Each of these bond enthalpies is an average enthalpy, measured from a series of 
similar molecules. Values of AH^ E for, say, C-H bonds in hydrocarbons are likely 
to be fairly similar, as shown by the values in Table 3.3. The bond energies of C-H 
bonds will differ (sometimes quite markedly) in more exceptional molecules, such as 
those bearing ionic charges, e.g. carbocations. AH^ E values differ for the OH bond 
in an alcohol, in a carboxylic acid and in a phenol. 

These energies relate to bond rearrangement in gaseous molecules, but calculations 
are often performed for reactions of condensed phases, by combining the enthalpies 
of vaporization, sublimation, etc. We can calculate a value without further correction 
if a crude value of AH T is sufficient, or we do not know the enthalpies of phase 



Worked Example 3.13 Use the bond enthalpies in Table 3.3 to calculate the enthalpy 
of burning methane (Equation (3.28)). Assume all processes occur in the gas phase. 

Strategy. We start by writing a list of the bonds that break and form. 

Broken: 4 x C-H and 2 x 0=0 
Formed: 2 x C=0 and 4 x O-H 


AH; = -[2 x A< E(C=0) + 4 x A< E(0 _ H) ] - [4 x A/f* E(C _ H) 

Inserting values of AH^ E from Table 3.3 into Equation (3.32): 

AH* = -[(2 x 803) + (4 x 464)] 

_ [(4 x 414) + (2 x 498)] klmoP 1 
AH* = -(3462 - 2652) kJmof 1 

+ 2x AH 


Reminder: all &H* E 
values are positive 
because they relate to 
dissociation of bonds. 


AH* = -810kJmol" 

which is similar to the value in Worked Example 3.11, but less exothermic. 


Calculations with bond enthalpies AH BE tend to be relatively inaccurate because each 
energy is an average. As a simple example, consider the sequential dissociation of 

(1) NH3 dissociates to form NH" and H", and re- 
quires an energy of 449 kj mol~ ' . 

(2) NHj dissociates to form NH" and H", and re- 
quires an energy of 384 kJmol -1 . 

The symbol V means 
a radical species, i.e. 
with unpaired elec- 

(3) NH* dissociates to form N" and H*, and requires an energy of 339 kJmol . 

The variations in AH m are clearly huge, so we usually work with an average bond 
enthalpy, which is sometimes written as BE or AH — The average bond enthalpy for 


the three processes above is 390.9 kJmol - . 



3.3 Indirect measurement of enthalpy 

How do we make "industrial alcohol'? 

Enthalpy Cycles from Hess's Law 

Industrial alcohol is an impure form of ethanol made by hydrolysing ethene, 
CH 2 =CH 2 : 


We pass ethene and water (as a vapour) at high pressure over a suitable catalyst, 
causing water to add across the double bond of the ethene molecule. The industrial 
alcohol is somewhat impure because it contains trace quantities of ethylene glycol 
(1,2-dihydroxyethane, III), which is toxic to humans. It also contains unreacted water, 
and some dissolved ethene. 

We may rephrase 
Hess's law, saying The 
standard enthalpy of an 
overall reaction is the 
sum of the standard 
enthalpies of the indi- 
vidual reactions into 
which the reaction may 
be divided'. 






But what is the enthalpy of the hydration reaction in Equation 
(3.33)? We first met Hess's law on p. 98. We now rephrase it by 
saying 'The standard enthalpy of an overall reaction is the sum of 
the standard enthalpies of the individual reactions into which the 
reaction may be divided.' 

Accordingly, we can obtain the enthalpy of reaction by drawing 
a Hess cycle, or we can obtain it algebraically. In this example, we 
will use the cycle method. 

Worked Example 3.14 What is the enthalpy change AH r of the reaction in Equation 

We start by looking up the enthalpies of formation AH t for ethene, 
ethanol and water. Values are readily found in books of data; Table 3.1 
contains a suitable selection. 

Notice that each of 
these formation reac- 
tions is highly exother- 
mic, explaining why 
energy is needed to 
obtain the pure ele- 

A// f(1) [CH 2 =CH 2 ] = -52 kJmol -1 

A# f(2) [CH 3 CH 2 OH] = -235 kJmol" 

A// f( 3)[H 2 0] = -286 kJmol" 


(We have numbered these three (1) to (3) simply to avoid the necessity of rewriting 
the equations.) 

To obtain the enthalpy of forming ethanol, we first draw a cycle. It is usual to start 
by writing the reaction of interest along the top, and the elements parallel, along the 
bottom. Remember, the value of A// r is our ultimate goal. 

Step 1 


CH 2 = CH 2 + H 2 -* CH 3 CH 2 OH Reaction of interest 

2C + 3H 2 + i-0 2 Elements 

The next three stages are inserting the three enthalpies AHf (1) to (3). Starting on 
the left-hand side, we insert AHf^y. 

AH r 

CH 2 =CH 2 + H 2 ► CH 3 CH 2 OH 

Step 2 

A// f (i) 

2C + 3H 2 + i0 2 

We then put in the enthalpy of forming water, AHf^y. 

AH r 

CH 2 =CH 2 + H 2 CH 3 CH 2 OH 

Step 3 

^H f(1) \ AH^ 

2C + 3H 2 + \0 2 

And finally we position the enthalpy of forming the product ethanol, A//f( 2 ): 

AH r 

CH 2 =CH 2 + H 2 ► CH 3 CH 2 OH 

Step 4 

AH f (i) \ ZiH f(3 ) \ / AH i( 2) 

2C + 3H 2 + i-0 2 

We can now determine the value of A// r . Notice how we only draw one arrow per 
reaction. The rules are as follows: 



We are only allowed 

to make a choice of 

route like this 


enthalpy is a 



(1) We wish to go from the left-hand side of the reaction to the 
right-hand side. We can either follow the arrow labelled 
AH V , or we pass via the elements (along the bottom line) 
and thence back up to the ethanol. 

(2) If we go along an arrow in the same direction as the arrow 
is pointing, then we use the value of AH as it is written. 

(3) If we have to go along an arrow, but in the opposite direction to the 

direction in which it points, then we multiply the value of AH by '— 1'. 

In the example here, to go from the left-hand side to the right-hand side via the 
elements, we need to go along two arrows AHf^ and AHfQ) in the opposite directions 
to the arrows, so we multiply the respective values of AH and multiply each by —1. 
We then go along the arrow AHfQ), but this time we move in the same direction as 
the arrow, so we leave the sign of the enthalpy unaltered. 

And then we tie the threads together and say: 

Note there are three 
arrows, so there are 
three AH terms within 
AH r . 

AH r = (-1 x AH m ) + (-1 x AH m ) + AH i(2) 
Inserting values into this equation: 


AH T = (-1 x -52 kJmol" 1 ) + (-1 x -286 kJmol _1 ) + (-235 kJmol -1 ) 
AH T = 52 kJmol -1 + 286 kJmol -1 + (-235 kJinoP 1 ) 

AH r = 103 kJmol" 1 

We obtained this value of AH r knowing the other enthalpies in the cycle, and 
remembering that enthalpy is a state function. Experimentally, the value of AH T = 
99 kJmol -1 , so this indirect measurement with Hess's law provides relatively good 

Sometimes, these cycles are considerably harder than the example here. In such 
cases, it is usual to write out a cycle for each reaction, and then use the results from 
each cycle to compile another, bigger cycle. 

How does an 'anti-smoking pipe' work? 

Hess's Law Cycles with Enthalpies of Combustion 

Smoking causes severe damage to the heart, lungs and respiratory system. The tobacco 
in a cigarette or cigar is a naturally occurring substance, and principally comprises 
the elements carbon, oxygen, hydrogen and nitrogen. 

Unfortunately, because the tobacco is contained within the bowl of a pipe or a paper 
wrapper, complete combustion is rare, meaning that the oxidation is incomplete. One 


of the worse side effects of incomplete combustion during smoking is the formation 
of carbon monoxide (CO) in relatively large quantities. Gaseous CO is highly toxic, 
and forms an irreversible complex with haemoglobin in the blood. This complex helps 
explain why people who smoke are often breathless. 

A simple way of overcoming the toxic effects of CO is to oxidize it before the 
smoker inhales the tobacco smoke. This is where the 'anti-smoking' pipe works. (In 
fact, the name is a misnomer: it does not stop someone smoking, but merely makes 
the smoke less toxic.) The cigarette is inserted into one end of a long, hollow tube 
(see Figure 3.7) and the smoker inhales from the other. Along the tube's length are 
a series of small holes. As the smoker inhales, oxygen enters the holes, mixes with 
the CO and combines chemically with it according to 

CO (g) + \0 2 ► C0 2(g) (3.34) 

The C02( g ) produced is considerably less toxic than CO( g ), thereby averting at least 
one aspect of tobacco poisoning. 

We might wonder: What is the enthalpy change of forming the CO in Equation 
(3.34)? It is relatively easy to make CO in the laboratory (for example by dehydrating 
formic acid with concentrated sulphuric acid), so the enthalpy of oxidizing CO to CO2 
is readily determined. Similarly, it is easy to determine the enthalpy of formation 
of CO2, by burning elemental carbon; but it is almost impossible to determined 
AH C for the reaction C + 5O2 —*■ CO, because the pressure of oxygen in a bomb 
calorimeter is so high that all the carbon is oxidized directly to CO2 rather than CO. 
Therefore, we will employ Hess's law once more, but this time employing enthalpies 
of combustion AH C . 

The enthalpies of combustion of carbon and CO are obtained 
readily from books of data. We can readily find out the following 
from such data books or Table 3.2: 

A// c( i)[C (s) + 2( g) ► C0 2( g)] = -393.5 kJmor 

A// c(2) [CO (g) + ±0 2(g) ► C0 2(g) ] = -283.0 kJmol" 


The enthalpy AH C(1) is 
huge, and helps explain 
why we employ coke 
and coal to warm a 
house; this reaction 
occurs when a coal fire 

Again, we have numbered the enthalpies, to save time. 

Once more, we start by drawing a Hess-law cycle with the elements at the bottom 
of the page. This time, it is not convenient to write the reaction of interest along the 




Small holes to allow in oxygen 

Figure 3.7 An anti-smoking device: the cigarette is inserted into the wider end. Partially oxidized 
carbon monoxide combines chemically with oxygen inside the device after leaving the end of the 
cigarette but before entering the smoker's mouth; the oxygen necessary to effect this oxidation 
enters the device through the small circular holes positioned along its length 


top, so we have drawn it on the left as AH r . 

CO + -L0 2 C0 2 Compounds 

Step 1 

C + O2 Constituent elements 

Next we insert the enthalpies for the reactions of interest; we first insert AH c ^y. 

CO + \0 2 C0 2 

c + o 2 

and finally, we insert AH C (2): 
. AH C(2) 

co + ^o 2 co 2 

Step 3 

/\M r \ / /iti Q n\ 

C + 2 

We want to calculate a value of AH T . Employing the same laws as before, we see 
that we can either go along the arrow for AH r directly, or along A// C (i) in the same 
direction as the arrow (so we do not change its sign), then along the arrow AH C ^)- 
Concerning this last arrow, we go in the opposite direction to the arrow, so we 
multiply its value by —1. 

The value of AH V is given as 

A// r = AH cm + (-1 x AH C(2) ) 

Inserting values: 

AH r = -393.5 kJmol _1 + (+283 kJmol -1 ) 


AH r = -110.5 kJmol" 1 

SAQ 3.10 Calcite and aragonite are both forms of calcium carbonate, 
CaC03. Calcite converts to form aragonite. If AHf (ca | Cite) = -1206.92 kJ 


mor 1 and Atf f & (aragonite) = 
for the transition process: 

1207.13 kJ mol 1 , calculate the value of A/-/ r 

CaC03( S , calcite) > CaC03( S , aragoni 



Why does dissolving a salt in water liberate heat? 

The 'lattice enthalpy' is 
defined as the standard 
change in enthalpy 
when a solid sub- 
stance is converted 
from solid to form 
gaseous constituent 
ions. Accordingly, val- 
ues of AH ( iattice) are 
always positive. 

Hess's Law Applied to Ions: Constructing Born-Haber Cycles 

Dissolving an ionic salt in water often liberates energy. For example, 32.8 kJmol -1 
of energy are released when 1 mol of potassium nitrate dissolves in water. Energy is 
released, as experienced by the test tube getting warmer. 

Before we dissolved the salt in water, the ions within the crys- 
tal were held together by strong electrostatic interactions, which 
obeyed Coulomb's law (see p. 313). We call the energetic sum 
of these interactions the lattice enthalpy (see p. 124). We need 
to overcome the lattice enthalpy if the salt is to dissolve. Stated 
another way, salts like magnesium sulphate are effectively insolu- 
ble in water because water, as a solvent, is unable to overcome the 
lattice enthalpy. 

But what is the magnitude of the lattice enthalpy? We cannot 
measure it directly experimentally, so we measure it indirectly, with 
a Hess's law energy cycle. The first scientists to determine lattice 
enthalpies this way were the German scientists Born and Haber: 
we construct a Born-Haber cycle, which is a form of Hess's- 
law cycle. 

Before we start, we perform a thought experiment; and, for con- 
venience, we will consider making 1 mol of sodium chloride at 
25 °C. There are two possible ways to generate 1 mol of gaseous 
Na + and Cl~ ions: we could start with 1 mol of solid NaCl and 
vaporize it: the energy needed is A// ( ^ ttice) . Alternatively, we could 
start with 1 mol of sodium chloride and convert it back to the ele- 
ments (1 mol of metallic sodium and 0.5 mol of elemental chlorine gas (for which 
the energy is — A// f °) and, then vaporize the elements one at a time, and ionize each 
in the gas phase. The energies needed to effect ionization are / for the sodium and 
£(ea) for the chlorine. 

In practice, we do not perform these two experiments because we can calculate a 
value of lattice enthalpy A //(lattice) with an energy cycle. Next, we appreciate how 
generating ions from metallic sodium and elemental chlorine involves several pro- 
cesses. If we first consider the sodium, we must: (i) convert it from its solid state to 
gaseous atoms (for which the energy is A// ( ^ ublimation) ); (ii) convert the gaseous atoms 
to gaseous cations (for which the energy is the ionization energy /). We next consider 
the chlorine, which is already a gas, so we do not need to volatilize it. But: (i) we 
must cleave each diatomic molecule to form atoms (for which the energy is AH^ E ); 

It is common to see 
values of AH ( i at t iC e) 
called 'lattice energy'. 
Strictly, this latter term 
is only correct when the 
temperature T is K. 



(ii) ionize the gaseous atoms of chlorine to form anions (for which the energy is the 
electron affinity £( ea )). Finally, we need to account for the way the sodium chloride 
forms from elemental sodium and chlorine, so the cycle must also include A// f s . 

Worked Example 3.15 What is the lattice enthalpy A//(i att i ce ) 
25 °C? 

of sodium chloride at 

Strategy. (1) We start by compiling data from tables. (2) We construct an energy cycle. 
(3) Conceptually, we equate two energies: we say the lattice enthalpy is the same as the 
sum of a series of enthalpies that describe our converting solid NaCl first to the respective 
elements and thence the respective gas-phase ions. 
(1) We compile the enthalpies: 

These energies are 
huge. Much of this 
energy is incorporated 
into the lattice; other- 
wise, the value of AH* 
would be massive. 

for sodium chloride 

Na (s) + ±Cl 2(g) 
for the sodium 
Na (s) 
Na (g) 

-* NaCl (s 

> Na (g) 

► Na+ (g) + e" 




Aff f * = 

-411.15 kJmol" 


= 107.32 kJmol" 

502.04 kJ mol" 

for the chlorine 

ci (g 

-> ci; 

C1 ( g ) + e ~ 

► cr 

5 A// BE 

121.68 kJmol" 1 (i.e. half of 243.36 kJmor 1 ) 



-354.81 kJmol" 

(2) We construct the appropriate energy cycle; see Figure 3.8. For simplicity, it is usual 
to draw the cycle with positive enthalpies going up and negative enthalpies going down. 
(3) We obtain a value of A H, ]mi . , equating it to the energy needed 

to convert solid NaCl to its elements and thence the gaseous ions. We 

construct the sum: 

We multiply AH f w by 
'-1' because we con- 
sider the reverse pro- 
cess to formation. (We 
travel in the opposite 
direction to the arrow 
representing &Hf in 
Figure 3.8.) 



= -AH t + AH { 

(sublimation) + 7 (Na) + J A// BE + £ (ea) 

Inserting values: 

A/ Vtice) = -(-411.153) + 107.32 + 502.04 
+ 121.676 + (-354.81) kJ moP 1 


^(Lice) = 787.38 kJmor 

SAQ 3.11 Calculate the enthalpy of formation AH* for calcium fluoride. 


lattice) = "2600 kJ mor 



= 178 kJmol -1 ;/ 

l(Ca^Ca + ) 



1 mol gaseous Na + and CI ions 


v (ea) 


1 V 






1 mol elements (in standard 

Figure 3.8 Born-Haber cycle constructed to obtain the lattice enthalpy AH f 


of sodium chlo- 

ride. All arrows pointing up represent endothermic processes and arrows pointing down represent 
exothermic processes (the figure is not drawn to scale) 

= 596kJmor 1 and I 2(Ca+ _^ Ca 2 +) = 1152 kJ mor 1 ; AWg E = 157 kJ mor 1 
and f ( ea) = -334 kJ mol" 1 . 

Why does our mouth feel cold after eating 

Enthalpy of Solution 

Natural peppermint contains several components that, if ingested, lead to a cold sen- 
sation in the mouth. The best known and best understood is (— )-menthol (IV), which 
is the dominant component of the peppermint oil extracted from Mentha piperiia and 
M. arvensia. 

CH 3 


CH3 CH3 



The cause of the cooling sensation is the unusually positive enthalpy of solution. 

Most values of AH, 


are positive, particularly for simple inorganic solutes. 

Pure IV is a solid at s.t.p. Dissolving IV in the mouth disrupts its molecular 
structure, especially the breaking of the hydrogen bonds associated with the hydroxy 1 
group. These bonds break concurrently with new hydrogen bonds forming with the 
water of the saliva. We require energy to break the existing bonds, and liberate energy 
as new bonds form. Energetically, dissolving (— )-menthol is seen to be endothermic, 
meaning we require energy. This energy comes from the mouth and, as we saw earlier, 
the macroscopic manifestation of a lower microscopic energy is a lower temperature. 
Our mouth feels cold. 

The other substance sometimes added to foodstuffs to cause cooling of the mouth 
is xylitol (V). It is added as a solid to some sweets, chewing gum, toothpastes and 
mouth-wash solutions. 




Measuring values of Ar^ solution 

with a 

It is quite difficult to measure an accurate enthalpy of solution A//, solutk , ni 
calorimeter, but we can measure it indirectly. Consider the example of sodium chlo- 
ride, NaCl. The ions in solid NaCl are held together in a tight array by strong ionic 
bonds. While dissolving in water, the ionic bonds holding the constituent ions of 
Na + and Cl~ in place break, and new bonds form between the ions and molecules 
of water to yield hydmted species. Most simple ions are surrounded with six water 
molecules, like the [Na(H20)6] + ion (VI). Exceptions include the proton with four 
water molecules (see p. 235) and lanthanide ions with eight. 

The positive charge 
does not reside on 
the central sodium 
alone. Some charge 
is distributed over the 
whole ion. 

OH 2 
H 2 0,, I ^OH 2 

' Na'~ 
H 2 0*^ [ ^OH 2 

OH 2 


Each hydration bond is partially ionic and partially covalent. 
Each oxygen atom (from the water molecules) donates a small 
amount of charge to the central sodium; hence the ionicity. The 
orbitals also overlap to impart covalency to the bond. 


Energy is needed to break the ionic bonds in the solid salt and energy is liberated 
forming hydration complexes like VI. We also break some of the natural hydrogen 
bonds in the water. The overall change in enthalpy is termed the enthalpy of solu- 
tion, A// ( ^ olmion) . Typical values are —207 kJmol - for nitric acid; 34 kJmol - for 
potassium nitrate and —65.5 kJmol -1 for silver chloride. 

One of the most sensitive ways of determining a value of AH? Mion) is to measure 
the temperature T at which a salt dissolves completely as a function of its solubility 
s. A plot of In s (as y) against 1 4- T (as x) is usually linear. We obtain a value 
of A// (solution N by multiplying the gradient of the graph by —R, where R is the gas 
constant (as described in Chapter 5, p. 210). 

low does a camper's ^emergency heat stick' work? 

Enthalpies of Complexation 

'Exposure' is a condi- 
tion of being exposed 
to the elements, lead- 
ing to hypothermia, 
and can lead to death. 

A camper is in great danger of exposure if alone on the moor or in 

the desert when night falls and the weather becomes very cold. If 

a camper has no additional heating, and knows that exposure is not 

far off, then he can employ an 'emergency heat stick'. The stick 

is long and thin. One of its ends contains a vial of water and, at 

the other, a salt such as anhydrous copper sulphate, CUSO4. Both 

compartments are housed within a thin-walled glass tube, itself encased in plastic. 

Bending the stick breaks the glass, allowing the water to come into contact with the 

copper sulphate and effect the following hydration reaction: 

CuS0 4(s) + 5H 2 0(i) ► CuS0 4 • 5H 2 (S) (3.36) 

The reaction in Equation (3.36) is highly exothermic and releases 134 kJmol -1 of 
energy. The camper is kept warm by this heat. The reaction in Equation (3.36) 
involves complexation. In this example, we could also call it 'hydration' or 'adding 
water of crystallization' . We will call the energy released the 'energy of complexation' 

^ ^(complexation) • 

Heat is liberated when adding water to anhydrous copper sulphate because a new 
crystal lattice forms in response to strong, new bonds forming between the water and 
Cu + and S0 4 ~ ions. As corroborative evidence of a change in the crystal structure, 
note how 'anhydrous' copper sulphate is off-white but the pentahydrate is blue. 


Reaction spontaneity and 
the direction of 
thermodynamic change 


We start by introducing the concept of entropy S to explain why some reactions 
occur spontaneously, without needing additional energy, yet others do not. The sign 
of A S for a thermodynamic universe must be positive for spontaneity. We explore 
the temperature dependence of A S. 

In the following sections, we introduce the concept of a thermodynamic universe 
(i.e. a system plus its surroundings). For a reaction to occur spontaneously in a 
system, we require the change in Gibbs function G to be negative. We then explore 
the thermodynamic behaviour of G as a function of pressure, temperature and reaction 

Finally, we investigate the relationship between AG and the respective equilibrium 
constant K, and outline the temperature interdependence of AG and K. 

4.1 The direction of physicochemical 
change: entropy 

Why does the colour spread when placing a drop 
of dye in a saucer of clean water? 

Reaction spontaneity and the direction of change 


However gently a drop of dye solution is added to a saucer of clean, pure water, 
the colour of the dye soon spreads into uncoloured regions of the water. This mixing 
occurs inevitably without warming or any kind of external agitation - the painter with 
watercolour would find his art impossible without this effect. Such mixing continues 



Mixing occurs spon- 
taneously, but we 
never see the reverse 
process, with dye sud- 
denly concentrating 
into a coloured blob 
surrounded by clear, 
uncoloured water. 

A reaction is 'spon- 
taneous' if it occurs 
without any additional 
energy input. 

It used to be thought 
reactions were spon- 
taneous if AH was 
negative. This sim- 
plistic idea is incorrect. 

until the composition of the solution in the saucer is homogeneous, 
with the mixing complete. We never see the reverse process, with 
dye suddenly concentrating into a coloured blob surrounded by 
clear, uncoloured water. 

In previous chapters we looked at the way heat travels from hot 
to cold, as described by the so called 'minus-oneth' law of ther- 
modynamics, and the way net movements of heat cease at thermal 
equilibrium (as described by the zeroth law). Although this transfer 
of heat energy was quantified within the context of the first law, we 
have not so far been able to describe why such chemical systems 
occur. Thermodynamic changes only ever proceed spontaneously 
in one direction, but not the other. Why the difference? 

In everyday life, we say the diffusion of a dye 'just happens' 
but, as scientists, we say the process is spontaneous. In years past, 
it was thought that all spontaneous reactions were exothermic, with 
non-spontaneous reactions being endothermic. There are now many 
exceptions to this overly simplistic rule; thus, we can confidently 
say that the sign of AH does not dictate whether the reaction is 
spontaneous or not, so we need a more sophisticated way of looking 
at the problem of spontaneity. 

When we spill a bowl of sugar, why do the grains go 
everywhere and cause such a mess? 

Changes in the extent of disorder 

The granules of spilt 
sugar have a range of 

Surely everyone has dropped a bowl of sugar, flour or salt, and caused a mess! The 
powder from the container spreads everywhere, and seems to cover the maximum 
area possible. Spatial distribution of the sugar granules ensures a 
range of energies; so, for example, some particles reside on higher 
surfaces than others, thereby creating a range of potential energies. 
And some granules travel faster than others, ensuring a spread of 
kinetic energies. 

The mess caused by dropping sugar reflects the way nature always seeks to max- 
imize disorder. Both examples so far, of dye diffusing in water and sugar causing 
a mess, demonstrate the achievement of greater disorder. But if we are specific, we 
should note how it is the energetic disorder that is maximized spontaneously. 

It is easy to create disorder; it is difficult to create order. It requires effort to clean up 
the sugar when re-establishing order, showing in effect how reversing a spontaneous 
process requires an input of energy. This is why the converse situation - dropping 
a mess of sugar grains and creating a neat package of sugar - does not happen 
spontaneously in nature. 



Why, when one end of the bath is hot and the other 
cold, do the temperatures equalize? 


The word 'entropy' 
comes from the Greek 
en tropa, meaning 
'in change' or 'during 

Entropy and the second law of thermodynamics 

Quite often, when running a bath, the water is initially quite cold. After the hot 
water from the tank has had time to travel through the pipes, the water from the tap 
is hot. As a result, one end of the bath is hotter than the other. But a short time 
later, the temperature of the water is the same throughout the bath, with the hot 
end cooler and the cold end warmer. Temperature equilibration occurs even without 
stirring. Why? 

We saw in Chapter 1 how the simplest way to gauge how much energy a molecule 
possesses is to look at its temperature. We deduce through a reasoning process such as 
this that molecules of water at the cold end of the bath have less energy than molecules 
at the hot end. Next, by combining the minus-oneth and zeroth laws of thermody- 
namics, we say that energy (in the form of heat) is transferred from molecules of 
water at the hot end of the bath to molecules at the cold end. Energy transfers until 
equilibrium is reached. All energy changes are adiabatic if the bath 
is lagged (to prevent energy loss), in accordance with the first law 
of thermodynamics. 

As no chemical reactions occur, we note how these thermo- 
dynamic changes are purely physical. But since no bonds form or 
break, what is the impetus - the cause - of the transfer of energy? 
We have already seen the way processes occur with an attendant 
increase in disorder. We now introduce the concept of entropy. The 
extent of energetic disorder is given the name entropy (and has the 
symbol S). A bigger value of S corresponds to a greater extent of 
energetic disorder. 

We now introduce the second law of thermodynamics : a physic- 
ochemical process only occurs spontaneously if accompanied by 
an increase in the entropy S. By corollary, a non-spontaneous 
process - one that we can force to occur by externally adding 
energy - would proceed concurrently with a decrease in the ener- 
getic disorder. 

We can often think of entropy merely in terms of spatial disorder, 
like the example of the sugar grains above; but the entropy of a 
substance is properly the extent of energetic disorder. Molecules of 
hot and cold water in a bath exchange energy in order to maximize 
the randomness of their energies. 

Figure 4.1 depicts a graph of the number of water molecules 
having the energy E (as y) against the energy of the water mole- 
cules E (as x). Trace (a) in Figure 4.1 shows the distribution of 
energies in a bath where half the molecules have one energy while 
the other half has a different energy, which explains why the graph 
contains two peaks. A distribution of energies soon forms as energy 

The 'second law of ther- 
modynamics' says a 
process occurs sponta- 
neously only when the 
concomitant energetic 
disorder increases. We 
can usually approxi- 
mate, and talk in terms 
of 'disorder' alone. 

The energy is trans- 
ferred via random, 
inelastic collisions bet- 
ween the molecules of 
water. Such molecular 
movement is some- 
times called Brownian 
motion; see p. 139. 


Colder end 
a of bath 



Hotter end 
of bath 

Energy of water molecules £ 



Energy of water molecules E 


Figure 4.1 Graph of the number of water molecules of energy E against energy, (a) Soon after 
running the bath, so one end is hotter and the other cooler; and (b) after thermal equilibration. The 
(average) energy at the peak relates to the macroscopic temperature 

is transferred from one set of water molecules to other. Trace (b) in Figure 4.1 shows 
the distribution of energies after equilibration. In other words, the energetic disorder 
S increases. The reading on a thermometer placed in the bath will represent an 
average energy. 

The spread of energies in Figure 4.1 is a direct indication of entropy, with a wider 
spread indicating a greater entropy. Such energetic disorder is the consequence of 
having a range of energies. The spread widens spontaneously; an example of a non- 
spontaneous process would be the reverse process, with the molecules in a bath at, 
say, 50 °C suddenly reverting to one having a temperature of 30 °C at one end and a 
temperature of 70 °C at the other. 

The German scientist Rudolf Clausius (1822-1888) was the 
first to understand the underlying physicochemical principles dic- 
tating reaction spontaneity. His early work aimed to understand 
the sky's blue colour, the red colours seen at sunrise and sun- 
set, and the polarization of light. Like so many of the 'greats' 
of early thermodynamics, he was a mathematician. He was inter- 
ested in engines, and was determined to improve the efficiency of 
steam-powered devices, such as pumping engines used to remove 
water from mines, and locomotives on the railways. Clausius was 
the first to introduce entropy as a new variable into physics and 

In the thermodynamic 
sense, an 'engine' is a 
device or machine for 
converting energy into 
work. Clausius himself 
wanted to devise an effi- 
cient machine to con- 
vert heat energy (from 
a fuel) into mechani- 
cal work. 



Why does a room containing oranges acquire 
their aroma? 

Spontaneity and the sign of AS 

When a bowl containing fresh oranges is placed on the dining room 
table, the room acquires their fragrance within a few hours. The 
organic substance we smell after its release from the oranges is the 
organic terpene (+)-limonene (I), each molecule of which is small 
and relatively non-polar. I readily evaporates at room temperature 
to form a vapour. 

We sometimes say 
these compounds 

The process we detect when we note the intensifying smell of the oranges, is: 






Gaseous materials 
have greater entropy 
than their respective 

so the concentration of volatile limonene in the gas phase increases with time. But 
why does it evaporate in this way? 

Liquids can flow (and hence transfer energy by inelastic colli- 
sions), so they will have a distribution of energies. Molecules in 
the liquid state possess a certain extent of energetic disorder and, 
therefore, have a certain extent of entropy S. By contrast, molecules 
in the gas phase have a greater freedom to move than do liquids, 
because there is a greater scope for physical movement: restrictions 
arising from hydrogen bonds or other physicochemical interactions are absent, and 
the large distances between each molecule allow for wider variations in speed, and 
hence in energy. Gas molecules, therefore, have greater entropy than do the liquids 
from which they derive. We deduce the simple result S( g ) > S(i). 

We could obtain this result more rigorously. We have met the symbol 'A' several 
times already, and recall its definition 'final state minus initial state', so the change 
in entropy AS for any process is given by the simple equation 




nal state) 

^(initial state) 


If the final disorder of a spontaneous process is greater than 
the initial disorder, then we appreciate from Equation (4.2) how a 
spontaneous process is accompanied by AS of positive sign. This 
will remain our working definition of spontaneity. 

Ultimately, the sign of AS explains why the smell of the oranges increases with time. 

A spontaneous process 
is accompanied by a 
positive value of AS. 



Worked Example 4.1 Show mathematically how the entropy of a gas is higher than 
the entropy of its respective liquid. 

If S (fina i state) is 5(g) and S (Mtia i state) is S(i) , then AS — 5(g) - S (l) . Because the volatilizing 
of the compound is spontaneous, the sign of A S must be positive. 
The only way to make AS positive is when S (g ) > Sq). 

Why do damp clothes become dry when hung outside? 

Reaction spontaneity by inspection 

Everyone knows damp clothes become dry when hung outside on the washing line. 
Any residual water is lost by evaporation from the cloth. In fact, moisture evap- 
orates even if the damp clothes hang limp in the absence of a breeze. The water 
spontaneously leaves the fabric to effect the physicochemical process H2O0) — »■ 
H 2 (g ). 

The loss of water occurs during drying in order to increase the 
overall amount of entropy, because molecules of gaseous water 
have a greater energy than do molecules of liquid, merely as a 
result of being gas and liquid respectively. In summary, we could 
have employed our working definition of entropy (above), which 
leads to a prediction of the clothes becoming dry, given time, 
as a result of the requirement to increase the entropy. Inspection 
alone allows us a shrewd guess at whether a process will occur 
spontaneously or not. 

We obtain the energy 
for evaporating the 
water by lowering the 
internal energy of the 
garment fibres, so the 
clothes feel cool to the 
touch when dry. 

Worked Example 4.2 By inspection alone, decide whether the sublimation of iodine 
(Equation (4.3)) will occur spontaneously or not: 


-> I 



Molecules in the gas phase have more entropy than molecules in the liquid phase; and 
molecules in the liquid phase have more entropy than molecules in the solid state. As an 
excellent generalization, the relative order of the entropies is given by 

This argument says 
nothing about the rare 
of sublimation. In fact, 
we do not see subli- 
mation occurring sig- 
nificantly at room tem- 
perature because it is 
so slow. 

5(g) » 5(1) > 5 (s 



The product of sublimation is a gas, and the precursor is a solid. 
Clearly, the product has greater entropy than the starting material, 
so A S increases during sublimation. The process is spontaneous 
because A S is positive. 

In a bottle of iodine, the space above the solid I2 always shows 
a slight purple hue, indicating the presence of iodine vapour. 



SAQ 4.1 By inspection alone, decide whether the condensation of water, 

H 2 


H 2 ( i) is spontaneous or not. 

Worked Example 4.3 Now consider the chemical process 

SOCl 2 (o + H 2 


2HCl (g) + SO 



The reaction occurs spontaneously in the laboratory without recourse 
to heating or catalysis. The sight of 'smoke' above a beaker of SOCl2(i) 
is ample proof of reaction spontaneity. 

We see that the reaction in Equation (4.5) consumes 1 mol of gas 
(i.e. water vapour) and 1 mol of liquid, and generates 3 mol of gas. 
There is a small change in the number of moles: principally, the 
amount of gas increases. As was seen above, the entropy of a gas 
is greater than its respective liquid, so we see a net increase in the 
entropy of the reaction, making AS positive. 

This increase in the 
entropy S of a gas 
explains why an open 
beaker of thionyl chlo- 
ride SOCI 2 in the lab- 
oratory appears to be 

Worked Example 4.4 By inspection alone, decide whether the formation of ammonia 
by the Haber process (Equation (4.6)) is spontaneous or not. 

N 2 ( g ) + 3H 2(g) > 2NH 



All the species in Equation (4.6) are gases, so we cannot use the simple method of looking 
to see the respective phases of reactants and products (cf. Equation (4.4)). 

But we notice the consumption of 4 mol of reactant to form 2 mol of product. As a 
crude generalization, then, we start by saying, '4 mol of energetic disorder are consumed 
during the process and 2 mol of energetic disorder are formed'. Next, with Equation (4.1) 
before us, we suggest the overall, crude entropy change AS is roughly —2 mol of disorder 
per mole of reaction, so the amount of disorder decreases. We suspect the process will 
not be spontaneous, because AS is negative. 

In fact, we require heating to produce ammonia by the Haber pro- 
cess, so the reaction is definitely not spontaneous. 

SAQ 4.2 By inspection alone, decide whether the oxi- 
dation of sulphur dioxide is thermodynamically sponta- 
neous or not. The stoichiometry of the reaction is 7202(g) 

+ S02(g) -» S03(g). 

The word 'stoichiom- 
etry' comes from the 
Greek stoicheion, mean- 
ing 'a part'. 

Worked Example 4.5 By inspection alone, decide whether the reaction Cl2( g ) + F 2(g ) 
2FCl( g ) should occur spontaneously. 

Occasionally we need to be far subtler when we look at reaction spontaneity. The 
reaction here involves two molecules of diatomic gas reacting to form two molecules of 
a different diatomic gas. Also, there is no phase change during reaction, nor any change 
in the numbers of molecules, so any change in the overall entropy is likely to be slight. 


Before we address this reaction, we need to emphasize how all these are equilibrium 
reactions: at completion, the reaction vessel contains product as well as unconsumed 
reactants. In consequence, there is a mixture at the completion of the reaction. 

This change in A S arises from the mixing of the elements between the two reacting 
species: before reaction, all atoms of chlorine were bonded only to other chlorine atoms 
in elemental CI2. By contrast, after the reaction has commenced, a choice arises with 
some chlorine atoms bonded to other chlorine atoms (unreacted CI2) and others attached 
to fluorine in the product, FC1. 

In fact, the experimental value of AS is very small and positive. 


Entropy as 'the arrow of time' 

The idea that entropy is continually increasing led many philosophers to call entropy 
'the arrow of time' . The argument goes something like this. From the Clausius equality 
(see p. 142), entropy is the ratio of a body's energy to its temperature. Entropy is 
generally understood to signify an inherent tendency towards disorganization. 

It has been claimed that the second law means that the universe as a whole must 
tend inexorably towards a state of maximum entropy. By an analogy with a closed 
system, the entire universe must eventually end up in a state of equilibrium, with the 
same temperature everywhere. The stars will run out of fuel. All life will cease. The 
universe will slowly peter out in a featureless expanse of nothingness. It will suffer a 
'heat death'. 

The idea was hugely influential. For example, it inspired the poet T. S. Eliot to write 
his poem The Hollow Men with perhaps his most famous lines 

This the way the world ends 
not with a bang but a whimper. 

He wrote this in 1925. Eliot's poem, in turn, inspired others. In 1927, the astronomer 
Sir Arthur Eddington said that if entropy was always increasing, then we can know the 
direction in which time moves by looking at the direction in which it increases. The 
phrase 'entropy is the arrow of time' gripped the popular imagination, although it is 
rather meaningless. 

In 1928, the English scientist and idealist Sir James Jean revived the old 'heat death' 
argument, augmented with elements from Einstein's relativity theory: since matter and 
energy are equivalents, he claimed, the universe must finally end up in the complete 
conversion of matter into energy: 

The second law of thermodynamics compels materials in the universe to move 
ever in the same direction along the same road which ends only in death and 



The extent of solute 
disorder decreases dur- 
ing crystallization. 

Why does crystallization of a solute occur? 

Thermodynamic systems and universes 

Atoms or ions of solute leave solution during the process of crystallization to form a 
regular repeat lattice. 

The extent of solute disorder is high before crystallization, be- 
cause each ion or molecule resides in solution, and thereby expe- 
riences the same freedom as a molecule of liquid. Conversely, 
the extent of disorder after crystallization will inevitably be much 
smaller, since solute is incorporated within a solid comprising a 
regular repeat lattice. 

The value of AS can only be negative because the symbol 'A' means 'final state 
minus initial state', and the extent of disorder during crystallization clearly follows the 
order 'solute disorder (initial) > solute disorder (nna i)'. We see how the extent of disorder 
in the solute decreases during crystallization in consequence of forming a lattice and, 
therefore, do not expect crystallization to be a spontaneous process. 

But crystallization does occur, causing us to ask, 'Why does 
crystallization occur even though AS for the process is negative?' 
To answer this question, we must consider all energetic consid- 
erations occurring during the process of crystallization, possibly 
including phenomena not directly related to the actual processes 
inside the beaker. 

Before crystallization, each particle of solute is solvated. As 
a simple example, a chloride ion in water is attached to six water molecules, as 
[Cl(H20)fi]~. Being bound to a solute species limits the freedom of solvent molecules, 
that is, when compared with free, unbound solvent. 

Crystallization releases these six waters of solvation; see Figure 4.2: 

A process is thermody- 
namically spontaneous 
only if the'overall' value 
of AS is positive. 

[Cl(H 2 0) 6 r 



(in solid lattice) ~T~ t>ri2U(f r ee, not solvating) 


J 2 ° H2 ° 
H 2 0— + Na— H 2 

H2(/ A 2 o 

Mobile aquo ion 

Na +6H 2 0(f re e) 


Ion immobilized 
within a 3-D 
repeat lattice 

Mobile water 

Figure 4.2 Schematic representation of a crystallization process. Each solvated ion, here Na + , 
releases six waters of solvation while incorporating into its crystal lattice. The overall entropy of 
the thermodynamic universe increases by this means 



The word 'universe' in 
this context is com- 
pletely different from 
a 'universe' in astron- 
omy, so the two should 
not be confused. A 
thermodynamic 'uni- 
verse' comprises both 
a 'system' and its 'sur- 

We define a thermo- 
dynamic universe as 
'that volume large 
enough to enclose 
all the changes'; the 
size of the surround- 
ings depends on the 

After their release from solvating this chloride ion, each water 
molecule has as much energetic disorder as did the whole chlo- 
ride ion complex. Therefore, we expect a sizeable increase in the 
entropy of the solvent during crystallization because many water 
molecules are released. 

When we look at the spontaneity of the crystallization pro- 
cess, we need to consider two entropy terms: (i) the solute (which 
decreases during crystallization) and (ii) the concurrent increase as 
solvent is freed. In summary, the entropy of the solute decreases 
while the entropy of the solvent increases. 

The crystallization process involves a system (which we are inter- 
ested in) and the surroundings. In terms of the component entropies 
in this example, we say AS( system ) is the entropy of the solute crys- 
tallizing and that A»S( Sun -oundings) represents the entropy change of 
the solvent molecules released. 

We call the sum of the system and its surroundings the ther- 
modynamic universe (see Figure 4.3). A thermodynamic universe 
is described as 'that volume large enough to enclose all the ther- 
modynamic changes'. The entropy change of the thermodynamic 
universe during crystallization is A5( to tai)> which equates to 

L-*-^ (total) — ^ ^(system) ~f~ ^^(surroundings) 


The value of A System) is negative in the example of crystallization. Accordingly, the 
value of AS(surroundings) must be so much larger than AS( S y Ste m) that AS( to tai) becomes 
positive. The crystallization is therefore spontaneous. 


Figure 4.3 We call the sum of the system and its surroundings the 'thermodynamic universe'. 
Energy is exchanged between the system and its surroundings; no energy is exchanged beyond 
the surrounds, i.e. outside the boundaries of the thermodynamic universe. Hence, the definition 'a 
universe is that volume large enough to enclose all the thermodynamic changes' 



This result of A5( to tai) being positive helps explain how consid- 
ering the entropy of a system's surroundings can obviate the appar- 
ent problems caused by only considering the processes occurring 
within a thermodynamic system. It also explains why crystallization 
is energetically feasible. 

The concept of a thermodynamic system is essentially macro- 
scopic, and assumes the participation of large numbers of 
molecules. Indeed, the word 'system' derives from the Greek 
susiema, meaning to set up, to stay together. 

As a new criterion for 
reaction spontaneity, 
we say AS (to tai) must be 
positive. We must con- 
sider the surroundings 
if we are to understand 
how the overall extent 
of energetic disorder 
increases during a 

4.2 The temperature dependence of entropy 

Why do dust particles move more quickly by Brownian 
motion in warm water? 

Entropy is a function of temperature 

Brownian motion is the random movement of small, solid particles sitting on the 
surface of water. They are held in position by the surface tension y of the meniscus. 
When looking at the dust under a microscope, the dust particles appear to 'dance' 
and move randomly. But when the water is warmed, the particles, be they chalk or 
house dust, move faster than on cold water. 

The cause of the Brownian motion is movement of water mole- 
cules, several hundred of which 'hold' on to the underside of each 
dust particle by surface tension. These water molecules move and 
jostle continually as a consequence of their own internal energy. 
Warming the water increases the internal energy, itself causing the 
molecules to move faster than if the water was cool. 

The faster molecules exhibit a greater randomness in their mo- 
tion than do slower molecules, as witnessed by the dust particles, 
which we see 'dancing' more erratically. The Brownian motion is 
more extreme. The enhanced randomness is a consequence of the water molecules 
having higher entropy at the higher temperature. Entropy is a function of temperature. 

'Brownian motion' is a 
macroscopic observa- 
tion of entropy. 

Entropy is a function of 

Why does the jam of a jam tart burn more than does 
the pastry? 

Relationship between entropy and heat capacity 

When biting into a freshly baked jam tart, the jam burns the tongue but the pastry 
(which is at the same temperature) causes relatively little harm. The reason why the 
jam is more likely to burn is its higher 'heat capacity' C. 



The heat capacity 
39 JK _1 mol 


Sections 3.1 and 3.2 describe heat capacity and explain how it 
may be determined at constant pressure C p or at constant volume 
Cy Most chemists need to make calculations with C p , which repre- 
sents the amount of energy (in the form of heat) that can be stored 
within a substance - the measurement having been performed at 
constant pressure p. For example, the heat capacity of solid water 
(ice) is 39 J K _1 mol -1 . The value of C p for liquid water is higher, 
at 75 JK _1 mol~ , so we store more energy in liquid water than when it is solid; 
stated another way, we need to add more energy to H20(i) if its temperature is to 
increase. C p for steam (H20( g )) is 34 JK _1 mol -1 . C p for solid sucrose (II) - a major 
component of any jam - is significantly higher at 425 JK _1 mol~ . 

tells us that adding 39 J 
of energy increases the 
temperature of 1 mol 
of water by 1 K. 

H OH/ 
H \/ \/ CH 2 OH 

The heat capacity of a liquid is always greater than the heat capacity of the respec- 
tive solid because the liquid, having a greater amount of energetic disorder, has a 
greater entropy according to 

AS = S 2 - Si 

f" C T 



More energy is 'stored' within a liquid than in its respective solid, as gauged by 
the relative values of C p implied by the connection between the heat capacity and 
entropy S (of a pure material). This is to be expected from everyday experience: 
to continue with our simplistic example, when a freshly baked jam tart is removed 
from the oven, the jam burns the mouth and not the pastry, because the (liquid) 
jam holds much more energy, i.e. has a higher C p than does the solid pastry, even 
though the two are at the same temperature. The jam, in cooling to the same tem- 
perature as the tongue, gives out more energy. The tongue cannot absorb all of this 
energy; the energy that is not absorbed causes other processes in the mouth, and hence 
the burn. 



This argument is oversimplified because it is expressed in terms of jam: 

(1) Jam, in comprising mainly water and sugar, will contain more moles per gram 
than does the pastry, which contains fats and polysaccharides, such as starch in 
the flour. Jam can, therefore, be considered to contain more energy per gram 
from a molar point of view, without even considering its liquid state. 

(2) The jam is more likely to stick to the skin than does the pastry (because it is 
sticky liquid), thereby maximizing the possibility of heat transferring to the 
skin; the pastry is flaky and/or dusty, and will exhibit a lower efficiency in 
transferring energy. 

Worked Example 4.6 Calculate the entropy change AS caused by heating 1 mol of 
sucrose from 360 K to 400 K, which is hot enough to badly burn the mouth. Take C p — 
425JK~ 1 mor 1 . 

Because the value of C p has a constant value, we place it outside the integral, which 
allows us to rewrite Equation (4.9), saying 

AS = C p ln(^) (4.10) 


We insert data into Equation (4.10) to obtain 

400 K 
360 K 

AS = 425 JKT'mor 1 x In 


AS = 425 JKT 1 mof 1 x ln(l.ll) 

AS = 44.8 JKT'mor 1 

SAQ 4.3 We want to warm the ice in a freezer from a temperature of 
-15 °C to 0°C. Calculate the change in entropy caused by the warming 
(assuming no melting occurs). Take C p for ice as 39 J K _1 mol -1 . [Hint: 
remember to convert to K from °C] 


C p is not independent of temperature, but varies slightly. For this reason, the approach 
here is only valid for relatively narrow temperature ranges of, say, 30 K. When determin- 
ing AS over wider temperature ranges, we can perform a calculation with Equation (4.9) 


provided that we know the way C p varies with temperature, expressed as a mathematical 
power series in T . For example, C p for liquid chloroform CHCI3 is 

CyjKT'mor 1 =91.47 + 7.5 x 10~ 2 T 

Alternatively, because Equation (4.9) has the form of an integral, we could plot a 
graph of C p -7- T (as y) against T (as x) and determine the area beneath the curve. We 
would need to follow this approach if C p -7- T was so complicated a function of T that 
we could not describe it mathematically. 

Justification Box 4.1 

Entropy is the ratio of a body's energy to its temperature according to the Clausius 

equality (as defined in the next section). For a reversible process, the change in entropy 

is defined by 

dS=y (4.11) 

where q is the change in heat and T is the thermodynamic temperature. Multiplying 
the right-hand side of Equation (4.11) by AT /AT (which clearly equals one), yields 

do dr 
dS= — x — (4.12) 

T dr 

If no expansion work is done, we can safely assume that q = H. Substituting H for q, 
and rearranging slightly yields 

/&H\ 1 

dS = x -dr (4.13) 

\dT J T 

where the term in brackets is simply C p . We write 

dS=-?-dT (4.14) 

Solution of Equation (4.14) takes two forms: (a) the case where C p is considered not 
to depend on temperature (i.e. determining the value of A S over a limited range of 
temperatures) and (b) the more realistic case where C p is recognized as having a finite 
temperature dependence. 

(a) C p is independent of temperature (over small temperature ranges). 

f 2 dS = C p f 

J Si Jt 

dr (4.15) 

r, 1 


AS = S 2 -Si = Cp[lnT]| (4.16) 


and hence 

AS = 



(b) C p is not independent of temperature 
ranges). We employ a similar approach to that above 
into the integral, yielding 


, except 


that C p 


is incorporated 

/ dS 

= C'^dT 


which, on integration, yields Equation 


Worked Example 4.7 What is the increase in entropy when warming 1 mol of chlo- 
roform (III) from 240 K to 330 K? Take the value of C p for chloroform from the Aside 
box on p. 142. 




We start with Equation (4.9), retaining the position of C p within the 
integral; inserting values: 

AS — Sao K — ^240 


/■330 K 

K= / 

J 240 K 

91.47 + 7.5 x 10~ 2 T 



/■330 K 
^240 K 


+ 7.5 x 10" 2 d7; 

Performing the integration, we obtain 

AS = [91.47 In T]^ 7.5 x 10 '\T\^£ 
Then, we insert the variables: 

AS = 91.47 In 

330 K 

The T' on top and 
bottom cancel in the 
second term within the 

+ 7.5 x 10 _2 (330 K-240 K) 

to yield 


,240 K, 
AS = (29.13 + 6.75) J KT 1 moP 1 

AS = 35.9 J K" 1 mol" 1 


SAQ 4.4 1 mol of oxygen is warmed from 300 K to 350 K. Calculate the 
associated rise in entropy AS if C p( o 2 )/J K _1 moT 1 = 25.8 + 1.2 x 10 _2 7/K. 

4.3 Introducing the Gibbs function 

Why is burning hydrogen gas in air (to form liquid 
water) a spontaneous reaction? 

Reaction spontaneity in a system 

The 'twin' subscript 
of 'I and g' arises 
because the reaction 
in Equation (4.19) is so 
exothermic that most 
of the water product 
will be steam. 

Equation (4.19) describes the reaction occurring when hydrogen 
gas is burnt in air: 

02(g) + 2H 2(g) ► 2H 2 (1 and g) (4.19) 

We notice straightaway how the number of moles decreases from 
three to two during the reaction, so a consideration of the system 
alone suggests a non-spontaneous reaction. There may also be a 
concurrent phase change from gas to liquid during the reaction, 
which confirms our original diagnosis: we expect AS to be negative, and so we 
predict a non-spontaneous reaction. 

But after a moment's reflection, we remember that one of the simplest tests for 
hydrogen gas generation in a test tube is to place a lighted splint nearby, and hear the 
'pop' sound of an explosion, i.e. the reaction in Equation (4.19) occurs spontaneously. 
The 'system' in this example comprises the volume within which chemicals com- 
bine. The 'surroundings' are the volume of air around the reaction vessel or flame; 
because of the explosive nature of reaction, we expect this volume to be huge. The 
surrounding air absorbs the energy liberated during the reaction; in this example, the 
energy is manifested as heat and sound. For example, the entropy of the air increases 
as it warms up. In fact, AS( surroim dings) is sufficiently large and positive that the value 
of ASftotai) is positive despite the value of AS( systsm ) being negative. So we can now 
explain why reactions such as that in Equation (4.19) are spontaneous, although at 
first sight we might predict otherwise. 

But, as chemists, we usually want to make quantitative predictions, which 
are clearly impossible here unless we can precisely determine the magnitude of 
AS( Sun - oun dings)> i- e - quantify the influence of the surroundings on the reaction, which 
is usually not a trivial problem. 

How does a reflux condenser work? 

Quantifying the changes in a system: the Gibbs function 

All preparative chemists are familiar with the familiar Liebig condenser, which we 
position on top of a refluxing flask to prevent the flask boiling dry. The evaporating 



solvent rises up the interior passage of the condenser from the flask, cools and thence 
condenses (Equation (4.20)) as it touches the inner surface of the condenser. Con- 
densed liquid then trickles back into the flask beneath. 



solvent(i) + energy 


The energy is transferred to the glass inner surface of the condenser. We maintain 
a cool temperature inside the condenser by running a constant flow of water through 
the condenser's jacketed sleeve. The solvent releases a large amount of heat energy 
as it converts back to liquid, which passes to the water circulating within the jacket, 
and is then swept away. 

Addition of heat energy to the flask causes several physicochemi- 
cal changes. Firstly, energy allows the chemical reaction to proceed, 
but energy is also consumed in order to convert the liquid solvent 
into gas. An 'audit' of this energy is difficult, because so much of 
the energy is lost to the escaping solvent and thence to the sur- 
rounding water. It would be totally impossible to account for all 
the energy changes without also including the surroundings as well 
as the system. 

So we see how the heater beneath the flask needs to provide 
energy to enable the reaction to proceed (which is what we want to 
happen) in addition to providing the energy to change the surround- 
ings, causing the evaporation of the solvent, the extent of which 
we do not usually want to quantify, even if we could. In short, 
we need a simple means of taking account of all the surroundings 
without, for example, having to assess their spatial extent. From 
the second law of thermodynamics, we write 

The 'Gibbs function' G 
is named after Josiah 
Willard Gibbs (1839- 
1903), a humble Amer- 
ican who contributed 
to most areas of phys- 
ical chemistry. He also 
had a delightful sense 
of humour: 'A math- 
ematician may say 
anything he pleases, 
but a physicist must be 
at least partially sane'. 








(see Justification Box 4.2) where the H, T and S terms have their 
usual definitions, as above, and G is the 'Gibbs function'. G is 
important because its value depends only on the system and not on 
the surroundings. By convention, a positive value of AH denotes 
an enthalpy absorbed by the system. 

AH here is simply the energy given out by the system, i.e. by the 
reaction, or taken into it during endothermic reactions. This energy 
transfer affects the energy of the surroundings, which respectively 
absorb or receive energy from the reaction. And the change in the 
energy of the surroundings causes changes in the entropy of the 
surroundings. In effect, we can devise a 'words-only' definition 
of the Gibbs function, saying it represents 'The energy available 
for reaction (i.e. the net energy), after adjusting for the entropy 
changes of the surroundings'. 

As well as calling G 
the Gibbs function, it is 
often called the 'Gibbs 
energy' or (incorrectly) 
'free energy'. 

The Gibbs function is 
the energy available for 
reaction after adjusting 
for the entropy changes 
of the surroundings. 


Justification Box 4.2 

The total change in entropy is A5'( tota i), which must be positive for a spontaneous 
process. From Equation (4.8), we say 

A O (total) = Ao (system) ~r ^ ^(surroundings) - > " 

We usually know a value for A5( sy stem) from tables. Almost universally, we do not know 

a Value for A S(surroundings) • 

The Clausius equality says that a microscopic process is at equilibrium if dS = dq/T 

where q is the heat change and T is the thermody- 
namic temperature (in kelvin). Similarly, for a macro- 
scopic process, AS = Aq/T. In a chemical reaction, 
the heat energy emitted is, in fact, the enthalpy change 
of reaction A //(system) , and the energy gained by the 
surroundings of the reaction vessel will therefore be 

-A //( Sys tem)- Accordingly, the Value of A Surroundings) 

is -AH 4- T. 

Rewriting Equation (4.8) by substituting for 

A ^(surroundings) gives 

This sign change occur- 
ring here follows since 
energy is absorbed 
by the surroundings if 
energy has been emit- 
ted by the reaction, 
and vice versa. 


AS(total) = AS( S y S tem) — (4.22a) 

The right-hand side must be positive if the process is spontaneous, so 

A^tem) - Aif T em) > (4.22b i 

^ -^(system) 

o > Ajf(system) - AS, 

rp (system) 

Multiplying throughout by T gives 

> AZ/( S ystem) — TASfsystem) (4.23) 

So the compound variable A//( Sys tem) — T AS( Sys tem) must be negative if a process is 

This compound variable occurs so often in chemistry that we will give it a symbol 
of its own: G, which we call the Gibbs function. Accordingly, a spontaneous process 
in a system is characterized by saying, 

> AG(system) (4.24) 

In words, the Gibbs energy must be negative if a change occurs spontaneously. 



The sign of AG 

Equation (4.24) in Justification Box 4.2 shows clearly that a pro- 
cess only occurs spontaneously within a system if the change in 
Gibbs function is negative, even if the sign of AS( S y Ste m) is slightly 
negative or if A//( Syste m) is slightly positive. Analysing the reaction 
in terms of our new variable AG represents a great advance: pre- 
viously, we could predict spontaneity if we knew that A5 , ( to tai) was 
positive - which we now realize is not necessarily a useful crite- 
rion, since we rarely know a value for AS( sun - oim di ng s)- It is clear 
from Equation (4.21) that all three variables, G, H and S, each 
relate to the system alone, so we can calculate the value of AG by 
looking up values of A S and AH from tables, and without needing 
to consider the surroundings in a quantitative way. 

A process occurring in a system is spontaneous if AG is nega- 
tive, and it is not spontaneous if AG is positive, regardless of the 
sign of A5( syste m)- The size of AG (which is negative) is maxi- 
mized for those processes and reactions for which AS is positive 
and which are exothermic, with a negative value of AH. 

Worked Example 4.8 Methanol (IV) can be prepared in the 
gas phase by reacting carbon monoxide with hydrogen, according 
to Equation (4.25). Is the reaction feasible at 298 K if A// e = 

The Gibbs energy must 
be negative if a change 
occurs spontaneously. 

We see the analytical 
power of AG when we 
realize how its value 
does not depend on 
the thermodynamic 
properties of the sur- 
roundings, but only on 
the system. 

90.7 kJmor 1 and AS* = -219 JKT 1 mol 

A process occurring in a 
system is spontaneous 
if AG is negative, and is 
not spontaneous when 
AG is positive. 

CO(g) + 2H 2( g) 

CH 3 OH 





We shall use Equation (4.21), AG 9 = A// s - TAS i 
values (and remembering to convert from kJ to J): 



= (-90700 J mol" 

l ) - (298 K 



= (-90700 + 65 262) JmoP 1 

= -25.4 kJ mol -1 

-219 J K" 1 mol -1 ) 

This value of AG e is negative, so the reaction will indeed be 
spontaneous in this example. This is an example of where 

The Gibbs function is 
a function of state, 
so values of AG G 
obtained with the van't 
Hoff isotherm (see 
p. 162) and routes 
such as Hess's law 
cycles are identical. 



the negative value of AH overcomes the unfavourable positive — AS term. In 
fact, although the reaction is thermodynamically feasible, the rate of reaction (see 
Chapter 8) is so small that we need to heat the reaction vessel strongly to about 550 K 
to generate significant quantities of product to make the reaction viable. 

Some zoologists be- 
lieve this reaction 
inspired the myth of 
fire-breathing dragons: 
some large tropical 
lizards have glands 
beside their mouths to 
produce both H 2 S and 
S0 2 - Under advanta- 
geous conditions, the 
resultant sulphur reacts 
so vigorously with the 
air that flames form as 
a self-defence mecha- 

SAQ 4.5 Consider the reaction 2H 2 S ( g) + SC>2(g) -> 
2H 2 ( g) + 3S (S) , where all species are gaseous except the 
sulphur. Calculate AG? for this reaction at 298 K with the 
thermodynamic data below: 

AH?/kl mol -1 
AS? /J KT 1 mor 1 

H 2 S 

S0 2 

H 2 



70.1 31.9 

[Hint: it is generally easier first to determine values of AH r 
and AS r by constructing separate Hess's-law-type cycles.] 

SAQ 4.6 The thermodynamic quantities of charge- 
transfer complex formation for the reaction 

methyl viologen + hydroquinone 

charge-transfer complex (4.26) 


are AH r = -22.6 kJ mor 1 and AS r = -62.1 J K _i mol" 1 at 298 K. Such val- 
ues have been described as 'typical of weak charge-transfer complex 
interactions'. Calculate the value of AG r . 

4.4 The effect of pressure on thermodynamic 

How much energy is needed? 

The Gibbs-Duhem equation 

'How much energy is needed?' is a pointless question. It is too imprecise to be useful 
to anyone. The amount of energy needed will depend on how much material we wish 
to investigate. It also depends on whether we wish to perform a chemical reaction 
or a physical change, such as compression. We cannot answer the question until we 
redefine it. 

The total amount we need to pay when purchasing goods at a shop depends both 
on the identity of the items we buy and how many of each. When buying sweets and 



apples, the total price will depend on the price of each item, and the amounts of each 
that we purchase. We could write it as 

d(money) = (price of item 1 x number of item 1) 
+ (price of item 2 x number of item 2) 


While more mathematical in form, we could have rewritten Equa- 
tion (4.27) 

3 (money) 3 (money) 

d(money) = — — x JV(1) + — — x N(2) + ■■■ 

J 3(1) 3(2) 

where N is merely the number of item (1) or item (2), and each 
bracket represents the price of each item: it is the amount of money 
per item. An equation like Equation (4.28) is called a total differ- 

In a similar way, we say that the value of the Gibbs function 
changes in response to changes in pressure and temperature. We 
write this as 

G = f( P , T) (4.29) 

and say G is a function of pressure and temperature. 

So, what is the change in G for a single, pure substance as the 
temperature and pressure are altered? A mathematician would start 
answering this question by writing out the total differential of G: 

dG = 

dG\ (dG 





We must use the sym- 
bol 3 ('curly d') in a dif- 
ferential when several 
terms are changing. 
The term in the first 
bracket is the rate of 
change of one variable 
when all other variables 
are constant. 

The value of G for 
a single, pure mate- 
rial is a function of 
both its temperature 
and pressure. 

The small subscripted 
p on the first bracket 
tells us the differential 
must be obtained at 
constant pressure. The 
subscripted T indicates 
constant temperature. 

which should remind us of Equation (4.28). The first term on the 

right of Equation (4.30) is the change in G per unit change in 

pressure, and the subsequent dp term accounts for the actual change in pressure. 

The second bracket on the right-hand side is the change in G per unit change in 

temperature, and the final dT term accounts for the actual change in temperature. 

Equation (4.30) certainly looks horrible, but in fact it's simply a statement of the 
obvious - and is directly analogous to the prices of apples and sweets we started by 
talking about, cf. Equation (4.27). 

We derived Equation (4.30) from first principles, using pure mathematics. An alter- 
native approach is to prepare a similar equation algebraically. The result of the 
algebraic derivation is the Gibbs-Duhem equation: 

dG = Vdp- SdT 




Justification Box 4.3 

We start with Equation (4.21) for a single phase, and write G = H — TS . Its differen- 

tial is 

dG = dH -TdS- SdT (4.32) 

From Chapter 2, we recall that H = U + pV , the differential of which is 

dH = dU + pdV + Vdp (4.33) 
Substituting for dH in Equation (4.32) with the expres- 

Equation (4.35) com- 
bines the first and 

second laws of ther- 

sion for dH in Equation (4.33), we obtain 

modynamics: it derives 

from Equation (3.5) 

dG = (dU + pdV +Vdp)-TdS- SdT (4.34) 

and says, in effect, 

till = dq + tiw. The ptiV 

For a closed system (i.e. one in which no expansion 

term relates to expan- 
sion work and the TtiS 

work is possible) 

term relates to the 
adiabatic transfer of 

dU = TdS-pdV (4.35) 

heat energy. 

Substituting for the dU term in Equation (4.34) with 
the expression for dU in Equation (4.35) yields 

dG = (T dS - p dV) + p dV + V dp - T dS - S dT 

The Gibbs-Duhem 

equation is also com- 
monly (mis-)spelt 
'Gibbs-Duheme '. 


The TdS and p dV terms will cancel, leaving the 
Gibbs-Duhem equation, Equation (4.31). 

We now come to the exciting part. By comparing the total differential of Equa- 
tion (4.30) with the Gibbs-Duhem equation in Equation (4.31) we can see a pat- 
tern emerge: 

dG = 
dG = 




dp + 




So, by direct analogy, comparing one equation with the other, we can say 


V = 








These two equations 
are known as the 
Maxwell relations. 

Equations (4.37) and (4.38) are known as the Maxwell relations. 
The second Maxwell relation (Equation (4.38)) may remind us of 
the form of the Clausius equality (see p. 142). Although the first Maxwell relation 
(Equation (4.37)) is not intuitively obvious, it will be of enormous help later when 
we look at the changes in G as a function of pressure. 

Why does a vacuum l suck'? 

The value of G as a function of pressure 

Consider two flasks of gas connected by a small tube. Imagine also that a tap separates 
them, as seen by the schematic illustration in Figure 4.4. One flask contains hydrogen 
gas at high pressure p, for example at 2 atm. The other has such a low pressure of 
hydrogen that it will be called a vacuum. 

As soon as the tap is opened, molecules of hydrogen move spon- 
taneously from the high-pressure flask to the vacuum flask. The 
movement of gas is usually so rapid that it makes a 'slurp' sound, 
which is why we often say the vacuum 'sucks'. 

Redistributing the hydrogen gas between the two flasks is essen- 
tially the same phenomenon as a dye diffusing, as we discussed 
at the start of this chapter: the redistribution is thermodynamically 
favourable because it increases the entropy, so A S is positive. 

We see how the spontaneous movement of gas always occurs 
from high pressure to low pressure, and also explains why a balloon 
will deflate or pop on its own, but work is needed to blow up 
the balloon or inflate a bicycle tyre (i.e. inflating a tyre is not 

The old dictum, 'nature 
abhors a vacuum' is not 
just an old wives tale, it 
is also a manifestation 
of the second law of 

Gases move spon- 
taneously from high 
pressure to low. 



Figure 4.4 Two flasks are connected by a tap. One contains gas at high pressure. As soon as the 
tap separating the two flasks is opened, molecules of gas move spontaneously from the flask under 
higher pressure to the flask at lower pressure. (The intensity of the shading represents the pressure 
of the gas) 


Why do we sneeze? 

The change in Gibbs function during gas movement: gas 
molecules move from high pressure to low 

When we sneeze, the gases contained in the lungs are ejected through the throat so 
violently that they can move at extraordinary speeds of well over a hundred miles 
an hour. One of the obvious reasons for a sneeze is to expel germs, dust, etc. in the 
nose - which is why a sneeze can be so messy. 

But a sneeze is much more sophisticated than mere germ removal: the pressure of 
the expelled gas is quite high because of its speed. Having left the mouth, a partial 
vacuum is left near the back of the throat as a result of the Bernoulli effect described 
below. Having a partial vacuum at the back of the throat is thermodynamically unsta- 
ble, since the pressure in the nose is sure to be higher. As in the example above, 
gas from a region of high pressure (the nose) will be sucked into the region of low 
pressure (the back of the mouth) to equalize them. The nose is unblocked during this 
process of pressure equilibration, so one of the major reasons why we sneeze is to 
unblock the nose. 

It is relatively easy to unblock a nose by blowing, but a sneeze is a superb means 
for unblocking the nose from the opposite direction. 


The Bernoulli effect 

Hold two corners of a piece of file paper along its narrow side. It will droop under its 
own weight because of the Earth's gravitational pull acting on it. But the paper will 
rise and stand out almost horizontally when we blow gently over its upper surface, as 
if by magic (see Figure 4.5). The paper droops before blowing. Blowing induces an 
additional force on the paper to counteract the force of gravity. 

The air pressure above the upper side of the paper decreases because the air moves 
over its surface faster than the air stream running past the paper's underside. The 

Blow over face of 
paper from here 

Partial vaccum here 
causes paper to lift 


Figure 4.5 The Bernoulli effect occurs when the flow of fluid over one face of a body is 
greater than over another, leading to pressure inequalities. Try it: (a) hold a pace of paper in 
both hands, and feel it sag under its own weight, (b) Blow over the paper's upper surface, 
and see it lift 



disparity in air speed leads to a difference in pressure. In effect, a partial vacuum forms 
above the paper, which 'sucks' the paper upwards. This is known as the Bernoulli effect. 
A similar effect enables an aeroplane to fly: the curve on a plane's wing is carefully 
designed such that the pressure above the wing is less than that below. The air flows over 
the upper face of the wing with an increased speed, leading to a decrease in pressure. 
Because the upward thrust on the underside of the wing is great (because of the induced 
vacuum), it counterbalances the downward force due to gravity, allowing the plane to 
stay airborne. 

How does a laboratory water pump work? 

Gibbs function of pressure change 

The water pump is another example of the Bernoulli effect, and is 
an everyday piece of equipment in most laboratories, for example 
being used during Biichner filtration. It comprises a piece of rubber 
tubing to connect the flask to be evacuated to a pump. Inside the 
pump, a rapid flow of water past one end of a small aperture inside 
the head decreases the pressure of the adjacent gas, so the pressure 
inside the pump soon decreases. 

Gas passes from the flask to the pump where the pressure is 
lower. The change in Gibbs function associated with these pressure 
changes is given by 

AG = RT lnf-^-) 

V /? (initial) / 


where the AG term represents the change in G per mole of gas. We 
will say here that gas enters the pump at pressure p( nn ai) from a flask 
initially at pressure /7 (in itiai)- Accordingly, since p (final) < p(Mm) 
and the term in brackets is clearly less than one, the logarithm term 
is negative. AG is thus negative, showing that gas movement from 
a higher pressure p(Mtiai) to a lower pressure P( nn ai) is spontaneous. 
It should be clear from Equation (4.39) that gas movement in 
the opposite direction, from low pressure (/?( nn ai)) to high (/"(initial)) 
would cause AG to be positive, thereby explaining why the process 
of gas going from low pressure to high never occurs naturally. 
Stated another way, compression can only occur if energy is put 
into the system; so, compression involves work, which explains 
why pumping up a car tyre is difficult, yet the tyre will deflate of 
its own accord if punctured. 

These highly oversim- 
plified explanations 
ignore the effects of 
turbulent flow, and the 
formation of vortices. 

The minimum pres- 
sure achievable with 
a water pump equals 
the vapour pressure of 
water, and has a value 
of about 28 mmHg. 

AG is negative for the 
physical process of gas 
moving from higher to 
lower pressure. 

Since Equation (4.39) 
relies on a ratio of pres- 
sures, we say that a 
gas moves from 'higher' 
to 'lower', rather than 
'high' to 'low'. 

Compression involves 



Equation (4.39) in- 
volves a ratio of pres- 
sures, so, although 
mmHg (millimetres of 
mercury) is not an SI 
unit of pressure, we 
are permitted to use 
it here. 

The pressure of vapour 
above a boiling liquid 
is the same as the 
atmospheric pressure. 

Worked Example 4.9 The pressure inside a water pump is the same 
as the vapour pressure of water (28 mmHg). The pressure of gas inside 
a flask is the same as atmospheric pressure (760 mmHg). What is 
the change in Gibbs function per mole of gas that moves? Take T — 
298 K. 

Inserting values into Equation (4.39) yields 

, i 28 mmHg 

AG = 8.414 JKT 1 moH 1 x 298 K x In ' 

760 mmHg 

AG = 2477 JmoP 1 x ln(3.68 x 10~ 2 ) 
AG = 2477 JmoP 1 x (-3.301) 
AG = -8.2kJmol _1 

SAQ 4.7 A flask of methyl-ethyl ether (V) is being evaporated. Its boiling 
temperature is 298 K (the same as room temperature) so the vapour 
pressure of ether above the liquid is the same as atmospheric pressure, 
i.e. at 100 kPa. The source of the vacuum is a water pump, so the pressure 
is the vapour pressure of water, 28 mmHg. 

Cri2 ,Cri3 

CH 3 ^ ^O 

(1) Convert the vacuum pressure P( V acuum) into an SI pressure, 
remembering that 1 atm = 101 325 kPa = 760 mmHg. 

(2) What is the molar change in Gibbs function that occurs when 
ether vapour is removed, i.e. when ether vapour goes from the 
flask at p s into the water pump at p (va cuum)? 

Justification Box 4.4 

We have already obtained the first Maxwell relation (Equation (4.37)) by comparing the 
Gibbs-Duhem equation with the total differential: 

— = V 


The ideal-gas equation says pV = nRT ', or, using a 
molar volume for the gas (Equation (1.13)): 

We obtain the molar 
volume l/m as V h- n. 

pV m 



Substituting for V m from Equation (4.37) into Equation (1.13) 


"(f) = Rr 


And separation of the variables gives 

dG 1 
= RT- 

dp p 



dG = RT-dp 



Integration, taking G, at p<MM) and G 2 at P( nna i), yields 

G 2 G X = RT lnf /?(final) ) 

V P (initial) / 


Or, more conveniently, if G 2 is the final value of G and G\ 

the initial value of G, then 

the change in Gibbs function is 

AG = RT ln( — ) 

i.e. Equation (4.39). 


In practice, it is often found that compressing or de- 
compressing a gas does not follow closely to the ideal- 
gas equation, particularly at high p or low T, as exem- 
plified by the need for equations such as the van der 
Waals equation or a virial expression. The equation 
above is a good approximation, though. 

A more thorough treatment takes one of two courses: 

The fugacity f can be 
regarded as an 'effec- 
tive' pressure. The 
'fugacity coefficient' y 
represents the devia- 
tion from ideality. The 
value of y tends to one 
asp tends to zero. 

(1) Utilize the concept of virial coefficients; see 

p. 57. 
(2) Use fugacity instead of pressure. 

Fugacity / is defined as 

f = pxy (4.44) 

The word 'fugacity' 
comes from the Latin 
fugere, which means 
'elusive' or 'difficult to 
capture'. The modern 
word 'fugitive' comes 
from the same source. 



where p is conventional pressure and y is the fugacity coefficient representing the 
deviation from ideality. Values of y can be measured or calculated. 
We employ both concepts to compensate for gas non-ideality. 

4.5 Thermodynamics and the extent 
of reaction 

Incomplete reactions and extent of reaction 

As chemists, we should perhaps re-cast the question 'Why is a 'weak' acid weak?' by 
asking 'How does the change in Gibbs function relate to the proportion of reactants 
that convert during a reaction to form products?' 

An acid is defined as a proton donor within the Lowry-Br0nsted theory (see 
Chapter 6). Molecules of acid ionize in aqueous solution to form an anion and a 
proton, both of which are solvated. An acid such as ethanoic acid (VI) is said to 
be 'weak' if the extent to which it dissociates is incomplete; we call it 'strong' if 
ionization is complete (see Section 6.2). 

H O 

H O-H 


Ionization is, in fact, a chemical reaction because bonds break and form. Consider 
the following general ionization reaction: 

HA + H 2 ► H 3 + + A" 


We give the extent of the reaction in Equation (4.45) the Greek 
symbol £. It should be clear that § has a value of zero before 
the reaction commences. By convention, we say that f = 1 mol 
if the reaction goes to completion. The value of % can take any 
value between these two extremes, its value increasing as the reac- 
tion proceeds. A reaction going to completion only stops when no 
reactant remains, which we define as £ having a value of 1 mol, 
although such a situation is comparatively rare except in inorganic 
redox reactions. In fact, to an excellent approximation, all preparative organic reac- 
tions fail to reach completion, so < £ < 1. 

The value of £ only stops changing when the reaction stops, although the rate at 
which % changes belongs properly to the topic of kinetics (see Chapter 8). We say 

We give the Greek 
symbol £ Cxi') to the 
extent of reaction. 
f is commonly mis- 
pronounced as 'ex- 



it reaches its position of equilibrium, for which the value of £ has 
its equilibrium value §( e q). We propose perhaps the simplest of the 
many possible definitions of equilibrium: 'after an initial period of 
reaction, no further net changes in reaction composition occur'. 

So when we say that a carboxylic acid is weak, we mean that 
§(eq) is small. Note how, by saying that £( eq ) is small at equilibrium, 
we effectively imply that the extent of ionization is small because 
[H 3 + ] and [A - ] are both small. 

But we need to be careful when talking about the magnitudes 
of £. Consider the case of sodium ethanoate dissolved in dilute 
mineral acid: the reaction occurring is, in fact, the reverse of that 
in Equation (4.45), with a proton and carboxylate anion associat- 
ing to form undissociated acid. In this case, £ = 1 mol before the 
reaction occurs, and its value decreases as the reaction proceeds. 
In other words, we need to define our reaction before we can speak 
knowledgeably about it. We can now rewrite our question, asking 
'Why is £ <C 1 for a weak acid?' 

The standard Gibbs function change for reaction is AG e , and 
represents the energy available for reaction if 1 mol of reactants 
react until reaching equilibrium. Figure 4.6 relates AG and f , and 
clearly shows how the amount of energy available for reaction 
AG decreases during reaction (i.e. in going from left to right as % 
increases). Stated another way, the gradient of the curve is always 
negative before the position of equilibrium, so any increases in % 
cause the value of AG to become more negative. 

We assume such an 
equilibrium is fully 
reversible in the sense 
of being dynamic - the 
rate at which products 
form is equal and oppo- 
site to the rate at which 
reactants regenerate 
via a back reaction. 

Reminder: the energy 
released during reac- 
tion originates from 
the making and break- 
ing of bonds, and the 
rearrangement of sol- 
vent. The full amount 
of energy given out 
is AH e , but the net 
energy available is 
less that AH*, being 
AH* -TAS*. 



Extent of reaction f 

Figure 4.6 The value of the Gibbs function AG decreases as the extent of reaction £ until, at 
£ ( eq), there is no longer any energy available for reaction, and AG = 0. £ = represents no reaction 
and f = 1 mol represents complete reaction 


The amount of energy liberated per incremental increase in reaction is quite large 
at the start of reaction, but decreases until, at equilibrium, a tiny increase in the extent 
of reaction would not change AG( tota i). The graph has reached a minimum, so the 
gradient at the bottom of the trough is zero. 

The minimum in the graph of AG against § is the reaction's position of equilib- 
rium - we call it £( eq ). The maximum amount of energy has already been expended 
at equilibrium, so AG is zero. 

Any further reaction beyond £( eq ) would not only fail to liberate any further energy, 
but also would in fact consume energy (we would start to go 'uphill' on the right- 
hand side of the figure). Any further increment of reaction would be character- 
ized by AG > 0, implying a non-spontaneous process, which is why the reaction 

Stops at £ (eq) . 

Why does the pH of the weak acid remain constant? 

The law of mass action and equilibrium constants 

The amount of ethanoic acid existing as ionized ethanoate anion and solvated proton 
is always small (see p. 253). For that reason, the pH of a solution of weak acid is 
always higher than a solution of the same concentration of a strong acid. A naive 
view suggests that, given time, all the undissociated acid will manage to dissociate, 
with the dual effect of making the acid strong, and hence lowering the pH. 

We return to the graph in Figure 4.6 of Gibbs function (as y) against extent of 
reaction % (as x). At the position of the minimum, the amounts of free acid and 
ionized products remain constant because there is no longer any energy available for 
reaction, as explained in the example above. 

The fundamental law of chemical equilibrium is the law of mass action, formulated in 
1 864 by Cato Maximilian Guldberg and Peter Waage. It has since been redefined several 
times. Consider the equilibrium between the four chemical species A, B, C and D: 

a A + bB = cC + dD (4.46) 

where the respective stoichiometric numbers are —a, —b, c and d. The law of mass 
action states that, at equilibrium, the mathematical ratio of the concentrations of 
the two reactants [A]" x [B] h and the product of the two product concentrations 
[C] c x [D] d , is equal. We could, therefore, define one of two possible fractions: 

[Af[B]" [Cf[DY 

7 or (4.47) 

[C] e [D] rf [A] a [B]'' 

This ratio of concentrations is called an equilibrium constant, and is symbolized as K. 
The two ratios above are clearly related, with one being the reciprocal of the other. 
Ultimately, the choice of which of these two we prefer is arbitrary, and usually relates 
to the way we write Equation (4.46). In consequence, the way we write this ratio is 
dictated by the sub-discipline of chemistry we practice. For example, in acid-base 



chemistry (see Chapter 6) we write a dissociation constant, but in complexation equi- 
libria we write a formation constant. 

In fact, an equilibrium constant is only ever useful when we have carefully defined 
the chemical process to which it refers. 

The reaction quotient 

It is well known that few reactions (other than inorganic redox reactions) ever reach 
completion. The value of £( eq ) is always less than one. 
The quotient of products to reactants during a reaction is 


Y\ [products] v 
\\ [reactants] v 


The mathematical sym- 
bol n means the 'pi 
product', meaning the 
terms are multiplied 
together, so \\{2, 3, 4) 
is 2 x 3 x 4 = 24. 

which is sometimes called the reaction quotient. The values of v 

are the respective stoichiometric numbers. The mathematical value 

of Q increases continually during the course of reaction because 

of the way it relates to concentrations during reaction. Initially, 

the reaction commences with a value of Q = 0, because there is no product (so the 

numerator is zero). 


A statement such as 'K = 0.4 moldm" ' is wrong, although we find examples in a 
great number of references and textbooks. We ought, rather, to say K = 0.4 when the 
equilibrium constant is formulated (i) in terms of concentrations, and (ii) where each 
concentration is expressed in the reference units of mol dm" . Equilibrium constants 
such as K c or K p are mere numbers. 

ery decrease to zero. 

Relationships between K and AG e 

The voltage of a new torch battery (AA type) is about 1 .5 V. After 
the battery has powered the torch for some time, its voltage drops, 
which we see in practice as the light beam becoming dimmer. If 
further power is withdrawn indefinitely then the voltage from the 
battery eventually drops to zero, at which point we say the battery 
is 'dead' and throw it away. 

A battery is a device for converting chemical energy into electri- 
cal energy (see p. 344), so the discharging occurs as a consequence 
of chemical reactions inside the battery. The reaction is complete 

The battery produces 
'power' W (energy per 
unit time) by passing 
a current through a 
resistor. The resister in 
a torch is the bulb 



The relationship be- 
tween AG and volt- 
age is discussed in 
Section 7.1. 

when the battery is 'dead', i.e. the reaction has reached its equilib- 
rium extent of reaction £( eq ) . 

The battery voltage is proportional to the change in the Gibbs 
function associated with the battery reaction, call it A G (battery)- 
Therefore, we deduce that AG (battery) must decrease to zero because 
the battery voltage drops to zero. Figure 4.7 shows a graph of battery voltage (as y) 
against time of battery discharge (as x); the time of discharge is directly analogous 
to extent of reaction £. Figure 4.7 is remarkably similar to the graph of AG against 
£ in Figure 4.6. 

The relationship between the energy available for reaction AG r and the extent of 
reaction (expressed in terms of the reaction quotient Q) is given by 

AG r = AG? + RT In Q 


where AG is the energy available for reaction during chemical changes, and AG^ 

is the standard change of Gibbs function AG S , representing the change in Gibbs 

function from £ = to £ = £( eq) . 

Equation (4.49) describes the shape of the graph in Figure 4.6. 
Before we look at Equation (4.46) in any quantitative sense, we 
note that if RT In Q is smaller than AG?, then AG r is positive. 
The value of AG r only reaches zero when AG? is exactly the same 
as RT In Q. In other words, there is no energy available for reaction 

when AG r = 0: we say the system has 'reached equilibrium'. In fact, AG r = is 

one of the best definitions of equilibrium. 

In summary, the voltage of the battery drops to zero because the value of AG r is 

zero, which happened at f = £( eq) . 

AG r = 

is one 

of the 

best defi 

nitions of equi- 


E 0.4 

20 30 40 50 
Extent of discharge f 

Figure 4.7 Graph of battery emf (as y) against extent of discharge (as x). Note the remarkable 
similarity between this figure and the left-hand side of Figure 4.6, which is not coincidental because 
emf on AG, and extent of discharge is proportional to £. The trace represents the ninth discharge of 
a rechargeable lithium-graphite battery, constructed with a solid-state electrolyte of polyethylene 
glycol containing LiC104. The shakiness of the trace reflects the difficulty in obtaining a reversible 
measurement. Reprinted from S. Lemont and D. Billaud, Journal of Power Sources 1995; 54: 338. 
Copyright © 1995, with permission from Elsevier 



Justification Box 4.5 

Consider again the simple reaction of Equation (4.46): 

aA + bB = cC + dD 
We ascertain the Gibbs energy change for this reaction. We start by saying 

AG = 2 , V G (products) — / t V G (reactants) 

where v is the respective stoichiometric number; so 

AG = cG c + dG D -aG A -bG B (4.50) 

From an equation like Equation (4.43), G = 6 s + RT ln(p/p & ), so each G term in 
Equation (4.50) may be converted to a standard Gibbs function by inserting a term like 
Equation (4.43): 

AG = 

cG° + cRT In ( -^ j + dG^ + dRT In ( -^ 

-a^rin I — 

bGl -bRT\n( — 
. p^ J ' \/>° 

We can combine the G e terms as AG e by saying 

AG S = cG c + <f G D — «G A — bG 

So Equation (4.51) simplifies to become: 

AG = AG* + cRT\n{ —) + dRT in' /l! 





Then, using the laws of logarithms, we can simplify further: 

' (Pc/P*Y(PD/P*y l 

AG = AG" + RTln 




The bracketed term is the reaction quotient, expressed 
in terms of pressures, allowing us to rewrite the equation 
in a less intimidating form of Equation (4.49): 

AG r = AGf + RT\nQ 

A similar proof may be used to derive an expression 
relating to AG® and K c . 

We changed the posi- 
tioned of each stoi- 
chiometric number via 
the laws of logarithms, 
saying b x Ina = Ina 6 . 




A further complication arises from the AG term in Equation (4.49). The diagram above 
is clearer than the derivation: in reality, the differential quantity 3G/3f only corresponds 
to the change in Gibbs function AG under certain, well defined, and precisely controlled 
experimental conditions. 

This partial differential is called the reaction affinity in older texts and in newer texts 
is called the reaction free energy. 

Why does the concentration of product stop 

The van't Hoff isotherm 

Because we only ever 
write K (rather than 
Q) at equilibrium, it is 
tautologous but very 
common to see K writ- 
ten as K (eq) or K e . 

The descriptor 'iso- 
therm' derives from 
the Greek iso meaning 
'same' and thermos 
meaning 'temperature'. 

Jacobus van't Hoff 
was a Dutch scientist 
(1852-1911). Notice 
the peculiar arrange- 
ment of the apostro- 
phe, and small and 
capital letters in his 

It would be beneficial if we could increase the yield of a chem- 
ical reaction by just leaving it to react longer. Unfortunately, the 
concentrations of reactant and product remain constant at the end 
of a reaction. In other words, the reaction quotient has reached a 
constant value. 

At equilibrium, when the reaction stops, we give the reaction 
quotient the special name of equilibrium constant, and re-symbolize 
it with the letter K. The values of K and Q are exactly the 
same at equilibrium when the reaction stops. The value of Q is 
always smaller than K before equilibrium is reached, because some 
product has yet to form. In other words, before equilibrium, the 
top line of Equation (4.48) is artificially small and the bottom is 
artificially big. 

Q and K only have the same value when the reaction has reached 
equilibrium, i.e. when AG r = 0. At this extent of reaction, the rela- 
tionship between £ and AG S is given by the van't Hoff isotherm: 




where R and T have their usual thermodynamics meanings. The 
equation shows the relationship between AG° and K, indicat- 
ing that these two parameters are interconvertible when the tem- 
perature is held constant. 

SAQ 4.8 Show that the van't Hoff isotherm is dimensionally self- 

Worked Example 4.10 Consider the dissociation of ethanoic (acetic) acid in water to 
form a solvated proton and a solvated ethanoate anion, CH3COOH + H2O -¥ CH3COO 



+ H30 + . This reaction has an equilibrium constant K of about 2 x 
10~ 5 at room temperature (298 K) when formulated in the usual units 
of concentration (mol dm~ ). What is the associated change in Gibbs 
function of this reaction? 

Inserting values into the van't Hoff isotherm (Equation (4.55)): 

AG* = -8.314 JKT 1 moP 1 x 298 K x ln(2 x 1(T 5 ) 
AG S = -2478 Jmor 1 x -10.8 


+26 811 Jmor 


AG* = +26.8 kJmor 1 
Note how AG 9 is positive here. We say it is endogenic. 

The correct use of the 
van't Hoff isotherm 
necessitates using the 
thermodynamic tem- 
perature (expressed 
in kelvin). 

A process occurring 
with a negative value 
of AG is said to be 
exogenic. A process 
occurring with a posi- 
tive value of AG is said 
to be endogenic. 

Justification Box 4.6 

We start with Equation 


AG = AG* + RT In Q 

At equilibrium, the value of AG is zero. Also, the value of Q is 

called K: 

= AG* + RTlnK 


Subtracting the '— RTlnK' term from both sides yields the 
(Equation (4.55)): 

AG* = -RTlnK 

van't Hoff isotherm 

This derivation proves that equilibrium constants do exist. The value of AG^ 
on T , so the value of K should be independent of the total pressure. 


We sometimes want to know the value of K from a value of AG*, in which case 
we employ a rearranged form of the isotherm: 

K = exp 



so a small change in the Gibbs function means a small value of K. Therefore, a weak 
acid is weak simply because AG S is small. 



Care: we must always 
convert from kJ to J 
before calculating with 
Equation (4.57). 

Worked Example 4.11 Consider the reaction between ethanol and 
ethanoic acid to form a sweet- smelling ester and water: 

CH 2 CH 2 OH + CH3COOH ► CH 2 CH 2 C0 2 CH 3 + H 2 (4.58) 

What is the equilibrium constant K at room temperature (298 K) if 
the associated change in Gibbs function is exogenic at —3.4 kJmol -1 ? 

Care: if we calculate 
a value of K that is 
extremely close to 
one, almost certainly 
we forgot to convert 
from kJ to J, mak- 
ing the fraction in the 
bracket a thousand 
times too small. 

Inserting values into eq. (4.57): 

K — exp 

-(-3400 J moP 1 ) 

SJWJKT'mor 1 x 298 K 
K = exp(+ 1.372) 
K = 3.95 

A value of K greater than one corresponds to a negative value of 
AG S , so the esterincation reaction is spontaneous and does occur 
to some extent without adding addition energy, e.g. by heating. 

A few values of AG 9 are summarized as a function of K in 
Table 4.1 and values of K as a function of AG 9 are listed in 
Table 4.2. Clearly, K becomes larger as AG becomes more neg- 
ative. Conversely, AG** is positive if K is less than one. 

Justification Box 4.7 

We start with Equation (4.55): 

AG e = -RTlnK 

Both sides are divided by — RT , yielding 

/-AG & \ 
\ RT ) 


Then we take the exponential of both sides to generate Equation (4.57). 

Table 4.1 The relationship between AG S and equilib- 
rium constant K: values of AG ^ as a function of K 

We see from Table 4.1 
that every decade 
increase in K causes 
AG° to become more 
negative by 5.7 kJ mor 1 
per tenfold increase 
in K. 


AG^/kJ mok 


l() 2 
10 3 
10 4 

10- 1 
10- 2 
10- 3 





+ 11.4 

+ 17.1 


Table 4.2 The relationship between AG e 
and equilibrium constant K: values of K as 
a function of AG S 

AG e 


moP 1 







-10 2 

3.38 x 10 17 

-10 3 


+ 1 


+ 10 


+ 10 2 

2.96 x 10~ 18 

SAQ 4.9 What is the value of AC corresponding to AGf 98 K = -12 kJ mo 

Why do chicken eggs have thinner shells 
in the summer? 

The effect of altering the concentration on £ 

Egg shells are made of calcium carbonate, CaCC>3. The chicken ingeniously makes 
shells for its eggs by a process involving carbon dioxide dissolved in its blood, 
yielding carbonate ions which combine chemically with calcium ions. An equilibrium 
is soon established between these ions and solid chalk, according to 

Ca 2+ (aq) + C0 2 - 3( a q ) = CaC0 3(s , she ii) (4.60) 

Unfortunately, chickens have no sweat glands, so they cannot perspire. To dissipate 
any excess body heat during the warm summer months, they must pant just like a 
dog. Panting increases the amount of carbon dioxide exhaled, itself decreasing the 
concentration of CO2 in a chicken's blood. The smaller concentration [C0 3 7 J during 
the warm summer causes the reaction in Equation (4.60) to shift further toward the 
left-hand side than in the cooler winter, i.e. the amount of chalk formed decreases. 
The end result is a thinner eggshell. 

Chicken farmers solve the problem of thin shells by carbonating the chickens' 
drinking water in the summer. We may never know what inspired the first farmer to 
follow this route, but any physical chemist could have solved this problem by first 
writing the equilibrium constant K for Equation (4.60): 

[CaC0 3(s )] 

^(shell formation) = — ^ 2 - , ( 4 " 61 ) 

|Ca (aq)J|C0 3(aq) J 

The value of ^( S h e ii formation) will not change provided the temperature is fixed. There- 
fore, we see that if the concentration of carbonate ions (see the bottom line) falls then 


the amount of chalk on the top line must also fall. These changes must occur in tandem 
if K is to remain constant. In other words, decreasing the amount of CO2 in a chicken's 
blood means less chalk is available for shell production. Conversely, the same reason- 
ing suggests that increasing the concentration of carbonate - by adding carbonated 
water to the chicken's drink - will increases the bottom line of Equation (4.61), and 
the chalk term on the top increases to maintain a constant value of K. 

Le Chatelier's principle 

Arguments of this type illustrate Le Chatelier's principle, which was formulated in 
1888. It says: 

Le Chatelier's prin- 
ciple is named after 
Henri Louis le Chatelier 
(1850-1937). He also 
spelt his first name the 
English way, as'Henry'. 

Any system in stable chemical equilibrium, subjected to the influence 
of an external cause which tends to change either its temperature or 
its condensation (pressure, concentration, number of molecules in unit 
volume), either as a whole or in some of its parts, can only undergo such 
internal modifications as would, if produced alone, bring about a change 
of temperature or of condensation of opposite sign to that resulting from 
the external cause. 

The principle represents a kind of 'chemical inertia', seeking to minimize the changes 
of the system. It has been summarized as, 'if a constraint is applied to a system in 
equilibrium, then the change that occurs is such that it tends to annul the constraint'. 
It is most readily seen in practice when: 

(1) The pressure in a closed system is increased (at fixed temperature) and 
shifts the equilibrium in the direction that decreases the system's volume, 
i.e. to decrease the change in pressure. 

(2) The temperature in a closed system is altered (at fixed pressure), and the 
equilibrium shifts in such a direction that the system absorbs heat from its 
surroundings to minimize the change in energy. 

4.6 The effect of temperature on 
thermodynamic variables 

Why does egg white denature when cooked but 
remain liquid at room temperature? 

Effects of temperature on AG & : the Gibbs-Helmholtz equation 

Boiling an egg causes the transparent and gelatinous albumen ('egg white') to modify 
chemically, causing it to become a white, opaque solid. Like all chemical reactions, 



denaturing involves the rearrangement of bonds - in this case, of 
hydrogen bonds. For convenience, in the discussion below we say 
that bonds change from a spatial arrangement termed '1' to a dif- 
ferent spatial arrangement '2'. 

From everyday experience, we know that an egg will not dena- 
ture at room temperature, however long it is left. We are not saying 
here that the egg denatures at an almost infinitesimal rate, so the 
lack of reaction at room temperature is not a kinetic phenomenon; 
rather, we see that denaturation is energetically non-spontaneous 
at one temperature (25 °C), and only becomes spontaneous as the 
temperature is raised above a certain threshold temperature, which 
we will call 7( cr jticai) (about 70 °C for an egg). 

The sign of the Gibbs function determines reaction spontaneity, 
so a reaction will occur if AG e is negative and will not occur if 
AG 9 is positive. When the reaction is 'poised' at r( cr iti ca i) between 
spontaneity and non-spontaneity, the value of AG S = 0. 

The changes to AG e with temperature may be quantified with 
the Gibbs-Helmholtz equation: 







where AG 9 is the change in Gibbs function at the temperature Tj_ 
and AGf is the change in Gibbs function at temperature T\. Note 
how values of T must be expressed in terms of thermodynamic 
temperatures. AH is the standard enthalpy of the chemical pro- 
cess or reaction, as determined experimentally by calorimetry or 
calculated via a Hess's-law-type cycle. 

The value of AH V for denaturing egg white is likely to be 
quite small, since it merely involves changes in hydrogen bonds. 
For the purposes of this calculation, we say A// r has a value of 
35 kJmol" 1 . 

Additionally, we can propose an equilibrium constant of reaction, 
although we must call it apseudo constant ^( pS eudo) because we can- 
not in reality determine its value. We need a value of A"(p Seu do) in 
order to describe the way hydrogen bonds change position during 
denaturing. We say that ^( pS eudo, i) relates to hydrogen bonds in the 
pre-reaction position '1' (i.e. prior to denaturing) and ^( pseu do, 2) 
relates to the number of hydrogen bonds reoriented in the post- 
reaction position '2' (i.e. after denaturing). We will say here that 
^(pseudo) is '1/10' before the denaturing reaction, i.e. before boiling 
the egg at 298 K. From the van't Hoff isotherm, ^"(pseudo) equates 
to a Gibbs function change of +5.7 kJmol -1 . 

This argument here has 
been oversimplified 
because the reaction 
is thermodynamically 
irreversible - after all, 
you cannot 'unboil an 

Some reactions 








at others. 

The temperature de- 
pendence of the Gibbs 
function change is 
described quantita- 
tively by the Gibbs- 
Helmholtz equation. 

The word 'pseudo' 
derives from the Greek 
stem pseudes meaning 
'falsehood', which is 
often taken to mean 
having an appearance 
that belies the actual 
nature of a thing. 

Denaturing albumen 
is an 'irreversible' 
process, yet the deriva- 
tions below assume 
thermodynamic re- 
versibility. In fact, 
complete reversibility 
is rarely essential; try 
to avoid making calcu- 
lations if a significant 
extent of irreversibility 
is apparent. 


Worked Example 4.12 The white of an egg denatures while immersed in water boiling 
at its normal boiling temperature of 373 K. What is the value of AG at this higher 
temperature? Take AH — 35 kJmol - . 

It does not matter 
which temperature is 
chosen as T± and which 
as 7" 2 so long as 7"i 
relates to AGi and T 2 
relates to AG 2 . 

The value of AG S at 298 K is +5.7 kJmol , the positive sign 
explaining the lack of a spontaneous reaction. Inserting values into 
the Gibbs-Helmholtz equation, Equation (4.62), yields 

AG373 k 5700 J mor 1 , / 1 1 

+35 000 J moP 

373 K 298 K V 373 K 298 K 

Note that T is a thermodynamic temperature, and is cited in kelvin. All energies have 
been converted from kJmoP 1 to Jmol" . 

P^ = (19.13 jr'mor')+ (35 000 Jmol" 1 ) x (-6.75 x 10" 4 K" 1 ) 

373 K 

A C & 

^Ji = (19.13 jk" 1 mor 1 ) - (23.62 JKT 1 moP 1 ) 

373 K 

A ^373 K ^_ 44JK -l mol -l 

373 K 



AG 3 e 73 K = -4.4 JK -1 mol -1 x 373 K 

AG 373 K = —1.67 kJmol 

AG 373 K has a negative value, implying that the reaction at this new, elevated temperature 
is now spontaneous. In summary, the Gibbs-Helmholtz equation quantifies a qualitative 
observation: the reaction to denature egg white is not spontaneous at room temperature, 
but it is spontaneous at elevated temperatures, e.g. when the egg is boiled in water. 

SAQ 4.10 Consider the reaction in Equation (4.63), which occurs wholly 
in the gas phase: 

NH 3 + |0 2 >NO + |H 2 (4.63) 

The value of AG? for this reaction is -239.9 kJ moT 1 at 298 K. If the 
enthalpy change of reaction A/-/ r = -406.9 kJ mor 1 , then 

(1) Calculate the associated entropy change for the reaction in 
Equation (4.63), and comment on its sign. 

(2) What is the value of the Gibbs function for this reaction when the 
temperature is increased by a further 34 K? 



Justification Box 4.8 

We will use the quantity 'G -r T' for the purposes of 
this derivation. Its differential is obtained by use of the 
product rule. In general terms, for a compound function 
ab, i.e. a function of the type y = f(a, b): 




do da 

= ax hDX — 

dx dx 

so here 

d(G H- T) 


1 dG / 

= - x \-Gx\- 

T dT I 


The function G- 


occurs so 





we call it 

the Planck 


All standard signs © 
have been omitted for 


Note that the term is —S, so the equation becomes 

dr H 

d(G -r T) _ S G 
df ~ ~T ~ f 1 


Recalling the now-familiar relationship G = H — TS , we may substitute for the —S 
term by saying 

G- H 

-S = 

d(G^T) (G-H\ 1 G 


T T 2 

The term in brackets on the right-hand side is then split up; so 

d(G -r T) 






T 2 

T 2 



On the right-hand side, the first and third terms cancel, yielding 
d(G H- T) 




Writing the equation in this way tells us that if we 
know the enthalpy of the system, we also know the 
temperature dependence of G 4- T . Separating the vari- 
ables and defining G\ as the Gibbs function change at 
T\ and similarly as the value of G2 at T 2 , yields 

/ d(G/T) = H - — 



This derivation as- 
sumes that both H 
and S are tempera- 
ture invariant - a safe 
assumption if the vari- 
ation between 7~i and 
7" 2 is small (say, 40 K 
or less). 




G 2 /T 2 

r 1 1 

T 2 

J . 

= H 


J - 



So, for a single chemical: 

G 1 _G 1 = H n_l\ 
T 2 T { \T 2 TJ 


And, for a chemical reaction we have Equation (4.62): 

AG 2 AG] / 1 1 \ 

= AH { 

T 2 T x \T 2 Tj 

We call this final equation the Gibbs-Helmholtz equation. 

At what temperature will the egg start to denature? 

Reactions 'poised' at the critical temperature 

Care: the nomenclature 
7"(criticai) is employed 
in many other areas 
of physical chemistry 
(e.g. see pp. 50 and 

The reaction is 'poised' 
at the critical tempera- 
ture with AG= 0. 

If AG 9 goes from positive to negative as the temperature alters, 
then clearly the value of AG 9 will transiently be zero at one unique 
temperature. At this 'point of reaction spontaneity', the value of 
AG 9 = 0. We often call this the 'critical temperature' T^criticai) - 

The value of T( cr iticai)» i- e - the temperature when the reaction first 
becomes thermodynamically feasible, can be determined approxi- 
mately from 


Critical) = — (4.74) 

Worked Example 4.13 

egg albumen 'poised' ? 


At what temperature is the denaturation of 

We will employ the thermodynamic data from Worked Example 4.12. Inserting 
values into Equation (4.74): 

This method yields only 
an approximate value 
of 7" (cr iticai) because AS 
and AH are themselves 
functions of tempera- 


35 000 J mol" 

' (critical) 

AS 98.3 JK" 1 mol" 

356 K or 83 C 

We deduce that an egg will start denaturing above about T = 
83 °C, confirming what every cook knows, that an egg cooks in 
boiling water but not in water that is merely 'hot'. 


SAQ 4.11 Water and carbon do not react at room temperature. Above 
what temperature is it feasible to prepare synthesis gas (a mixture of CO 
and H2)? The reaction is 


2H 2 



(g) "I" H 2 (g) 


Take AH & = 132 kJ mol" 1 and AS* = 134 J K _1 mol -1 . 

Justification Box 4.9 

We start with Equation (4.21): 

AG e =0= AH - TAS 

When just 'poised', the value of AG* is equal to zero. Accordingly, = 

= AH 

= TAS. 

Rearranging slightly, we obtain 



which, after dividing both sides by 'AS', yields Equation (4.74). 

Why does recrystallization work? 

The effect of temperature on K: the van't Hoff isochore 

To purify a freshly prepared sample, the preparative chemist will crystallize then 
recrystallize the compound until convinced it is pure. To recrystallize, we first dissolve 
the compound in hot solvent. The solubility s of the compound depends on the 
temperature T. The value of s is high at high temperature, but it decreases at lower 
temperatures until the solubility limit is first reached and then surpassed, and solute 
precipitates from solution (hopefully) to yield crystals. 

The solubility s relates to a special equilibrium constant we call 
the 'solubility product' K s , defined by 

K s = [solute] 



We say the value of 
[solute]( S ) = 1 because 
its activity is unity; see 
Section 7.3. 

The [solute] term may, in fact, comprise several component parts 
if the solute is ionic, or precipitation involves agglomeration. This 
equilibrium constant is not written as a fraction because the 'effec- 
tive concentration' of the undissolved solute [solute] ( s ) can be taken 
to be unity. 

Like all equilibrium constants, the magnitude of the equilibrium 
constant K s depends quite strongly on temperature, according to 

The word 'isochore' 
implies constant pres- 
sure, since iso is Greek 
for 'same' and the root 
chore means pressure. 



the van't Hoff isochore: 









where R is the gas constant, AH, t , is the change in enthalpy associated with 
crystallization, and the two temperatures are expressed in kelvin, i.e. thermodynamic 

Worked Example 4.14 The solubility s of potassium nitrate is 140 g per 100 g of water 
at 70.9 °C, which decreases to 63.6 g per 100 g of water at 39.9 °C. Calculate the enthalpy 
of crystallization, AH, t . 

Strategy. For convenience, we will call the higher temperature Ti and the lower temper- 
ature T\. (1) The van't Hoff isochore, Equation (4.78), is written in terms of a ratio, so 
we do not need the absolute values. In other words, in this example, we can employ the 
solubilities s without further manipulation. We can dispense with the units of s for the 
same reason. (2) We convert the two temperatures to kelvin, for the van't Hoff isochore 
requires thermodynamic temperatures, so T-i — 343.9 K and T\ — 312.0 K. (3) We insert 
values into the van't Hoff isochore (Equation (4.78)): 

Note how the two 
minus signs on the 
right will cancel. 





8.314 JKT'mol 



343.9 K 312.0 K 

ln(2.20) = 



8.314 JK" 1 mol" 
In 2.20 = 0.7889 

x (-2.973 x 10" 4 K _1 ) 

We then divide both sides by '2.973 x 10 4 K , so: 




Only when the differ- 
ence between T 2 and 
7"i is less than ca. 
40 K can we assume 
the reaction enthalpy 
AH 9 is independent of 
temperature. We oth- 
erwise correct for the 
temperature depen- 
dence of AH e with 
the Kirch hoff equation 
(Equation (3.19)). 

2.973 x 10- 4 K" 

8.314 JKT'mor 1 

The term on the left equals 2.654 x 10 3 K. Multiplying both sides by 
R then yields: 

A/7* ryst) = 2.654 x 10 3 K x 8.314 JK ' mol" 1 

Atf ( : ry st) = 22.1 kJ mol" 1 

SAQ 4.12 The simple aldehyde ethanal (VII) reacts with 
the di-alcohol ethylene glycol (VIII) to form a cyclic 
acetal (IX): 



CH 3 ^ 


= + 


XH 2 

,CH 2 


CHo 0-CH 2 

H 0-CH 2 

Calculate the enthalpy change AH & for the reaction if the equilibrium constant for 
the reaction halves when the temperature is raised from 300 K to 340 K. 

Justification Box 4.10 

We start with the Gibbs-Helmholtz equation (Equation (4.62)): 


AG7 „/l 1 
1 = Afl e 


T 2 Ti 

Each value of AG e can be converted to an equilibrium constant via the van't Hoff 

isotherm AG* 

-RTlnK, Equation (4.55). We say, AGf = -RT { \nK { at 7i; and 


-RT 2 \nK 2 at T 2 . 

Substituting for each value of AG S yields 
-RT 2 lnK 2 -RTilnKi 

AH i 


T 2 Ti \T 2 : 

which, after cancelling the T\ and T 2 terms on the right-hand side, simplifies to 

{-R\nK 2 )-(-R\nK0 MI»[±--±- 

J 2 1\ 



Next, we divide throughout by '-R' to yield 

In K 2 - In Ki = 


1 1 

¥ 2 ~Y t 


and, by use of the laws of logarithms, In a — \nb = In (a -=- b), so the left-hand side of 
the equation may be grouped, to generate the van't Hoff isochore. Note: it is common 
(but incorrect) to see A// e written without its plimsoll sign ' e '. 

The isochore, Equation (4.81), was derived from the integrated 
form of the Gibbs-Helmholtz equation. It is readily shown that the 
van't Hoff isochore can be rewritten in a slightly different form, as: 



talk about 


isochore' when we 

mean the 

'van't Hoff 







x ^+ 






which is known as either the linear or graphical form of the equation. By analogy 
with the equation for a straight line (y = mx + c), a plot of In K (as y) against 1 4- T 
(as x) should be linear, with a gradient of — AH®/R. 

Worked Example 4.15 The isomerization of 1-butene (X) to form trans -2-butsne. (XI). 
The equilibrium constants of reaction are given below. Determine the enthalpy of reaction 
AH® using a suitable graphical method. 











Strategy. To obtain AH®: (1) we plot a graph of In K (as y) against values of 1 4- T 
(as x) according to Equation (4.82); (2) we determine the gradient; and (3) multiply the 
gradient by — R. 

(1) The graph is depicted in Figure 4.8. (2) The best gradient is 1415 K. (3) AH® — 
'gradient x — R', so 

AH® = 1415 K x (-8.314 JK" 1 mol -1 ) 


-11.8 kJmob 

SAQ 4.13 The following data refer to the chemical reaction between 
ethanoic acid and glucose. Obtain AH® from the data using a suitable 
graphical method. (Hint: remember to convert all temperatures to kelvin.) 

7/ °C 



21.2 x 

10 4 


15.1 x 

10 4 


8.56 x 

10 4 


5.46 x 

10 4 







Figure 4.8 The equilibrium constant K for the isomerization of 1-butene depends on the temper- 
ature: van't Hoff isochore plot of In K (as y) against 1 -h T (as x) from which a value of A// e 
can be calculated as 'gradient x — R' 


Writing table headers 

Each term in the table for SAQ 4.13 has been multiplied by 10 4 , which is repetitious, 
takes up extra space, and makes the table look messy and cumbersome. 

Most of the time, we try to avoid writing tables in this way, by incorporating the 
common factor of 'xlO 4 ' into the header. We accomplish this by making use of the 
quantity calculus concept (see p. 13). Consider the second value of K. The table says 
that, at 36 °C, K = 15.1 x 10 4 . If we divide both sides of this little equation by 10 4 , 
we obtain, K -r 10 4 = 15.1. This equation is completely correct, but is more usually 
written as 10~ 4 K = 15.1. 

According, we might rewrite the first two lines of the table as: 


lO" 4 ^ 



The style of this latter version is wholly correct and probably more popular than the 
style we started with. Don't be fooled: a common mistake is to look at the table heading 
and say, 'we need to multiply K by 10~ 4 '. It has been already! 


Phase equilibria 


Phase equilibrium describes the way phases (such as solid, liquid and/or gas) co-exist 
at some temperatures and pressure, but interchange at others. 

First, the criteria for phase equilibria are discussed in terms of single-component 
systems. Then, when the ground rules are in place, multi-component systems are 
discussed in terms of partition, distillation and mixing. 

The chapter also outlines the criteria for equilibrium in terms of the Gibbs function 
and chemical potential, together with the criteria for spontaneity. 

5.1 Energetic introduction to phase equilibria 

Why does an ice cube melt in the mouth? 

Introduction to phase equilibria 

The temperature of the mouth is about 37 °C, so an overly simple explanation of 
why ice melts in the mouth is to say that the mouth is warmer than the transition 
temperature T( me \ t ). And, being warmer, the mouth supplies energy to the immobilized 
water molecules, thereby allowing them to break free from those bonds that hold them 
rigid. In this process, solid H2O turns to liquid H2O - the ice melts. 

Incidentally, this argument also explains why the mouth feels cold after the ice has 
melted, since the energy necessary to melt the ice comes entirely from the mouth. In 
consequence, the mouth has less energy after the melting than before; this statement 
is wholly in accord with the zeroth law of thermodynamics, since heat energy travels 
from the hot mouth to the cold ice. Furthermore, if the mouth is considered as an 
adiabatic chamber (see p. 89), then the only way for the energy to be found for 
melting is for the temperature of the mouth to fall. 



A phase is a compo- 
nent within a system, 
existing in a precisely 
defined physical state, 
e.g. gas, liquid, or a 
solid that has a single 
crystallographic form. 

Further thermodynamic background: terminology 

In the thermodynamic sense, a phase is defined as part of a chemical system in 
which all the material has the same composition and state. Appropriately, the word 
comes from the Greek phasis, meaning 'appearance'. Ice, water and steam are the 
three simple phases of H2O. Indeed, for almost all matter, the three 
simple phases are solid, liquid and gas, although we must note that 
there may be many different solid phases possible since H20( s ) 
can adopt several different crystallographic forms. As a related 
example, the two stable phases of solid sulphur are its monoclinic 
and orthorhombic crystal forms. 

Ice is a solid form of water, and is its only stable form below 
°C. The liquid form of H2O is the only stable form in the tem- 
perature range < T < 100 °C. Above 100 °C, the normal, stable 
phase is gaseous water, 'steam'. Water's normal melting temperature T( me i t ) is 0°C 
(273.15 K). The word 'normal' in this context implies 'at standard pressure p & ' . The 
pressure p & has a value of 10 5 Pa. This temperature T( me it) is often called the melting 
point because water and ice coexist indefinitely at this temperature and pressure, but 
at no other temperature can they coexist. We say they reside together at equilibrium. 
To melt the ice, an amount of energy equal to A// (melt) must be 
added to overcome those forces that promote the water adopting 
a solid-state structure. Such forces will include hydrogen bonds. 
Re-cooling the melted water to re-solidify it back to ice involves 
the same amount of energy, but this time energy is liberated, so 
. The freezing process is often called fusion. 

Concerning transi- 





phases '1' 


'2', Hess's Law 



AH ( i-2) = 

-1 X 





= -AH, 


(Strictly, we ought to define the energy by saying that no pres- 
sure-volume work is performed during the melting and freezing 
processes, and that the melting and freezing processes occur without any changes in 

Table 5.1 gives a few everyday examples of phase changes, together with some 
useful vocabulary. 

Two or more phases can coexist indefinitely provided that we maintain certain 
conditions of temperature T and pressure p. The normal boiling temperature of water 
is 100 °C, because this is the only temperature (at p = p & ) at which both liquid and 

Table 5.1 Summary of terms used to describe phase changes 

Phase transition 

Name of transition 

Everyday examples 

Solid — >• gas 
Solid -*■ liquid 
Liquid — ► gas 
Liquid — >• solid 
Gas — y liquid 
Gas — y solid 



Boiling or vaporization 

Freezing, solidification or fusion 

Condensation or liquification 


'Smoke' formed from dry ice 

Melting of snow or ice 

Steam formed by a kettle 

Ice cubes formed in a fridge; hail 

Formation of dew or rain 

Formation of frost 



A phase diagram is a 
graph showing values 
of applied pressure and 
temperature at which 
equilibrium exists. 

gaseous H2O coexist at equilibrium. Note that this equilibrium is dynamic, because 
as liquid is converted to gas an equal amount of gas is also converted back to liquid. 

However, the values of pressure and temperature at equilibrium 
depend on each other; so, if we change the pressure, then the tem- 
perature of equilibrium shifts accordingly (as discussed further in 
Section 5.2). If we plotted all the experimental values of pressure 
and temperature at which equilibrium exists, to see the way they 
affect the equilibrium changes, then we obtain a graph called a 
phase diagram, which looks something like the schematic graph in 
Figure 5.1. 

We call each solid line in this graph a phase boundary. If the val- 
ues of p and T lie on a phase boundary, then equilibrium between 
two phases is guaranteed. There are three common phase bound- 
aries: liquid-solid, liquid-gas and solid-gas. The line separating 
the regions labelled 'solid' and 'liquid', for example, represents 
values of pressure and temperature at which these two phases coex- 
ist - a line sometimes called the 'melting-point phase boundary'. 

The point where the three lines join is called the triple point, 
because three phases coexist at this single value of p and T. The 
triple point for water occurs at T = 273.16 K (i.e. at 0.01 °C) and 
p = 610 Pa (0.006/5°). We will discuss the critical point later. 

Only a single phase is stable if the applied pressure and tem- 
perature do not lie on a phase boundary, i.e. in one of the areas 
between the phase boundaries. For example, common sense tells is 
that on a warm and sunny summer's day, and at normal pressure, 
the only stable phase of H2O is liquid water. These conditions of 
p and T are indicated on the figure as point 'D'. 

A phase boundary is a 
line on a phase diagram 
representing values of 
applied pressure and 
temperature at which 
equilibrium exists. 

The triple point on a 
phase diagram rep- 
resents the value of 
pressure and temper- 
ature at which three 
phases coexist at equi- 

.2 0.006 p*= 

273.16 K 

Temperature T 

Figure 5.1 Schematic phase diagram showing pressures and temperatures at which two phases 
are at equilibrium. Phase boundary (a) represents the equilibrium between steam and ice; boundary 
(b) represents equilibrium between water and ice; and boundary (c) represents equilibrium between 
water and steam. The point D represents p and T on a warm, sunny day. Inset: warming an ice 
cube from —5 °C to the mouth at 37 °C at constant pressure causes the stable phase to convert from 
solid to liquid. The phase change occurs at 0°C at p e 



When labelling a phase 
diagram, recall how the 
only stable phase at 
high pressure and low 
temperature is a solid; 
a gas is most stable 
at low pressure and 
high temperature. The 
phase within the crook 
of the 'Y' is therefore 
a liquid. 

We can predict whether an ice cube will melt just by looking 
carefully at the phase diagram. As an example, suppose we take 
an ice cube from a freezer at —5 °C and put it straightaway in 
our mouth at a temperature of 37 °C (see the inset to Figure 5.1). 
The temperature of the ice cube is initially cooler than that of the 
mouth. The ice cube, therefore, will warm up as a consequence 
of the zeroth law of thermodynamics (see p. 8) until it reaches the 
temperature of the mouth. Only then will it attain equilibrium. But, 
as the temperature of the ice cube rises, it crosses the phase bound- 
ary, as represented by the bold horizontal arrow, and undergoes a 
phase transition from solid to liquid. 

We know from Hess's law (see p. 98) that it is often useful to 
consider (mentally) a physical or chemical change by dissecting it 
into its component parts. Accordingly, we will consider the melting of the ice cube 
as comprising two processes: warming from —5 °C to 37 °C, and subsequent melting 
at 37 °C. During warming, the water crosses the phase boundary, implying that it 
changes from being a stable solid (when below 0°C) to being an unstable solid 
(above °C). Having reached the temperature of the mouth at 37 °C, the solid ice 
converts to its stable phase (water) in order to regain stability, i.e. the ice cube melts 
in the mouth. (It would be more realistic to consider three pro- 
cesses: warming to 0°C, melting at constant temperature, then 
warming from to 37 °C.) 

Although the situation with melting in two stages appears a little 
artificial, we ought to remind ourselves that the phase diagram is 
made up of thermodynamic data alone. In other words, it is possible 
to see liquid water at 105 °C, but it would be a metastable phase, 
i.e. it would not last long! 

The Greek root meta 
means 'adjacent to' or 
'near to'. Something 
metastable is almost 
stable . .. but not quite. 


The arguments in this example are somewhat simplified. 

Remember that the phase diagram's y-axis is the applied pressure. At room tempera- 
ture and pressure, liquid water evaporates as a consequence of entropy (e.g. see p. 134). 
For this reason, both liquid and vapour are apparent even at s.t.p. The pressure of the 
vapour is known as the saturated vapour pressure (s.v.p.), and can be quite high. 

The s.v.p. is not an applied pressure, so its magnitude is generally quite low. The 
s.v.p. of water will certainly be lower than atmospheric pressure. The s.v.p. increases 
with temperature until, at the boiling temperature, it equals the atmospheric pressure. 
One definition of boiling says that the s.v.p. equals the applied pressure. 

The arguments in this section ignore the saturated vapour pressure. 




y does water placed in a freezer become ice? 

Spontaneity of phase changes 

It will be useful to concentrate on the diagram in Figure 5.2 when considering why 
a 'phase change' occurs spontaneously. We recall from Chapter 4 that one of the 
simplest tests of whether a thermodynamic event can occur is to ascertain whether 
the value of AG is negative (in which case the change is indeed spontaneous) or 
positive (when the change is not spontaneous). 

The graph in Figure 5.2 shows the molar Gibbs function G m as a function of 
temperature. (G m decreases with temperature because of increasing entropy.) The 
value of G m for ice follows the line on the left-hand side of the graph; the line in 
the centre of the graph gives values of G m for liquid water; and the line on the 
right represents G m for gaseous water, i.e. steam. We now consider the process of 
an ice cube being warmed from below T( me i t ) to above it. The molar Gibbs functions 
of water and ice become comparable when the temperature reaches T( me \ t j . At T( me i t ) 
itself, the two values of G m are the same - which is one definition of equilibrium. 
The two values diverge once more above 7(meit)- 

Below 7( me i t ), the two values of G m are different, implying that the two forms of 
water are energetically different. It should be clear that if one energy is lower than 
the other, then the lower energy form is the stablest; in this case, 
the liquid water has a higher value of G m and is less stable than 
solid ice (see the heavy vertical arrow, inset to Figure 5.2). Liquid 
water, therefore, is energetically unfavourable, and for that reason it 
is unstable. To attain stability, the liquid water must release energy 
and, in the process, undergo a phase change from liquid to solid, 
i.e. it freezes. 

Remember how the 
symbol A means 'final 
state minus initial 
state', so AG m = 

G m (final state) — 
Gm (initial state)' 


Temperature T 

Figure 5.2 Graph of molar Gibbs function G m as a function of temperature. Inset: at temperatures 
below r (rae i t ) the phase transition from liquid to solid involves a negative change in Gibbs function, 
so it is spontaneous 



These arguments rep- 
resent a simple ex- 
ample of phase equi- 
libria. This branch of 
thermodynamics tells 
us about the direction 
of change, but says 
nothing about the rare 
at which such changes 

temperature if 

It should be clear from the graph in Figure 5.2 that AG m is 
negative (as required for a spontaneous change) only if the final 
state is solid ice and the initial state is liquid water. This sign of 
AG m is all that is needed to explain why liquid water freezes at 
temperatures below 7( me i t ). 

Conversely, if an ice cube is warmed beyond T (me i t ) to the tem- 
perature of the mouth at 37 °C, now it is the solid water that 
has excess energy; to stabilize it relative to liquid water at 37 °C 
requires a different phase change to occur, this time from ice to 
liquid water. This argument again relies on the relative magnitudes 
of the molar Gibbs function, so AG m is only negative at this higher 
the final state is liquid and the initial state is solid. 

Why was Napoleon's Russian campaign 
such a disaster? 

Solid-state phase transitions 

A large number of French soldiers froze to death in the winter of 1812 within a 
matter of weeks of their emperor Napoleon Bonaparte leading them into Russia. The 
loss of manpower was one of the principal reasons why Napoleon withdrew from the 
outskirts of Moscow, and hence lost his Russian campaign. 

But why was so ruthless a general and so obsessively careful a tactician as Napoleon 
foolhardy enough to lead an unprepared army into the frozen wastes of Russia? In 
fact, he thought he was prepared, and his troops were originally well clothed with 
thick winter coats. The only problem was that, so the story goes, he chose at the last 
moment to replace the brass of the soldiers' buttons with tin, to save money. 

Metallic tin has many allotropic forms: rhombic white tin (also called /3-tin) is 
stable at temperatures above 13 °C, whereas the stable form at lower temperatures is 
cubic grey tin (also called a-tin). A transition such as tin (w hit e ) —*■ tm (grey) is called a 
solid-state phase transition. 

Figure 5.3 shows the phase diagram of tin, and clearly shows the transition from 
t m (white) to tin( grey ). Unfortunately, the tin allotropes have very 

The transition from 
white tin to grey was 
first noted in Europe 
during the Middle Ages, 
e.g. as the pipes of 
cathedral organs dis- 
integrated, but the 
process was thought 
to be the work of 
the devil. 

different densities p, so p^n, grey) =5.8 gem -3 but p (tin> white ) = 
7.3 gem -3 . The difference in p during the transition from white to 
grey tin causes such an unbearable mechanical stress that the metal 
often cracks and turns to dust - a phenomenon sometimes called 
'tin disease' or 'tin pest'. 

The air temperature when Napoleon entered Russia was appar- 
ently as low as —35 °C, so the soldiers' tin buttons converted from 
white to grey tin and, concurrently, disintegrated into powder. So, 
if this story is true, then Napoleon's troops froze to death because 
they lacked effective coat fastenings. Other common metals, such 






Solid \ 
(grey) \ 



Solid N. 
(white) \ 
tin \ 

Liquid tin 




_^-"-^ Tin vapour 

200 400 600 800 


1000 1200 

Figure 5.3 Phase diagram of tin computed from thermodynamic data, showing the transition 
from grey tin from white tin at temperatures below 13 °C. Note the logarithmic y-axis. At 
/> e , Tf w hite^ grey) = 13 °C, and T( me ] t) = 231. 9°C. (Figure constructed from data published in Tin 
and its Alloys and Compounds, B. T. K. Barry and C. J. Thwaits, Ellis Horwood, Chichester, 1983) 

as copper or zinc, and alloys such as brass, do not undergo phase changes of this sort, 
implying that the troops could have survived but for Napoleon's last-minute change 
of button material. 

The kinetics of phase changes 

Like all spontaneous changes, the rate at which the two forms of tin interconvert is a 
function of temperature. Napoleon's troops would have survived if they had entered 
Russia in the summer or autumn, when the air temperature is similar to the phase- 
transition temperature. The rate of conversion would have been slower in the autumn, 
even if the air temperature had been slightly less than T( trans j t j on ) - after all, the tin 
coating of a can of beans does not disintegrate while sitting in a cool cupboard! The 
conversion is only rapid enough to noticeably destroy the integrity of the buttons when 
the air temperature is much lower than ^transition) » i.e. when the difference between 
r (air) and r (transition) is large. 

Phase changes involving liquids and gases are generally fast, owing to the high 
mobility of the molecules. Conversely, while phase changes such as tin( wmte ) — »■ 
i m (grey) can and do occur in the solid state, the reaction is usually very much slower 
because it must occur wholly in the solid state, often causing any thermodynamic 
instabilities to remain 'locked in'; as an example, it is clear from the phase diagram 
of carbon in Figure 5.4 that graphite is the stable form of carbon (cf. p. 109), yet the 

phase change carbon 




is so slow that a significant extent of 

conversion requires millions of years. 
We consider chemical kinetics further in Chapter 8. 








10 6 - 


10 4 - 

10 2 _ 




I i 




Figure 5.4 The phase diagram of carbon showing the two solid-state extremes of diamond and 
graphite. Graphite is the thermodynamically stable form of carbon at room temperature and pressure, 
but the rate of the transition C(diamond) — * C( grap hite) is virtually infinitesimal 

5.2 Pressure and temperature changes 
with a single-component system: 
qualitative discussion 

How is the ' Smoke' in horror films made? 

Effect of temperature on a phase change: sublimation 

Horror films commonly show scenes depicting smoke or fog billowing about the 
screen during the 'spooky' bits. Similarly, smoke is also popular during pop concerts, 
perhaps to distract the fans from something occurring on or off 

Dry ice is solid car 
bon dioxide. 

stage. In both cases, it is the adding of dry ice to water that produces 

the 'smoke'. 

Dry ice is carbon dioxide (CO2) in its solid phase. We call it 
'dry' because it is wholly liquid-free at p & : such solid CO2 looks similar to normal ice 
(solid water), but it 'melts' without leaving a puddle. We say it sublimes, i.e. undergoes 
a phase change involving direct conversion from solid to gas, without liquid forming 
as an intermediate phase. CC>2(i) can only be formed at extreme pressures. 

Solid CO2 is slightly denser than water, so it sinks when placed in a bucket of 
water. The water is likely to have a temperature of 20 °C or so at room temperature, 
while typically the dry ice has a maximum temperature of ca — 78 °C (195 K). The 
stable phase at the temperature of the water is therefore gaseous CC>2- We should 
understand that the C02( s ) is thermodynamically unstable, causing the phase transition 


C02( g ) on immersion in the water. 



-78.2 56.6 31 

Temperature 77°C 

Figure 5.5 Phase diagram of a system that sublimes at room temperature: phase diagram of carbon 
dioxide. (Note that the y-'dxis here is logarithmic) 

Incidentally, the water in the bucket is essential for generating the effect of theatrical 
'smoke' because the large volumes of C02( g ) entrap minute particles of water (which 
forms a colloid; see Chapter 10.2). This colloidal water is visible because it creates 
the same atmospheric condition known as fog, which is opaque. 

Look at the phase diagram of CO2 in Figure 5.5, which is clearly similar in general 
form to the schematic phase diagram in Figure 5.1. A closer inspection shows that 
some features are different. Firstly, notice that the phase boundary between solid and 
liquid now has a positive gradient; in fact, water is almost unique in having a negative 
gradient for this line (see Section 5.1). Secondly, the conditions of room temperature 
(T = 298 K and p = p & ) relate to conditions of the solid-gas phase boundary rather 
than the liquid-gas phase boundary. 

By drawing a horizontal line across the figure at p = p B , we see how the line cuts 
the solid-gas phase boundary at — 78.2°C. Below this temperature, the stable form 
of CO2 is solid dry ice, and C02( g ) is the stable form above it. Liquid CO2 is never 
the stable form at p & ; in fact, Figure 5.5 shows that CO20) will not form at pressures 
below 5.1 x p e . In other words, liquid CO2 is never seen naturally on Earth; which 
explains why dry ice sublimes rather than melts under s.t.p. conditions. 

How does freeze-drying work? 

Effect of pressure change on a phase change 

Packets of instant coffee proudly proclaim that the product has been 'freeze-dried'. 
In practice, beans of coffee are ground, boiled in water and filtered to remove the 
depleted grounds. This process yields conventional 'fresh' coffee, as characterized 
by its usual colour and attractive smell. Finally, water is removed from the coffee 
solution to prepare granules of 'instant' coffee. 

In principle, we could remove the water from the coffee by just boiling it off, to 
leave a solid residue as a form of 'instant coffee'. In fact, some early varieties of 
instant coffee were made in just this way, but the flavour was generally unpleasant as 



a result of charring during prolonged heating. Clearly, a better method of removing 
the water was required. 

We now look at the phase diagram of water in Figure 5.6, which will help us follow 
the modern method of removing the water from coffee to yield anhydrous granules. 
A low temperature is desirable to avoid charring the coffee. Water vapour can be 
removed from the coffee solution at any temperature, because liquids are always sur- 
rounded by their respective vapour. The pressure of the vapour is the saturated vapour 
pressure, s.v.p. The water is removed faster when the applied pressure decreases. 
Again, a higher temperature increases the rate at which the vapour is removed. The 
fastest possible rate occurs when the solution boils at a temperature we call r^oii). 
Figure 5.6 shows the way in which the boiling temperature alters, 
with boiling becoming easier as the applied pressure decreases or 
the temperature increases, and suggests that the coffee solution will 
boil at a lower temperature when warmed in a partial vacuum. At 

Freeze-drying is a 
layman's description, 
and acknowledges that 
external conditions 
may alter the 
conditions of a phase 
change, i.e. the drying 
process (removal of 
water) occurs at a 
temperature lower than 

a pressure of about -j4j x p , water is removed from the coffee 

by warming it to temperatures of about 30 C, when it boils. We 
see that the coffee is dried and yet is never subjected to a high 
temperature for long periods of time. 

It is clear that decreasing the external pressure makes boiling 
easier. It is quite possible to remove the water from coffee at or 
near its freezing temperature - which explains the original name 
of freeze-drying. 

In many laboratories, a nomograph (see Figure 5.7) is pinned to 
the wall behind a rotary evaporator. A nomograph allows for a simple estimate of 
the boiling temperature as a function of pressure. Typically, pressure is expressed in 
the old-fashioned units of atmospheres (atm) or millimetres of mercury (mmHg). 
1 atm = 760 mmHg. (The curvature of the nomograph is a consequence of the 
mathematical nature of the way pressure and temperature are related; see Section 5.2). 


Temperature T 

Figure 5.6 Freeze-drying works by decreasing the pressure, and causing a phase change; at higher 
pressure, the stable form of water is liquid, but the stable form at lower pressures is vapour. 
Consequently, water (as vapour) leaves a sample when placed in a vacuum or low-pressure chamber: 
we say the sample is 'freeze-dried' 


"C °F 




boiling point 


Boiling point 
corrected to 
760 mmHg 




Figure 5.7 A typical nomograph for estimating the temperature at which a pure liquid boils when 
the pressure is decreased 

This is how a boiling temperature at reduced pressure is estimated with a nomo- 
graph: place a straight ruler against the applied pressure as indicated on the curved 
right-hand scale (c). The ruler must also pass through the 'normal' boiling temperature 
on the middle scale (b). The reduced-pressure boiling temperature is then read off 
the left-hand scale (a). As an example, if the normal boiling temperature is 200 °C, 
then the reduced boiling temperature may be halved to 100 °C if the applied pressure 
is approximately 20 mmHg. 

SAQ 5.1 A liquid has a normal boiling temperature of 140 °C. Use the 
nomograph to estimate the applied pressure needed to decrease the 
boiling temperature to 90°C. 



How does a rotary evaporator work? 

Thermodynamics of phase changes 

Rotary evaporators are a common feature in most undergraduate laboratories. Their 
primary purpose is to remove solvent following a reflux, perhaps before crystallization 
of a reaction product. 

To operate the evaporator, we place the reaction solution in a round-bottomed flask 
while the pressure inside the evaporator is decreased to about ^ x p & . The flask 
is then rotated. The solvent evaporates more easily at this low pressure than at p & . 
The solvent removed under vacuum is trapped by a condenser and collected for easy 
re-use, or disposal in an environmentally sensitive way. 

But molecules need energy if they are to leave the solution during boiling. The 
energy comes from the solution. The temperature of the solution would decrease 
rapidly if no external supply of energy was available, as a reflection of its depleted 
energy content (see p. 33). In fact, the solution would freeze during evaporation, so 
the rotating bulb is typically immersed in a bath of warm water. 

An atmosphere of vapour always resides above a liquid, whether 
the liquid is pure, part of a mixture, or has solute dissolved within 
it. We saw on p. 180 how the pressure of this gaseous phase is 
called its saturation vapour pressure, s.v.p. The s.v.p. increases with 
increased temperature until, at the boiling point r(t,oii), it equals the 
external pressure above the liquid. Evaporation occurs at tempera- 
tures below T(boii), and only above this temperature will the s.v.p. 
exceed p & . The applied pressure in a rotary evaporator is less than 
p & , so the s.v.p. of the solvent can exceed the applied pressure 
(and allow the liquid to boil) at pressures lower than p e . 

We see this phenomenon in a different way when we look back at 

the phase diagram in Figure 5.6. The stable phase is liquid before 

applying a vacuum. After turning on the water pump, to decrease 

the applied pressure, the s.v.p. exceeds /?( ap piied), and the solvent 

boils at a lower pressure. The bold arrow again indicates how a 

phase change occurs during a depression of the external pressure. 

We see how decreasing the pressure causes boiling of the solvent 

at a lower temperature than at its normal boiling temperature, i.e. 

if the external pressure were p & . Such a vacuum distillation is 

desirable for a preparative organic chemist, because a lower boiling 

temperature decreases the extent to which the compounds degrade. 

Coffee, for example, itself does not evaporate even at low pressure, since it is 

a solid. Solids are generally much less volatile than liquids, owing to the stronger 

interactions between the particles. In consequence, the vapour pressure of a solid is 

several orders of magnitude smaller than that above a liquid. 

Strictly, the term s.v.p. 
applies to pure liq- 
uids. By using the term 
s.v.p., we are implying 
that all other com- 
ponents are wholly 
involatile, and the 
s.v.p. relates only to 
the solvent. 

Normal in the con- 
text of phase equilibria 
means 'performed at a 
pressure of 1 bar, p s '. 

The rotary evaporator 
is a simple example of 
a vacuum distillation. 



How is coffee decaffeinated? 

Critical and supercritical fluids 

We continue our theme of 'coffee'. Most coffees contain a large amount of the 
heterocyclic stimulant caffeine (I). Some people prefer to decrease the amounts of 
caffeine they ingest for health reasons, or they simply do not like to consume it at 
all, and they ask for decaffeinated coffee instead. 

H 3 C 

The modern method of removing I from coffee resembles the operation of a coffee 
percolator, in which the water-soluble chemicals giving flavour, colour and aroma 
are leached from the ground-up coffee during constant irrigation with a stream of 
boiling water. 

Figure 5.8 shows such a system: we call it a Soxhlet apparatus. Solvent is passed 
continually through a porous cup holding the ground coffee. The solvent removes the 
caffeine and trickles through the holes at the bottom of the cup, i.e. as a solution of 
caffeine. The solvent is then recycled: solvent at the bottom of the flask evaporates 
to form a gas, which condenses at the top of the column. This pure, clean solvent 
then irrigates the coffee a second time, and a third time, etc., until all the caffeine 
has been removed. 

Water is a good choice of solvent in a standard kitchen percolator because it removes 
all the water-soluble components from the coffee - hence the flavour. Clearly, how- 
ever, a different solvent is required if only the caffeine is to be removed. Such a 
solvent must be cheap, have a low boiling point to prevent charring of the coffee 
and, most importantly, should leave no toxic residues. The presence of any residue 
would be unsatisfactory to a customer, since it would almost certainly leave a taste; 
and there are also health and safety implications when residues persist. 

The preferred solvent is supercritical CO2. The reasons for this 
choice are many and various. Firstly, the CO2 is not hot (CO2 first 
becomes critical at 31 °C and 73 atm pressure; see Figure 5.5), so 
no charring of the coffee occurs during decaffeination. Furthermore, 
at such a low temperature, all the components within the coffee that 
impart the flavour and aroma remain within the solid coffee - try 
soaking coffee beans in cold water and see how the water tastes afterwards! Caffeine 
is removed while retaining a full flavour. 

C0 2 is supercritical 
at temperatures and 
pressures above the 
critical point. 




Cool, pure 


beans to be 


Hot, pure 

Reservoir, where 


caffeine accumulates 

| Heat 

Figure 5.8 Coffee is decaffeinated by constantly irrigating the ground beans with supercritical 
carbon dioxide: schematic representation of a Soxhlet apparatus for removing caffeine from coffee 

Secondly, solid CO2 is relatively cheap. Finally, after caffeine removal, any occluded 
CO2 will vaporize from the coffee without the need to heat it or employ expensive 
vacuum technology. Again, we retain the volatile essential oils of the coffee. Even 
if some CO2 were to persist within the coffee granules, it is chemically inert, has no 
taste and would be released rapidly as soon as boiling water was added to the solid, 
decaffeinated coffee. 

What is a critical or supercritical fluid? 

We look once more at the phase diagram of CO2 in Figure 5.5. The simplest way of 
obtaining the data needed to construct such a figure would be to take a sample of 
CO2 and determine those temperatures and pressures at which the liquid, solid and 
gaseous phases coexist at equilibrium. (An appropriate apparatus 
involves a robust container having an observation window to allow 
us to observe the meniscus.) We then plot these values of p (as 
'y') against T (as 'x'). 

We first looked at criti- 
cal fluids on p. 50. 



It is impossible to dis- 
tinguish between the 
liquid and gaseous 
phases of C0 2 at tem- 
peratures and pres- 
sures at and above the 
critical point. 

Let us consider more closely what happens as the conditions become more extreme 
inside the observation can. As heating proceeds, so the amount of CO20) convert- 
ing to form gas increases. Accordingly, the amount of CO2 within the gaseous 
phase increases, which will cause the density p of the vapour to increase. Con- 
versely, if we consider the liquid, at no time does its density alter 
appreciably, even though its volume decreases as a result of liquid 
forming vapour. 

From a consideration of the relative densities, we expect the 
liquid phase to reside at the bottom of the container, with the 
less-dense gaseous phase 'floating' above it. The 'critical' point is 
reached when the density of the gas has increased until it becomes 
the same as that of the liquid. In consequence, there is now no 
longer a lighter and a heavier phase, because P(Uq U id) = P(vapour)- 
Accordingly, we no longer see a meniscus separating liquid at the 
bottom of the container and vapour above it: it is impossible to see 
a clear distinction between the liquid and gas components. We say 
that the CO2 is critical. 

Further heating or additional increases in pressure generate 
supercritical CO2. The pressure and temperature at which the fluid 
first becomes critical are respectively termed r( cr iticai) and /? (critical) • 
Table 5.2 contains a few examples of ^(critical) and p (critical)- 

The inability to distinguish liquid from gaseous CO2 explains 
why we describe critical and supercritical systems as fluids - they 
are neither liquid nor gas. 

It is impossible to distinguish between the liquid and gaseous 
phases of CO2 at and above the critical point, which explains 
why a phase diagram has no phase boundary at temperatures and 
pressures above ^critical). The formation of a critical fluid has an 
unusual corollary: at temperatures above ^critical), we cannot 
cause the liquid and gaseous phases to separate by decreasing or 
increasing the pressure alone. The critical temperature, therefore, 
represents the maximum values of p and T at which liquification 

The intensive prop- 
erties of the liquid 
and gas (density, heat 
capacity, etc.) become 
equal at the criti- 
cal point, which is 
the highest temper- 
ature and pressure at 
which both the liquid 
and gaseous phases 
of a given compound 
can coexist. 

Table 5.2 Critical constants T^ticd) an d /?(criticai) for 
some common elements and bi-element compounds 

IUPAC defines super- 
critical chromatogra- 
phy as a separation 
technique in which 
the mobile phase is 
kept above (or rel- 
atively close to) its 
critical temperature 
and pressure. 


^(critical) /**■ 


H 2 






o 2 



Cl 2 



C0 2 



S0 2 



H 2 



NH 3 





of the gas is possible. We say that there cannot be any C02(i) at temperatures above 

-» (critical) ■ 

Furthermore, supercritical CO2 does not behave as merely a mixture of liquid 
and gaseous CO2, but often exhibits an exceptional ability to solvate molecules in 
a specific way. The removal of caffeine from coffee relies on the chromatographic 
separation of caffeine and the other organic substances in a coffee bean; supercritical 
fluid chromatography is a growing and exciting branch of chemistry. 

5.3 Quantitative effects of pressure 
and temperature change 
for a single-component system 




y is ice so slippery 

The coefficient of fric- 
tion [i (also called 
'friction factor') is the 
quotient of the fric- 
tional force and the 
normal force. In other 
words, when we apply a 
force, is there a resis- 
tance to movement 
or not? 

Effect of p and T on the position of a solid -liquid equilibrium 

We say something is 'as slippery as an ice rink' if it is has a tiny 
coefficient of friction, and we cannot get a grip underfoot. This is 
odd because the coefficient of friction /x for ice is quite high - try 
dragging a fingernail along the surface of some ice fresh from the 
ice box. It requires quite a lot of effort (and hence work) for a 
body to move over the surface of ice. 

At first sight, these facts appear to represent a contradiction in 
terms. In fact, the reason why it is so easy to slip on ice is that ice 
usually has a thin layer of liquid water covering its surface: it is 
this water-ice combination that is treacherous and slippery. 
But why does any water form on the ice if the weather is 
sufficiently cold for water to have frozen to form ice? Consider the ice directly 
beneath the blade on a skater's ice-shoe in Figure 5.9: the edge of the blade is 
so sharp that an enormous pressure is exerted on the ice, as indicated by the grey 

We now look at the phase diagram for water in Figure 5.10. 
Ice melts at 0°C if the pressure is p s (as represented by 7\ and 
Px respectively on the figure). If the pressure exerted on the ice 
increases to P2, then the freezing temperature decreases to T2. (The 
freezing temperature decreases in response to the negative slope 
of the liquid-solid phase boundary (see the inset to Figure 5.10), 
which is most unusual; virtually all other substances show a posi- 
tive slope of dp/dT.) 
If the temperature Ti is lower than the freezing temperature of the water - and it 
usually is - then some of the ice converts to form liquid water; squeezing decreases 
the freezing temperature of the water. The water-on-ice beneath the skater's blade is 
slippery enough to allow effortless skating. 

The sign of dp/dT for 
the liquid-solid line 
on a phase diagram 
is almost always posi- 
tive. Water is the only 
common exception. 


Skater's blade 

Water formed by pressure 

Figure 5.9 Skaters apply an enormous pressure beneath the blades of their skates. This pressure 
causes solid ice to melt and form liquid water 


Figure 5.10 Phase diagram of water. Inset: applying a high pressure from p\ (here p e ) to p 2 
causes the melting temperature of the ice to decrease from temperature T\ (here 0°C) to T 2 

What is ' black ice'? 

The Clapeyron equation 

We give the name 'black ice' to the phenomenon of invisible ice on a road. In 
practice, anything applying a pressure to solid ice will cause a similar depression of 
the freezing temperature to that of the skater, so a car or heavy vehicle travelling 
over ice will also cause a momentary melting of the ice beneath its wheels. This 
water-on-ice causes the car to skid - often uncontrollably - and leads to many deaths 
every year. Such ice is particularly dangerous: whereas an ice skater wants the ice to 
be slippery, a driver does not. 



We move from the qualitative argument that T( me \ t ) decreases as p increases, and 
next look for a quantitative measure of the changes in melt temperature with pressure. 
We will employ the Clapeywn equation: 




In fact, it does not 
matter whether AH 
relates to the direction 
of change of solid -> 
liquid or of liquid -> 
solid, provided that AV 
relates to the same 
direction of change. 

The molar change in 
volume AV m has units 
of m 3 mol -1 . Values 
typically lie in the 
range irr 5 -lCT 6 m 3 
mol -1 . 

The minus sign of AV m 
reflects the way water 
expands on freez- 
ing. This expansion 
explains why a car 
radiator cracks in cold 
weather (if it contains 
no 'de-icer'): the water 
freezes and, in expand- 
ing, exerts a huge a 
pressure on the metal. 

where T is the normal melting temperature, AT is the change in 
the melting temperature caused by changing the applied pressure 
by an amount of dp (in SI units of pascals), where 1 Pa is the 
pressure exerted by a force of 1 N over an area of 1 m 2 . AH is 
the enthalpy change associated with the melting of water and A V 
is the change in volume on melting. Strictly, both AH and AV are 
molar quantities, and are often written as AH m and A V m , although 
the 'm' is frequently omitted. 

The molar change in volume A V m has SI units of m 3 mol - . 
We should note how these volumes are molar volumes, so they 
refer to 1 mol of material, explaining why AV is always very 
small. The value of AV m is usually about 10~ 6 to 10~ 5 m 3 mol _1 
in magnitude, equating to 1 to 10 cm 3 mol -1 respectively. 

We recall from Chapter 1 how the symbol A means 'final state 
minus initial state', so a positive value of AVm during melting 
(which is y m (ii qu id> — V m (solid)) tells us that the liquid has a slightly 
larger volume than the solid from which it came. AV m (melt) is 
positive in the overwhelming majority of cases, but for water 
A V m (melt) = — 1.6 x 10~ 6 m 3 mol~ . This minus sign is extremely 
unusual: it means that ice is less dense than water. This explains 
why an iceberg floats in water, yet most solids sink when immersed 
in their respective liquid phases. 

The enthalpy A^ elt) is the energy required to melt 1 mol of 
material at constant pressure. We need to be careful when obtain- 
ing data from tables, because many books cite the enthalpy of 
fusion, which is the energy released during the opposite process of 
solidification. We do not need to worry, though, because we know 

from Hess's law that AH, 




The molar enthalpy 

of melting water is +6.0 kJmol 

Care: following Hess's 
law, we say: 




(fusion) ' 

Worked Example 5.1 Consider a car weighing 1000 kg (about 
2200 lbs) parked on a sheet of ice at 273.15 K. Take the area under 
wheels in contact with the ice as 100 cm 2 i.e. 10~ 2 m 2 . What is the 

'(melt) — 

new melting temperature of the ice - call it T^nai)? Take AH*. 

6.0 kJmol and water AV, 

m (melt) 

= -1.6 x 10 -6 m J mor 



Strategy. (1) Calculate the pressure exerted and hence the pressure change. (2) Insert 
values into the Clapeyron equation (Equation (5.1)). 

(1) The pressure exerted by the car is given by the equation 
'force 4- area'. The force is simply the car weight expressed 
in newtons (N): force = 10000 N, so we calculate the 
pressure exerted by the wheels as 10 000 N -f- 10~ 2 m 2 , 
which is 10 6 Pa. We see how a car exerts the astonishing 
pressure beneath its wheels of 10 6 Pa (about 10 bar). 

(2) Before inserting values into the Clapeyron equation, we 
rearrange it slightly, first by multiplying both sides by 
T AV , then dividing both sides by A// e , to give 




a mass 


1 kg 

has a weight 


2. exerts a 

force) of 


mately 10 N. 

dr = 


dp TAV 

10 6 Pax 273.15Kx (-1.6 x 10 

6 ) m 3 mol 

Next, we recall that the symbol 
Accordingly, we say 

6.0 x 10 3 Jmor 

dT = -0.07 K 

'A' means 'final state — initial state' 

AT = T, 


' (initial) 

Notice how the free- 
zing temperature of 
water decreases when 
a pressure is applied. 
This decrease is directly 
attributable to the 
minus sign of AV. 

where the temperatures relate to the melting of ice. The normal melting temperature of 
ice r(i n j t i a i) is 273.15 K. The final temperature T(fiDai) of the ice with the car resting on it 
is obtained by rearranging, and saying 

AT + r^tiai) = -0.07 K + 273.15 K 
%nal) = 273.08 K 

The new melting temperature of the ice T^.^ is 273.08 K. Note how we performed 
this calculation with the car parked and immobile on the ice. When driving rather than 
parked, the pressure exerted beneath its wheels is actually considerably greater. Since 
Equation (5.1) suggests that dp oc dT, the change in freezing temperature dT will be 
proportionately larger (perhaps as much as —3 K), so there will be a layer of water on 
the surface of the ice even if the ambient temperature is —3 °C. Drive 

with care! 

SAQ 5.2 Paraffin wax has a normal melting temperature 
7~(meit) of 320 K. The temperature of equilibrium is raised 

Care: the word 'normal' 
here is code: it means 
, atp = p°'. 


by 1.2 K if the pressure is increased fivefold. Calculate AV m for the wax as 
it melts. Take AW (me it) = 8.064 kJ mol" 1 . 

Justification Box 5.1 

Consider two phases (call them 1 and 2) that reside together in thermodynamic equi- 
librium. We can apply the Gibbs-Duhem equation (Equation 4.31) for each of the two 
phases, 1 and 2. 

For phase 1 : dG a) = (V m( i) dp) - (S m(i) dT) (5.2) 

For phase 2 : dG (2 ) = (V m(2 ) dp) - (S m(2) dT) (5.3) 

where the subscripts 'm' imply molar quantities, i.e. per mole of substance in each phase. 

Now, because equilibrium exists between the two phases 1 and 2, the dG term in 
each equation must be the same. If they were different, then the change from phase 1 to 
phase 2 (Gp) — G(i)) would not be zero at all points; but at equilibrium, the value of 
AG will be zero, which occurs when G(2) = G(i). In fact, along the line of the phase 
boundary we say dG(i) = dG(2). 

In consequence, we may equate the two equations, saying: 

(V m( i) x dp) - (5 m( i) x dT) = (V m{2 ) x dp) - (5 m(2) x dT) 
Factorizing will group together the two S and V terms to yield 

(5 m (2) - Sm(l)) dT = (Vm(2) - V m (i)) dp 

which, after a little rearranging, becomes 

dp A5 m (i_ > .2) 

dT AV, 



Finally, since AG & = Afl* - TAS* (Equation (4.21)), and since AG S = at 


T = AS m( i^ 2 ) 

Inserting this relationship into Equation (5.4) yields the Clapeyron equation in its famil- 
iar form. 

dp _ A #m(1^2) 

dT ~ TAV m(l ^ 2) 

In fact, Equation (5.4) is also called the Clapeyron equation. This equation holds for 
phase changes between any two phases and, at heart, quantitatively defines the phase 


boundaries of a phase diagram. For example: 

for the liquid — > solid line 

for the vapour — > solid line 

dp AH, 

m (l->s) 


m (l->-s) 

d/> AH m ( v -«) 

d7 TA'V, 

v *m (v-*-s) 

for the vapour — > liquid line — = 

dp AH* (V _ 

dr taLv, 

v *m (v->-s) 

Approximations to the Clapeyron equation 

We need to exercise a little caution with our terminology: we performed the calculation 

in Worked Example 5.1 with Equation (5.1) as written, but we should have written Ap 

rather than dp because 10 6 Pa is a very large change in pressure. 

Similarly, the resultant change in temperature should have been 

written as AT rather than dT, although 0.07 K is not large. To 

accommodate these larger changes in p and T, we ought to be 

rewrite Equation (5.1) in the related form: 



The 'd' in Equation 
(5.1) means an infini- 
tesimal change, 
whereas the 'A' symbol 
here means a large, 
macroscopic, change. 

We are permitted to assume that dp is directly proportional to dT because AH and 
A V are regarded as constants, although even a casual inspection of a phase diagram 
shows how curved the solid-gas and liquid-gas phase boundaries are. Such curvature 
clearly indicates that the Clapeyron equation fails to work except 
over extremely limited ranges of p and T . Why? 

We assumed in Justification Box 5.1 that A// (melt) is indepen- 
dent of temperature and pressure, which is not quite true, although 
the dependence is usually sufficiently slight that we can legiti- 
mately ignore it. For accurate work, we need to recall the Kirchhoff 
equation (Equation (3.19)) to correct for changes in AH. 

Also, we saw on p. 23 how Boyle's Law relates the volume of 
a gas to changes in the applied pressure. Similar expressions apply 
for liquids and solids (although such phases are usually much less 
compressible than gases). Furthermore, we assumed in the deriva- 
tion of Equation (5.1) that AV m does not depend on the pressure 
changes, which implies that the volumes of liquid and solid phases 
each change by an identical amount during compression. This 
approximation is only good when (1) the pressure change is not 
extreme, and (2) we are considering equilibria for the solid-liquid 

The Clapeyron equa- 
tion fails to work for 
phase changes involv- 
ing gases, except 
over extremely limited 
ranges of p and T. 

It is preferable to 
analyse the equilib- 
ria of gases in terms 
of the related Clau- 
sius-Clapeyron equa- 
tion; see Equation (5.5). 


phase boundary, which describes melting and solidification. For these reasons, the 
Clapeyron equation is most effective when dp is relatively small, i.e. 2-10 atm 
at most. 

The worst deviations from the Clapeyron equation occur when one of the phases 
is a gas. This occurs because the volume of a gas depends strongly on temperature, 
whereas the volume of a liquid or solid does not. Accordingly, the value of A V m is 
not independent of temperature when the equilibrium involves a gas. 

Why does deflating the tyres on a car improve its 
road-holding on ice? 

The Clapeyron equation, continued 

We saw from the Clapeyron equation, Equation (5.1), how the decrease in freezing 
temperature AT is proportional to the applied pressure dp, so one of the easiest ways 
of avoiding the lethal conversion of solid ice forming liquid water is to apply a smaller 
pressure - which will decrease AT in direct proportion. 

The pressure change dp is caused by the additional weight of, 
for example, a car, lorry or ice skater, travelling over the surface 
of the ice. We recall our definition of 'pressure' as 'force -r area'. 
There is rarely a straightforward way of decreasing the weight of 
a person or car exerting the force, so the best way to decrease 
the pressure is to apply the same force but over a larger area. 
An elementary example will suffice: cutting with a sharp knife 
is easier than with a blunt one, because the active area along 
the knife-edge is greater when the knife is blunt, thus causing p 
to decrease. 
In a similar way, if we deflate slightly the tyres on a car, we see the tyre bulge 
a little, causing it to 'sag', with more of the tyre in contact with the road surface. 
So, although the weight of the car does not alter, deflating the tyre increases the area 
over which its weight (i.e. force) is exerted, with the result that we proportionately 
decrease the pressure. 

In summary, we see that letting out some air from a car tyre decreases the value of 
dp, with the result that the change in melting temperature dT of the ice, as calculated 
with the Clapeyron equation (Equation (5.1)), also decreases, thereby making driving 
on ice much safer. 

The pressure beneath 
the blades of an ice- 
skater's shoe is enor- 
mous - maybe as much 
as 100 atm when the 
skater twists and turns 
at speed. 

SAQ 5.3 A man is determined not to slip on the ice, so instead of wearing 
skates of area 10 cm 2 he now wears snow shoes, with the underside of 
each sole having an extremely large area to spread his 100 kg mass 
(equating to 1000 N). If the area of each snow shoe is 0.5 m 2 , what is the 
depression of the freezing temperature of the ice caused by his walking 
over it? 
Use the thermodynamic data for water given in Worked Example 5.1. 




If water behaved in a similar fashion to most other materials and possessed a positive 
value of AV m , then water would spontaneously freeze when pressure was applied, 
rather than solid ice melting under pressure. Furthermore, a positive value of AV m 
would instantly remove the problems discussed above, caused by vehicles travelling 
over 'black' ice, because the ice would remain solid under pressure; and remember that 
the slipperiness occurs because liquid water forms on top of solid ice. 

Unfortunately, a different problem would present itself if AV m was positive! If AV m 
was positive, then Equation (5.1) shows that applying a pressure to liquid water would 
convert it to ice, even at temperatures slightly higher than °C, which provides a 
different source of black ice. 

Why does a pressure cooker work? 

The Clausius-Clapeyron equation 

A pressure cooker is a sealed saucepan in which food cooks faster than it does in a 
simple saucepan - where 'simple', in this context means a saucepan that is open to 
the air. A pressure cooker is heated on top of a cooker or hob in the conventional way 
but, as the water inside it boils, the formation of steam rapidly causes the internal 
pressure to increase within its sealed cavity; see Figure 5.11. The internal pressure 
inside a good-quality pressure cooker can be as high 6 atm. 

The phase diagram in Figure 5.12 highlights the pressure- temp- 
erature behaviour of the boiling (gas-liquid) equilibrium. The 
normal boiling temperature 7(b ii) of water is 100 °C, but 7(boii) 
increases at higher pressures and decreases if the pressure de- 
creases. As a simple example, a glass of water would boil instantly at the cold 
temperature of 3 K in the hard vacuum of deep space. The inset to Figure 5.12 

Remember that all 
equilibria are dynamic. 





Figure 5.11 A pressure cooker enables food to cook fast because its internal pressure is high, 
which elevates the temperature at which food cooks 



2 p^ 





The normal boiling point of water 



100 °C 

Temperature T 

Figure 5.12 Phase diagram to show how a pressure cooker works. Inset: applying a high pressure 
from p & to p2 causes the boiling temperature of the water to increase from temperature 100 °C 

to T 2 

shows why the 
consequence of 

The Clausius -Clapey- 
ron equation quantifies 
the way a boiling tem- 
perature changes as a 
function of the applied 
pressure. At the boil- 
ing points of 7"i and 
T 2 , the external pres- 
sures Pi and pi are the 
same as the respective 
vapour pressures. 

water inside the pressure cooker boils at a higher temperature as a 

the pan's large internal pressure. 

Having qualitatively discussed the way a pressure cooker facil- 
itates rapid cooking, we now turn to a quantitative discussion. 
The Clapeyron equation, Equation (5.1), would lead us to suppose 
that dp oc AT , but the liquid -gas phase boundary in Figure 5.12 
is clearly curved, implying deviations from the equation. There- 
fore, we require a new version of the Clapeyron equation, adapted 
to cope with the large volume change of a gas. To this end, we 
introduce the Clausius- Clapeyron equation: 


p 2 at T 2 
Pi at T x 





~T 2 



It does not matter 
which of the values we 
choose as '1' and 
'2' provided that 7~i 
relates to pi and T 2 
relates to p 2 . It is per- 
missible to swap 7~i 
for 7" 2 and p\ for p 2 
simultaneously, which 
amounts to multiply- 
ing both sides of the 
equation by '-!'. 

where R is the familiar gas constant, and AH? n) is the enthalpy 

of vaporization. AHf [X) is always positive because energy must be 
put in to a liquid if it is to boil. T 2 here is the boiling temperature 
when the applied pressure is p 2 , whereas changing the pressure to 
p\ will cause the liquid to boil at a different temperature, T\. 

We need to understand that the Clausius -Clapeyron equation is 
really just a special case of the Clapeyron equation, and relates to 
phase changes in which one of the phases is a gas. 

Worked Example 5.2 What is the boiling temperature of pure water 
inside a pressure cooker? Let 7i be the normal boiling temperature 
r (boi i) of water (i.e. 100 °C, 373 K, at p°) and let p 2 of 6 x p & be the 
pressure inside the pan. The enthalpy of boiling water is 50.0 kJmol - . 



In this example, it is simpler to insert values into Equation (5.5) and 
to rearrange later. Inserting values gives 


6x p* 

-50 000 J mo!" 
8.314 JK-'mol" 



373 K 

We can omit the units of the two pressures on the left-hand side 
because Equation (5.5) is written as a ratio, so the units cancel: we 
require only a relative change in pressure. 

Notice how the ratio 
within the bracket on 
the left-hand side of the 
equation permits us 
to dispense with abso- 
lute pressures. 

In 6.0= -6104 Kx 



T 2 373 K 

where In 6.0 has a value of —1.79. Next, we rearrange slightly by dividing both sides by 
6104 K, to yield: 

-1.79 1 




6104 K T 2 373 K 



= -2.98 x 10" 4 K _1 + 
T 2 373 K 

— = 2.38 x 10" 3 K" 1 

T 2 

We obtain the temperature at which water boils by taking the reciprocal of both side. 
T 2 , is 420 K, or 147 °C at a pressure of 6 x p & , which is much higher than the normal 
boiling temperature of 100 °C. 

SAQ 5.4 A mountaineer climbs Mount Everest and wishes to make a 
strong cup of tea. He boils his kettle, but the final drink tastes lousy 
because the water boiled at too low a temperature, itself because the 
pressure at the top of the mountain is only 0.4 xp e . Again taking the 
enthalpy of boiling the water to be 50 kJmor 1 and the normal boiling 
temperature of water to be 373 K, calculate the temperature of the water 
as it boils at the top of the mountain. 

The form of the Clausius-Clapeyron equation in Equation (5.5) is called the inte- 
grated form. If pressures are known for more than two temperatures, an alternative 
form may be employed: 






+ constant 




so a graph of the form 'y = mx + c' is obtained by plotting In p (as 'y') against 
1/r (as 'x'). The gradient of this Clapeyron graph is '— AH,. ir . 4- R', so we obtain 



We employ the inte- 
grated form of the 
equation when we 
know two tempera- 
tures and pressures, 
and the graphical form 
for three or more. 


as 'gradient x — 1 x R\ 

The intercept of a Clapeyron graph is not useful; its value may 
best be thought of as the pressure exerted by water boiling at infinite 
temperature. This alternative of the Clausius-Clapeyron equation 
is sometimes referred to as the linear (or graphical) form. 

Worked Example 5.3 The Clausius-Clapeyron equation need not 
apply merely to boiling (liquid-gas) equilibria, it also describes sub- 
limation equilibria (gas-solid). 

Consider the following thermodynamic data, which concern the sub- 
limation of iodine: 

-* (sublimation) / -^ 
P(h)/ Pa 

270 280 290 300 310 320 330 340 

50 133 334 787 1755 3722 7542 14 659 

We obtain the value 
of AH as 'gradientx 
-1 xR'. 

Figure 5.13 shows a plot of In p (h) (as 'y') against 1 / r (sublimation) (as 

V). The enthalpy AH* 


is obtained via the gradient of the 

graph 62 kJ mol (note the positive sign). 




Figure 5.13 The linear form of the Clausius-Clapeyron equation: a graph of In p (as 'y') against 
1/r (as 'x') should be linear with a slope of —AH 




Why does food cook faster at higher pressures? 

The process of cooking involves a complicated series of chemical reactions, each of 
which proceeds with a rate constant of k. When boiling an egg, for example, the rate- 
limiting process is denaturation of the proteins from which albumen is made. Such 
denaturation has an activation energy E a of about 40 kj mol~ . 
The rate constant of reaction varies with tempera- 

We consider the Arrhe- 
nius equation in appro- 
priate detail in Chap- 
ter 8. 

ture, with k increasing as the temperature increases, k 
is a function of T according to the well-known Arrhe- 
nius equation: 

\katTj R \T 2 TJ 

We saw in Worked Example 5.2 how the temperature of the boiling water increases 
from 100 °C to 147 °C in a pressure cooker. A simple calculation with the Arrhenius 
equation (Equation (5.7)) shows that the rate constant of cooking increases by a little 
over fourfold at the higher temperature inside a pressure cooker. 

Boiling an egg takes about 4 min at 100 °C, so boiling an egg in a pressure cooker 
takes about 1 min. 

Justification Box 5.2 

The Clapeyron equation, Equation (5.1), yields a quantitative description of a phase 
boundary on a phase diagram. Equation (5.1) works quite well for the liquid- solid 
phase boundary, but if the equilibrium is boiling or sublimation - both of which involve 
a gaseous phase - then the Clapeyron equation is a poor predictor. 

For simplicity, we will suppose the phase change is the boiling of a liquid: liquid — > 
gas. We must make three assumptions if we are to derive a variant that can accommodate 
the large changes in the volume of a gas: 

Assumption 1: we assume the enthalpy of the phase change is independent of tem- 
perature and pressure. This assumption is good over limited ranges of both p and T, 
although note how the Kirchhoff equation (Equation (3.19)) quantifies changes in AH . 

Assumption 2: we assume the gas is perfect, i.e. it obeys the ideal-gas equation, 
Equation (1.13), so 

pV = nRT or pV m = RT 

where V m is the molar volume of the gas. 



Assumption 3: AV m is the molar change in volume during the phase change. The 

value of A V m = V m ( g ) — V m (i) , where V m (i) is typically 20 cm mol 
22.4 dm 3 mol" 1 (at s.t.p.), i.e. 22400 cm 3 mol"* 
between V m ( g ) and V m (i), we assume that AV m 

and V, 


In response to the vast discrepancy 
i.e. that V m (i) is negligible by 



comparison. This third approximation is generally good, and will only break down at 
very low temperatures. 

First, we rewrite the Clapeyron equation in response to approximation 2: 



dT TV, 

mi f i 

Next, since we assume the gas is ideal, we can substitute for the V m term via the 
ideal-gas equation, and say V m = RT -4- p: 

dp_ = Afl^ x _p_ 
dT T K RT 

Next, we multiply together the two T terms, rearrange and separate the variables, to 

1 . AH* 1 


— — x — dT 
R T 2 

We place the'AHZ + R' 
term outside the right- 
hand integral because 
its value is constant. 

Integrating with the limits p2 at T2 and p\ at T\ gives 

f' 2 1 AH* f Tl 1 

/ -dp = —^ —dT 

J P , p R Jt, t 2 

Subsequent integration yields 



-| ? 2 

-I T, 

Next, we insert limits: 


AH* ( 1 1 \ 

And, finally, we group together the two logarithmic terms to yield the Clausius- Clap- 
eyron equation: 

> 2 \ AH*/\ 1 

Pi) R \T 2 Ti 




5.4 Phase equilibria involving two-component 
systems: partition 

Why does a fizzy drink lose its fizz and go flat? 

Equilibrium constants of partition 

Drinks such as lemonade, orangeade or coke contain dissolved CO2 gas. As soon as 
the drink enters the warm interior of the mouth, CO2 comes out of solution, imparting 
a sensation we say is 'fizzy'. 

The CO2 is pumped into the drink at the relatively high pressure of about 3 bar. 
After sealing the bottle, equilibrium soon forms between the gaseous CO2 in the space 
above the drink and the CO2 dissolved in the liquid drink (Figure 5.14). We say the 
CO2 is partitioned between the gas and liquid phases. 

The proportions of CO2 in the space above the liquid and in the liquid are fixed 
according to an equilibrium constant, which we call the partition constant: 



amount of CO2 in phase 1 
amount of CO2 in phase 2 

We need to note how the identities of phases 1 and 2 must be 
defined before K can be cited. We need to be aware that ^(partition) 
is only ever useful if the identities of phases 1 and 2 are defined. 
On opening the drink bottle we hear a hissing sound, which 
occurs because the pressure of the escaping CO2 gas above the 
liquid is greater than the atmospheric pressure. We saw in Chapter 4 
that the molar change in Gibbs function for movement of a gas is 
given by 

AG = RTln(-^^-) 

V /'(initial)/ 


This equilibrium con- 
stant is often incor- 
rectly called a 'partition 
function' - which is in 
fact a term from statis- 
tical mechanics. 


Carbon dioxide 
in the gas phase 

Carbon dioxide 
dissolved in solution 

Figure 5.14 In a bottle of fizzy drink, carbon dioxide is partitioned between the gas and the 
solution phases 



The value of AG is only ever negative, as required by a thermodynamically sponta- 
neous process, if the initial pressure p (initial) is greater than the final pressure p(finai)> 
i.e. the fraction is less than one. In other words, Equation (5.9) shows why AG is 
negative only if the pressure of the CO2 in the space above the liquid has a pressure 
that is greater than p & . 

We disrupted the equilibrium in the bottle when we allowed out much of the CO2 
gas that formerly resided within the space above the liquid; conversely, the CO2 
dissolved in the liquid remains in solution. 

After drinking a mouthful of the drink, we screw on the bottle top to stop any 
more CO2 being lost, and come back to the bottle later when a thirst returns. The 
CO2 re-equilibrates rapidly, with some of the CO2 in the liquid phase passing to the 
gaseous phase. Movement of CO2 occurs in order to maintain the constant value of 
^(partition): we call it 're-partitioning' . 

Although the value of ^(partition) does not alter, the amount of 
CO2 in each of the phases has decreased because some of the CO2 
was lost on opening the bottle. The liquid, therefore, contains less 
CO2 than before, which is why it is perceived to be less fizzy. And 
after opening the bottle several times, and losing gaseous CO2 each 
time, the overall amount of CO2 in the liquid is so depleted that the 
drink no longer sparkles, which is when we say it has 'gone flat'. 

A fizzy drink goes 'flat' 
after opening it sev- 
eral times because the 
water is depleted of 
C0 2 . 

Worked Example 5.4 A bottle of fizzy pop contains CO2. What are the relative amounts 
of CO2 in the water and air if ^( part i t i on ) = 4? 

Firstly, we need to note that stating a value of ^"(partition) is useless unless we know how 
the equilibrium constant A"( part ition) was written, i.e. which of the phases '1' and '2' in 
Equation (5.8) is the air and which is the liquid? 

In fact, most of the CO2 resides in the liquid, so Equation (5.8) would be written as 


(partition) — 

concentration of CO2 in the drink 
concentration of CO2 in the air above the liquid 

A bottle of fizzy drink 
going flat is a fairly triv- 
ial example of partition, 
but the principle is vital 
to processes such as 
reactions in two-phase 
media or the operation 
of a high-performance 
liquid chromatography 

This partition constant has a value of 4, which means that four times 
as much CO2 resides in the drink as in the liquid of the space above 
the drink. Stated another way, four-fifths of the CO2 is in the gas 
phase and one-fifth is in solution (in the drink). 

SAQ 5.5 An aqueous solution of sucrose is prepared. It 
is shaken with an equal volume of pure chloroform. The 
two solutions do not mix. The sucrose partitions between 
the two solutions, and is more soluble in the water. The 
value of /((partition) for this water-chloroform system is 5.3. 
What percentage of the sucrose resides in the chloro- 



How does a separating funnel work? 

Partition as a function of solvent 

The operation of a separating funnel depends on partition. A solvent contains some 
solute. A different solvent, which is immiscible with the first, contains no compound. 
Because the two solvents are immiscible - which means they do 
not mix - the separating funnel will show two distinct layers (see 
Figure 5.15). After shaking the funnel vigorously, and allowing its 
contents to settle, some of the solute will have partitioned between 
the two solvents, with some sample passing from the solution into 
the previously pure solvent 1. 

We usually repeat this procedure two or three times during the 
practice of solvent extraction, and separate the two layers after each 
vigorous shake (we call this procedure 'running off the heavier, 
lower layer of liquid). Several extractions are needed because ^(partition) is usually 
quite small, which implies that only a fraction of the solute is removed from the 
solution during each partition cycle. 

Immiscible solutions 
do not mix. The words 
'miscible' and its con- 
verse 'immiscible' 
derive from the Latin 
word miscere, meaning 
'to mix'. 

Solvent 1 

Meniscus (across which 
partition occurs) 

Solvent 2 

Figure 5.15 A separating funnel is a good example of partition: solute is partitioned between two 
immiscible liquids 




The reason we need to shake the two solutions together when partitioning is because 
the solute only passes from one solvent to the other across the interface between them, 
i.e. across the meniscus. 

The meniscus is quite small if the funnel is kept still, and partitioning is slow. 
Conversely, shaking the funnel generates a large number of small globules of solvent, 
which greatly increases the 'active' surface area of the meniscus. Therefore, we shake 
the funnel to increase the rate of partitioning. 

A fish would not be 
able to 'breathe' in 
water if it contained no 
oxygen gas. 

Why is an ice cube only misty at its centre? 

The temperature dependence of partition 

Most ice cubes look misty at their centre, but are otherwise quite clear. The ice from 
which the ice cubes are made is usually obtained from the tap, so it contains dissolved 
impurities such as chlorine (to ensure its sterility) and gases from the atmosphere. The 
mist at the centre of the ice cube comprises millions of minute air bubbles containing 
these gases, principally nitrogen and oxygen. 

Gaseous oxygen readily partitions with oxygen dissolved in solu- 
tion, in much the same way as the partitioning of CO2 in the 
fizzy-drink example above. The exact amount of oxygen in solu- 
tion depends on the value of ^(partition), which itself depends on the 

Tap water is always saturated with oxygen, the amount depend- 
ing on the temperature. The maximum concentration of oxygen in 
water - about 0.02 moldm -3 - occurs at a temperature of 3 °C. 
The amount of oxygen dissolved in water will decrease below 
this temperature, since ^partition) decreases. Accordingly, much dis- 
solved oxygen is expelled from solution as the water freezes, merely 
to keep track of the constant decreasing value of ^(partition)- 
The tap water in the ice tray of our fridge undergoes some interesting phase changes 
during freezing. Even cold water straight from a tap is warmer than the air within a 
freezer. Water is a poor thermal conductor and does not freeze evenly, i.e. all at once; 
rather, it freezes progressively. The first part of the water to freeze is that adjacent to 
the freezer atmosphere; this outer layer of ice gradually becomes thicker with time, 
causing the amount of liquid water at the cube's core to decrease during freezing. 

But ice cannot contain much dissolved oxygen, so air is expelled from solution 
each time an increment of water freezes. This oxygen enters any liquid water nearby, 
which clearly resides near the centre of the cube. We see how the oxygen from the 
water concentrates progressively near the cube centre during freezing. 

Eventually, all the oxygen formerly in the water resides in a small volume of water 
near the cube centre. Finally, as the freezing process nears its completion and even 

Like all other equilib- 
rium constants, the 

value Of /((partition) 

depends strongly on 



this last portion solidifies, the amount of oxygen in solution exceeds ^(partition) and 
leaves solution as gaseous oxygen. It is this expelled oxygen we see as tiny bubbles 
of gas. 


Zone refining is a technique for decreasing the level of impurities in some metals, alloys, 
semiconductors, and other materials; this is particularly so for doped semiconductors, in 
which the amount of an impurity must be known and carefully controlled. The technique 
relies on the impurities being more soluble in a molten sample (like oxygen in water, 
as noted above) than in the solid state. 

To exploit this observation, a cylindrical bar of material is passed slowly through 
an induction heater and a narrow molten 'zone' is moved along its length. This causes 
the impurities to segregate at one end of the bar and super-pure material at the other. 
In general, the impurities move in the same direction as the molten zone moves if the 
impurities lower the melting point of the material (see p. 212). 

How does recrystallization work? 

Partition and the solubility product 

We say the solution is saturated if solute is partitioned between a liquid-phase solution 
and undissolved, solid material (Figure 5.16). In other words, the solution contains 
as much solute as is feasible, thermodynamically, while the remainder remains as 
solid. The best way to tell whether a solution is saturated, therefore, is to look for 
undissolved solid. If ^(partition) is small then we say that not much of the solute resides 
in solution, so most of the salt remains as solid - we say the salt is not very soluble. 
Conversely, most, if not all, of the salt enters solution if ^(partition) is large. 

Like all equilibrium constants, the value of ^(partition) depends on temperature, 
sometimes strongly so. It also depends on the solvent polarity. For example, ^T(partition) 

Solution saturated 
with solute 

Solid crystals 
of solute 

Figure 5.16 In a saturated solution, the solute is partitioned between the solid state and solute 
in solution 



of sodium chloride (NaCl) in water is large, so a saturated solution has a concentration 
of about 4 moldm -3 ; a saturated solution of NaCl in ethanol contains less than 
0.01 moldm -3 of solute. 

An alternative way of expressing the partition constant of a spar- 
ingly soluble salt is to define its 'solubility product' K sp (also called 
the 'solubility constant' K s ). K s is defined as the product of the ion 
activities of an ionic solute in its saturated solution, each raised to 
its stoichiometric number V;. K s is expressed with due reference to 
the dissociation equilibria involved and the ions present. 
We saw above how the extent of partition is temperature depen- 
dent; in that example, excess air was expelled from solution during freezing, since 
the solubility of air was exceeded in a cold freezer box, and the gas left solution in 

Strictly, we should 
speak in terms of ionic 
activities rather than 
concentrations; see 
p. 312 ff. 

order for the value of K 


to be maintained. 

Like all equilibrium constants, ^(partition) is a function of temper- 
ature, thereby allowing the preparative chemist to recrystallize a 
freshly made compound. In practice, we dissolve the compound in 
a solvent that is sufficiently hot so that ^(partition) is large, as shown 
by the high solubility. Conversely, ^(partition) decreases so much on 
cooling that much of the solute undergoes a phase change from 
the solution phase to solid in order to maintain the new, lower 
value of X" (partition) • The preparative chemist delights in the way 
that the precipitated solid retrieved is generally purer than that initially added to the 
hot solvent. 

The energy necessary to dissolve 1 mol of solute is called the 'enthalpy of solution' 
(cf. p. 125). A value of AH can be estimated by analysing the solubility 

The improved purity 
of precipitated solute 
implies that /C (partition) 
for the impurities is 
different from that for 
the major solute. 



s of a solute (which is clearly a function of K (partition)) with temperature T. 

The value of ^(partition) changes with temperature; the temperature dependence of 
an equilibrium constant is given by the van't Hoff isochore: 


^(partition)2 \ AH, 


(partition) 1 


* (solution) / ^ 




AW (s iution) is sometimes 
called 'heat of solu- 
tion', particularly in 
older books. The word 
'heat' here can mislead, 
and tempts us to ignore 
the possibility of pres- 
sure-volume work. 

so an approximate value of AH^ olutio . may be obtained from the 
gradient of a graph of an isochore plot of In s (as 'y') against 
1/r (as 'x'). Since s increases with increased T, we predict that 
A Solution) wil1 be Positive. 

Worked Example 5.5 Calculate the enthalpy of solution AH, olution) 
from the following solubilities 5 of potassium nitrate as a function of 
temperature T. Values of s were obtained from solubility experiments. 

77K 354 347.6 342 334 329 322 319 317 

slg per 100 g of water 140.0 117.0 100.0 79.8 68.7 54.6 49.4 46.1 







Figure 5.17 The solubility s of a partially soluble salt is related to the equilibrium constant 
^(partition) an d obeys the van't Hoff isochore, so a plot of In s (as 'y') against \/T (as 'x') should 
be linear, with a slope of '— A//- luti . -=- R' . Note how the temperature is expressed in kelvin; a 
graph drawn with temperatures expressed in Celsius would have produced a curved plot. The label 
KIT on the x-axis comes from \/T -f- 1/K 

The solubility s is a function of A" (partition) ; so, from Equation (5.10), a plot of In s 
(as 'v') against \IT (as 'x') yields the straight-line graph in Figure 5.17. A value of 



= 34 kJmol is obtained by multiplying the gradient '—1 x R' . 

Why are some eggshells brown and some white? 

Partition between two solid solutes 

The major component within an eggshell is calcium carbonate (chalk). Binders and 
pigments make up the remainder of the eggshell mass, accounting for about 2-5%. 

Before a hen lays its egg, the shell forms inside its body via a complicated series 
of precipitation reactions, the precursor for each being water soluble. Sometimes the 
hen's diet includes highly coloured compounds, such as corn husk. The chemicals 
forming the colour co-precipitate with the calcium carbonate of the shell during shell 
formation, which we see as the egg shell's colour. Any substantial change in the hen's 
diet causes a different combination of chemicals to precipitate during shell formation, 
explaining why we see differently coloured eggs. 

In summary, this simple example illustrates the partition of solutes during pre- 
cipitation: the colour of an egg shell results from the partitioning of chemicals, 
some coloured, between the growing shell and the gut of the hen during shell 


5.5 Phase equilibria and colligative properties 

Why does a mixed-melting-point determination work? 

Effects of impurity on phase equilibria 
in a two-component system 

The best 'fail-safe' way of telling whether a freshly prepared compound is identical 
to a sample prepared previously is to perform a mixed-melting-point experiment. 

In practice, we take two samples: the first comprises material 

A 'mixed melting point 
is the only absolutely 
fail-safe way of deter- 
mining the purity of 
a sample. 

whose origin and purity we know is good. The second is fresh 
from the laboratory bench: it may be pure and identical to the 
first sample, pure but a different compound, or impure, i.e. a mix- 
ture. We take the melting point of each separately, and call them 
respectively T (mdu pure) and T imAt , unknown)- We know for sure that 
the samples are different if these two melting temperatures differ. 
Ambiguity remains, though. What if the melting temperatures are the same but, by 
some strange coincidence, the new sample is different from the pure sample but has 
the same melting temperature? We therefore determine the melting temperature of a 
mixture. We mix some of the material known to be pure into the sample of unknown 
compound. If the two melting points are still the same then the two materials are 
indeed identical. But any decrease in T( me i ti i mpure ) means they are not the same. The 
value of 7( me i t m i xtU re) will always be lower than 7( me i t] pure ) if the two samples are 
different, as evidenced by the decrease in 7( me i t ). We call it a depression of melting 
point (or depression of freezing point). 

Introduction to colligative properties: chemical potential 

The depression of a melting point is one of the simplest manifestations of a colliga- 
tive property. Other everyday examples include pressure, osmotic pressure, vapour 
pressure and elevation of boiling point. 

'Colligative properties' 
depend on the num- 
ber, rather than the 
nature, of the chem- 
ical particles (atoms 
or molecules) under 

For simplicity, we will start by thinking of one compound as the 
'host' with the other is a 'contaminant'. We find experimentally 
that the magnitude of the depression AT depends only on the 
amount of contaminant added to the host and not on the identity of 
the compounds involved - this is a general finding when working 
with colligative properties. A simple example will demonstrate how 
this finding can occur: consider a gas at room temperature. The 
ideal-gas equation (Equation (1.13)) says pV = nRT, and holds 
reasonably well under s.t.p. conditions. The equation makes it clear that the pressure 
p depends only on n, V and T, where V and T are thermodynamic variables, and 
n relates to the number of the particles but does not depend on the chemical nature 
of the compounds from which the gas is made. Therefore, we see how pressure is a 
colligative property within the above definition. 



Earlier, on p. 181, we looked at the phase changes of a single-component system 
(our examples included the melting of an ice cube) in terms of changes in the molar 
Gibbs function AG m . In a similar manner, we now look at changes in the Gibbs 
function for each component within the mixture; and because several components 
participate, we need to consider more variables, to describe both the host and the 

We are now in a position to understand why the melting point 
of a mixture is lower than that of the pure host. Previously, when 
we considered the melting of a simple single-component system, 
we framed our thinking in terms of the molar Gibbs function G m - 
In a similar way, we now look at the molar Gibbs function of each 
component i within a mixture. Component i could be a contami- 
nant. But because i is only one part of a system, we call the value of G m for material 
i the partial molar Gibbs function. The partial molar Gibbs function is also called 
the chemical potential, and is symbolized with the Greek letter mu, /x. 

We define the 'mole fraction' x, as the number of moles of component i expressed 
as a proportion of the total number of moles present: 

For a pure substance, 
the chemical potential 
11 is merely another 
name for the molar 
Gibbs function. 

number of moles of component i 

Xi = 

total number of moles 

The value of /x, - the molar Gibbs function of the contaminant - 
decreases as x, decreases. In fact, the chemical potential //, of 
the contaminant is a function of its mole fraction within the host, 
according to Equation (5.11): 


fit = [i: + RT \r\Xi 


where x,- is the mole fraction of the species i, and fx i is its 
standard chemical potential. Equation (5.12) should remind us of 
Equation (4.49), which relates AG and AG e . 

Notice that the mole fraction x has a maximum value of unity. 
The value of x decreases as the proportion of contaminant in- 
creases. Since the logarithm of a number less than one is always 
negative, we see how the RT In x, term on the right-hand side of 
Equation (5.12) is zero for a pure material (implying /x, = fx t ). At 
x, < 1, causing the term RT In x, to be negative. In other words, 
will always decrease from a maximum value of \x i as the amount 
in creases. 

Figure 5.18 depicts graphically the relationship in Equa- 
tion (5.12), and shows the partial molar Gibbs function of the host 
material as a function of temperature. We first consider the heavy 
bold lines, which relate to a pure host material, i.e. before con- 
tamination. The figure clearly shows two bold lines, one each for 
the material when solid and another at higher temperatures for the 

The mole fraction x of 
the host DEcreases as 
the amount of contam- 
inant INcreases. The 
sum of all the mole 
fractions must always 
equal one; and the 
mole fraction of a pure 
material is also one. 

Strictly, Equation (5.12) 
relates to an ideal 
mixture at constant p 
and T. 

all other times, 
the value of /x 
of contaminant 

Remember: in this type 
of graph, the lines 
for solid and liquid 
intersect at the melting 







\. Liquid 




en as 

E E 

_ © 

(0 J= 



' (melt) mixture ' (melt) pure solvent (pure) 

Temperature T 

Figure 5.18 Adding a chemical to a host (mixing) causes its chemical potential [i to decrease, 
thereby explaining why a melting-point temperature is a good test of purity. The heavy solid 
lines represent the chemical potential of the pure material and the thin lines are those of the host 
containing impurities 

respective liquid. In fact, when we remember that the chemical potential for a pure 
material is the same as the molar Gibbs function, we see how this graph (the bold 
line for the pure host material) is identical to Figure 5.2. And we recall from the start 
of this chapter how the lines representing G m for solid and G m for liquid intersect 
at the melting temperature, because liquid and solid are in equilibrium at r( me i t ), i.e. 

G m (li qu id) = G m ( so lid) at T( me i t ). 

We look once more at Figure 5.18, but this time we concentrate on the thinner 
lines. These lines are seen to be parallel to the bold lines, but have been displaced 
down the page. These thin lines represent the values of G m of the host within the 
mixture (i.e. the once pure material following contamination). The line for the solid 
mixture has been displaced to a lesser extent than the line for the 
liquid, simply because the Gibbs function for liquid phases is more 
sensitive to contamination. 

The vertical difference between the upper bold line (representing 
/x^) and the lower thin line (which is /x) arises from Eq. (5.12): 
it is a direct consequence of mixing. In fact, the mathematical 
composition of Eq. (5.12) dictates that we draw the line for an 
impure material (when x, < 1) lower on the page than the line for 
the pure material. 

It is now time to draw all the threads together, and look at 
the temperature at which the thin lines intersect. It is clear from 
Figure 5.18 that the intersection temperature for the mixture occurs 
at a cooler temperature than that for the pure material, showing 
why the melting point temperature for a mixture is depressed rel- 
ative to a pure compound. The depression of freezing point is a 
direct consequence of chemical potentials as defined in Equation 

As the mole fraction of 
contaminant increases 
(as Xj gets larger), so 
we are forced to draw 
the line progressively 
lower down the figure. 

A mixed-melting-point 
experiment is an ideal 
test of a material's 
purity since 7" (me i t) 
never drops unless the 
compound is impure. 


We now see why the melting-point temperature decreases following contamination, 
when its mole fraction deviates from unity. Conversely, the mole fraction does not 
change at all if the two components within the mixed-melting-point experiment are 
the same, in which T( me it) remains the same. 

Justification Box 5.3 

When we formulated the total differential of G (Equation (4.30)) in Chapter 4, we only 
considered the case of a pure substance, saying 

/3G\ /3G\ 

We assumed then the only variables were temperature and pressure. We must now 
rewrite Equation (4.30), but we add another variable, the amount of substance «, in a 

We append an additional subscript to this expression for dG to emphasize that we refer 
to the material i within a mixture. As written, Equation (5.13) could refer to either the 
host or the contaminant - so long as we define which is i . 

The term 3G,/3«j occurs so often in second law of thermodynamics that it has its 
own name: the 'chemical potential' /x, which is defined more formally as 

V 9 ".' / p,T,nj 

where the subscripts to the bracket indicate that the variables p, T ', and the amounts of 
all other components tij in the mixture, each remain constant. The chemical potential 
is therefore seen to be the slope on a graph of Gibbs function G (as 'y') against the 
amount of substance m, (as 'x'); see Figure 5.19. In general, the chemical potential 
varies with composition, according to Equation (5.12). 

The chemical potential fi can be thought of as the constant of proportionality between 
a change in the amount of a species and the resultant change in the Gibbs function of 
a system. 

The way we wrote 3G in Equation (5.13) suggests the chemical potential /x is the 
Gibbs function of 1 mol of species i mixed into an infinite amount of host material. 
For example, if we dissolve 1 mol of sugar in a roomful of tea then the increase in 
Gibbs function is /X( SUg ar)- An alternative way to think of the chemical potential /x is to 
consider dissolving an infinitesimal amount of chemical i in 1 mol of host. 



Composition n, 

Figure 5.19 The chemical potential //,- (the partial molar Gibbs function) of a species in a 
mixture is obtained as the slope of a graph of Gibbs function G as a function of composition 

We need to employ 'mental acrobatics' of this type merely to ensure that our definition 
of ji is watertight - the overall composition of the mixture cannot be allowed to change 

How did the Victorians make ice cream? 

Cryoscopy and the depression of freezing point 

The people of London and Paris in Victorian times (the second half of the nineteenth 
century) were always keen to experience the latest fad or novelty, just like many rich 
and prosperous people today. And one of their favourite 'new inventions' was ice 
cream and sorbets made of frozen fruit. 

The ice cream was made this way: the fruit and/or cream to be frozen is packed 
into a small tub and suspended in an ice bath. Rock salt is then added to the ice, 
which depresses its freezing temperature (in effect causing the ice to melt). Energy 
is needed to melt the ice. A//( me i t ) = 6.0 kJmol - for pure water. This energy comes 
from the fruit and cream in the tub. As energy from the cream and fruit passes through 
the tub wall to the ice, it freezes. Again, we see how a body's temperature is a good 
gauge of its internal energy (see p. 34). 

The first satisfactory theory to explain how this cooling process works was that of 
Francois -Marie Raoult, in 1878. Though forgotten now, Raoult already knew 'Blag- 
den's law': a dissolved substance lowers the freezing point of a 
solvent in direct proportion to the concentration of the solute. In 
practice, this law was interpreted by saying that an ice-brine mix- 
ture (made with five cups of ice to one of rock salt) had a freezing 
point at about —2.7 °C. Adding too much salt caused the tempera- 
ture to fall too far and too fast, causing the outside of the ice 

Dissolving a solute in 
a solvent causes a 
depression of freezing 
point, in the same way 
as mixing solids. 



cream to freeze prematurely while the core remained liquid. Adding 
too little salt meant that the ice did not melt, or remained at a tem- 
perature close to °C, so the cream and fruit juices remained liquid. 
This depression of the freezing point occurs in just the same 
way as the lower melting point of an impure sample, as discussed 
previously. This determination of the depression of the freezing 
point is termed crysoscopy. 

The word 'cryoscopy' 
comes from the Greek 
kryos, which literally 
means 'frost'. 

Why boil vegetables in salted water? 

Ebullioscopy and the elevation of boiling point 

We often boil vegetables in salted water (the concentration of table salt is usually 
in the range 0.01-0.05 moldm -3 ). The salt makes the food taste nicer, although we 
should wash off any excess salt water if we wish to maintain a healthy blood pressure. 

But salted water boils at a higher temperature than does pure 
water, so the food cooks more quickly. (We saw on p. 203 how 
a hotter temperature promotes faster cooking.) The salt causes an 
elevation of boiling point, which is another colligative property. We 
call the determination of such an elevation ebullioscopy. 

Look at Figure 5.20, the left-hand side of which should remind us 
of Figure 5.18; it has two intersection points. At the low-tempera- 
ture end of the graph, we see again why the French ice-cream 
makers added salt to the ice, to depress its freezing point. But, 
when we look at the right-hand side of the figure, we see a second 
intersection, this time between the lines for liquid and gas: the temperature at which 
the lines intersect gives us the boiling point T(boii). 

The word 'ebullioscopy' 
comes from the Latin 
(e)bulirre, meaning 
'bubbles' or 'bubbly'. 
In a related way, we 
say that someone is 
'ebullient' if they have a 
'bubbly' personality. 

Freezing point 
of water + solute 

Freezing point Boiling point Boiling point of 

of pure water °* P ure water water + solute 

Figure 5.20 Salt in water causes the water to boil at a higher temperature and freeze at a lower 
temperature; adding a solute to a solvent decreases the chemical potential ji of the solvent. The 
bold lines represent pure water and the thinner lines represent water-containing solute 



The figure shows how adding salt to the water has caused both the lines for liquid 
and for gas to drop down the page, thus causing the intersection temperature to change. 
Therefore, a second consequence of adding salt to water, in addition to changing its 
chemical potential, is to change the temperature at which boiling occurs. Note that 
the boiling temperature is raised, relative to that of pure water. 

Why does the ice on a path melt when sprinkled 
with salt? 

Quantitative cryoscopy 

The ice on a path or road is slippery and dangerous, as we saw when considering 
black ice and ice skaters. One of the simplest ways to make a road or path safer is to 
sprinkle salt on it, which causes the ice to melt. In practice, rock salt is preferred to 
table salt, because it is cheap (it does not need to be purified) and because its coarse 
grains lend additional grip underfoot, even before the salt has dissolved fully. 

The depression of freezing temperature occurs because ions from the salt enter the 
lattice of the solid ice. The contaminated ice melts at a lower temperature than does 
pure ice, and so the freezing point decreases. Even at temperatures below the normal 
melting temperatures of pure ice, salted water remains a liquid - which explains why 
the path or road is safer. 

We must appreciate, however, that no chemical reaction occurs 
between the salt and the water; more or less, any ionic salt, when 
put on ice, will therefore cause it to melt. The chemical identity of 
the salt is irrelevant - it need not be sodium chloride at all. What 
matters is the amount of the salt added to the ice, which relates 
eventually to the mole fraction of salt. So, what is the magnitude 
of the freezing-point depression? 

Let the depression of the freezing point be AT, the magnitude 
of which depends entirely on the amount of solute in the solvent. 
Re-interpreting Blagden's law gives 

The 'molaLity' m is the 
number of moles of 
solute dissolved per 
unit mass of solvent; 
'molaRity' (note the 
different spelling) is 
the number of moles 
of solute dissolved per 
unit volume. 

We prefer 'molaL- 
ity' m to 'molaRity' 
(i.e. concentration c) 
because the volume 
of a liquid or solution 
changes with temper- 
ature, whereas that 
of a mass does not. 
Accordingly, molal- 
ity is temperature 
independent whereas 
concentration is not. 

AT oc molality 


The amount is measured in terms of the molality of the solute. 
Molality (note the spelling) is defined as the amount of solute 
dissolved per unit mass of solvent: 

molality, m 

moles of solute 
mass of solvent 


where the number of moles of solute is equal to 'mass of solute 4- 
molar mass of solute'. The proportionality constant in Equation 
(5.15) is the cryoscopic constant AT(cryoscopic) ■ Table 5.3 contains a 
few typical values of ^(cryoscopic)* from which it can be seen that 


Table 5.3 Sample values of boiling and freezing points, and cryoscopic and ebullioscopic constants 




**■ (cryoscopic) 


point/ °C 

point/ °C 

/K kg mof 1 

/KkgrnoP 1 

Acetic acid 




















Carbon disulphide 





Carbon tetrachloride 




















Ethyl acetate 





Ethyl ether 










Methyl acetate 







































camphor as a solvent causes the largest depression. Note that K has the units of 
K kgmol -1 , whereas mass and molar mass are both expressed with the units in units 
of grammes, so any combination of Equations (5.15) and (5.16) requires a correction 
term of 1000 gkg -1 . Accordingly, Equation (5.15) becomes 

(mass of solute \ 1 

— ; ;— ; — x ;— ; — < 5 - 17 ) 
molar mass of solute/ mass of solvent 

where the term in parentheses is n, the number of moles of solute. 

Worked Example 5.6 10 g of pure sodium chloride is dissolved in 1000 g of water. 
By how much is the freezing temperature depressed from its normal melting temperature 
of T — 273.15 K? Take A"( cryoscop i C ) from Table 5.3 as 1.86 Kkgmol -1 . 

Inserting values into Equation (5.17) yields 

AT = 1.86 KkgrnoP 1 x lOOOgkg" 1 x 

10e 1 

58.5 gmol -1 1000 g 
so AT = 0.32 K 



This value of AT repre- 
sents the depression of 
the freezing tempera- 
ture, so it is negative 

showing that the water will freeze at the lower temperature of 
(273.16-0.32) K. 

SAQ 5.6 Pure water has a normal freezing point 
of 273.15 K. What will be the new normal freezing point 
of water if 11 g of KCI is dissolved in 0.9 dm 3 of water? 
The cryoscopic constant of water is 1.86 Kkg 1 mol -1 ; assume the density 
of water is 1 g cm' 3 , i.e. molality and molarity are the same. 

An almost identical equation relates the elevation of boiling point to the molality: 


"(elevation) = -K(ebuiiioscopic) x 1000 x molality of the salt (5.18) 

where -K"( e buiiioscopic) relates to the elevation of boiling temperature. Table 5.3 con- 
tains a few sample values of ^(ebuilioscopic) • It can be seen from the relative val- 
ues of AT(ebuiiioscopic) and AT( C ryoscopic) in Table 5.3 that dissolving a solute in a sol- 
vent has a more pronounced effect on the freezing temperature than on the boiling 


The ice on a car windscreen will also melt when squirted with de-icer. Similarly, we add 
anti- freeze to the water circulating in a car radiator to prevent it freezing; the radiator 
would probably crack on freezing without it; see the note on p. 194. 

Windscreen de-icer and engine anti-freeze both depress the freezing point of water 
via the same principle as rock salt depressing the temperature at which ice freezes on a 
road. The active ingredient in these cryoscopic products is ethylene glycol (II), which 
is more environmentally friendly than rock salt. It has two physicochemical advantages 
over rock salt: (1) being liquid, it can more readily enter between the microscopic 
crystals of solid ice, thereby speeding up the process of cryoscopic melting; (2) rock 
salt is impure, whereas II is pure, so we need less II to effect the same depression of 
freezing point. 

CH2 CH2 

/ \ 



Ethylene glycol is also less destructive to the paintwork of a car than rock salt is, 
but it is toxic to humans. 



5.6 Phase equilibria involving 
vapour pressure 

Why does petrol sometimes have a strong smell 
and sometimes not? 

Dalton's law 

The acrid smell of petrol on a station forecourt is sometimes overpoweringly strong, 
yet at other times it is so weak as to be almost absent. The smell is usually stronger 
on a still day with no wind, and inspection shows that someone has spilled some 
petrol on the ground nearby. At the other extreme, the smell is weaker when there 
is a breeze, which either blows away the spilt liquid or merely dilutes the petrol in 
the air. 

The subjective experience of how strong a smell is relates to the amount of petrol 
in the air; and the amount is directly proportional to the pressure of gaseous petrol. 
We call this pressure of petrol the 'partial pressure' /?( pe troi)- 

And if several gases exist together, which is the case for petrol in air, then the total 
pressure equals the sum of the partial pressures according to Dalton's law: 

P (total) 



In the case of a petrol smell near a station forecourt, the smell is strong when the 
partial pressure of the petrol vapour is large, and it is slight when p( pe troi) is small. 

These differences in /?( pe troi) need not mean any difference in the overall pressure 
P (total) » merely that the composition of the gaseous mixture we breathe is variable. 

SAQ 5.7 What is the total pressure of 10 g of nitrogen gas and 15 g 
of methane at 298 K, and what is the partial pressure of nitrogen in the 
mixture? [Hint: you must first calculate the number of moles involved.] 

Justification Box 5.4 

The total number of moles equals the sum of its constituents, so 

"(total) = «A + «B + • • ■ 

The ideal-gas equation (Equation (1.13)) says pV = nRT; thus P( pe troi)^ = n(petroi)RT, 

SO M(petrol) = P(petrol) V ~ RT . 

Accordingly, in a mixture of gases such as petrol, oxygen and nitrogen: 

P(total)V _ ,P(petrol)V P (oxygen) V p (nitrogen) V 






We can 

cancel the 


constant R, the volume and 




all constant, 

to yield 

P(total) = 

P(petrol) t P(oxygen 

~r P (nitrogen) 

which is 

Dalton 's 




How do anaesthetics work? 


Gases dissolving in liquids: Henry's law 

'Anaesthesia' is the 
science of making 
someone unconscious. 
The word comes from 
the Greek aesthesis, 
meaning sensation 
(from which we get 
the modern English 
word 'aesthetic', i.e. to 
please the sensations). 
The initial s ana' makes 
the word negative, i.e. 
without sensation. 

An anaesthetist administers chemicals such as halothane (III) to a 
patient before and during an operation to promote unconsciousness. 
Medical procedures such as operations would be impossible for 
the surgeon if the patient were awake and could move; and they 
would also be traumatic for a patient who was aware of what the 
surgery entailed. 



F H 


A really deep, chem- 
ically induced sleep 
is termed 'narcosis', 
from the Greek narke, 
meaning 'numbness'. 
Similarly, we similarly 
call a class-A drug a 

Henry's law is named 
after William Henry 
(1775-1836), and says 
that the amount of gas 
dissolved in a liquid or 
solid is in direct pro- 
portion to the partial 
pressure of the gas. 

Although the topic of anaesthesia is hugely complicated, it is 
clear that the physiological effect of the compounds depends on 
their entrapment in the blood. Once dissolved, the compounds pass 
to the brain where they promote their narcotic effects. It is now 
clear that the best anaesthetics dissolve in the lipids from which 
cell membranes are generally made. The anaesthetic probably alters 
the properties of the cell membranes, altering the rates at which 
neurotransmitters enter and leave the cell. 

A really deep 'sleep' requires a large amount of anaesthetic and a 
shallower sleep requires less material. A trained anaesthetist knows 
just how much anaesthetic to administer to induce the correct depth 
of sleep, and achieves this by varying the relative pressures of the 
gases breathed by the patient. 

In effect, the anaesthetist relies on Henry's law, which states 
that the equilibrium amount of gas that dissolves in a liquid is 
proportional to the mole fraction of the gas above the liquid. Henry 
published his studies in 1803, and showed how the amount of gas 
dissolved in a liquid is directly proportional to the pressure (or 



Table 5.4 Henry's law constants 
&h for gases in water at 25 °C 

Gas & H /moldm~ 3 bar -1 

C0 2 

o 2 

CH 4 

N 2 

3.38 x 1(T 2 
1.28 x 1(T 3 
1.34 x 1(T 3 
6.48 x 1(T 4 

partial pressure) of the gas above it. Stated in another form, Henry's 
law says: 

[z'(soln)] = knpi 


where pt is the partial pressure of the gas i, and [z'( SO in)] is the 
concentration of the material i in solution. The constant of propor- 
tionality ku is the respective value of Henry's constant for the gas, 
which relates to the solubility of the gas in the medium of choice. 
Table 5.4 lists a few Henry's law constants, which relate to the 
solubility of gases in water. 

Worked Example 5.7 What is the concentration of molecular oxy- 
gen in water at 25 °C? The atmosphere above the water has a pressure 
of 10 5 Pa and contains 21 per cent of oxygen. 

Strategy. (1) We calculate the partial pressure of oxygen /?(o,)- (2) We 
calculate the concentration [02( aq )] using Henry's law, Equation 

(5.20), [0 2 (aq)] = P(0 2 ) X £ H (0 2 )- 

One of the simplest 
ways of removing 
gaseous oxygen from 
water is to bubble 
nitrogen gas through 
it (a process called 

Strictly, Henry's law 
only holds for dilute 
systems, typically in 
the mole-fraction range 
0-2 per cent. The law 
tends to break down 
as the mole fraction 
x increases. 

(1) From the partial of oxygen p(o 2 ) — X{0 2 ) x the total pressure P( to tai)> where x is 
the mole fraction: 

p ( 2 ) = 0.21 x 10 5 Pa 


2.1 x 10 4 Pa or 0.21 bar 


To obtain the concentration of oxygen, we insert values into Henry's law, 
Equation (5.20): 



[0 2( aq)] = 0.21 x p a x 1.28 x 10 _J moldnT^ bar" 


[0 2 (aq)] = 2.69 x 10~ 4 moldm" 3 

We need to be aware that ku is an equilibrium constant, so 
its value depends strongly on temperature. For example, at 35 °C, 
water only accommodates 7.03 mg of oxygen per litre, which ex- 
plains why fish in warm water sometimes die from oxygen 

This relatively high 
concentration of oxy- 
gen helps explain why 
fish can survive in 



How do carbon monoxide sensors work? 

Henry's law and solid-state systems 

Small, portable sensors are now available to monitor the air we breathe for such 
toxins as carbon monoxide, CO. As soon as the air contains more than a critical 
concentration of CO, the sensor alerts the householder, who then opens a window or 
identifies the source of the gas. 

At the 'heart' of the sensor is a slab of doped transition-metal oxide. Its mode 
of operation is to detect the concentration of CO within the oxide slab, which is in 
direct proportion to the concentration of CO gas in the air surrounding it, according 
to Henry's law. 

A small voltage is applied across the metal oxide. When it contains no CO, the 
electrical conductivity of the oxide is quite poor, so the current through the sensor is 
minute (we argue this corollary from Ohm's law). But increasing the concentration of 
CO in the air causes a proportionate increase in the amount of CO incorporating into 
the solid oxide, which has a profound influence on electrical conductivity through the 
slab, causing the current through the slab to increase dramatically. A microchip within 

the sensor continually monitors the current. As soon as the current 

increases above its minimum permissible level, the alarm sounds. 
So, in summary, CO gas partitions between the air and carefully 

formulated solid oxides. Henry's law dictates the amount of CO in 

the oxide. 

In general, Henry's 
law only applies over 
relatively small ranges 
of gas pressure. 

Why does green petrol smell different from 
leaded petrol? 

Effects of amount of material on vapour pressure 

Petrol is only useful in 
a car engine because it 
is volatile. 

A car engine requires petrol as its source of fuel. Such petrol has 

a low boiling temperature of about 60 °C. Being so volatile, the 

liquid petrol is always surrounded with petrol vapour. We say it has 

a high vapour pressure (also called 'saturated vapour pressure'), 

which explains why we smell it so readily. 

Once started, the engine carburettor squirts a mixture of air and volatile petrol 

into a hot engine cylinder, where the mixture is ignited with a spark. The resultant 

explosion (we call it 'firing') provides the ultimate source of kinetic energy to propel 

the car. 

A car engine typically requires four cylinders, which fire in a carefully synchro- 
nized manner. Unfortunately, these explosions sometimes occur prematurely, before 



the spark has been applied, so the explosions cease to be synchro- 
nized. It is clearly undesirable for a cylinder to fire out of sequence, 
since the kinetic energy is supplied in a jerky, irreproducible man- 
ner. The engine sounds dreadful, hence the word 'knock'. 

Modern petrol contains small amounts of additives to inhibit this 
knocking. 'Leaded' petrol, for example, contains the organometallic 
compound lead tetraethyl, PbEt4. Although PbEt4 is excellent at 
stopping knocking, the lead by-products are toxic. In fact, most 
EU countries now ban PbEt/t. 

So-called 'green' petrol is a preferred alternative to leaded petrol: 
it contains about 3 per cent of the aromatic hydrocarbon benzene 
(CgHg, IV) as an additive, the benzene acting as a lead -free alter- 
native to PbEt 4 as an 'anti-knocking' compound. 

We experience knock- 
ing (which we collo- 
quially call 'pinking') 
when explosions within 
a car engine are not 

Lead tetraethyl is the 
most widely made 
organometallic com- 
pound in the world. 
It is toxic, and killed 
over 40 chemical work- 
ers during its early 

The PbEt4 in petrol does not smell much because it is not volatile. By contrast, 
benzene is much more volatile - almost as volatile as petrol. The vapour above 'green' 
petrol, therefore, contains quite a high proportion of benzene (as detected by its 
cloying, sweet smell) as well as gaseous petrol. That is why green petrol has a 
sweeter smell than petrol on its own. 

Why do some brands of ' green' petrol smell different 
from others? 


Rao u It's law 

The 'petrol' we buy comprises a mixture of naturally occurring 
hydrocarbons, a principal component of which is octane; but the 
mixture also contains a small amount of benzene. Some brands of 
petrol contain more benzene than others, both because of varia- 
tions in the conditions with which the crude oil is distilled into 
fractions, and also variations in the reservoir from which the crude 
oil is obtained. The proportion varies quite widely: the average is 
presently about 3 per cent. 

Petrol containing a lot of benzene smells more strongly of ben- 
zene than petrol containing less of it. In fact, the intensity of 
the smell is in direct proportion to the amount of benzene in the 
petrol: at equilibrium, the pressure of vapour above a liquid mixture 

In the countries of 
North America, petrol is 
often called 'gas', which 
is short for gasoline'. 

Raoult's law is merely a 
special form of Henry's 



depends on the liquid's composition, according to Raoult's law. 

P (benzene) — P (benzene) X (benzene) 


Raoult's law states 
that (at constant tem- 
perature) the partial 
pressure of component 
/ in the vapour residing 
at equilibrium above 
a liquid is proportional 
to the mole fraction 
x, of component in 
the liquid. 

where X(b en zene) is the mole fraction of the benzene in the liquid. 
If we assume that liquid benzene and petrol have the same den- 
sities (which is entirely reasonable), then petrol containing 3 per 
cent of benzene represents a mole fraction X(benzene) = 0.03; the 
mole fraction of the petrol in the liquid mixture is therefore 0.97 
(or 97 per cent). The vapour above the petrol mixture will also 
be a mixture, containing some of each hydrocarbon in the petrol. 
We call the pressure due to the benzene component its partial pres- 
sure ^(benzene)- The constant of proportionality in Equation (5.21) is 
P (benzene) ' which represents the pressure of gaseous benzene above 
pure (i.e. unmixed) liquid benzene. 

Calculations with Raoult's law 

If a two-component 
system of A and B 
forms an ideal mixture, 
then we can calcu- 
late xa if we know xb 
because x A + x B = 1, so 
x B = (l-x A ). 

If we know the mole fraction of a liquid i (via Equation (5.11)) 
and the vapour pressures of the pure liquids p, , then we can ascer- 
tain the total vapour pressure of the gaseous mixture hovering at 
equilibrium above the liquid. 

The intensity of the benzene smell is proportional to the amount 
of benzene in the vapour, /? (benzene)- According to Equation (5.21), 
P(benzene) is a simple function of how much benzene resides within 
the liquid petrol mixture. Figure 5.21 shows a graph of the partial 
pressures of benzene and octane above a mixture of the two liq- 
uids. (For convenience, we assume here that the mixture comprises only these two 

The extreme mole fractions, and 1, at either end of the graph relate to pure petrol 
(x = 0) and pure benzene (x = 1) respectively. The mole fractions between these 
values represent mixtures of the two. The solid, bold line represents the total mole 
fraction while the dashed lines represent the vapour pressures of the two constituent 
vapours. It is clear that the sum of the two dashed lines equals the bold line, and 
represents another way of saying Dalton's law: the total vapour pressure above a 
mixture of liquids is the sum of the individual vapour pressures. 

Benzene is more vola- 
tile than bromobenzene 
because its vapour 
pressure is higher. 

Worked Example 5.8 The two liquids benzene and bromobenzene 
are mixed intimately at 298 K. At equilibrium, the pressures of the 
gases above beakers of the pure liquids are 100.1 kPa and 60.4 kPa 
respectively. What is the vapour pressure above the mixture if 3 mol 
of benzene are mixed with 4 mol of bromobenzene? 








70 000 - 

60 000 - 

50 000 - 

40 000 - 

30 000 - 

20 000 - 

10 000- 



Total vapour pressure^----^'''^ ..O 

^^-^^ ,o"" 

^^ a'' 

^ ,.--'' Vapour pressu 













,.0 * of benzene 

Vapour pressure h*"~* 

of octane "^^^ 






i i i i 
0.2 0.4 0.6 0.8 


Mole fraction of benzene 

Figure 5.21 Petrol ('gasoline') is a mixture of liquid hydrocarbons. The partial pressure of ben- 
zene is nearly twice that of octane, making it much more volatile. The bold line represents the 
total pressure of vapour above a basin of petrol, and comprises the sum of two partial pressures: 
benzene (open circles) and octane (filled circles). Each partial pressure is proportional to the mole 
fraction of the respective liquid in the petrol mixture 

From Dalton's law, the total vapour pressure is simply the sum of the individual vapour 

/'(total) — "(benzene) ' /'(bromobenzene) 

so, from Raoult's law, these partial pressures may be obtained by 
substituting each p term with p i x x-, : 

/"(total) - (/? (benzene) X ^(benzene)) + (P 

(bromobenzene) X -'•(bromobenzene)) 


Care: do not confuse 
p & (the standard pres- 
sure of 10 5 Pa) withpf , 
the vapour pressure of 
pure /". 

We know from the question that there are 7 mol of liquid. We obtain the respective mole 
fractions x from Equation (5.11): the mole fraction of benzene is | and the mole fraction 
of bromobenzene is ^ . 
Substituting values of x-, and p t into Equation (5.22) yields the total pressure /? (total) as 

/'(total) = (100.1 kPa x f) + (60.4 kPa x f) 
Pdotai) = (42.9 kPa) + (34.5 kPa) 


P (total) = 77.4 kPa 



An ideal mixture com- 
prises a pair (or more) 
of liquids that obey 
Raoult's law. 

Because these two liquids, when mixed, obey Raoult's law, we 
say they form an ideal mixture. In fact, relatively few pairs of 
liquids form ideal mixtures: a few examples include benzene and 
bromobenzene, benzene and toluene, bromobenzene and chloroben- 
zene, re-pentane and i-pentane. Note how each set represents a pair 
of liquids showing a significant extent of similarity. 

SAQ 5.8 Benzene and toluene form an ideal mixture, i.e. they obey 
Raoult's law. At 20 °C, the pressure p e of benzene and toluene are 0.747 x 
p^ and 0.223 xp e respectively. What is the pressure above a mixture of 
these two liquids that contains 12 mol% of benzene? 

Worked Example 5.9 (Continuing from Worked Example 5.8.) What are the mole frac- 
tions of benzene and bromobenzene in the vapour! 

From the definition of mole fraction x in Equation (5.11) above, we say 

moles of benzene in the vapour 

■^(benzene, vapour) 

total number of moles in the vapour phase 

The numbers of moles «, are directly proportional to the partial pressures /?, if we assume 
that each vapour behaves as an ideal gas (we assume here that T, R and V are constant). 
Accordingly, we can say 


pressure of benzene 

Note how the units 
cancel to yield a dimen- 
sionless mole fraction. 

total pressure 
Substituting numbers from Worked Example 5.8: 

42.9 kPa 


77.4 kPa 

•^-(benzene) — U.JJ4- 

The mole fraction of benzene in the vapour is 0.554, so it contains 55.4 
per cent benzene. The remainder of the vapour comprises the second 
component bromobenzene, so the vapour contains (100 — 55.4)% = 
44.6% of bromobenzene. 

Note how the liquid comprises 43 per cent benzene and 57 per 
cent bromobenzene, but the vapour contains proportionately more of 
the volatile benzene. We should expect the vapour to be richer in the 
more volatile component. 

SAQ 5.9 Continuing with the system in SAQ 5.8, what is the mole fraction 
of toluene in the vapour above the mixture? 

In fact, most liquid mixtures do not obey Raoult's law particularly well, owing to 
molecular interactions. 

We need four mole 
fractions to define this 
two-component sys- 
tem - two for the liquid 
phases and two for the 
vapour phases. 



Why does a cup of hot coffee yield more steam than 
above a cup of boiling water at the same temperature? 

The rate of steam pro- 
duction decreases with 
time as the water cools 
down because energy 
is lost from the cup as 
water molecules enter 
the gas phase. 

The effects of poor mixing (immiscibility) 

Prepare two cups: put boiling water into one and boiling coffee in the other. The 
temperature of each is the same because the water comes from the same kettle, yet 
the amount of steam coming from the coffee is seen to be greater. (We obtain a better 
view of the steam by placing both cups on a sunny window sill, and looking at the 
shadows cast on the opposing wall as the light passes through the vapour as it rises 
from the cups.) 

When performing this little experiment, we will probably notice 
how the steam above the coffee has an extremely strong smell of 
coffee, although the smell dissipates rapidly as the rate of steam 
production decreases. 

This experiment is a simple example of steam distillation. 
Adding steam promotes the volatilization of otherwise non-volatile 
components, simplifying their extraction. For simplicity, we will 
say that the smell derives from a single sweet-smelling chemical 
'coffee'. Coffee and water are not wholly miscible, with some of 
the essential oils from the coffee existing as tiny globules - we call the mixture a 
colloid (see Chapter 10). We have generated a two-phase system. Both phases, the 
water and the coffee, are saturated with each other. In fact, these globules would 
cause strong coffee to appear slightly misty, but for its strong colour blocking all 
light. We never see phase separation in the coffee cup, with a layer of oil floating 
above a layer of water, because the coffee's concentration is never high enough. 

We say a pure liquid boils when its vapour pressure equals the 
external, atmospheric pressure (see p. 188). Similarly, when boiling 
a mixture, boiling occurs when the sum of the partial pressures 
(P(water) + /'(coffee)) equals p* . It is for this reason that the steam 
above the coffee cup smells strongly of coffee, because the vapour 
contains the essential oils (e.g. esters) that impart the smell. But the 
water generates steam at a pressure of p & when the water added 
to the cup is boiling, so the partial pressure of the coffee P( co ttee) 
is additional. For this reason, we produce more steam than above 
the cup containing only water. 

The boiling of such a 
mixture requires the 
sum of the pressures, 
not just the pressure 
of one component, to 
equal p e . 

How are essential oils for aromatherapy extracted 
from plants? 

Steam distillation 


The 'essential oils' of a plant or crop usually comprise a mixture of esters. At its 
simplest, the oils are extracted from a plant by distillation, as employed in a standard 



undergraduate laboratory. Since plants contain such a small amount of this precious 
oil, a ton of plant may be needed to produce a single fluid ounce. Some flowers, such 
as jasmine or tuberose, contain very small amounts of essential oil, and the petals 
are very temperature sensitive, so heating them would destroy the blossoms before 
releasing the essential oils. 

To add to the cost further, many of these compounds are rather sensitive to tem- 
perature and would decompose before vaporizing. For example, oil of cloves (from 
Eugenia caryophyllata) is rich in the phenol eugenol (V), which has a boiling point 
of 250 °C). We cannot extract the oils via a conventional distillation apparatus. 

Heat-sensitive or 
compounds are purified 
by steam distillation 
at temperatures 
considerably lower 
than their usual boiling 

Solvent extraction of 
essential oils tends to 
generate material that 
is contaminated with 
solvent (and cannot be 
sold); and mechanical 
pressing of a plant usu- 
ally generates too poor 
a yield to be economi- 
cally viable. 

The most common method of extracting essential oils is steam 
distillation. The plant is first crushed mechanically, to ensure a 
high surface area, and placed in a closed still. High-pressure steam 
is forced through the still, with the plant pulp becoming hot as the 
steam yields its heat of vaporization (see p. 79). The steam forces 
the microscopic pockets holding the essential oils to open and to 
release their contents. Tiny droplets of essential oil evaporate and 
mix in the gas-phase mixture with the steam. The mixture is then 
swept through the still before condensing in a similar manner to a 
conventional distillation. 

Such 'steam heating' is even, and avoids the risk of overheating 
and decomposition that can occur in hot spots when external heat- 
ing is used. The steam condenses back into water and the droplets 
coagulate to form liquid oil. Esters and essential oils do not mix 
with water, so phase separation occurs on cooling, and we see 
a layer of oil forming above a layer of condensed water. The 
oil is decanted or skimmed off the surface of the water, dried, 
and packaged. 

The only practical problem encountered when collecting organic 
compounds by steam distillation is that liquids of low volatility 
will usually distil slowly, since the proportion of compound in the 
vapour is proportional to the vapour pressure, according to 



/'(water) "(water) 



In practice, we force water vapour (steam) at high pressure through the clove pulp to 
obtain a significant partial pressure of eugenol (V). 

Justification Box 5.5 

When considering the theory behind steam distillation, we start with the ideal-gas 
equation (Equation (1.13)), pV = nRT . We will consider two components: oil and 
water. For the oil, we say P( i\)V = n^n^RT, and for the water /'(water)V' = "(water) RT . 
Dividing the two equations by R and V (which are both constant) yields 

/'(oil) = "(oil) x T for the oil 

P(water) = "(water) x T for the water 

We then divide each pressure by the respective number of moles «, , to obtain 

P(oii) -r "(oil) = T for the oil 
P(water) ^- "(water) = T for the water 

The temperature of the two materials will be T, which is the same for each as they are 
in thermal equilibrium. We therefore equate the two expressions, saying 

/'(oil) -H "(oil) = P(water) "T "(water) 

Dividing both sides by />( wa ter) and multiplying both sides by n( aj yields Equation (5.23): 

/>(oil) "(oil) 

P (water) "(water) 

so we see how the percentage of each constituent in the vapour depends only on its 
vapour pressure at the distillation temperature. 

To extract a relatively involatile oil such as eugenol (V) without charring requires a 
high pressure of steam, although the steam will not be hotter than 100 °C, so we generate 
a mixture of vapours at a temperature lower than that of the less volatile component. 


Acids and Bases 


Equilibria involving acids and bases are discussed from within the Lowry-Br0nsted 
theory, which defines an acid as a proton donor and a base as a proton acceptor (or 
'abstracter'). The additional concept of pH is then introduced. 'Strong' and 'weak' 
acids are discussed in terms of the acidity constant K a , and then conjugate acids and 
bases are identified. 

Acid -base buffers comprise both a weak acid or base and its respective salt. Cal- 
culations with buffers employing the Henderson- Has selbach equation are introduced 
and evaluated, thereby allowing the calculation of the pH of a buffer. Next, titrations 
and pH indicators are discussed, and their modes of action placed into context. 

6.1 Properties of Lowry-Bronsted acids 
and bases 

Why does vinegar taste sour? 

The Lowry-Br0nsted theory of acids 

We instantly experience a sour, bitter taste when consuming anything containing 
vinegar. The component within the vinegar causing the sensation is ethanoic acid, 
CH3COOH (I) (also called 'acetic acid' in industry). Vinegar con- 
tains between 10 and 15% by volume of ethanoic acid, the remain- 
der being water (85-90%) and small amounts of other components 
such as caramel, which are added to impart extra flavour. 

The Latin word for 
'sour' is acidus. 

H — C- 



O — H 



The German chemist 
Liebig, in 1838, was 
the first to suggest 
mobile, replaceable, 
hydrogen atoms being 
responsible for acidic 
properties. Arrhenius 
extended the idea in 
1887, when he said the 
hydrogen existed as a 

The ethanoic acid molecule is essentially covalent, explaining 
why it is liquid when pure at room temperature. Nevertheless, the 
molecule is charged, with the O-H bond characterized by a high 
percentage of ionic character. Because water is so polar a sol- 
vent, it strongly solvates any solute dissolved within it. In aqueous 
solutions, water molecules strongly solvate the oxygen- and proton- 
containing ends of the O-H bond, causing the bond to break in a 
significant proportion of the ethanoic acid molecules, according to 
the following simplistic reaction: 

CH 3 COOH (aq) 

CH 3 COO" 


+ H" 1 



The O-H bond in an 
acid is sometimes said 
to be 'labile', since it is 
so easily broken. The 
word derives from the 
Latin labi, to lapse (i.e. 
to change). 

The 'Lowry-Br0nsted 
theory' says an acid is 
a proton donor. 

Acid property 

We say the acid dissociates. The bare proton is very small, and has 
a large charge density, causing it to attract the negative end of the 
water dipole. The proton produced by Equation (6.1) is, therefore, 
hydrated in aqueous solutions, and is more accurately represented 
by saying H+ (aq) . 

We see how solvated protons impart the subjective impression 
of a sour, bitter flavour to the ethanoic acid in vinegar. In fact, not 
only the sour flavour, but also the majority of the properties we 
typically associate with an acid (see Table 6.1) can be attributed to 
an acidic material forming one or more solvated protons H + (aq ) in 

This classification of an acid is called the Lowry-Br0nsted theory 
after the two scientists who (independently) proposed this definition 
of an acid in 1923. More succinctly, their theory says an acid is 

Table 6.1 Typical properties of Lowry-Br0nsted acids 

Example from everyday life 

Acids dissolve a metal to form a salt plus 

Acids dissolve a metal oxide to form a salt 

and water 

Acids react with metal carbonates to form a 
salt and carbon dioxide 

Acids are corrosive 

Acids react with a base to form a salt and 
water ('neutralization') 

Metallic sodium reacts with water, and 
'fizzes' as hydrogen gas evolves 

The ability of vinegar to clean tarnished 
silver by dissolving away the coloured 
coating of Ag 2 

The fizzing sensation in the mouth when 
eating sherbet (saliva is acidic, with a pH 
of 6.5); sherbet generally contains an 
organic acid, such as malic or ascorbic 

Teeth decay after eating sugar, and one of the 
first metabolites from sugar is lactic acid 

Rubbing a dock leaf (which contains an 
organic base) on the site of a nettle sting 
(which contains acid) will neutralize the 
acid and relieve the pain 


a substance capable of donating a proton. Therefore, we describe ethanoic acid as a 
'Lowry-Br0nsted acid'. 

Care: the symbol of 
conductivity is not K 
but the Greek letter 
kappa, k. 

Why is it dangerous to allow water near an electrical 
appliance, if water is an insulator? 

The solvated proton 

We all need to know that electricity and water are an extremely dangerous combi- 
nation, and explains why we are taught never to sit in the bath at the same time as 
shaving with a plug-in razor or drying our hair. Electrocution is almost inevitable, 
and is often fatal. 

The quantity we call 'electricity' is a manifestation of charge Q 
passing through a suitable conductor. The electrical conductivity k 
of water (be it dishwater, rainwater, bathwater, etc.) must be rela- 
tively high because electricity can readily conduct through water. 
Nevertheless, the value of k for water is so low that we class water 
as an insulator. Surely there is a contradiction here? 

'Super-pure water' has been distilled several times, and is indeed an insulator: 
its conductivity k is low at 6.2 x 10 -8 Scm -1 at 298 K, and lies midway between 
classic insulators such as Teflon, with a conductivity of about 10 -15 Scm -1 , and 
semiconductors such as doped silicon, for which k = 10 -2 Scm -1 . The conductivity 
of metallic copper is as high as 10 6 Scm -1 . 

The value of k cited above was for super-pure water, i.e. the 
product of multiple distillations, but 'normal' water from a tap 
will inevitably contain solutes (hence the 'furring' inside a kettle 
or pipe). Inorganic solutes are generally ionic salts; most organic 
solutes are not ionic. The conductivity of super-pure water is low 
because the molecules of water are almost exclusively covalent, 
with the extent of ionicity being very slight. But as soon as a salt 
dissolves, the extent of ionicity in the water increases dramatically, 
causing more extensive water dissociation. 

Ignoring for the moment the solute in solution, the water dis- 
sociation involves the splitting of water itself in a process called 
autoprotolysis. The reaction is usually represented as 

Pure water is a mixture 
of three components: 
H 2 0, and its two dis- 
sociation products, the 
solvated proton (H 3 + ) 
and the hydroxide ion 

2H 2 

H 3 cr 


+ OH" 



where the H30 + ( aq ) species is often called a solvated proton. It also 
has the names hydroxonium ion and hydronium ion. The complex 
ion H30 + ( aq ) is a more accurate representation of the proton respon- 
sible for acidic behaviour than the simplistic 'H + (aq )' we wrote in 
Equation (6.1). Note how the left-hand side of Equation (6.2) is 
covalent and the right-hand side is ionic. 

It is safer in many 
instances to assume 
the solvated proton 
has the formula unit 
[H(H 2 0) 4 ]+, with four 
water molecules ar- 
ranged tetrahedrally 
around a central pro- 
ton, the proton being 
stabilized by a lone 
pair from each oxygen 



Dissolving a solute generally shifts the reaction in Equation (6.2) from left to 
right, thereby increasing the concentration of ionic species in solution. This increased 
number of ions causes the conductivity k of water to increase, thereby making it a 
fatally efficient conductor of charge. 

Why is bottled water ^neutral'? 


The word 'criterion' is 
used of a principle or 
thing we choose to use 
as a standard when 
judging a situation. 
The plural or crite- 
rion is 'criteria', not 

'Autoprotolysis' comes 
from proto- indicating 
the proton, and lysis, 
which is a Greek root 
meaning 'to cleave 
or split'. The prefix 
auto means 'by self 
or 'without external 

The labels of many cosmetic products, as well as those on most 
bottles of drinking water, emphasize how the product is 'neutral', 
implying how it is neither acidic nor alkaline. This stipulation is 
deemed to show how healthy the water is. But how do they know? 
And, furthermore, what is their criterion for testing? 

A better way of defining 'neutral' is to say equal numbers of 
protons and hydroxide ions reside in solution (both types of ion 
being solvated). How does this situation arise? Autoprotolysis, as 
mentioned above, represents the .^//-production of protons, which 
is achieved by the splitting of water according to Equation (6.2). 
It is clear from Equation (6.2) how the consequence of such auto- 
protolytic splitting is a solution with equal numbers of protons and 
hydroxide ions. 

When water contains no dissolved solutes, the concentrations of 
the solvated protons and the hydroxide ions are equal. Accord- 
ingly, from our definition of 'neutral' above, we see why pure 
water should always be neutral, since [H30 + ( aq )] = [OH~( aq )]. 

As with all physicochemical processes, the extent of Equation 
(6.2) may be quantified by an equilibrium constant K. We call it 
the autoprotolysis constant, as defined by 

K = 

[H3O ( aq )][OH ( aq )] 

[H 2 0] 2 


The water term in the denominator of Equation (6.3) is always large when com- 
pared with the other two concentrations on the top, so we say it remains constant. 
This assumption explains why it is rare to see the autoprotolysis constant written as 
Equation (6.3). Rather, we usually rewrite it as 


[H3O (aq)][OH ( aq )] 


where K in Equation (6.3) = K w x [H2O] 2 in Equation (6.4). We will only employ 
Equation (6.4) from now on. We call K w the autoprotolysis constant or ionic product 
of water. 


Table 6.2 Values of the autoprotolysis constant A" w as a function of 

Temperature 77 °C 18 25 34 50 

K w x 10 14 0.12 0.61 1.04 2.05 5.66 

Source: Physical Chemistry, W. J. Moore (4th Edn), Longmans, London, 1962, 
p. 365. 

The value of K w is 1.04 x 10~ 14 at 298 K when expressed in 

concentration units of mol dm -3 . Like all equilibrium constants, its 
value depends on the temperature. Table 6.2 lists a few values of 
K w as a function of temperature. Note how K w increases slightly 
as the temperature increases. 

It should now be clear from Equation (6.4) how water splits 
(dissociates) to form equal number of protons and hydroxide ions, 
hence its neutrality, allowing us to calculate the numbers of each 
from the value of K w . 

Note how K„ has units. 

K„ is often re-ex- 
pressed as p/C w , where 
the 'p' is a mathe- 
matical operator (see 
p. 246). p/C w has a 
value of 14 at 298 K. 

Worked Example 6.1 What is the concentration of the solvated proton in super-pure 

Since [H30 + ( aq )] = [OH~( aq )], we could rewrite Equation (6.4) as 

K w — [H 3 (aq )] 

Taking the square root of both sides of this expression, we obtain 

[H 3 + (aq) ]/mol dm" 3 = /I^ = [10" 14 ] 1/2 


[H 3 + (aq) ] = 1(T 7 mol dm" 3 

The concentration of solvated protons in super-pure water is clearly very small. 

What is 'acid rain'? 


Acid rain is one of the worst manifestations of the damage we, as humans, inflict on 
our planet. Chemicals combine with elemental oxygen during the burning of fossil 
fuels, trees and rubbish to generate large amounts of 'acidic oxides' such as nitrogen 
monoxide (NO), carbon dioxide (CO2) and sulphur dioxide (SO2). 

Natural coal and oil contain many compounds of nitrogen. One of the worst products 
of their combustion is the acidic oxide of nitrogen, NO. At once, we are startled by 
this terminology, because the Lowry-Br0nsted definition of an acid involves the 
release of a proton, yet nitrogen monoxide NO has no proton to give. 



As long ago as the 
18th century, French 
chemists appreciated 
how burning elemen- 
tal carbon, nitrogen 
or sulphur generated 
compounds which, 
when dissolved in 
water, yielded an acidic 

The nitric acid in acid 
rain forms by a more 
complicated mecha- 
nism: 4NO (g) + 2H 2 ( D 

+ 2( g) ► 4HN0 3 (aq) 

'Hydrolysis' means to 
split water, the word 
coming from the two 
Greek roots hydro 
meaning water, and 
lysis meaning 'to cleave 
or split'. 

To understand the acidity of pollutants such as NO and CO2, we 
need to appreciate how the gas does not so much dissolve in water 
as react with it, according to 

C0 2(g) + 2H 2 O fl ) ► HCO- 3 (a q) + H 3 H 



Carbonic acid, H2C03( aq ), never exists as a pure compound; it only 
exists as a species in aqueous solution, where it dissociates in 
just the same way as ethanoic acid in Equation (6.1) to form a 
solvated proton and the HCCC -, ion. Note how we form a sol- 
vated proton H30 + ( aq ) by splitting a molecule of water, rather 
than merely donating a proton. Carbonic acid is, nevertheless, a 
Lowry-Br0nsted acid. 

The carbonic acid produced in Equation (6.5) is a proton donor, 
so the solution contains more solvated protons than hydroxide 
ions, resulting in rain that is (overall) an acid. To make the risk 
of pollution worse, 'acid rain' in fact contains a mixture of sev- 
eral water-borne acids, principally nitric acid, HNO3 (from nitrous 
oxide in water), and sulphurous acid, H2SO3 (an aqueous solution 
of sulphur dioxide). 

In summary, we see how the concentrations of H3O" 1 " and OH~ 
are the same if water contains no dissolved solutes, but dissolving 
a solute such as NO increases the concentration of H3O" 1 "; in a 
similar way, the concentration of OH~ will increase if the water 
contains any species capable of consuming protons. 

It is time to introduce a few new words. We say carbonic acid 
forms by hydrolysis, i.e. by splitting a molecule of water. We 
describe the extent of hydrolysis in Equation (6.5) by the following 
equilibrium constant: 


[HCQ3-][H 3 + ] 
[C0 2 ][H 2 0] 2 


Care: the values of K 
from these equations 
are only meaningful 
for concentrations at 

We sometimes call Equation (6.6) the hydrolysis constant of carbon 
dioxide. In fact, the water term in the 'denominator' (the bottom 
line) is so large compared with all the other terms that it remains 
essentially constant. Therefore, we write Equation (6.6) in a differ- 
ent form: 

[HC0 3 -][H 3 0+] 


[C0 2 ] 


Note how the two K terms, K in Equation (6.6) and K' in Equation 
(6.7), will have different values. 



Why does cutting an onion make us cry? 

Other aqueous acids in the environment 

The reason why our eyes weep copiously when peeling an onion is because the onion 
contains minute pockets of sulphur trioxide, S03( g ). Cutting the onion releases this 
gas. A mammalian eye is covered with a thin film of water-based liquid ('tears') to 
minimize friction with the eyelid. The tears occur in response to SO3 dissolving in 
this layer of water to form sulphuric acid: 

S0 3(g ) + H 2 0(1) 

H 2 SO 



The sulphuric acid produced dissociates in the water to form SO4 2 and two protons. 
The eyes sting as a direct consequence of contact with this acid. 

Why does splashing the hands 

with sodium hydroxide solution make 

them feel x soapy'? 

Proton abstraction 

Sodium hydroxide in solution dissociates to yield solvated cations 
and anions, Na + and the hydroxide ion 0H~ respectively: 





+ 0H" 



The solvated hydroxide ion in Equation (6.9) is formed in addi- 
tion to the hydroxide ions produced during water autoprotolysis, 
so there are more hydroxide ions in solution than solvated pro- 
tons, yielding excess hydroxide in solution. We say the solution 
is alkaline. As an alternative name, we say hydroxide is a base 
(see p. 241). 

Oils in the skin react readily with the hydroxide ions via the 
same chemical process occurring when spray-on oven cleaner 'eats' 
into the grime in an oven, reacting principally by the OH~( aq ) 
ion consuming protons. Let us start, for example, with a molecule 
possessing a proton capable of being donated; call it HA, where 
'A' is merely an anion of some sort. This proton must be labile. 
The hydroxide ion removes this labile proton to generate water, 
according to 

HA + OH" 


> A (aq) + H 2 


This proton-removing ability characterizes the reactions of hy- 
droxide ions in aqueous solutions, and of bases in general. We 

Care: It 


a common 



call the 

Ol-T ion 



It is not: 


hydroxyl is 




bound - 


group, for 



an alcohol. 

Fullers' earth is a type 
of clay named after a 
fuller, whose job was to 
clean cloth, e.g. strip- 
ping wool of its grease. 
Fullers' earth removes 
oils and grease from 
cloth because of its 
alkalinity, just like an 
oven cleaner solution. 

The Lowry-Br0nsted 
theory says a base is a 
proton remover. 



A basic chemical con- 
sumes protons. 

go further by defining hydroxide as a base because it reacts with 
(i.e. consumes) labile protons. And any chemical capable of remov- 
ing protons is said to be basic. 



The word 'saponify' 
comes from the Latin 
sapo, meaning soap. 

Hydroxide ions react to split ('hydrolyse') natural esters 
in the skin to form glycerol (II) and palmitic or stearic 
acid - a reaction called saponification. Palmitic and stea- 
ric acids subsequently react with the base to form the 

respective long-chain carboxy late anions - which is soap. 


H2C j CH2 

OH ° H OH 

These cleansing properties of bases were appreciated in antiquity. For example, in a 
portion of the Bible probably written in about 1200 BC, a character called Job declares his 
desire to be clean, saying, 'If I washed myself with soap and snow, and my hands with 
washing soda . . .' (snow was thought to be especially pure and soda (Na2CC>3 • IOH2O) 
is alkaline and has long been used as a soap). This quote may be found in full in the 
Bible, see Job 9:30. 

The Jewish Prophet Jeremiah writing in about 700 BC says much the same thing: 
look at Jeremiah 2:21-23 in the Hebrew Bible. 

Why is aqueous ammonia alkaline'-. 

Lowry-Br0nsted bases 

All aqueous solutions naturally contain hydroxide ions in consequence of the auto- 
protolytic reaction in Equation (6.2). As we have seen, there will be equal numbers 
of solvated protons and solvated hydroxide ions unless we add an acid or base to 
it. A solution containing more solvated protons than hydroxide ions is said to be an 
'acid' within the Lowry-Br0nsted theory, and a solution comprising more hydroxide 
ions than solvated protons is said to be a base. 



But a word of caution: species other than metal hydroxides can 
act as bases. Ammonia is such an example, since it can abstract 
protons in aqueous solution according to 



+ H 2 ► NH" 


+ OH" 



We say a proton is 
abstracted when re- 
moved selectively. 
Similarly, we call a 
selective summary or 
precis of a piece of 
prose 'an abstract'. 

To abstract a proton is to remove only the proton. The substan- 
tial extent of dissociation in Equation (6.11) helps explain why 
'aqueous ammonia' is more properly called 'ammonium hydrox- 
ide', NH4OH. We generate the solvated hydroxide ion OH~( aq ) by abstracting a proton 
from water. The OH~( aq ) ion in Equation (6.1 1) is chemically and physically identical 
to the solvated hydroxide ion generated by dissolving NaOH or KOH in water. 

Why is there no vinegar in crisps of salt 
and vinegar flavour? 

Conjugate acids and bases 

Potato crisps come in many flavours, perhaps the most popular being 'salt and vine- 
gar'. Curiously, a quick glance at the packet's list of ingredients reveals how the 
crisps contain unhealthy amounts of salt, but no vinegar (ethanoic acid) at all. In fact, 
the manufacturer dusts the crisps with powdered sodium ethanoate (NaCC^CHs), 
because 'real' vinegar would soon make the crisps limp and soggy. Inside the mouth, 
acid from the saliva reacts with the ethanoate anion to form ethanoic acid: 

CH 3 CO- aq) +H 3 0^ 


-> CH 3 C0 2 H (aq) + H 2 


This reaction proceeds inside the mouth, rapidly reaching its position of equilibrium 
and allowing the ethanoic acid to impart its distinctive vinegary flavour. 

The solvated proton on the left of Equation (6.12) acts as an 
acid, since it donates a proton at the same time as the ethanoate 
ion behaves as a base, because it accepts a proton. To complicate 
the situation, the reaction is one half of a dynamic equilibrium, 
i.e. it proceeds in both the forward and backward directions. In the 
backward direction, we notice how this time the ethanoic acid acts 
as an acid and the water acts as a base. 

The reaction in Equation (6.12) illustrates the coexistence of two 
acids and two bases. We say the ethanoate ion and ethanoic acid 
represent a conjugate pair, and the solvated proton and the water 
form a second conjugate pair. Within the ethanoic-ethanoate pair, 
the ethanoic acid is the conjugate acid and the ethanoate anion 
is the conjugate base. Similarly, H3O" 1 " is a conjugate acid to the 

The word 'conjugate' 
comes from the Latin 
conjugare, meaning 
'to yoke together' 
(the prefix con means 
'together' and jugare 
is 'to yoke'). Simi- 
larly, the English word 
'conjugal' relates to 
marriage and concerns 
the joining of husband 
and wife. 



conjugate base of H2O. Other examples of conjugate acid-base pairs include nitric 
acid and nitrate ion, and ammonium ion and ammonia (the acid being cited first in 
each case). 

We must treat with caution one further aspect of the Br0nsted theory: multiple 
proton-donation reactions. Consider the example of the bicarbonate ion HC03~ in 
water. When titrating bicarbonate with a base such as hydroxide, the ion behaves as 
an acid to form the carbonate anion and water: 

A substance like bicar- 
bonate, which can react 
as either an acid or as 
a base, is said to be 
amphoteric. The word 
comes from the Greek 
amphoteros, meaning 

HCO3" + OH" 

C0 3 2 " + H 2 


But, conversely, when titrating ions with an acid, the bicarbonate 
behaves as a base, losing its proton to form carbonic acid: 

HC0 3 "+H 3 H 

-> H2CO3 + H 2 


We see how the same ion acts as an acid or as a base, depending 
on the other reagents in solution. We say the bicarbonate ion is 
amphoteric, since it reacts either as an acid or as a base. 

SAQ 6.1 Consider the following pairs, and for each decide which is 
the conjugate acid and which the base: (a) carbonate and bicarbonate; 
(b) H 2 EDTA 2 - and H3EDTA ; (c) HN0 2 and N0 2 . 


Related models of acids and bases 

The concept of acid and base can be generalized in several ways. In liquid ammonia, for 
example, the ammonium and amide ions (NH4 + and NH2~ respectively) coexist. The 
roles of these ions are directly comparable with H30 + and OH~ in water. In ammonia, 
the species NH4CI and NaNH2 can be considered to be the respective acid and base 
conjugates, just as HC1 and NaOH are an acid-base pair in water. This solvent-based 
classification of acids and bases derived from Franklin, in 1905. His ideas are worth 
careful thought, although we no longer use his terminology. 

Br0nsted's definition of acids and bases (see p. 234 and 240) emphasizes the com- 
plementary nature of acids and bases, but it is broader than Franklin's model because 
it does not require a solvent, and can even be applied to gas-phase reactions, e.g. 

HCl( g ) + NH : 


NH 4 C1 (S) . 

How did soldiers avoid chlorine gas poisoning 
at the Second Battle of Ypres? 

Neutralization reactions with acids and bases 

The bloody Second Battle of Ypres was fought in France on 22 April 1915, and was 
the first time in modern warfare when poison gases were employed. At a crucial 



stage in the battle, the German forces filled the air above the enemy trenches with 
chlorine gas. 

Elemental chlorine CI2 dissolves slightly in water, and hydrolyses 
some of the water to yield hypochlorous acid, HOC1, according to 

Cl2(aq) + H2O 


> HCbaal + HOC1 




The reaction in Equation (6.15) occurs readily in the lungs and 
eyes (the sensitive tissues of which are lined with water) to cause 
irreparable damage. Troops exposed to chlorine apparently experi- 
enced a particularly slow and nasty death. 

The German troops did not advance, because they were not sure 
if the gas masks issued to their own troops could withstand the 
chlorine. They were also deterred by the incursion of a Canadian 
regiment. But one of the young Canadian soldiers knew a little 
chemistry: sniffing the gas, he guessed its identity correctly, and 
ordered the soldiers to cover their faces with handkerchiefs (or 
bandages) soaked in their own urine. The idea spread quickly, and 
the Canadians, together with two Yorkshire territorial battalions, 
were able to push back the German troops. 

One of the major constituents of urine is the di-amine, urea 
(III). Each amine group in urea should remind us of ammonia in 
Equation (6.11). Solutions of urea in water are basic because the 
two amine moieties each abstract a proton from water, to generate 
an ammonium salt and a hydroxide ion: 





Hypochlorous acid, 
HOCI, is one of the 
active components in 
household bleach. 

After this battle, both 
sides showed reluc- 
tance to employ poi- 
sonous gases again, 
being afraid it would 
drift back and poi- 
son their own troops. 
Cl 2 gas also caused 
extensive corrosion 
of rifles and artillery 
breech blocks, making 
them unusable. 

Reminder, to a chem- 
ist, the word basic does 
not mean 'elementary' 
or 'fundamental', but 
'proton abstracting'. 



+ 2H 2 

NH 2 



NH 3 



The two OH~ ions formed during Equation (6.16) explain why aqueous solutions of 
urea are alkaline. 

As we saw above, chlorine forms hypochlorous acid, HOCI. The hydroxide ions 
generated from urea react with the hypochlorous acid in a typical acid-base reaction, 



to form a salt and water: 

HOCl( a q) + OH ( aq ) 



+ H 2 


where C10~ is the hypochlorite ion. 

Equation (6.17) is an example of a neutralization reaction, a topic we discuss in 
more depth in Section 6.3. 

How is sherbet made? 

Effervescence and reactions of acids 

Malic acid (IV) occurs 
naturally, and is the 
cause of the sharp taste 
in over-ripe apples. 

Sherbet and sweets yielding a fizzy sensation in the mouth gen- 
erally contain two components, an acid and a simple carbonate 
or bicarbonate. A typical reaction of an acid with a carbonate is 
effervescence: the generation of gaseous carbon dioxide. In a well- 
known brand of British 'fizzy lolly', the base is sodium bicarbonate 
and the acid is malic acid (IV). Ascorbic acid (vitamin C) is another 

common acid included within sherbet. 


X H 




The acid and the bicarbonate dissolve in saliva as soon as the 'fizzy lolly' is placed 
in the mouth. If we abbreviate the malic acid to HM (M being the maliate anion), 
the 'fizzing' reaction in the mouth is described by 

HM (aq) + HCO 


> M" (aq) + C0 2(g) f +H 2 



where in this case the subscript 'aq' means aqueous saliva. The subjective sensation 

of 'fizz' derives from the evolution of gaseous carbon dioxide. 

This reaction between an acid and a carbonate is one of the oldest 
chemical reactions known to man. For example, it says in a portion 
of the Hebrew Bible written about 1000 bc, '. . . like vinegar poured 
on soda, is one who sings songs to a heavy heart', i.e. inappropriate 
and lighthearted singing can lead to a dramatic response ! The quote 
may be found in full in the book of Proverbs 25:20. 

Soda is an 

old fash- 

ioned name for sodium 


Na 2 C0 3 - 

10H 2 O. 

Ossified means 'to 
make rigid and hard.' 
The word comes from 
the Latin ossis, mean- 
ing bone. 

Why do steps made of limestone 
sometimes feel slippery? 

The typical reactions of an alkali 

Limestone is an ossified form of calcium carbonate, CaC03. Lime- 
stone surfaces soon become slippery and can be quite dangerous if 



Table 6.3 Typical properties of Lowry-Br0nsted bases 

Base property 

Example from everyday life 

Bases react with an acid to form a salt and 
water ('neutralization') 

Bases react with esters to form an alcohol 

and carboxylic acid ('saponification') 
Bases can be corrosive 

Rubbing a dock leaf (which contains an 
organic base) on the site of a nettle sting 
(which contains acid) will neutralize the 
acid and relieve the pain 

Aqueous solutions of base feel 'soapy' to the 

Oven cleaner comprising caustic soda 

(NaOH) can cause severe burns to the skin 

pools of stagnant rainwater collect. Although calcium carbonate is essentially insoluble 
in water, minute amounts do dissolve to form a dilute solution, which is alkaline. 

The alkali in these water pools reacts with organic matter such as algae and moss 
growing on the stone. The most common of these reactions is saponification (see 
p. 240), which causes naturally occurring esters to split, to form the respective car- 
boxylic acid and an alcohol. Once formed, this carboxylic acid 
reacts with more alkaline rainwater to form a metal carboxylate, 
according to 

2RCOOH (aa) + CaCO 



> Ca z+ (RCOO-) 2(aq) 

+ C0 2 ( aq ) + H 2 


The carbon dioxide generated by Equation (6.19) generally remains 
in solution as carbonic acid, although the rainwater can look a little 
cloudy because minute bubbles form. 

Like most other metal carboxylates, the calcium carboxylate 
(Ca(RCOO) 2 ) formed during Equation (6.19) readily forms a 
'soap', the name arising since its aqueous solutions feel slippery 
and soapy to the touch. The other commonly encountered metal 
carboxylates are the major components of household soap, which 
is typically a mixture of potassium stearate and potassium palmitate 
(the salts of stearic and palmitic acids). 

In summary, steps of limestone become slippery because the 
stagnant water on their surface is alkaline, thereby generating a 
solution of an organic soap. Other typical properties of bases and 
alkalis are listed in Table 6.3. 

Metal carboxylates are 
called soaps because 
they saponify oils in 
the skin and decrease 
the surface tension y of 
water, which makes the 
surfaces more slippery. 

The serial TV programs 
known as 'soap operas' 
earned their name in 
the USA at a time when 
much of a program- 
maker's funding came 
from adverts for house- 
hold soap. 

Why is the acid in a car battery more corrosive 
than vinegar? 


Car batteries generally contain sulphuric acid at a concentration of about 10 mol dm -3 . 
It is extremely corrosive, and can generate horrific chemical burns. By contrast, the 



Care: the 'H' in pH 
derives from the sym- 
bol for hydrogen, and 
is always given a big 
letter. The 'p' is a math- 
ematical operator, and 
is always small. 

concentration of the solvated protons in vinegar lies in the range 
10- 4 -10" 5 mol dm" 3 . 

Between these two acids, there is up to a million-fold differ- 
ence in the number of solvated protons per litre. We cannot cope 
with the unwieldy magnitude of this difference and tend to talk 
instead in terms of the logarithm of the concentration. To this end, 
we introduce a new concept: the pH. This is defined mathemati- 
cally as 'minus the logarithm (to the base ten) of the hydrogen ion 
concentration' : 

An acid's pH is defined 
as minus the loga- 
rithm (to the base ten) 
of the hydrogen ion 

pH = -log 10 [H+/mol dm" 3 ] 


The 'p' in Equation 
(6.20) is the mathe- 
matical operator 
'-log 10 ' of something. 
pH means we have 
applied the operator 'p' 
to [H + ]. The p is short 
for potenz, German for 

The concentrations of bench acids in an undergraduate labora- 
tory are generally less than 1 mol dm~ , so by corollary the minus 
sign to Equation (6.20) suggests we generally work with positive 
values of pH. Only if the solution has a concentration greater than 
1 mol dm -3 will the pH be negative. Contrary to popular belief, 
a negative pH is not impossible. (Try inserting a concentration of 
2.0 mol dm -3 into Equation (6.20) and see what happens!) 

Notice how we generally infer the solvated proton, H3O" 1 ", each 
time we write a concentration as [H + ], which helps explain why 
the concept of pH is rarely useful when considering acids dis- 
solved in non-aqueous solvents. When comparing the battery acid 
with the bench acid, we say that the battery acid has a lower pH 
than does the bench acid, because the number of solvated protons 
is greater and, therefore, it is more acidic. Figure 6.1 shows the 
relationship between the concentration of the solvated protons and 
pH. We now appreciate why the pH increases as the concentration 

Apart from the convenience of the logarithmically compressed 

scale, the concept of pH remains popular because one of the most 

popular methods of measuring the acidity of an aqueous solution is the glass electrode 

(see p. 336), the measurement of which is directly proportional to pH, rather than to 

[H 3 0+]. 

We need to introduce a word of caution. Most modern calculators cite an answer 
with as many as ten significant figures, but we do not know the concentration to more 
than two or three significant figures. In a related way, we note how the pH of blood 
is routinely measured to within 0.001 of a pH unit, but most chemical applications 

The lower the pH, the 
more concentrated the 

[H 3 + ]/mol dm" 3 10 1 1CT 1 1CT 2 10 -3 . . . 1CT 7 1CT 10 1Cr 11 

pH -10 1 2 3 ... 7 ... 10 11 

Figure 6.1 The relationship between concentrations of strong acids and the solution pH 



do not require us to cite the pH to more than 0.05 of a pH unit. In fact, we can rarely 
cite pH to a greater precision than 0.01 for most biological applications. 

Worked Example 6.2 What is the pH of bench nitric acid having a 
concentration of 0.25 moldnT 3 ? 

Inserting values into Equation (6.20): 

pH = -log[0.25 moldm~ 3 /mol dm" 3 ] = 0.6 

The acid has a pH of 0.6. 

SAQ 6.2 What is the pH of hydrochloric acid having a 
concentration of 0.2 moldrrT 3 ? 

SAQ 6.3 To highlight the point made above concerning 
numbers of significant figures, determine the (negligible) 
difference in [H 3 + ] between two acid solutions, one hav- 
ing a pH of 6.31 and the other 6.32. 

Sometimes we know the pH and wish to know the concentration 
of the solvated protons. Hence, we need to rewrite Equation (6.20), 
making [H + ] the subject, to obtain 

The concentration of 
the solvated protons in 
Equation (6.18) needs 
to be expressed in the 
familiar (but non-SI) 
units of mol dm 3 ; the 
SI unit of concentration 
is mol rrr 3 . 

We divide the concen- 
tration by its units to 
yield a dimensionless 

[H+/mol dm" 3 ] = 10 _pH 


We are assuming the 
concentration of the 
proton H + is the same 
as the concentration of 

H3O ( aq). 

Worked Example 6.3 What is the concentration of nitric acid having a pH of 3.5? 
Inserting values into Equation (6.21): 

[H+] = 10 



[H+] = 3.16 x 10~ 4 mol dm" 3 
SAQ 6.4 What is the concentration of nitric acid of pH = 2.2? 

Occasionally, we can merely look at a pH and say straightaway 
what is its concentration. If the pH is a whole number - call it 
x - the concentration will take the form 1 x 10 - * mol dm -3 . As 
examples, if the pH is 6, the concentration is 10~ 6 mol dm -3 ; if 
the concentration is 10~ 3 mol dm -3 then the pH is 3, and so on. 

Sometimes, we refer to 
a whole number as an 

Worked Example 6.4 Without using a calculator, what is the concentration of HNO3 
solution if its pH is 4? 


If the pH is x, then the concentration will be lO - * moldirT , so the acid concentration 
is 10~ 4 moldirT 3 . 

SAQ 6.5 Without using a calculator, what is the pH of hydrochloric acid 
of concentration 10~ 5 moldm" 3 ? 



We encounter problems when it becomes necessary to take the logarithm of a con- 
centration (which has units), since it contravenes one of the laws of mathematics. To 
overcome this problem, we implicitly employ a 'dodge' by rewriting the equation as 

, .'[H 3 + ] 
pH = - log 10 

where the term c e is the standard state, which generally has the value of 1 moldm -3 . 
The c° term is introduced merely to allow the units within the bracketed term above 
to cancel. Throughout this chapter, concentrations will be employed relative to this 
standard, thereby obviating the problem inherent with concentrations having units. 

S0renson introduced 
this definition of pH in 

Justification Box 6.1 

The definition of pH is given in Equation (6.20) as 
P H=-log 10 [H+] 

First we multiply both sides by '— 1': 

- P H = log 10 [H+] 

and then take the antilog, to expose the concentration term. Correctly, the function 
'antilog' means a mathematical operator, which performs the opposite job to the original 
function. The opposite function to log is a type of exponential. As the log is written in 
base 10, so the exponent must also be in base 10. 

If y = log| JC, then x = 10 > . This way we obtain Equation (6.21). 

Sometimes we need to know the pH of basic solutions. 



PK W = 

-log 10 K w 

pH = 

-log 10 [H 3 O+ q) ] 

pOH = 

-log 10 [OH- (aq) ] 

Worked Example 6.5 What is the pH of a solution of sodium hydroxide of concentra- 
tion 0.02 moldirT 3 ? Assume the temperature is 298 K. 

At first sight, this problem appears to be identical to those in previ- 
ous Worked Examples, but we soon appreciate how it is complicated 
because we need first to calculate the concentration of the free protons 
before we can convert to a pH. However, if we know the concentration 
of the alkali, we can calculate the pH thus: 

pK w = pH + pOH (6.22) 

where pA" w and pH have their usual definitions, and we define pOH as 

pOH=-log 10 [Oir] (6.23) 

Inserting the concentration [NaOH] = 0.02 mol dm - into Equation 
(6.22) yields the value of pOH = 2. The value of pA" w at 298 K is 14 
(see Table 6.2). Therefore: 

pH = pK w - pOH 

pH = 14 - 2 

pH= 12 

SAQ 6.6 What is the pH of a solution of potassium hydroxide of concen- 
tration 6 x 10 3 mol dm 3 . Again, assume the temperature is 298 K. 

We must look up the 
value of K w from Table 
6.2 if the temperature 
differs from 298 K, 
and then calculate a 
different value using 
Equation (6.26). 

Justification Box 6.2 

Water dissociates to form ions according to Equation (6.2). The ionic 
concentrations is the autoprotolysis constant K w , according to Equation 
Taking logarithms of Equation (6.4) yields 



of the 

log 10 K w = log 10 [H 3 O + ] + log 10 [OH-] 


Next, we multiply each term by '— 1' to yield 

- log 10 K w = - log 10 [H 3 O + ] - log 10 [OH- 



The term '— 
according to 

log 10 [H3O + ]' is the solution pH and the term 
pK w = - log 10 K w 

'- lo gio 

K w ' is 


We give the 
tion (6.22). 

name pOH to the third term '— log 10 [OH - ]'. We thereby obtain 




Why do equimolar solutions of sulphuric acid 
and nitric acid have different pHs? 

Mono-, di- and tri- basic acids 

Equimolar means 'of 
equal molarity', so 
equimolar solutions 
have the same concen- 

Nitric acid, HNO3, readily dissolves in water, where it dissociates 
according to 

HN0 3 (a q) +H 2 ► lH 3 0+ (aq) +NO 



The stoichiometry illustrates how each formula unit generates a sin- 
gle solvated proton. By contrast, sulphuric acid, H2SO4, dissociates 
in solution according to 

H 2 S0 4( a q ) + 2H 2 

2H,0 H 


+ SO 



so each formula unit of sulphuric acid generates two solvated protons. In other words, 
each mole of nitric acid generates only 1 mol of solvated protons but each mole of 
sulphuric acid generates 2 mol of solvated protons. We say nitric acid is a mono-protic 
acid and sulphuric acid is a di-protic acid. Tri-protic acids are rare. Fully protonated 
ethylene diamine tetra-acetic acid H4EDTA (V) is a tetra-protic acid. 



N-CH 2 -CH 2 -N 





Equation (6.27) demonstrated how the concentration of the solvated protons equates 
to the concentration of a mono-protic acid from which it derived; but, from Equation 
(6.28), the concentration of the solvated protons will be twice the concentration if the 
parent acid is di-protic. These different stoichiometries affect the pH, as demonstrated 
now by Worked Examples 6.6 and 6.7. 

Worked Example 6.6 Nitric acid of concentration 0.01 mol dm" is dissolved in water. 
What is its pH? 

Since one solvated proton is formed per molecule of acid, the concentration [H + ( aq )] is 
also 0.01 mol dm -3 . 

The pH of this acidic solution is obtained by inserting values into Equation (6.20): 

pH = -log 10 [0.01] 
pH = 2 



Worked Example 6.7 What is the pH of sulphuric acid having the same concentration 
in water as the nitric acid in Worked Example 6.6? 

This time, two solvated protons are formed per molecule of acid, so the concentration of 


will be 0.02 mol dm 

The pH of this acidic solution is obtained by inserting values into Equation (6.20): 

P H=-log 10 [2x0.01] 
pH= 1.68 

A pH electrode immersed in turn into these two solutions would 
register a different pH despite the concentrations of the parent acids 
being the same. 

We need to be careful with these calculations, because the extent 
of dissociation may also differ; see p. 255 ff. 

pH electrodes and pH 
meters are discussed in 
Chapter 7. 

All neutral solutions 
have a pH of 7 at 
298 K. 

What is the pH of a y neutral' solution? 

pH and neutrality 

A medicine or skin lotion is often described as 'pH neutral' as though it was obvi- 
ously a good thing. A solution is defined as neutral if it contains neither an excess 
of solvated protons nor an excess of hydroxide ions. Equation (6.4) tells us the auto- 
protolysis constant K w of super-pure water (water containing no 
additional solute) is 10 -14 (mol dm -3 ) 2 . Furthermore, we saw in 
Worked Example 6.1 how the concentration of the solvated protons 
was 10" 7 mol dm" 3 at 298 K. 

By considering both the definition of pH in Equation (6.20) 
and the concentration of the solvated protons from Worked 
Example 6.1, we see how a sample of super-pure water - which is 
neutral - has a pH of 7 at 298 K. We now go further and say all 
neutral solutions have a pH of 7. By corollary, we need to appre- 
ciate how an acidic solution always has a pH less than 7. If the pH 
is exactly 7, then the solution is neutral. 

The pH of a Lowry-Br0nsted acid decreases as its concentration 
increases. Bench nitric acid of concentration 1 mol dm -3 has a 
pH = 0. An acid of higher concentration will, therefore, have a 
negative pH (the occasions when we need to employ such solutions 
are, thankfully, rare). 

What do we mean when we say blood plasma 
has a l pHof7.4'? 

The pH of alkaline solutions 

Table 6.4 lists the pH of many natural substances, and suggests human blood plasma, 
for example, should have a pH in the range 7.3-7.5. The pH of many natural 


maximum pH 


an acid 

will be just less 



at 298 K. 

The pH of a Lowry- 
Brpnsted acid DEcre- 
ases as its concentra- 
tion INcreases. 



Table 6.4 The pHs of naturally occurring substances, listed in order 
of decreasing acidity 

Human body 

Common foodstuffs 

Gastric juices 








Duodenal contents 


















Maple syrup 


Spinal fluid 


Tap water 


Blood plasma 


Egg white (fresh) 


The word 'product' 
is used here in its 
mathematical sense of 
'multiplied by'. 

Source: Handbook of Chemistry and Physics (66th Edition), R. C. Weast (ed.), 
CRC Press, Boca Raton, Florida, 1985, page D-146. 

substances is higher than 7, so we cannot call them either 'acidic' or 'neutral'. The 
pHs range from 1.8 for limes (which explains why they taste so sour) to 7.8 or so 
for fresh egg white (albumen). 

Blood 'plasma' is that part of the blood remaining after removal 
of the haemoglobin cells that impart a characteristic 'blood-red' 
colour. According to Table 6.4, most people's plasma has a pH 
in the range 7.3-7.5. So, what is the concentration of solvated 
protons in such plasma? We met the autoprotolysis constant A" w 
in Equation (6.4). Although we discussed it in terms of super-pure 
water, curiously the relationship still applies to any aqueous system. The product of 
the concentrations of solvated protons and hydroxide ions is always 10~ 14 at 298 K. 
If Equation (6.4) applies although the water contains dissolved solute, then we can 
calculate the concentration of solvated protons and the concentration of hydroxide 
ions, and hence ascertain what a pH of more than '7' actually means. 

Worked Example 6.8 What is the concentration [OH~( aq) ] in blood plasma of pH 7.4? 

Answer strategy. (1) We first calculate the concentration of solvated protons from the pH, 
via Equation (6.20). (2) Second, we compare the concentration [H30 + ( aq )] with that of a 
neutral solution, via Equation (6.4). 

(1) The concentration [H 3 + (aq )] is obtained from Equation (6.20). Inserting values: 

[H 3 H 




[H 3 + (aq) ] = 4x 10~ 8 moldm" 3 

(2) We see how the concentration obtained in part (1) is in fact less than the 1 x 
10~ 7 moldnT we saw for pure water, as calculated in Worked Example 6.1. 



Equation (6.4) says A" w = [H30 + ( aq )] x [OH ( aq )], where the product of the two 
concentration has a value of 10~ 14 (moldm~ ) 2 at 298 K. Knowing the values of K w and 
[H30 + ( aq )], we can calculate the concentration of hydroxide ions in the blood plasma. 

Rearranging to make [OH~( aq) ] the subject, we obtain 

Inserting values, 




[H 3 0+ (aq) ] 


[OH- (aq) ] 

10" 14 (mol dm" 3 ) 
4 x 10" 8 mol dm" 


[OH" (aq) ] = 2.5 x 1(T 7 mol dm" 

Aqueous solutions in 
which the concen- 
tration of hydroxide 
exceeds the concentra- 
tion of solvated protons 
show a pH higher 
than 7. 

We see how the pH of blood plasma is higher than 7, so the con- 
centration of hydroxide ions exceeds their concentration in super- 
pure water. We derive the generalization: aqueous solutions in 
which the concentration of hydroxide is greater than the concen- 
tration of solvated protons have a pH higher than 7. The pH is 
lower than 7 if the concentration of hydroxide is less than the 
concentration [H30 + ( aq )]. 

SAQ 6.7 A solution of ammonia in water has a pH of 9. Without using a 
calculator, what is the concentration of solvated protons and hence what 
is the concentration of hydroxide ions? 

SAQ 6.8 What is the pH of sodium hydroxide solution of concentration 
lO" 2 moldrrr 3 ? 

6.2 "Strong' and "weak' acids and bases 

Why is a nettle sting more painful than a bur 
from ethanoic acid? 

Introducing 'strong' and 'weak' acids 

Brushing against a common nettle Urtica dioica can cause a pain- 
ful sting. The active component in a nettle sting is methanoic acid 
(VI), also called 'formic acid'. The sting of a nettle also contains 
natural additives to ensure that the methanoic acid stays on the 
skin, thereby maximizing the damage to its sensitive underlying 
tissue known as the epidermis. 

The sting of a red ant 
also contains meth- 
anoic acid. 



The word 'epidermis' 
derives from two Greek 
words, derma meaning 
skin, and epi meaning 
'at', 'at the base of, or 
'in additional to'. The 
same root epi occurs 
in 'epidural', a form of 
pain relief in which an 
injection is made at 
the base of the dura, 
located in the spine. 

We say an acid is 
strong if the extent 
of its ionization is 
high, and weak if the 
extent of its ionization 
is small. 

Care: do not confuse 
the words strong and 
weak acids with every- 
day usage, where we 
usually say something 
is 'strong' if its con- 
centration is large, and 
'weak' if its concentra- 
tion is small. 



The chemical structures of I and VI reveal the strong similarities 
between ethanoic and methanoic acids, yet the smaller molecule is 
considerably nastier to the skin. Why? Methanoic acid dissociates 

in water to form the solvated methanoate anion HCOO 



a solvated proton in a directly analogous fashion to ethanoic acid 
dissolving in water; Equation (6.1). In methanoic acid of concen- 
tration 0.01 moldm -3 , about 0.14 per cent of the molecules have 
dissociated to yield a solvated proton. By contrast, in ethanoic acid 
of the same concentration, only 0.04 per cent of the molecules 
have dissociated. We say the methanoic acid is a stronger acid 
than ethanoic since it yields more protons per mole. Conversely, 
ethanoic acid is weaker. 

We might rephrase this statement, and say an acid is strong if its 
extent of ionization is high, and weak if the extent of ionization is 
small. Within this latter definition, both I and VI are weak acids. 

In summary, the word 'acid' is better applied to methanoic acid 
than to ethanoic acid, since it is more acidic, and so methanoic acid 
in a nettle sting is more able to damage the skin than the ethanoic 
acid in vinegar. 

But we need to be careful. In everyday usage, we say often some- 
thing is 'strong' when we mean its concentration is large; similarly, 
we say something is 'weak' if its concentration is small. As a good 
example, when a strong cup of tea has a dark brown colour (because 
the compounds imparting a colour are concentrated) we say the tea 
is 'strong'. To a chemist, the words 'strong' and 'weak' relate only 
to the extent of ionic dissociation. 

To a chemist, the 
words 'strong' and 
'weak' relate only to 
the extent of ionic dis 

Why is 'carbolic acid' not in fact an acid? 

Acidity constants 

'Carbolic acid' is the old-fashioned name for hydroxybenzene 
(VII), otherwise known as phenol. It was first used as an anti- 
septic to prevent the infection of post-operative wounds. The British surgeon Joseph 
(later 'Lord') Lister (1827-1912) discovered these antiseptic qualities in 1867 while 
working as Professor of Medicine at the Glasgow Royal Infirmary. He squirted a 



dilute aqueous solution of VII directly onto a post-operative wound 
and found that the phenol killed all the bacteria, thereby yielding 
the first reliable antiseptic in an era when medical science was in 
its infancy. 


The antibacterial properties of VII are no longer utilized in 
modern hospitals because more potent antiseptics have now been 
formulated. But its memory persists in the continued use of 'car- 
bolic soap', which contains small amounts of phenol. 

Phenol in water is relatively reactive, thereby explaining its 
potency against bacteria. But phenol dissolved in water contains 
relatively few solvated protons, so it is not particularly acidic. But 
its old name is carbolic acid\ 

Phenol (VII) can dissociate according to 

PhOH (aq) + H 2 


->PhO- (aq) + H 3 4 



where Ph is a phenyl ring and PhO ( aq ) is the phenolate anion. An 
equilibrium constant may be written to describe this reaction: 

K = 

[PhO"(a q )][H 3 + (aq) ] 
[PhOH (aq) ][H 2 0] 


In fact, the water term in the denominator remains essentially con- 
stant, since it is always huge compared with all the other terms. 
Accordingly, we usually write a slightly altered version of K, cross- 
multiplying both sides of the equation with the concentration of 
water to yield 

[PhO- (aq) ][H 3 + (aq) ] 


[PhOH (aq) ] 


The resultant (modified) equilibrium constant is called the acidity 
constant of phenol, and has the new symbol K a , which has a value 
is 10~ 10 for phenol. K & is also called the acid constant, the acid 
dissociation constant or just the dissociation constant. The value 
of K a for phenol is clearly tiny, and quantifies just how small the 
extent is to which it dissociates to form a solvated proton. 

Phenol is a rare exam- 
ple of a stable enol 
(pronounced 'ene-ol'), 
with a hydroxyl bonded 
to a C=C bond. Most 
enols tautomerize to 
form a ketone. 

The word 'antiseptic' 
comes from the Latin 
prefix anti meaning 
'before' or 'against', 
and 'septic' comes 
from the Latin sep- 
tis, meaning a bacterial 
infection. An 'antisep- 
tic', therefore, prevents 
the processes or sub- 
stances causing an 

Historically, carbolic 
acid was so called 
because solid phenol 
causes nasty chemical 
burns to the skin. The 
root carbo comes from 
the French for 'coal'. 

Care: Do not con- 
fuse 'Ph' (a common 
abbreviation for a 
phenyl ring) with 'pH' 
(which is a mathemat- 
ical operator meaning 
-log 10 [H + (aq) ]). 

K a for phenol is 10 10 
when expressing the 
concentrations with the 
units of mol dm 3 . 



A strong acid has 


large value 

of K a , 


a weak acid 

has a 


value of K a . 

The value of K a for ethanoic acid is a hundred thousand times larger at 1.8 x 10~ 5 , 
and K a for methanoic acid is ten times larger still, at 1.8 x 10~ 4 ; so methanoic acid 
generates more solvated protons per mole of acid than either phenol or ethanoic acid. 
We discover the relative differences in K a when walking in the country, for a nettle 
can give a nasty sting (i.e. a chemical burn) but vinegar does not burn the skin. We 
say methanoic acid is a stronger acid than ethanoic acid because its value of K a is 
larger. A mole of phenol yields few protons, so we say it is a weak acid, because its 
value of K a is tiny. 

These descriptions of 'strong' and 'weak' acid are no longer 
subjective, but depend on the magnitude of K a : a strong acid has 
a large value of K a and a weak acid has a low value of K a . Stated 
another way, the position of the acid-dissociation equilibrium lies 
close to the reactants for a weak acid but close to the products for 
a strong acid, as shown schematically in Figure 6.2. 

Carboxylic acids such as ethanoic acid are generally weak be- 
cause their values of K a are small (although see p. 261). By con- 
trast, so called mineral acids such as sulphuric or nitric are classed 
as strong because their respective values of K a are large. Although 
there is little consensus, a simplistic rule suggests we class an acid 
as weak if its value of K a drops below about 10~ 3 . The acid is 
strong if K a > 10~ 3 . 

Table 6.5 contains a selection of K a values. Acids characterized 
by large values of K a are stronger than those with smaller values of 
K a . Each K a value in Table 6.5 was obtained at 298 K. Being an 
equilibrium constant, we anticipate temperature-dependent values 
of K a , with K a generally increasing slightly as T increases. 

A crude generalization 
suggests that inorganic 
acids are strong and 
organic acids are weak. 

The values of K a gen- 
erally increase with 
increasing tempera- 
ture, causing the acid to 
be stronger at high T. 


Reactants, i.e. HA + H 2 Products, i.e. H 3 + + A" 

Extent of reaction f 

Figure 6.2 Graph of Gibbs function G (as 'y') against the extent of reaction % (as 'x'). The 
minimum of the graph corresponds to the position of equilibrium: the position of equilibrium for 
a weak acid, such as ethanoic acid, lies near the un-ionized reactants; the position of equilibrium 
for a strong acid, like sulphuric acid, lies near the ionized products 



Table 6.5 Acidity ('dissociation') constants K a for inorganic Lowry-Br0nsted acids in water 
at 298 K. Values of K & are dimensionless: all values presuppose equilibrium constants such as 
Equation (6.35), and were calculated with concentrations expressed in mol dnT 3 








Hypochlorous, HOC1 
Hydrochloric, HC1 
Nitrous, HN0 2 
Sulphuric, H2SO4 
Sulphurous, H2SO3 
Carbonic, H 2 C0 3 
Phosphoric, H3PO4 

4.0 x 10~ 8 

1.0 x 10 7 

4.6 x 10~ 4 

1.0 x 10 2 

1.4 x 10~ 2 

4.3 x 10~ 7 
7.53 x 10~ 3 

1.2 x 10~ 2 
1.02 x 10~ 7 
5.61 x 10-" 
6.23 x 10~ 8 

2.2 x 10- 

Table 6.6 As for Table 6.5, but for inorganic acids 
and showing the effects of various structural changes 


10 5 K, 

Effect of extent of halogenation 








23 200 

Effect of halide 









Effect of chain length 









Effect of substituent in benzoic acids 

C 6 H 5 COOH 


p-N0 2 -C 6 H 4 COOH 


p-CH 3 0-C 6 H 4 COOH 


p-NH 2 -C 6 H 4 COOH 


In summary, carbolic acid (phenol, VII) is an extremely weak 
acid because its value of K a is 10~ 10 , quantifying how small is the 
concentration of solvated protons in its solutions. 

Basicity constants 

Having categorized acids into 'strong' and 'weak' via the concept 
of acidity constants K a , we now look at the strengths of various 

The cause of phenol's 
corrosive properties 
does not relate to its 
ability to form solvated 
protons (as indicated 
by the value of K a ) but 
its ability to penetrate 
the skin and disrupt 
the chemical processes 
occurring within the 
epidermis, to painful 



bases. It is possible to write an equilibrium constant K to describe the hydrolysis of 
bases such as ammonia (see Equation (6.12)). We write the appropriate equilibrium 
constant in just the same way as we wrote an expression for K to describe the acidic 
behaviour of phenol: 

[NH 4 +][OH-] 


We sometimes call the 
equilibrium constant 
in Equation (6.33) a 
basicity constant, and 
symbolize it as K b . 

K = 

[NH 3 ][H 2 0] 

As with the expression in Equation (6.6), this equilibrium constant 
can be simplified by incorporating the water term into K, thereby 
yielding a new constant which we will call ^b, the basicity con- 


[NH 3 ] 

where K\, in Equation (6.33) is quite different from the K in Equation (6.32). The 
value of K\, for ammonia is 1.74 x 10~ 5 , which is quite small, causing us to say 
ammonia is a weak base. The value of K b for sodium hydroxide is much larger at 
0.6, so we say NaOH is a strong base. 

But, curiously, this new equilibrium constant K\, is redundant because we could 
have calculated its value from known values of K a according to 

K„ x Kh = K„ 


where K w is the autoprotolysis constant of water from p. 236. Older textbooks some- 
times cite values of K b , but we really do not need to employ two separate K constants. 

SAQ 6.9 What is the value of K a for the ammonium ion, NH 4 +? Take K^ 
from the paragraphs immediately above, and K w = 10~ 14 . 

Justification Box 6.3 

Consider a weak acid, HA 

, dissociating: HA — » H 3 + + A~ 

. Its 

acidity constant A" a is 

given by 


[A"][H 3 + ] 


and then consider a weak base (the 

conjugate of the weak 


forming a 


ion in solution, H2O + A~ 

-> OH" 

+ HA. Its basicity constant is given by 

K b 





Multiplying the 

expressions for K a and K\, yields 

K*XK h 

[A-][H 3 0+] 


The HA and A" 

terms clearly 

cancel to yield [H: 

+ ][OH"], 

which is 

K w . 

Why does carbonic acid behave as a mono-protic acid? 

Variations in the value of K a 

Carbonic acid, H2CO3, is naturally occurring, and forms when carbon dioxide from 
the air dissolves in water. From its formula, we expect it to be a di-protic acid, but 
it is generally classed as mono-protic. Why? 

In water at 298 K, the ionization reaction follows the equation 

H2C03( aq ) + H 2 

HC0 3 - (aq) +H 3 0^ 



The value of K a for the reaction in Equation (6.37) is 4.3 x 10~ 7 , so carbonic acid is 
certainly a very weak acid. The hydrogen carbonate anion HC03~ could dissociate 
further, according to 



+ H 2 ► CO^ aq) + H 3 



but its value of K a is low at 5.6 x 10 -11 , so we conclude that the HCO^~ ion is too 
weak an acid to shed its proton under normal conditions. Thus, carbonic acid has two 
protons: the loss of the first one is relatively easy, but the proportion of molecules 
losing both protons is truly minute. Only one of the protons is labile. 

This situation is relatively common. If we look, for example, at the values of 
K a in Table 6.5, we see that phosphoric acid is a strong acid insofar as the loss of 
the first proton occurs with K a = 7.5 x 10~ 3 , but the loss of the second proton, to 
form HP0 4 ~, is difficult, as characterized by K a = 6.2 x 10~ 8 . In other words, the 
dihydrogen phosphate anion H2PO4 is a very weak acid. And the hydrogen phosphate 
di-anion HP0 4 ~ has a low a value of K a = 2.2 x 10 -11 , causing us to say that the 
P0 4 ~ anion does not normally exist. Even the loss of the second proton of sulphuric 
acid is characterized by a modest value of K a = 10~ 2 . 

This formal definition of K a can be extended to multi-protic 
acids. We consider the dissociation to occur in a step-wise manner, 
the acid losing one proton at a time. Consider, for example, the 
two-proton donation reactions of sulphuric acid: 




> H + + HSO4 
■* H+ + SO* _ 

Multi-protic acids have 
a different value of K a 
for each proton dona- 
tion step, with the 
values of K a decreas- 
ing with each proton 
donation step. 



The subscript (1) tells 
us we are considering 
the first proton to be 

The equilibrium constant for the dissociation of H2SO4: 

[HS0 4 -][H+] 




Ka ( i) is always bigger 
than K a(2) . 



[H 2 S0 4 ] 


In fact, we can also extend this treatment to bases, looking at 
the step-base addition of protons. 

SAQ 6.10 The tetra-protic acid H 4 EDTA (V) has four possible proton 
equilibrium constants. Write an expression for each, for K a{1) to K a(4) . 

Why is an organic acid such as trichloroethanoic acid 
so strong? 

Effect of structure on the K a of a weak acid 

The value of K a for trichloroacetic acid CCI3COOH (VIII) is very large at 0.23. 
Indeed, it is stronger as an acid than the HSO4 ion - quite remarkable for an organic 







Let us return to the example of ethanoic acid (I). The principal structural difference 
between I and VIII is the way we replace each of the three methyl protons in ethanoic 
acid with chlorine atoms. 

The three methyl protons in I are slightly electropositive, implying that the central 
carbon of the -CH3 group bears a slight negative charge. This excess charge is not 
large, but it is sufficient to disrupt the position of the acid-dissociation equilibrium, as 
follows. Although the undissociated acid has no formal charge, the ethanoate anion 
has a full negative charge, which is located principally on the carboxyl end of the 
anion. It might be easier to think of this negative charge residing on just one of the 
oxygen atoms within the anion, but in fact both oxygen atoms and 
the central carbon each bear some of the charge. We say the charge 
is delocalized, according to structure IX, which is a more accurate 
representation of the carboxylate anion than merely -COO~. The 

Derealization is a 
means of stabilizing 
an ion. 



right-hand structure of IX is effectively a mixture of the two structures to its left. 
Note how the name resonance implies charge derealization. 






A double-headed arrow 
'+*' indicates reso- 

Ions are more likely to 
form if they are stable, 
and less likely to form 
if unstable. 

To reiterate, the hydrogen atoms in the methyl group are slightly electropositive, 
with each seeking to relocate their own small amounts of charge onto the central 
carboxyl carbon. In consequence, the ethanoic anion (cf. structure 
IX with R = CH3) has a central carbon bearing a larger negative 
charge - both from the ionization reaction but also from the hydro- 
gen atoms of the -CH3 group. In consequence, the central carbon 
of the ethanoate anion is slightly destabilized; and any chemical 
species is less likely to form if it is unstable. 

Next we look at the structure of trichloroethanoic acid (VIII). In contrast to the 
hydrogen atoms of ethanoic acid, the three chlorine atoms are powerfully electron 
withdrawing. The chlorine atoms cause extensive derealization of the negative charge 
on the C^COO - anion, with most of the negative charge absorbed by the three 
chlorine atoms and less on the oxygen atoms of the carboxyl. Such a relocation of 
charge stabilizes the anion; and any chemical species is more likely to form if it 
is stable. 

Statistically, we find fewer ethanoate anions than trichloroethanoate anions in the 
respective solutions of the two acids. And if there are fewer ethanoate anions in 
solution per mole of ethanoic acid, then there will be fewer solvated protons. In other 
words, the extent to which ethanoic acid dissociates is less than the corresponding 
extent for trichloroethanoic acid. I is a weak acid and VIII is strong; dipping a simple 
pH electrode into a solution of each of the two acids rapidly demonstrates this truth. 

This sort of derealization stabilizes the ion; in fact, the ClsCOO - anion is more 
stable than the parent molecule, CI3COOH. For this reason, the solvated anion resides 
in solution in preference to the acid. K a is therefore large, making trichloroacetic acid 
one of the strongest of the common organic acids. 

Trifluoroethanoic acid (probably better known as trifluoroocef/c acid, TFA) is 
stronger still, with a value of K a = 1.70. 

6.3 Titration analyses 

iy does a dock leaf bring relief after 
a nettle sting? 

Introducing titrations 

We first met nettle stings on p. 253, where methanoic ('formic') 
acid was identified as the active toxin causing the pain. Like its 

The common dock leaf, 
Rumex obtusifolia, and 
the yellow dock leaf, 
Rumex crispus, are in 
fact equally common. 



The naturally occurring 
substance histamine 
causes blood capillaries 
to dilate and smooth 
muscle to contract. 
Most cells release it 
in response to wound- 
ing, allergies, and most 
inflammatory condi- 
tions. Antihistamines 
block the production of 
this substance, thereby 
combating a painful 

structurally similar sister, ethanoic acid (I), methanoic acid disso- 
ciates in water to yield a solvated proton, H30 + ( aq ). 

Rubbing the site of the sting with a crushed dock leaf is a sim- 
ple yet rapid way of decreasing the extent of the pain. In common 
with many other weeds, the sap of a dock leaf contains a mix- 
ture of natural amines (e.g. urea (III) above), as well as natural 
antihistamines to help decrease any inflammation. The amines are 
solvated and, because the sap is water based, are alkaline. Being 
alkaline, these amines react with methanoic acid to yield a neutral 
salt, according to 





:n — R 





H-,N — R 


where R is the remainder of the amine molecule. We see how the 
process of pain removal involves a neutralization process. 

Notice how the lone 
pair on nitrogen of the 
amine attracts a proton 
from the carboxylic 

How do indigestion tablets won 

Calculations concerning neutralization 

Excess acid in the stomach is one of the major causes of indiges- 
tion, arising from a difficulty in digesting food. The usual cause 
of such indigestion is the stomach simply containing too much hydrochloric acid, or 
the stomach acid having too high a concentration (its pH should be about 3). These 
failures cause acid to remain even when all the food has been digested fully. The 
excess acid is not passive, but tends to digest the lining of the stomach to cause an 
ulcer, or reacts by alternative reaction routes, generally resulting in 'wind', the gases 
of which principally comprises methane. 

Most indigestion tablets are made of aluminium or magnesium 
hydroxides. The hydroxide in the tablet removes the excess stom- 
ach acid via a simple acid-base neutralization reaction: 

Some indigestion 
tablets contain chalk 
(CaC0 3 ) but the large 
volume of C0 2 pro- 
duced (cf. Equation 
(6.19)) can itself cause 



+ Al(OH) 3 (table t ) ► 3H 2 + A1C1 3 



The cause of the indigestion is removed because the acid is con- 
sumed. Solid (unreacted) aluminium hydroxide is relatively insolu- 
ble in the gut, and does not dissolve to generate an alkaline solution. 
Rather, the outer layer of the tablet dissolves slowly, with just sufficient entering solu- 
tion to neutralize the acid. Tablet dissolution stops when the neutralization reaction 
is complete. 



A similar process occurs when we spread a thick paste of zinc and 
castor oil on a baby's bottom each time we change its nappy. The 
'zinc' is in fact zinc oxide, ZnO, which, being amphoteric, reacts 
with the uric acid in the baby's urine, thereby neutralizing it. 

Worked Example 6.9 But how much stomach acid is neutralized 
by a single indigestion tablet? The tablet contains 0.01 mol of MOH, 
where 'M' is a monovalent metal and M + its cation. 

We first consider the reaction in the stomach, saying it proceeds with 
1 mol of hydrochloric acid reacting with 1 mol of alkali: 

Some campaigners 
believe the AICI 3 pro- 
duced by Equation 
(6.40) hastens the 
onset of Alzheimer's 
disease. Certainly, the 
brains of people with 
this nasty condition 
contain too much alu- 

MOH (s) + H 3 H 


► M+ (aq) + 2H 2 


M + is merely a cation. We say Equation (6.41) is a 1:1 reaction, occurring with a 1:1 
stoichiometry. Such a stoichiometry simplifies the calculation; the 3:1 stoichiometry in 
Equation (6.40) will be considered later. 

From the stoichiometry of Equation (6.41), we say the neutralization is complete 
after equal amounts of acid and alkali react. In other words, we neutralize an amount 
n of hydrochloric acid with exactly the same amount of metal hydroxide, i.e. with 
1 x 10" 2 mol. 

The tablet can neutralize 0.01 mol of stomach acid. 

This simple calculation illustrates the fundamental truth under- 
lying neutralization reactions: complete reaction requires equal 
amounts of acid and alkali. In fact, the primary purpose of a 
titration is to measure an unknown amount of a substance in 
a sample, as determined via a chemical reaction with a known 
amount of a suitable reagent. We perform the titration to ascer- 
tain when an equivalent amount of the reagent has been added to 
the sample. When the amount of acid and alkali are just equal, 
we have the equivalence point, from which we can determine the 
unknown amount. 

In a typical titration experiment, we start with a known volume 
of sample, call it V( sam pie)- If we know its concentration c (samp i e ), 
we also know the amount of it, as V( sam pie) x C( sam pie) • During the 
course of the titration, the unknown reagent is added to the solu- 
tion, usually drop wise, until the equivalence point is reached (e.g. 
determining the endpoint by adding an indicator; see p. 273ff). 
At equivalence, the amounts of known and unknown reagents are 
the same, so n( sam pie) = w (unknown)- Knowing the amount of sample 
and the volume of solution of the unknown, we can calculate the 
concentration of the unknown. 

The experimental tech- 
nique of measuring out 
the amount of acid and 
alkali needed for neu- 
tralization is termed a 

The amounts of acid 
and alkali are equal at 
the equivalence point. 
The linguistic similar- 
ity between these two 
words is no coinci- 

We need equal num- 
bers of moles of acid 
and alkali to effect 



Worked Example 6.10 The methanoic acid from a nettle sting is extracted into 50 cm 3 
of water and neutralized in the laboratory by titrating with sodium hydroxide solution. 
The concentration of NaOH is 0.010 mol dm -3 . The volume of NaOH solution needed to 
neutralize the acid is 34.2 cm 3 . What is the concentration c of the acid? 

Unlike the Worked Example 6.9, we do not know the number of moles n of either reactant, 
we only know the volumes of each. But we do know one of the concentrations. 

Answer strategy. (1) First, we calculate the amount n of hydroxide required to neutralize 
the acid. (2) We equate this amount n with the amount of acid neutralized by the alkali. 
(3) Knowing the amount of acid, we finally calculate its concentration. 

(1) To determine the amount of alkali, we first remember the defi- 
nition of concentration c as 

concentration, c ■■ 

amount, n 
volume, V 


We could have achiev- 
ed this conversion 
with quantity calcu- 
lus: knowing there are 
1000 cm 3 per dm 3 (so 
10 3 dm 3 cnrr 3 ). In SI 
units, we write the 
volume as 34.2 cm 3 x 
10- 3 dm 3 crrr 3 . The 
units of cm 3 and cnrr 3 
cancel to yield V = 
0.0342 dm 3 . 

Notice now the units of 
dm 3 and drrr 3 cancel 
out here. 

Notice how we con- 
verted the volume of 
acid solution (50 cm 3 ) 
to 0.05 dm 3 . 

and rearrange Equation (6.41) to make amount n the subject, i.e. 

n — c x V 

The volume V of alkali is 34.2 cm . As there are 1000 cm in a litre, 
this volume equates to (34.2 4- 1000) dm 3 = 0.0342 dm 3 . Accordingly 

n = 0.0342 dm 3 x 0.010 mol dm -3 

3.42 x 10" 4 mol 

(2) The reaction between the acid and alkali is a simple 1:1 reac- 
tion, so 3.42 x 10~ 4 mol of alkali reacts with exactly 3.42 x 10~ 4 mol 
of acid. 

(3) The concentration of the acid is given by Equation (6.42) again. 
Inserting values: 

concentration c 

3.42 x 10" 4 mol 

0.05 dm 3 
c = 6.8x 10~ 3 mol dm" 3 

After extraction, the concentration of the methanoic acid is 0.068 mol 
dm" 3 . 

An altogether simpler and quicker way of calculating the concentration of an acid 
during a titration is to employ the equation 

C(acid) X V(acid) = C(alkali) X V(alkali) (6.43) 

where the V terms are volumes of solution and the c terms are concentrations. 



Worked Example 6.11 A titration is performed with 25 cm of NaOH neutralizing 

29.4 cm 3 of nitric acid. The concentration of NaOH is 0.02 moldm 
concentration of the acid. 

. Calculate the 

We rearrange Equation (6.43), to make C( ac id) the subject: 

C(alkali) ^(alkali) 

C(acid) — ~ 


We then insert values into Equation (6.44): 

0.02 moldm 3 x 25cm 3 




29.4 cm 3 
0.017 moldm -3 

We obtain here a ratio 
of volumes (l/ (a ikaii) h- 
l/(add)), enabling us to 
cancel the units of the 
two volumes. Units 
are irrelevant if both 
volumes have the same 

Justification Box 6.4 

A definition of the point of 'neutralization' in words says, 'at the neutralization point, 
the number of moles of acid equals the number of moles of hydroxide'. We re-express 
the definition as 

«(acid) = "(alkali) (6.45) 

Next, from Equation (6.42), we recall how the concentration of a solution c when 
multiplied by its respective volume V equals the number of moles of solute: n = c x V. 
Clearly, «( ac ;d) = C( ac id) x V( ac id), and n(aikaii) = C( a ik a u) x V( a ikaii) • 

Accordingly, substituting for «( ac ;d) 
tion (6.43). 

and n 


into Equation (6.45) yields Equa- 

SAQ 6.11 What volume of NaOH (of concentration 0.07 moldm ) is 
required to neutralize 12 cm 3 of nitric acid of concentration 0.05 mol 
drrr 3 ? 


Equation (6.43) is a simplified version of a more general equation: 

C (alkali) ^(alkali) 

C(acid) = S - 




where s is the so-called stoichiometric ratio. 

For the calculation of a mono-protic acid with a mono-basic base, the stoichiometry 
is simply 1 : 1 because 1 mol of acid reacts with 1 mol of base. We say the stoichiometric 
ratio s = 1 . The value of s will be two if sulphuric acid reacts with NaOH since 2 mol 
of base are required to react fully with 1 mol of acid. For the reaction of NaOH with 
citric acid, s = 3; and s = 4 if the acid is H4EDTA. 



The value of s when Ca(OH) 2 reacts with HNO3 will be j, and the value when citric 
acid reacts with Ca(OH) 2 will be |. 


SAQ 6.12 What volume of Ca(OH) 2 (of concentration 0.20 moldrrr J ) 
is required to neutralize 50 cm 3 of nitric acid of concentration 0.10 mol 
dm" 3 ? 

An alternative way of determining the endpoint of a titration 
is to monitor the pH during a titration, and plot a graph of pH 
(as 'y') against volume V of alkali added (as 'x'). Typically, the 
concentration of the acid is unknown, but we know accurately the 
concentration of alkali. Figure 6.3 shows such as graph - we call 
it a pH curve - in schematic form. The shape is sigmoidal, with 
the pH changing very rapidly at the end point. 

In practice, we obtain the end point by extrapolating the two 
linear regions of the pH curve (the extrapolants should be parallel). 
A third parallel line is drawn, positioned exactly midway between 
the two extrapolants. The volume at which this third line crosses the 
pH curve indicates the end point. Knowing the volume V( en d p0 mt)> 

we can calculate the concentration of the acid via a calculation similar to Worked 

Example 6.11. 

Incidentally, the end point also represents the volume at which the pH changes most 

dramatically, i.e. the steepest portion of the graph. For this reason, we occasionally 

plot a different graph of gradient (as 'y') against volume V (as 'x'); see Figure 6.4. 

We obtain the gradient as 'ApH -j- AV. The end point in Figure 6.4 relates to the 

graph maximum. 

Sigmoidal literally 
means 'shaped like 
a Greek sigma g'. The 
name derives from the 
Greek word sigmoides, 
meaning 'sigma-like'. 
(There are two Greek 
letters called sigma, 
used differently in word 
construction. The other 
has the shape a.) 

Volume of alkali added V 

Figure 6.3 A schematic pH curve for the titration of a strong acid with a strong base. At the 
equivalence point, the amount of alkali added is the same as the amount of acid in solution 
initially, allowing for an accurate calculation of the acid's concentration. Note how the end 
point is determined by extrapolating the linear regions, and drawing a third parallel line between 



Volume of alkali added V 

Figure 6.4 A schematic of the first derivative of the pH curve in Figure 6.3. The end-point volume 
is determined as the volume at the peak. A first derivative plot such as this can yield a more accurate 
end point than drawing parallel lines on Figure 6.3 

6.4 pH buffers 

Why does the pH of blood not alter after eating pickle? 

Introduction to buffers 

A 'pickle' is a food preserved in vinegar (ethanoic acid). Pickles generally have a sharp, 
acidic flavour in consequence of the acid preservative. Many systems - especially living 
cells - require their pH to be maintained over a very restricted range in order to prevent 
catastrophic damage to the cell. Enzymes and proteins denature, for example, if the 
pH deviates by more than a fraction. Traces of the food we eat are readily detected in 
the blood quite soon after eating, so why does the concentration in the blood remain 
constant, rather than dropping substantially with the additional acid in our diet? 

Before we attempt an answer, look again at Figure 6.3, which clearly shows an 
almost invariant pH after adding a small volume of alkali. Similarly, at the right- 
hand side of the graph the pH does not vary much. We see an insensitivity of the 
solution pH to adding acid or alkali; only around the end point does the pH alter 
appreciably. The parts of the titration graph having an invariant pH are termed the 
buffer regions, and we call the attendant pH stabilization a buffer action. 

In a similar way, blood does not change its pH because it contains suitable concen- 
trations of carbonic acid and bicarbonate ion, which act as a buffer, as below. 

Why are some lakes more acidic than 

Buffer action 

Acid rain is the major cause of acidity in open-air lakes and ponds 
(see p. 237). Various natural oxides such as CO2 dissolve in water 

Pollutant gases include 
S0 2 , S0 3 , NO and N0 2 . 
It is now common to 
write SO x and NO x 
to indicate this vari- 
able valency within the 



Remember: 'weak' in 
this sense indicates the 
extent to which a weak 
acid dissociates, and 
does not relate to its 

to generate an acid, so the typical pH of normal rainwater is about 5.6; but rainwater 
becomes more acidic if pollutants, particularly SO v and NOjc, in addition to natural 
CO2, dissolve in the water. As an example, the average pH of rain in the eastern United 
States of America (which produces about one-quarter of the world's pollution) lies in 
the range 3.9-4.5. Over a continental landmass, the partial pressure of SO2 can be 
as high as 5 x 10~ 9 x p & , representing a truly massive amount of pollution. 

After rainfall, the pH of the water in some lakes does not change, 

whereas others rapidly become too acidic to sustain aquatic life. 

Why? The difference arises from the buffering action of the water. 

Some lakes resist gross change in pH because they contain other 

chemicals that are able to take up or release protons into solution 

following the addition of acid (in the rain). These chemicals in 

the lake help stabilize the water pH, to form a buffer. Look at 

Figure 6.5, which shows a pH curve for a weak acid titrated with 

an alkali. Figure 6.5 is clearly similar to Figure 6.3 after the end-point volume, but it 

has a much shallower curve at lower volumes. In fact, we occasionally have difficulty 

ascertaining a clear end point because the curvature is so pronounced. 

A buffer comprises (1) a weak acid and a salt of that acid, (2) a weak base and a 
salt of that base, or (3) it may contain an acid salt. We define an acid-base buffer as 
'a solution whose pH does not change after adding (small amounts 
of) a strong acid or base'. Sodium ascorbate is a favourite buffer 
in the food industry. 

We can think of water entering the lake in terms of a titration. 
A solution of alkali enters a fixed volume of acid: the alkaline 
solution is water entering from the lake's tributary rivers, and the 
acid is the lake, which contains the weak acid H2CO3 (carbonic 
acid) deriving from atmospheric carbon dioxide. The alkali in the 
tributary rivers is calcium hydroxide Ca(OH) 2 , which enters the 

A buffer is a solution 
of constant pH, which 
resists changes in pH 
following the addition 
of small amounts of 
acid or alkali. 


Half volume 
at the end point 

Volume of alkali added V 

Figure 6.5 A typical pH curve for the titration of carbonic acid (a weak acid) with a strong base. 
The concentration of H2CO3 and HCO^ are the same after adding half the neutralization volume 
of alkali. At this point, pH = pK 3 



water as the river passes over the limestone floor of river basins. 
Calcium hydroxide is a fairly strong base. 

Figure 6.5 shows a buffering action since the pH does not change 
particularly while adding alkali to the solution. In fact, as soon as 
the alkali mixes with the acid in the lake, its hydroxide ions are 
neutralized by reaction with solvated protons in the lake, thereby 
resisting changes in the pH. Figure 6.5 shows how little the lake pH 
changes; we term the relatively invariant range of constant pH the 
buffer region of the lake water. The mid pH of the buffer region cor- 
responds quite closely to the pK a of the weak acid (here H2CO3), 
where the pK a is a mathematical function of K a , as defined by 

Limestone or chalk dis- 
solve in water to a lim- 
ited extent. The CaC0 3 
decomposes naturally 
to form Ca(OH) 2 , 
thereby generating 
alkaline water. 

pK a = - log 10 K a 


The natural buffers in 
the lake 'mop up' any 
additional alkali enter- 
ing the lake from the 
tributary rivers, thereby 
restricting any changes 
to the pH. 

A buffer is only really 
effective at restrict- 
ing changes if the pH 
remains in the range 
p/C a ±l. 

As a good generalization, the buffer region extends over the range 
0fp^ a ±l. 

Only when all the acid in the lake has been consumed will the pH 
rise significantly. In fact, the end point of such a titration is gauged 
when the pH rises above pH 7, i.e. the pH of acid-base neutrality. 

The pH of the lake water fluctuates when not replenished by the 
alkaline river water in the tributary rivers. In fact, the pH of the 
lake water drops significantly each time it rains (i.e. when more 
H2CO3 enters the lake). If the amounts of acid and alkali in the 
water remain relatively low, then the slight fluctuations in water pH will not be great 
enough to kill life forms in the lake. 

The system above describes the addition of alkali to a lake containing a weak 
acid. The reverse process also occurs, with acid being added to a base, e.g. when the 
tributary rivers deliver acid rain to a lake and the lake basin is made of limestone or 
chalk. In such a case, the lake pH drops as the acid rain from the rivers depletes the 
amounts of natural Ca(OH) 2 dissolved in the lake. 

As a further permutation, adding a strong acid to a weak base also yields a buffer 
solution, this time with a buffer region centred on the pK a of the base. The pH at the 
end point will be lower than 7. 


Each species within a buffer solution participates in an equilibrium reaction, as char- 
acterized by an equilibrium constant K. Adding an acid (or base) to a buffer solution 
causes the equilibrium to shift, thereby preventing the number of 
protons from changing, itself preventing changes in the pH. The 
change in the reaction's position of equilibrium is another mani- 
festation of Le Chatelier's principle (see p. 166). 

One of the most common buffers in the laboratory is the so- 
called 'phosphate buffer', which has a pH of 7.0. It comprises salts 

A buffer is a solution 
of a weak acid mixed 
with its conjugate base, 
which restricts changes 
to the pH. 



of hydrogen phosphate and dihydrogen phosphate, in the following equilibrium: 

H 2 P0 4 2 - 

— > 


+ H+ 


conjugate acid 

conjugate base 

If this example were to proceed in the reverse direction, then the hydrogen phosphate 
(on the right) would be the base, since it receives a proton, and the dihydrogen 
phosphate (on the left) would be the conjugate the acid. 

The equilibrium constant of the reaction in Equation (6.48) is given by 

K„ = 

[H 2 P07] 


Notice how the equilibrium constant K in Equation (6.49) is also an acidity constant, 
hence the subscripted 'a'. The value of K remains constant provided the temperature 
is not altered. 

Now imagine adding some acid to the solution - either by mistake or deliberately. 
Clearly, the concentration of H + will increase. To prevent the value of K A changing, 
some of the hydrogen phosphate ions combine with the additional protons to form di- 
hydrogen phosphate (i.e. Equation (6.48) in reverse). The position of the equilibrium 
adjusts quickly and efficiently to 'mop up' the extra protons in the buffer solution. In 
summary, the pH is prevented from changing because protons are consumed. 

How do we make a 'constant-pH solution'? 

The Henderson - Hasselbach equation 

We often need to prepare a solution having a constant pH. Such solutions are vital in 
the cosmetics industry, as well as when making foodstuffs and in the more traditional 
experiments performed by the biologist and physical chemist. 

To make such a solution, we could calculate exactly how many moles of acid to add 
to water, but this method is generally difficult, since even small errors in weighing 
the acid can cause wide fluctuations in the pH. Furthermore, we cannot easily weigh 
out one of acid oxides such as NO. Anyway, the pH of a weak acid does not clearly 
follow the acid's concentration (see p. 254). 

The Henderson-Hasselbach equation, Equation (6.50), relates 

the pH of a buffer solution to the amounts of conjugate acid and 

conjugate base it contains: 

In some texts, Equa- 
tion (6.50) is called the 
HasselbaLch equation. 

pH = pK a + log 




We follow the usual pattern here by making a buffer with a weak acid HA and a 
solution of its conjugate base, such as the sodium salt of the respective anion, A - . 


We can prepare a buffer of almost any pH provided we know the pK a of the acid; and 
such values are easily calculated from the K d values in Table 6.5 and in most books 
of physical chemistry and Equation (6.50). We first choose a weak acid whose pK^ 
is relatively close to the buffer pH we want. We then need to measure out accurately 
the volume of acid and base solutions, as dictated by Equation (6.50). 

Worked Example 6.12 We need to prepare a buffer of pH 9.8 by mixing solutions of 
ammonia and ammonium chloride solution. What volumes of each are required? Take 
the K & of the ammonium ion as 6 x 10~ 10 . Assume the two solutions have the same 
concentration before mixing. 

Strategy: (1) We calculate the pAT a of the acid. (2) We identify which component is the 
acid and which the base. (3) And we calculate the proportions of each according to 
Equation (6.50). 

(1) From Equation (6.50), we define the pK. A as — log 10 K^. Inserting values, we obtain 
a p# a of 9.22. 

(2) The action of the buffer represents the balanced reaction, NH4CI — > NH3 + HC1, 
so NH4CI is the acid and NH3 is the base. 

(3) To calculate the ratio of acid to base, we insert values into Equation (6.50): 

, [NH 

= 9.2 + log 10 

0.6 = log 10 


[NH 3 ] 

Taking antilogs of both sides to remove the logarithm, we obtain 


,,,.„ _ [NH 3 ] 
" [NH+] 
[NH 3 ] 


= 4 

We are permitted to 
calculate a ratio like 
this if the concen- 
trations of acid and 
conjugate base are the 

So, we calculate the buffer requires four volumes of ammonia solution 
to one of ammonium (as the chloride salt, here). 

SAQ 6.13 What is the pH of ammonia-ammonium buffer if three vol- 
umes of NH4CI are added to two volumes of NH 3 ? 

We want a buffer solution because its pH stays constant after adding small amounts 
of acid or base. Consider the example of adding hydrochloric acid to a buffer, as 
described in the following Worked Example. 

Worked Example 6.13 Consider the so-called 'acetate buffer', made with equal vol- 
umes of sodium ethanoate and ethanoic acid solutions. The concentration of each solu- 
tion is 0.1 moldnT . A small volume (10 cm ) of strong acid (HC1 of concentration 
1 moldm^ 3 ) is added to a litre of this buffer. The pH before adding HC1 is 4.70. What 
is its new pH? 



Strategy: we first calculate the number of moles of hydrochloric 
acid added. Second, we calculate the new concentrations of ethanoic 
acid and ethanoate. And third, we employ the Henderson-Hasselbach 
equation once more. 

(1) 10 cm 3 represents one-hundredth of a litre. From 
Equation (6.42), the number of moles is 0.01 mol. 

(2) Before adding the hydrochloric acid, the concentrations 
of ethanoate and ethanoic acid are constant at 0.1 mol dm" 3 . The 
hydrochloric acid added reacts with the conjugate base in the buffer 

(the ethanoate anion) to form ethanoic acid. Accordingly, the concentration [CH^COO - ] 
decreases and the concentration [CH3COOH] increases. (We assume the reaction is 
quantitative.) Therefore, the concentration of ethanoate is (0.1 — 0.01) mol dnT = 
0.09 mol dm" 3 . The concentration of ethanoic acid is (0.1 + 0.01) mol dnT 3 = 
0.11 mol dm" 3 . 

(3) Inserting values into Equation (6.50): 

The acetate buffer is 
an extremely popu- 
lar choice in the food 
industry. The buffer 
might be described on 
a food packet as an 
acidity regulator. 

pH = 4.70 + log 10 




pH = 4.70 + log (0.818) 
pH = 4.70 + (-0.09) 

pH = 4.61 

So, we see how the pH shifts by less than one tenth of a pH unit after adding quite 
a lot of acid. Adding this same amount of HC1 to distilled water would change the 
pH from 7 to 2, a shift of five pH units. 

SAQ 6.14 Consider the ammonia -ammonium buffer in Worked Exam- 
ple 6.12. Starting with 1 dm 3 of buffer solution containing 0.05 mol dm" 3 
each of NH 3 and NH 4 CI, calculate the pH after adding 8 cm 3 of NaOH 
solution of concentration 0.1 mol dm" 3 . 

Justification Box 6.5 

We start by 


the equilibrium constant for a weak acid HA dissociating in water, 

HA + H 2 

-> H 3 

*" + A" , where each ion is solvated. The dissociation constant for 

the acid K a 

is given 

by Equation (6.35): 

„ [H 3 0+][A-] 

K a = 


where, as usual, we 

ignore the water term. Taking logarithms of 

Equation (6.35) yields 

v , [H 3 0+][A-] 

lOglQ ^a = lOg 10 




We can split the fraction term in Equation (6.51) by employing 

the laws of logarithms, 

to yield 

+ [A - ] 
logio K * = log 10 [H 3 CT] + log 10 


The term Tog| K^' should remind us of pA" a (Equation 6.52), and the term log 10 [H3O + ] 

will remind us of 

pH in Equation (6.20), so we rewrite Equation (6.52) as 

[A - ] 

p^ a = pH + log 10 


which, after a 

little rearranging, yields the Henderson- 



Equation (6.50). 

6.5 Acid- base indicators 

What is l the litmus test'? 

pH indicators 

Litmus is a naturally occurring substance obtained from lichen. It 
imparts an intense colour to aqueous solutions. In this sense, the 
indicator is a dye whose colour is sensitive to the solution pH. 
If the solution is rich in solvated protons (causing the pH to be 
less than 7) then litmus has an intense red colour. Conversely, a 
solution rich in hydroxide ions (with a pH greater than 7) causes 
the litmus to have a blue colour. 

To the practical chemist, the utility of litmus arises from the way 
its colour changes as a function of pH. Placing a single drop of 
litmus solution into a beaker of solution allows us an instant test of 
the acidity (or lack of it). It indicates whether the pH is less than 
7 (the litmus is red, so the solution is acidic), or the pH is greater 
than 7 (the litmus is blue, so the solution is alkaline). Accordingly, 
we call litmus a pH indicator. 

In practical terms, we generally employ litmus during a titration. 
The flask will contain a known volume of acid of unknown con- 
centration, and we add alkali from a burette. We know we have 
reached neutralization when the Litmus changes from red (acid 
in excess) and just starts changing to blue. We know the pH of 
the solution is exactly 7 when neutralization is complete, and then 
note the volume of the alkali, and perform a calculation similar to 
Worked Example 6.11. 

The great English scientist Robert Boyle (1627-1691) was the 
first to document the use of natural vegetable dyes as acid-base indicators. 

The name 'litmus' 
comes from the Old 
Norse litmosi, which 
derives from litr and 
mosi, meaning dye and 
moss respectively. 

Much of the litmus 
in a laboratory is 
pre-impregnated on 
dry paper. 

Litmus is an indicator. 
To avoid ambiguity, 
we shall call it an 
'acid -base indicator' 
or a "pH indicator'. 

Litmus often looks 
purple-grey at the 
neutralization point. 
This colour tells us we 
have a mixture of both 
the red and blue forms 
of litmus. 



Why do some hydrangea bushes look red and others 

The chemical basis of acid -base indicators 


The name 'hydrangea' 
derives from classi- 
cal Greek mythol- 
ogy, in which the 
'hydra' was a beast 
with many heads. 

Hydrangeas (genus Hydrangea) are beautiful bushy plants having 
multiple flower heads. In soils comprising much compost the flow- 
ers have a blue colour, but in soils with much lime or bone meal 
the heads are pink or even crimson-pink in colour. Very occa- 
sionally, the flowers are mauve. 'Lime' is the old-fashioned name 
for calcium oxide, and is alkaline; bone meal contains a lot of 
phosphate, which is also likely to raise the soil pH. The colour of 
the hydrangea is therefore an indication of the acid content of the 
soil: the flower of a hydrangea is blue in acidic soil because the 
plant sap is slightly acidic; red hydrangeas exist in alkaline soil 
because the sap transports alkali from the soil to the petals. The 
rare mauve hydrangea indicates a soil of neutral pH. We see how 
the chromophore in the flower is an acid -base indicator. 

The chromophore in hydrangeas is delphinidin (X), which is a 
member of the anthrocyanidin class of compounds. Compound X 
reminds us of phenol (VII), indicating that delphinidin is also a 
weak acid. In fact, all pH indicators are weak acids or weak bases, 
and the ability to change colour is a visible manifestation of the 
indicator's ability to undergo reversible changes in structure. In the 
laboratory, only a tiny amount of the pH indicator is added to the 
titration solution, so it is really just a probe of the solution pH. It 
does not participate in the acid -base reaction, except insofar as its 
own structure changes with the solution pH. 
As an example, whereas the anthracene-based core of molecular X is relatively inert, 

the side-chain 'X' is remarkably sensitive to the pH of its surroundings (principally, 

to the pH of the solution in which it dissolves). 

The word 'chromo- 
phore' comes from two 
Greek words, ^khro- 
mos' meaning colour 
and 'prtoro', which 
means 'to give' or 'to 
impart'. A chromophore 
is therefore a species 
imparting colour. 

All pH indicators are 
weak acids or weak 

Figure 6.6 shows the structure of the side substituent as a function of pH. 

The hydroxyl group placed para to the anthracene core is protonated in acidic 
solutions (i.e. when the hydrangea sap is slightly acidic). The proton is abstracted in 
alkaline sap, causing molecular rearrangement to form the quinone moiety. 






Figure 6.6 Anthrocyanidins impart colour to many natural substances, such as strawberries and 
cherries. The choice of side chains can cause a huge change in the anthrocyanidin's colour. If the 
side chain is pH sensitive then the anthrocyanidin acts as an acid-base indicator: structures of an 
anthrocyanidin at three pHs (red in high acidity and low pH, blue in low acidity and high pH and 
mauve in inter-midiate pHs) 

It is also astonishing how the rich blue of a cornflower (Centaurea cyanus) and 
the majestic red flame of the corn poppy (Papaver rheas) each derive from the same 
chromophore - again based on an anthrocyanidin. The pH of cornflower and poppy 
sap does not vary with soil composition, which explains why we see neither red 
cornflowers nor blue corn poppies. 


It is fascinating to appreciate the economy with which nature produces colours (ele- 
mentary colour theory is outlined in Chapter 9). The trihydroxyphenyl group of the 
anthrocyanidin (X) imparts a colour to both hydrangeas and delphiniums. The dihydrox- 
yphenyl group (XI) is remarkably similar, and imparts a red or blue colour to roses, 
cherries and blackberries. The singly hydroxylated phenyl ring in XII is the chromophore 
giving a red colour to raspberries, strawberries and geraniums, but it is not pH sensitive. 






Why does phenolphthalein indicator not turn red until 
pH 8.2? 

Which acid -base indicator to use? 

Litmus was probably the most popular choice of acid-base indicator, but it is not a 
good choice for colour-blind chemists. The use of phenolphthalein as an acid-base 
indicator comes a close second. Phenolphthalein (XIII) is another weak organic acid. 
It is not particularly water soluble, so we generally dissolve it in aqueous ethanol. 
The ethanol explains the pleasant, sweet smell of phenolphthalein solutions. 



Phenolphthalein is colourless and clear in acidic solutions, but imparts an intense 
puce pink colour in alkaline solutions of higher pH, with A( max ) = 552 nm. The 
coloured form of phenolphthalein contains a quinone moiety; in fact, any chromophore 
based on a quinone has a red colour. But if a solution is prepared at pH 7 (e.g. as 
determined with a pH meter), we find the phenolphthalein indicator is still colour- 
less, and the pink colour only appears when the pH reaches 8.2. Therefore, we have 
a problem: the indicator has not detected neutrality, since it changes colour at too 

Table 6.7 Some common pH indicators, their useful pH ranges and 
the changes in colour occurring as the pH increases. An increasing 
pH accompanies a decreasing concentration of the solvated proton 


pH range 

Colour change 

Methyl violet 


Yellow — ¥ blue 

Crystal violet 


Yellow -*■ blue 



Red -*■ blue 

Methyl orange 


Red — ► yellow 

Ethyl red 


Colourless — ► red 

Alizarin red S 


Yellow — ► red 



Colourless — > yellow 



Colourless — > pink 


high a value of pH. However, the graph in Figure 6.3 shows how the pH of the 
titration solution changes dramatically near the end point: in fact, only a tiny incre- 
mental addition of alkali solution is needed to substantially increase the solution pH 
by several pH units. In other words, a fraction of a drop of alkali solution is the only 
difference between pH 7 (at the true volume at neutralization) and pH 8 when the 
phenolphthalein changes from colourless to puce pink. 

Table 6.7 lists the pH changes for a series of common pH indicators. The colour 
changes occur over a wide range of pHs, the exact value depending on the indicator 
chosen. Methyl violet changes from yellow to blue as the pH increases between and 
1.6. At the opposite extreme, phenolphthalein responds to pH changes in the range 
8.2 to 10. 




This chapter commences by describing cells and redox chemistry. Faraday's laws of 
electrolysis describe the way that charge and current passage necessarily consume 
and produce redox materials. The properties of each component within a cell are 
described in terms of potential, current and composition. 

Next, the nature of half-cells is explained, together with the necessary thermody- 
namic backgrounds of the theory of activity and the Nernst equation. 

In the final sections, we introduce several key electrochemical applications, such 
as the pH electrode (a type of concentration cell), nerve cells (which rely on junction 
potentials) and batteries. 

7.1 Introduction to cells: terminology 
and background 

Why does putting aluminium foil in the mouth cause 


Introduction to electrochemistry 

Most people have at some time experienced a severe pain in their teeth after acci- 
dentally eating a piece of sweet wrapper. Those teeth that hurt are usually nowhere 
near the scrap of wrapper. The only people who escape this nasty sensation are those 
without metal fillings in their teeth. 

The type of sweet wrapper referred to here is generally made of aluminium metal, 
even if we call it 'silver paper'. Such aluminium dissolves readily in acidic, conductive 
electrolytes; and the pH of saliva is about 6.5-7.2. 

The dissolution of aluminium is an oxidative process, so it gener- 
ates several electrons. The resultant aluminium ions stay in solution 
next to the metal from which they came. We generate a redox cou- 
ple, which we define as 'two redox states of the same material'. 

Two redox states of 
the same material are 
called a redox couple. 



The word 'amalgam' 
probably comes from 
the Greek malagma 
meaning 'to make 
soft', because a metal 
becomes pliable when 
dissolved in mercury. 
Another English word 
from the same root is 

A cell comprises two 
or more half-cells in 
contact with a common 
electrolyte. The cell is 
the cause of the pain. 

While it feels as though all the mouth fills with this pain, in fact 
the pain only manifests itself through those teeth filled with metal, 
the metal being silver dissolved in mercury to form a solid - we 
call it a silver amalgam. Corrosion of the filling's surface causes it 
to bear a layer of oxidized silver, so the tooth filling also represents 
a redox couple, with silver and silver oxide coexisting. 

An electrochemical cell is defined as 'two or more half-cells in 
contact with a common electrolyte'. We see from this definition 
how a cell forms within the mouth, with aluminium as the more 
positive pole (the anode) and the fillings acting as the more negative 
pole (the cathode). Saliva completes this cell as an electrolyte. All 
the electrochemical processes occurring are contained within the 
boundaries of the cell. 

Oxidation proceeds at the anode of the cell according to 






+ 3e" 


Oxidation reactions 
occur at the anode. 

Reduction reactions 
occur at the cathode. 

and occurs concurrently with a reduction reaction at the cathode: 

Ag 2 (s) + 2e" ► 2Ag° (s) + O 2 " (7.2) 

The origins of the words 'anode' and 'cathode' tell us much. 
'Anode' comes from the Greek words ana, meaning 'up', and 
hodos means 'way' or 'route', so the anode is the electrode to which electrons travel 
from oxidation, travelling to higher energies (i.e. energetically 'uphill'). The word 
'cathode' comes from the Greek hodos (as above), and cat meaning 'descent'. The 
English word 'cascade' comes from this same source, so a cathode is the electrode 
to which the electrons travel (energetically downhill) during reduction. 

The oxidation and reduction reactions must occur concurrently because the electrons 
released by the dissolution of the aluminium are required for the reduction of the 
silver oxide layer on the surface of the filling. For this reason, we need to balance 
the two electrode reactions in Equations (7.1) and (7.2) to ensure the same number of 
electrons appear in each. The pain felt at the tooth's nerve is a response to this flow 
of electrons. The paths of electron flow are depicted schematically in Figure 7.1. 

Each electron has a 'charge' Q. When we quantify the number of electrons produced 
or consumed, we measure the overall charge flowing. Alternatively, we might measure 
the rate at which the electrons flow (how many flow per unit time, t): this rate is 
termed the current /. Equation (7.3) shows the relationship between current / and 
charge Q: 




So ultimately the pain we feel in our teeth comes from a flow of current. 



Aluminium metal 

Aluminium ions 


Silver metal ► Tooth interior 

and nerve 

- electrons 

+ electrons 

Silver ions 


('silver foil') 




(tooth filling) 

Figure 7.1 Schematic illustration of the electron cycles that ultimately cause a sensation of pain 
in the teeth in people who have metallic fillings and who have inadvertently eaten a piece of 
aluminium ('silver') foil, e.g. while eating sweets 

Why does an electric cattle prod cause pain? 

The magnitude of a current 

An electric cattle prod looks a little like a walking stick with an attached battery. A 
potential from the tip of the stick is applied to a cow's flank, and the induced current 
hurts the animal. The cow moves where prompted to avoid a reapplication of the 
pain, thereby simplifying the job of cowherding. 

Ohm's law says that applying a potential V across an electrical resistor R induces 
a proportional current /. We can state this relationship mathematically as 

V = IR 


It is reported that the great Victorian scientist Michael Faraday 
discovered a variation of Ohm's law some 20 years before Ohm 
himself published the law that today bears his name. Faraday, with 
his typical phlegm and ingenuity, grasped a resistor between his 
two hands, immersed them both in a bowl of tepid water, and 
applied a voltage between them. It hurt. He found the empirical 
relationship 'pain oc I x R\ 

The pain Faraday induced was in direct proportion to the magni- 
tude of the current passed. He discovered the principle underlying 
the action of an electric cattle prod. It is sobering to realize how 
Faraday's result is today employed through most of the world as 
the basis of torture. Despite the explicit banning of torture by the 
UN Charter on Human Rights (of 1948), it is common knowledge that giving an 
electric shock is one of the most effective known means of causing pain. 

In summary, the cattle prod causes pain because of the current formed in response 
to applying a voltage. 

The word 'empirical' 
implies a result derived 
from experiment rather 
than theory. Similarly, 
a chemist calculates a 
compound's 'empirical 
formula' while fully 
aware its value is based 
on experiment rather 
than theory. 



What is the simplest way to clean a tarnished silver 

Electrochemistry: the chemistry of electron transfer 


Oxidation is loss of 
electrons. Reduction is 
gain of electrons. 

Cutlery or ornaments made of silver tarnish and become black; 
this is a shame, because clean, shiny silver is very attractive. The 
'tarnish' comprises a thin layer of silver that has oxidized following 
contact with the air to form black silver(I) oxide: 

In fact, the tarnish on 
silver usually comprises 
both silver(I) oxide 
and a little silver(I) 

4Ag (s) + 2( g) ► 2Ag 2 (s) 


Such silver can be rather difficult to clean without abrasives 
(which wear away the metal). The following is a simple electro- 
chemical means of cleaning the silver: immerse the tarnished silver 
in a saucer of electrolyte, such as salt solution or vinegar, and wrap 
it in a piece of aluminium foil. Within a few minutes the silver is cleaner and bright, 
whereas the aluminium has lost some of its shininess. 

The shine from the aluminium is lost as atoms on the surface of the foil are 
oxidized to form Al 3+ ions (Equation (7.1)), which diffuse into solution. Because 
the aluminium touches the silver, the electrons generated by Equation (7.1) enter the 
silver and cause electro-reduction of the surface layer of Ag 2 (Equation (7.2)). 

In summary, we construct a simple electrochemical cell in which the silver to be 
cleaned is the cathode (Equation (7.2)) and aluminium foil as the anode supplies the 
electrons via Equation (7.1). 

The salt or vinegar acts as an electrolyte, and is needed since the product Al 3+ 
requires counter ions to ensure electro-neutrality (so aluminium ethanoate forms). 
The oxide ions combine with protons from the vinegar to form water. Figure 7.2 
illustrates these processes occurring in schematic form. 

Aluminium metal 

Aluminium ions 

- electrons 

Silver metal 

+ electrons 

Silver ions 

Aluminium foil 



Silver spoon 

Figure 7.2 Illustration of the electron cycles that allow for the trouble-free cleaning of silver: we 
immerse the tarnished silver in an electrolyte, such as vinegar, and touch the silver with aluminium 


And finally a word of caution: aluminium ethanoate is toxic, so wash the silver 
spoon thoroughly after removing its tarnish. 

How does Electrolysis' stop hair growth? 

Electrochemical reactions and electrolysis 

To many people, particularly the image conscious, electrolysis 
means removing hairs from the arms and legs - a practice some- 
times called 'electrology'. The purpose of such electrolysis is to 
remove the hair follicles temporarily, thereby avoiding the need to 
shave. We 'treat' each individual hair by inserting a tiny surgical 
'probe' (in reality, an electrode) into the hair follicle. Applying a 
voltage to the hair root for a fraction of a second kills the root. The 
electrolysed hair is then removed, and will not regrow for some 
time. This procedure is performed repeatedly until the desired area 
is cleared. But how does it work? 

A voltage is applied to the electrode: we say it is polarized. A 
current flows in response to the voltage, and electrons are consumed 
by electrochemical reactions around the electrodes. Electrolysis 
occurs. Each electron that flows through the electrode must be 
involved in a redox reaction, either oxidation or reduction. The 
electrons entering or leaving the electrode move as a result of 
reactions occurring in the immediate vicinity of the electrode. Con- 
versely, electrons can travel in the opposite direction (leaving the 
electrode) to facilitate reduction reactions. 

In summary, this form of electrolysis is effective because the 
charge passing through the electrode generates chemicals inside 
the hair follicle. The resultant trauma kills the hair root. A leg or 
arm treated in this way remains hairless until a new, healthy root 
regrows later in the previously damaged follicle. 

Electrochemical means 
the chemistry of the 

It is wrong to claim 
a permanent method 
of hair removal: after 
electrolysis, the hair 
will grow back, albeit 
thinner and finer. 

The word 'electroly- 
sis' derives from the 
Greek words lysis, 
meaning 'splitting' or 
'cleavage', and the 
root electro, meaning 
'charge' or 'electric- 
ity'. Strictly, then, 
electrolysis involves 
electrochemical bond 

Why power a car with a heavy-duty battery yet use a 
small battery in a watch? 

Faraday's laws of electrolysis 

A battery is a device for converting chemical energy into electrical 
energy. The amount of energy required by the user varies according 
to the application in mind. For example, a watch battery only pow- 
ers the tiny display on the face of a digital watch. For this purpose, 

Batteries are described 
in more detail in 
Section 7.7. 


Table 7.1 Faraday's laws of electrolysis 

Faraday' s first law 

The number of moles of a species formed at an electrode during electrolysis is proportional to the 
electrochemical charge passed: Q = I x t 

Faraday's second law 

A given charge liberates different species in the ratio of their relative formula masses, divided by 
the number of electrons in the electrode reaction 

it only needs to deliver a tiny current of a micro-amp or so. Conversely, a car battery 
(usually a 'lead-acid cell', as described on p. 347) is bulky and heavy because it 
must deliver a massive amount of electrical energy, particularly when starting the car. 
Other batteries generate currents of intermediate magnitude, such as those needed in 
torches, mobile phones and portable cassette and CD players. 

The amount of charge generated or consumed by a battery is in direct proportion 
to the number of electrons involved, according to Faraday 's laws, which are given in 
Table 7.1. Both electrons and ions possess charge. When a current is drawn through 
a cell, the charged electrons move through the conductive electrodes (as defined on 
p. 300) concurrently with charged ions moving through the electrolyte. The ions are 
anions (which bear a negative charge) and cations (which are positive). 

Underlying both of Faraday's laws lies the fundamental truth that each electron 
possesses the same charge. 

Worked Example 7.1 What is the charge on 1 mol of electrons? 

The charge e on a single electron is 1.6 x 10" 19 C and there are 
6.022 x 10 23 electrons per mole (the Avogadro number L), so the 
charge on a mole of electrons is given by the simple expression 

The coulomb, C, is the 
SI unit of charge. 

The charge on 1 mol of 
electrons is termed 'a 
faraday' F. 

charge on one electron = L x e (7.6) 

Inserting numbers into Equation (7.6), we obtain 

charge on 1 mole of electrons 

= 1.6 x 10" 19 C x 6.026 x 10 23 mol" 1 

We see that 1 mol has a charge of 96487 Cmol" . This quantity of 
charge is known as the 'Faraday' F. 

SAQ 7.1 An electrolysis needle (i.e. an electrode) delivers 1 nrmol of 
electrons to a hair root. How many faraday's of charge are consumed, and 
how many coulombs does it represent? 



Worked Example 7.2 How much silver is generated by reductively 
passing 1 F of charge through a silver-based watch battery? 

We will answer this question in terms Faraday's first law, which was 
first formulated in 1834 (see Table 7.1). 

Each electron is required to effect the reduction reaction Ag + (aq ) + 
e~ — > Ag (s) , so one electron generates one atom of Ag, and IF of 
charge (i.e. 1 mol of electrons) generates 1 mol of Ag atoms. The 
metal forms at the expense of 1 mol of Ag + ions. Similarly, 1 mol 
of electrons, if passed oxidatively, would generate 1 mol of Ag + ion 
from Ag metal. 

We see a direct proportionality between the charge passed and the 
amount of material formed during electrolysis, as predicted by Fara- 
day's first law. 

Silver has the symbol 
'Ag' because its Latin 
name is argentium, 
itself derived from the 
Greek for money, argu- 
rion. The Spanish col- 
onized parts of South 
America in the 16th 
century. They named 
it Argentina (dog-Latin 
for 'silver land') when 
they discovered its vast 
reserves of silver. 

SAQ 7.2 10~ 10 F of charge are inserted into a hair pore through a fine nee- 
dle electrode. Each electron generates one molecule of a chemical to poison 
the hair root. How much of the chemical is formed? 

Worked Example 7.3 A Daniell cell (see p. 345) is constructed, and IF of charge is 
passed through a solution containing copper(II) ions. What mass of copper is formed? 
Assume that the charge is only consumed during the reduction reaction, and is performed 
with 100 percent efficiency. 

We will answer this question by introducing Faraday's second law (see Table 7.1). 
The reduction of one copper(II) ion requires two electrons according to the reaction 

Cu z 


+ 2e~ 

-* Cu (s 



We need 2F of charge to generate 1 mol of Cu( S ), and IF of charge will form 0.5 mol 
of copper. 

1 mol of copper has a mass of 64 g, so 1 F generates (64 4- 2) g of copper. The mass 
of copper generated by passing 1 F is 32 g. 

SAQ 7.3 How much aluminium metal is formed by passing 2F of charge 
through a solution of Al 3+ ions? [Hint: assume the reaction at the electrode 

is AI J 



Al ( s).] 

low is coloured Canodized') aluminium produced? 

Currents generate chemicals: dynamic electrochemistry 

Saucepans and other household implements made of aluminium often have a brightly 
coloured, shiny coating. This outer layer comprises aluminium oxide incorporating a 
small amount of dye. 



AI2O3 is also called 

The layer is deposited with the saucepan immersed in a vat of 
dye solution (usually acidified to pH 1 or 2), and made the positive 
terminal of a cell. As the electrolysis proceeds, so the aluminium 
on the surface of the saucepan is oxidized: 

2A1 (S) + 3H 2 ► A1 2 3 ( S ) + 6H^ 



We say the dye occlu- 
des within a matrix of 
solid aluminium oxide. 

The aluminium is white and shiny before applying the potential. A critical potential 
exists below which no electro-oxidation will commence. At more extreme potentials, 
the surface atoms of the aluminium oxidize to form Al + ions, which combine with 
oxide ions from the water to form AI2O3. This electro-precipitation of solid alu- 
minium oxide is so rapid that molecules of dye get trapped within it, and hence its 
coloured aspect. 

The dye resides inside the layer of alumina. Its colour persists 
because it is protected from harmful UV light, as well as mechan- 
ical abrasion and chemical attack. 

But the chemical reaction forming this coloured layer of oxide 
represents only one part of the cell. A cell contains a minimum of 
two electrodes, so a cell comprises two reactions - we call them 
half-reactions: one describes the chemical changes at the positive 
electrode (the anode) and the other describes the changes that occur 
at the negative cathode. 
The same number of electrons conducts through (i.e. are conducted by) each of the 
two electrodes. If we think in terms of charge flowing per unit time, we would say 
the same 'current' / flows through each electrode. The electrons travel in opposite 
directions, insofar as they leave or enter an electrode, which explains why the current 
through the anode is oxidative and the current through the cathode is reductive. 
We say 

A anode) = — ^(cathode) (7.9) 

A cell must comprise a 
minimum of two elec- 

where the minus sign reminds us that the electrons either move in or out of the 

Because these two currents are equal (and opposite), the same amount of reaction 
will occur at either electrode. We see how an electrode reaction must also occur at 
the cathode as well as the desired oxidative formation of alumina at the anode. (The 
exact nature of the reaction at the anode will depend on factors such as the choice of 
electrode material.) 

How do we prevent the corrosion of an oil rig? 

Introduction to electrochemical equilibrium 

Oil rigs are often built to survive in some of the most inhospitable climates in the 
world. For example, the oil rigs in the North Sea between the UK and Scandinavia 



frequently withstand force- 10 gales. Having built the rig, we appreciate how impor- 
tant is the need of maximizing its lifetime. And one of the major limits to its life 
span is corrosion. 

Oil rigs are made of steel. The sea in which they stand contains vast quantities of 
dissolved salts such as sodium chloride, which is particularly 'aggressive' to ferrous 
metals. The corrosion reaction generally involves oxidative dissolution of the iron, to 
yield ferric salts, which dissolve in the sea: 



-> Fe 



+ 3e" 


In reality, several of 
the iron compounds 
are solid, such as rust. 
This clever method of 
averting corrosion can 
also arrest the corro- 
sion of rails and the 
undersides of boats. 

If left unchecked, dissolution would cause thinning and hence weakening of the legs 
on which the rig stands. 

One of the most ingenious ways in which corrosion is inhibited 
is to strap a power pack to each leg (just above the level of the 
sea) and apply a continuous reductive current. An electrode couple 
would form when a small portion of the iron oxidizes. The couple 
would itself set up a small voltage, itself promoting further disso- 
lution. The reductive current coming from the power pack reduces 
any ferric ions back to iron metal, which significantly decreases 
the rate at which the rig leg corrodes. 

Clearly, we want the net current at the iron to be zero (hence no 
overall reaction). The rate of corrosion would be enhanced if the 
power pack supplied an oxidative current, and wasteful side reac- 
tions involving the seawater itself would occur if the power pack 
produced a large reductive current. The net current through the 
iron can be positive, negative or zero, depending on the potential 
applied to the rig's leg. The conserver of the rig wants equilibrium, 
implying no change. 

All the discussions of electrochemistry so far in this chapter concern current - the 
flow of charged electrons. We call this branch of electrochemistry dynamic, implying 
that compositions change in response to the flow of electrons. Much of the time, 
however, we wish to perform electrochemical experiments at equilibrium. 

One of our best definitions of 'equilibrium electrochemistry' says 
the net current is zero; and from Faraday's laws (Table 7.1), a zero 
current means that no material is consumed and no products are 
formed at the electrode. 

But this equilibrium at the oil rig is dynamic: the phrase 
'dynamic equilibrium' implies that currents do pass, but the cur- 
rent of the forward reaction is equal and opposite to the current of the back reaction, 
according to 

' (forward, eq) = ' (backward, eq) \ ' • * *) 

The simplest definition 
of equilibrium in an 
electrochemistry cell is 
that no concentrations 

Electrochemical mea- 
surements at equilib- 
rium are made at zero 

and the overall (net) current is the sum of these two: 

J(net) = I* 


+ /, 





Equation (7.11) is important, since it emphasizes how currents flow even at equi- 

But the value of 7( net ) is only ever zero at equilibrium because /(forward) = — /(backward)* 
which can only happen at one particular energy, neither too reductive nor too oxidative. 
The voltage around the legs of the oil rig needs to be chosen carefully. 

What is a battery? 

The emf of cells 

A battery is a device 
for converting chemical 
energy into electrical 

The word 'cell' comes 
from the Latin for'smal 
room', which explains 
why a prisoner is kept 
in a 'cell'. 

A battery is an electrochemical cell, and is defined as 'a device 
comprising two or more redox couples' (where each couple com- 
prises two redox states of the same material). An oxidation reaction 
occurs at the negative pole of the battery in tandem with a reduc- 
tion reaction at the positive pole. Both reactions proceed with the 
passage of current. The two redox couples are separated physically 
by an electrolyte. 

The battery requires two redox couples because it is a cell. Each 
couple could be thought of as representing half of a complete cell. 
This sort of reasoning explains why the two redox couples are 
called half-cells. We could, therefore, redefine a cell as a device 
comprising two half-cells separated with an electrolyte. 
In practice, the voltage of a battery is measured when its two ends are connected 
to the two terminals of a voltmeter, one contact secured to the positive terminal of 
the battery and the other at the negative. But a voltmeter is a device to measure 
differences in potential, so we start to see how the 'voltage' cited on a battery label 
is simply the difference in potential between the two poles of the battery. 

While the voltage of the cell represents the potential difference 
between the two 'terminals' of the battery, in reality it relates to 
the separation in energy between the two half-cells. We call this 
separation the emf where the initials derive from the archaic phrase 
electromotive force. An emf is defined as always being positive. 

We have already seen from Faraday's laws how a zero current 
implies that no redox chemistry occurs. Accordingly, we stipulate 
that the meter must draw absolutely no current if we want to mea- 
sure the battery's emf at equilibrium. Henceforth, we will assume 
that all values of emf were determined at zero current. 

The cell's emf is a pri- 
mary physicochemical 
property, and is mea- 
sured with a voltmeter 
or potentiometer. 

An emf is always 
defined as being posi 


The term 


follows from the archaic term 'electromotive force'. Physicists prefer 

to call the 


a 'potential difference' or symbolize it as a 


Confusingly, potential also has the symbols of U, V and E 

, depending on 

the context. 



Why do hydrogen fuel cells sometimes *dry up'? 

Cells and half-cells 

Hydrogen fuel cells promise to fuel prototype cars in the near future. We define such 
a fuel cell as a machine for utilizing the energies of hydrogen and oxygen gases, 
hitherto separated, to yield a usable electric current without combustion or explosion. 
Unlike the simple batteries above, the oxygen and hydrogen gases fuelling these cells 
are transported from large, high-pressure tanks outside the cell. The gases then feed 
through separate pipes onto the opposing sides of a semi-permeable membrane (see 
Figure 7.3), the two sides of which are coated with a thin layer of platinum metal, 
and represent the anode and cathode of the fuel cell. This membrane helps explain 
why such cells are often called PEM fuel cells, where the acronym stands for 'proton 
exchange membrane'. 

When it reaches the polymer membrane, hydrogen gas is 
oxidized at the negative side of the cell (drawn on the left of Figure 
7.3), forming protons according to 



-* 2H+ + 2e" 



The subscripted 'Pt' helps emphasize how the two electrons con- 
duct away from the membrane through the thin layer of platinum 

The energy necessary 
to cleave the H-H bond 
is provided by the 
energy liberated when 
forming the two H-Pt 
bonds after molecular 


Hydrogen gas 

polymer - 


Oxygen gas 



Thin electrodes 
(layers of platinum) 

Figure 7.3 A hydrogen-oxygen fuel cell. The water formed at the cathode on the right-hand side 
of the cell condenses and collects at the bottom of the cell, and drains through a channel at the 
bottom right-hand side 



metal surrounding the electrolyte, and enter into the external circuit where they per- 
form work. The platinum also catalyses the dissociation of diatomic H2 gas to form 
reactive H" atoms. 

Once formed, the protons diffuse through the platinum layer and 
enter deep into the layer of semi-permeable membrane. They travel 
from the left-hand side of the membrane to its right extremity in 
response to a gradient in concentration. (Movement caused by a 
concentration gradient will remind us of dye diffusing through a 
saucer of water, as described on p. 129.) 

At the positive side of the cell (drawn here on the right), oxygen 
gas is reduced to oxide ions, according to 

The membrane, made 
from a perfluorinated 
polymer bearing sul- 
phonic acid groups, is 
known in the trade as 

Note: the electron 
count in Equations 
(7.13) and (7.14) 
should balance in 

02(g) + 4e" 





The electrons necessary to effect the reduction of gaseous oxy- 
gen come from the external circuit, and enter the oxygen half-cell 
through the layer of platinum coating the cathode, thereby explain- 
ing why the electrons in Equation (7.14) are subscripted with 'Pt'. 

The 2 ~ ions combine chemically with protons that have traversed the Nafion 
membrane and form water, which collects at the foot of the cell. 

Because the overall cell reaction is exothermic, the value of the cell emf decreases 
with increasing temperature, so the temperature is generally kept relatively low at 
about 200 °C. The cell emf is 1.23 V at this temperature. 

One of the main advantages of this hydrogen fuel cell is the rapid 
rate at which hydrogen is oxidized at the platinum surface. Most 
of the cell's operational difficulties relate to the oxygen side of the 
device. Firstly, the reduction of gaseous oxygen in Equation (7.14) 
is relatively slow, so the rate at which the cell operates is somewhat 
limited. But more serious is the way the cell requires a continual 
flow of gas, as below. The hydrogen half-cell comprises an elec- 
trode couple because two redox states of the same material coexist 
there (H2 and H + ). In a similar way, the oxygen half-cell also comprises an electrode 
couple, but this time of O2 and O ~. 

If the flow of oxygen falters, e.g. when the surface of the cathode is covered with 
water, then no gaseous O2 can reach the platinum outer layer. In response, firstly no 
electrons are consumed to yield oxide ions, and secondly the right-hand side of the 
cell 'floods' with the excess protons that have traversed the polymer membrane and 
not yet reacted with O 2- . Furthermore, without the reduction of oxygen, there is no 
redox couple at the cathode. The fuel cell ceases to operate, and can produce no more 
electrical energy. 

This simple example helps explain why a cell requires no fewer than two half- 
cells. A half-cell on its own cannot exchange electrons, and cannot truly be termed 
a cell. 

The rate of oxygen 
reduction can be accel- 
erated by finely divid- 
ing the platinum cata- 
lyst, thereby increasing 
its effective area. 



Why bother to draw cells? 

Cells and the 'cell schematic' 

Having denned a cell, we now want to know the best way of representing it. Undoubt- 
edly, the simplest is to draw it diagrammatically - no doubt each picture of a cell 
being a work of art in miniature. 

But drawing is laborious, so we generally employ a more sensible alternative: we 
write a cell schematic, which is a convenient abbreviation of a cell. It can be 'read' 
as though it was a cross-section, showing each interface and phase. It is, therefore, 
simply a shorthand way of saying which components are incorporated in the cell as 
cathode, anode, electrolyte, etc., and where they reside. 

Most people find that a correct understanding of how to write a cell schematic also 
helps them understand the way a cell works. Accordingly, Table 7.2 contains a series 
of simple rules for constructing the schematic. 

Worked Example 7.4 We construct a cell with the copper(II) | copper and zinc(II) | 
zinc redox couples, the copper couple being more positive than the zinc couple. What is 
the cell schematic? 

Answer strategy: we will work sequentially through the rules in Table 7.2. 

First, we note how the copper couple is the most positive, so we 
write it on the right. The zinc is, therefore, the more negative and 
we write it on the left. We commence the schematic by writing, 

ezn (s) (s) e. 

For convenience, we 
often omit the subscript 
descriptors and the 'e' 
and '©' signs. 

Table 7.2 Rules for constructing a cell schematic 

1. We always write the redox couple associated with the positive electrode on the 
right-hand side 

2. We write the redox couple associated with the negative electrode on the left-hand side 

3. We write a salt bridge as a double vertical line: || 

4. If one redox form is conductive and can function as an electrode, then we write it on one 
extremity of the schematic. 

5. We represent the phase boundary separating this electrode and the solution containing the 
other redox species by a single vertical line: a | 

6. If both redox states of a couple reside in the same solution (e.g. Pb + and Pb + ), then they 
share the same phase. Such a couple is written conventionally with the two redox states 
separated by only a comma: Pb 4+ , Pb 2+ 

7. Following from 6: we see that no electrode is in solution to measure the energy at 
equilibration of the two redox species. Therefore, we place an inert electrode in solution; 
almost universally, platinum is the choice 

a We write a single line | or, better, a dotted vertical line, if the salt bridge is replaced by a simple porous 



There is a phase bound- 
ary between the Zn 
and Zn 2+ because the 
Zn is solid but the Zn 2+ 
is dissolved within a 
liquid electrolyte. A 
similar boundary exists 
in the copper half-cell. 

An alternative way of 
looking at the schematic 
is to consider it as 
'the path taken by a 
charged particle dur- 
ing a walk from one 
electrode to the other'. 

To be a redox couple, the zinc ions will be in contact with the 
zinc electrode, which we write as Zn 2+ ( aq )|Zn( S ), the vertical line 
emphasizing that there is a phase boundary between them. We can 
write the other couple as Cu 2+ ( aq )|Cu (s ), with similar reasoning. Note 
that if the two electrodes are written at the extreme ends of the 
cell schematic, then the redox ionic states must be located some- 
where between them. The schematic now looks like ©Zn( S ) |Zn 2+ ( aq ) . . . 




|Cu (s 


Finally, we note that the two half-cells must 'communicate' some- 
how - they must be connected. It is common practice to assume that 
a salt bridge has been incorporated, unless stated otherwise, so we 
join the notations for the two half-cell with a double vertical line, as 

0Zn (s) I Zn 2+ (aq) 1 1 Cu 2+ (aq) | Cu (s) © . 

SAQ 7.4 Write the cell schematic for a ce 
the Fe 3+ c ~ ' — "■'"-'* ■"-* r ~- 2 

,Fe (positive) and Co^ 

Co (negative) couples. 

Worked Example 7.5 Write a cell schematic for a cell comprising 
the couples Br2, Br - and H + , H2. The bromine | bromide couple is 
the more positive. Assume that all solutions are aqueous. 

This is a more complicated cell, because we have to consider the involvement of more 
phases than in the previous example. 

Right-hand side: the bromine couple is the more positive couple, so we write it on the 
right of the schematic. Neither Br2 nor Br~ is metallic, so we need an inert electrode. By 
convention, we employ platinum if no other choice is stipulated. The electrode at the far 
right of the schematic is therefore Pt, as . . . Pt( S )©. 

Br2 and Br - are both soluble in water - indeed, they are mutually 
soluble, forming a single-phase solution. Being in the same phase, we 
cannot write a phase boundary (as either 'Br2|Br~' or as 'Br~|Br2'), 

We write the oxidized 
form first if both redox 
states reside in the 
same phase, sepa- 
rating them with a 

so we write it as 'Br2, Br~ (aq )'. Note how we write the oxidized form 
first and separate the two redox states with a comma. The right-hand 
side of the schematic is therefore 'Br2, Br~( aq )|Pt( S )©'. 

Left-hand side : neither gaseous hydrogen nor aqueous protonic solu- 
tions will conduct electrons, so again an inert electrode is required on 
the extremity of the schematic. We again choose platinum. The left-hand side of the cell 
is: ePt( S ). 

Hydrogen gas is in immediate contact with the platinum inert electrode. (We bub- 
ble it through an acidic solution.) Gas and solution are different phases, so we write the 
hydrogen couple as H2( g )|H + ( aq ), and the left-hand side of the schematic becomes '©Pt( S )| 
H2( g )|H + (aq) '. 



Finally, we join the two half-cells via a salt bridge, as 

Pt (s) | H 2(g) | H+ (aq) || Br 2 , Br" (aq) | Pt (s) © . 

From now on, we will omit both the symbols and ©, and merely 
assume that the right-hand side is the positive pole and the left-hand 
side the negative. 

As an extra check, 
we note how the salt 
bridge dips into the 
proton solution, so the 
term for H + needs to 
be written adjacent to 
the symbol Y- 

Why do digital watches lose time in the winter? 

The temperature dependence of emf 

A digital watch keeps time by applying a tiny potential (voltage) 
across a crystal of quartz, causing it to vibrate at a precise fre- 
quency of v cycles per second. The watch keeps time by counting 
off 1 s each time the quartz vibrates v times, explaining why the 
majority of the components within the watch comprise a count- 
ing mechanism. 

Unfortunately, the number of vibrations of the crystal per second 
is dictated by the potential applied to the quartz, so a larger voltage 
makes the frequency v increase, and a smaller voltage causes v 
to slow. For this reason, the potential of the watch battery must 
be constant. 

In Section 4.6, we saw how the value of AG is never indepen- 
dent of temperature, except in those rare cases when A5 , ( ce ii) = 0. 
Accordingly, the value of AG( ce ii) for a battery depends on whether 
someone is wearing the watch while playing outside in the cold 
snow or is sunbathing in the blistering heat of a tropical summer. 

And the emf of the watch battery is itself a function of the 
change in Gibbs function, AG( ce ii) according to 

AG ( C eii) = —nF x emf 


where F is the Faraday constant and n is the number of moles of 
electrons transferred per mole of reaction. The value of AG( ce ii) is 
negative if the reaction proceeds reversibly (see Section 4.3), so the 
emf is defined as positive to ensure that AG( ce ii) is always negative. 
In other words, the value of AG( ce ii) relates to the spontaneous 
cell reaction. 

So we understand that as the emf changes with temperature, so 
the quartz crystal vibrates at a different frequency - all because 

Frequency v has the SI 
unit of hertz (Hz). 1 Hz 
represents 1 cycle or 
vibration per second, 
so a frequency of v 
Hz means v cycles per 

The voltage from the 
battery induces a minute 
mechanical strain in 
the crystal, causing it 
to vibrate - a property 
known as the piezo- 
electric effect. 

Care: The output volt- 
age of a battery is only 
an emf when mea- 
sured at zero current, 
i.e. when not operating 
the watch. 



AG ( C eii) is a function of temperature. Ultimately, then, a digital watch loses time in 
the winter as a simple result of the cold. 

Worked Example 7.6 The emf of a typical 'lithium' watch battery (which is a cell) is 
3.0 V. What is AG ^11)? 

Look at the units and 
note how 1 J = 1 C x 

The number of electrons transferred n in a lithium battery is one, 
since the redox couple is Li + , Li. Inserting values into Equation 
(7.15) yields: 

AG ( 


-1 x 96485 CmoP 1 x 3.0 V 

AG (C eii) = -289 500 Jmol -1 = -290 kJmoi" 1 

Notice how the molar energy released by a simply battery is enormous. 

SAQ 7.5 A manganese dioxide battery has an emf of 1.5 V and n = 2. 
Calculate AG (C eii)- 

In the absence of any pressure- volume work, the value of AG( ce ii) is equal to the 
work needed to transfer charge from the negative end of the cell to the positive. In 
practice, AG( ce ii) equates to the amount of charge passed, i.e. the number of charged 
particles multiplied by the magnitude of that charge. 

hy Is a battery's potential no 

Non-equilibrium measurements of emf 

We define the emf 
as having a positive 
value and, strictly, it is 
always determined at 

The two currents (for 
anode and cathode) 
are generally differ- 
ent. Neither of them is 
related to potential in a 
linear way. 

A healthy battery for powering a Walkman or radio has a voltage 
of about 1.5 V. In the terminology of batteries, this value is called 
its open-circuit potential, but an electrochemist talking in terms of 
cells will call it the emf. This voltage is read on a voltmeter when 
we remove the battery from the device before measurement. But 
the voltage would be different if we had measured it while the 
battery was, for example, powering a torch. 

We perform work whenever we connect the two poles of the 
battery across a load. The 'load' in this respect might be a torch, 
calculator, car, phone or watch - anything which causes a current 
to pass. And this flow of current causes the voltage across the 
battery or cell to decrease; see Figure 7.4. We call this voltage the 
'voltage under load'. 

A similar graph to Figure 7.4 could have been drawn but with 
the x axis being the resistance between the two electrodes: if the 




Current drawn/A 

Figure 7.4 Schematic diagram showing how a cell's potential decreases with current. We call the 
cell potential the emf only when the current is zero 

resistance between the two electrodes is zero, which is clearly the case if they should 
touch, then the cell potential is zero - we say the cell has 'shorted'. 

Ohm's law, Equation (7.16), describes the difference between the emf and a voltage 
under load: 

V=IR (7.16) 

where / is the current flowing, R is the resistance of the load and V is the decrease 
in the voltage of the cell. When a current is drawn, the potential of the cell decreases 
by the amount V in Equation (7.16). We will call this new (smaller) voltage E(\ oad ), 
and its magnitude is given by 

£(k>ad) = emf - IR 

In summary, we say the voltage of a cell is the same as a cell's 
emf if determined at zero current. From Faraday's laws of elec- 
trolysis, this criterion implies that none of the compositions within 
the cell can change. In other words, a cell emf is an equilib- 
rium quantity. 

For this reason, it is not wise to speak of terms such as 'anode' 
of 'cathode' for a cell at equilibrium, because these terms relate 
to electrodes that give or receive charge during current flow; and 
our definition of equilibrium implies that no current does flows. 
We therefore adopt the convention: the terms 'anode' or 'cath- 
ode ' will no longer be employed in our treatment of equilibrium 
electrochemistry . 


The emf can only ever 
be determined at zero 

Why a 

battery's emf 

decreases permanently 

after a 

current has 


is explained on 

p. 328. 

What is a 'standard cell'? 

The thermodynamics of cells 

A standard cell produces a precise voltage and, before the advent of reliable volt- 
meters, was needed to calibrate medical and laboratory equipment. It is generally 
agreed that the first standard cell was the Clark cell (see p. 299), but the most popular 
was the Weston saturated cadmium cell, patented in 1893. 



Edward Weston (1850-1936) was a giant in the history of electrical measuring 
instruments. In the field of measurement, he developed three important components: 
the standard cell, the manganin resistor and the electrical indicating instrument. 

The main advantage of Weston's cell was its insensitivity to temperature, and the 
emf of almost 1 V: to be precise, 1.0183 V at 20 °C. It is usually constructed in an H- 
shaped glass vessel. One arm contains a cadmium amalgam electrode beneath a paste 
of hydrated cadmium sulphate (3CdSC>4 • 5H2O) cadmium sulphate and mercury(I) sul- 
phate. The other arm contains elemental mercury. Its schematic is Cd(Hg)|CdS04( aq ), 
Hg 2 S0 4 |Hg. 

The Weston saturated cadmium cell became the international 
standard for emf in 1911. Weston waived his patent rights shortly 
afterward to ensure that anyone was allowed to manufacture it. 

Weston's cell was much less temperature sensitive than the previ- 
ous standard, the Clark cell. We recall how the value of AG changes 
with temperature according to Equation (4.38). In a similar way, the 
value of AG ( C eii) for a cell relates to the entropy change AS( ce n) such 
that the change of emf with temperature follows 

The Clark cell was 
patented by Latimer 
Clark in the 1880s, and 
was the first standard 


the small 


'p' indi- 

cates that the quantity 

is measurec 

at con- 

stant pressure. It does 

not mean 'multiplied 


The temperature volt- 
age coefficient has 
several names: 'tem- 
perature coefficient', 
'voltage coefficient' or 
'temperature coef- 
ficient of voltage'. 
Table 7.3 contains a 
few values of 
(d (emO/dT). 

A 5, 






the value of (d(emf)/dT) p is virtually zero for the Weston cell. 

If we assume the differential (d (emf)IdT) is a constant, then 
Equation (7.18) has the form of a straight line, y = mx, and a 
graph of emf (as 'v') against T (as 'x') should be linear. Figure 7.5 
shows such a graph for the Clark cell, Hg|HgS0 4 , ZnS04(sat'd)| 
Zn. Its gradient represents the extent to which the cell emf varies 
with temperature, and is called the temperature voltage coefficient. 
The gradient may be either positive or negative depending on the 
cell, and typically has a magnitude in the range 10~ 5 to 10~ 4 
VK" 1 . We want a smaller value of (d (emf)IdT) if the emf is 
to be insensitive to temperature. 

Having determined the temperature dependence of emf as the 
gradient of a graph of emf against temperature, we obtain the value 
of AS( C eii) as 'gradient x n x F' . 

That this value of 
d(emf)/dT is neg- 
ative tells us that 
the emf DEcreases 
when the temperature 

Worked Example 7.7 The temperature voltage coefficient for a sim- 
ple alkaline torch battery is —6.0 x 10~ 4 VK - '. What is the entropy 
change associated with battery discharge? The number of electrons 
transferred in the cell reaction n = 1. 

Inserting values into Equation (7.18): 

A 5, 


= 2x 96485 CmoP 1 x -6.0 x 10" 4 VK" 

AS( C eii) = -116 J K" 1 mor 1 







Gradient = 'temperature 
voltage coefficient' 






300 310 

Temperature T/K 



Figure 7.5 Graph of cell emf against temperature for the Clark cell Hg|HgS0 4 , ZnS04(sat'd)|Zn. 
We call the gradient of this graph the 'temperature voltage coefficient' 

Table 7.3 Temperature voltage coefficients for various 
cells and half cells 

Cell 3 


Standard hydrogen electrode 
Clark standard cell 
Saturated calomel electrode 
Silver- silver chloride 
Silver- silver bromide 
Weston standard cell 

(by definition) 13 
6.0 x 1(T 4 
+7 x 1CT 4 
-8.655 x 1CT 5 
-4.99 x 1(T 4 
-5 x irr 5 

"Individual electrodes are cited with the SHE as the second elec- 
trode of the cell. 
b The potential of the SHE is defined as zero at all temperatures. 

SAQ 7.6 The emf of a lithium watch battery is exactly 3.000 V at 298 K, 
but the value decreases to 2.985 V at 270 K. Calculate the temperature 
voltage coefficient and hence the change in entropy AS (C eii) during cell 
discharge. (Take n = 1.) 

Occasionally, the temperature voltage coefficient is not expressed as a simple number, 
but as a power series in T (we generally call it a virial series, or expansion). For example, 
Equation (7.19) cites such a series for the cell Pt( s )|H2( g )|HBr( aq )|AgBr (s JAg (s) : 

emf IN = 0.071 31 - 4.99 x 10 _4 (r/K - 298) 

3.45 x 10 _6 (r/K - 298) 2 (7.19) 

We insert values of temperature T into the expression to obtain a value for emf. Values 
of AS( ce ii) are obtained by performing two calculations, inserting first one temperature 



7\ to obtain the emf at T\, and then a second T2 to obtain another value of emf. We 
calculate a value of (d(emf)/dT) using 

d(emf) emf at T2 — emf at T\ 
AT ~ T 2 -Ti 

The value of A5( ce ii) is then determined in the usual way via Equation (7.18) 


SAQ 7.7 Insert values of T = 310 K into Equation (7.19) to calculate the 
potential of the cell Pt (S )|H2(g)|HBr (aq) |AgBr (s) |Ag (s) . 

SAQ 7.8 Repeat the calculation in SAQ 7.7, this time with 7 = 360 K, 
and hence determine AS (ce ii). 

Justification Box 7.1 

The relationship between changes in Gibbs function and temperature (at constant pres- 
sure p) is defined using Equation (4.38): 


dT ), 

We know from Equation (7.15) that the change in AG( ce ii) with temperature is '— nF x 
emf . The entropy change of the cell AS( Ce ii) is then obtained by substituting for AG( ce ii) 
in Equation (7.18): 

-A5 ( . 


d(—nF x emf) 


Firstly, the two minus signs cancel; and, secondly, n and F are both constants. Taking 
them out of the differential yields Equation (7.18) in the form above. 

These values of AG, 
AH and AS relate to a 
complete cell, because 
thermodynamic data 
cannot be measured 
experimentally for half- 
cells alone. 

To obtain the change in enthalpy during the cell reaction, we 
recall from the second law of thermodynamics how AH = AG + 
TAS (Equation (4.21)). In this context, each term relates to the 
cell. We substitute for AG( ce ii) and A5( ce n) via Equations (7.15) 
and (7.18) respectively, to yield 



= — nF x emf + TnF 



so, knowing the emf as a function of temperature, we can readily obtain a value of 

A //(cell)- 



Worked Example 7.8 The Clark cell Zn|Zn 2+ , Hg 2 S04|Hg is often employed as a 
standard cell since its emf is known exactly as a function of temperature. The cell emf is 
1.423 V at 298 K and its temperature coefficient of voltage is —1.2 x 10~ 4 VK" 1 . What 
are AG( ce ii), AS( ce ii) and thence A//( ce u) at 298 K? 

Before we commence, we note that the spontaneous cell reaction is 

Zn + Hg 2 S0 4 + 7H 2 ► ZnS0 4 ■ 7H 2 + 2Hg° 

so the cell reaction is a two-electron process. 

Next, we recall from Equation (7.15) that AG( Ce u) = — nF x emf. Inserting values for 
the cell at 298 K gives 

AG (cell) = -2 x 96485 Cmol" 1 x 1.423 V 
AG ( C eii) = —275 kJnmT 1 

Then, from Equation (7.18), we recall that 

AS( C eii) = nF 


Inserting values: 

AS (ce ii) = 2x 96485 CmoP 1 x (-1.2 x 10" 4 VK~') 
A5 (ce ii) = -23.2 JK" 1 mor 1 

Finally, from Equation (4.21), we say that A//( ce ii) = AG (ce n) + TAS( ce \i). We again 
insert values: 



(-275 kJmol" 1 ) + (298 K x -23.2 JKT 1 moP 1 ) 




-282 kJ mor 

SAQ 7.9 A different cell has an emf of 1.100 V at 298 K. 
The temperature voltage coefficient is +0.35 mVK -1 . Cal- 
culate AG ( ceii), AS ( ceii) and hence AW (ce n) for the cell at 
298 K. Take n = 2. 

Remember: 5 mV (5 
m/7//-volts) is the same 
as 5 x 10- 3 V. 

Having performed a few calculations, we note how values of AG( ce ii) and for 
A //(cell) tend to be rather large. Selections of AG( ce ii) values are given in Table 7.4. 



Table 7.4 Table of values of AG (CE ii) as a function of 
emf and n 



mor 1 

AG ( 

C eii)/kJ mol ' 

emf IV 

(when n - 

= D 

(when n = 2) 
































While electrochemical methods are experimentally easy, 
the practical difficulties of obtaining accurate thermo- 
dynamic data are so severe that the experimentally deter- 
mined values of AG( Ce ii), A//( ce u) and A5( ce ii) can only 
be regarded as 'approximate' unless we perform a daunt- 
ing series of precautions. The two most common errors 
are: (1) allowing current passage to occur, causing the 
value of the cell emf to be too small; and (2) not per- 
forming the measurement reversibly. 

The most common fault under (2) is changing the 
temperature of the cell too fast, so the temperature inside the cell is not the same as the 
temperature of, for example, the water bath in which a thermometer is placed. 

None of these ther- 
modynamic equations 
is reliable if we fail 
to operate the cell 
reversibly, since the 
emf is no longer an 
exact thermodynamic 

Why aren't electrodes made from wood? 

Electrodes: redox, passive and amalgam 

Electrochemical cells comprise a minimum of two half-cells, the 
energetic separation between them being proportional to the cell 
emf. Since this energy is usually expressed as a voltage, we see 
that the energy needs to be measured electrically as a voltage. 

We have seen already how a cell's composition changes if a 
charge flows through it - we argued this phenomenon in terms 
of Faraday's laws. We cause electrochemical reactions to occur 
whenever a cell converts chemical energy into electrical energy. 

The impetus for elec- 
tronic motion is the 
chance to lose energy 
(see p. 60), so elec- 
trons move from high 
energy to low. 



Currents conduct 
through an electrode 
by means of electrons. 

A redox electrode acts 
as a reagent as well as 
an electron conductor, 
as the metal of an 
electrode can also be 
one component part of 
a redox couple. 

For this reason, we say a battery or cell discharges during operation, with each 
electron from the cell flowing from high energy to low. 

All electrochemical cells (including batteries) have two poles: 
one relates to the half-cell that is positively charged and the other 
relates to the negatively charged pole. Negatively charged electrons 
are produced at the anode as one of the products of the electro- 
chemical reaction occurring at there. But if the electrons are to 
move then we need something through which they can conduct to and from the 
terminals of the cell: we need an electrode. 

The phenomenon we call electricity comprises a flow of charged electrons. Wood 
is a poor conductor of electricity because electrons are inhibited from moving freely 
through it: we say the wood has a high electrical 'resistance' R. By 
contrast, most metals are good conductors of charge. We see how 
an electrode needs to be electrically conductive if the electrons are 
to move. 

Most electrodes are metallic. Sometimes the metal of an electrode 
can also be one component part of a redox couple. Good examples 
include metallic iron, copper, zinc, lead or tin. A tin electrode forms 
a couple when in contact with tin(IV) ions, etc. Such electrodes are 
called redox electrodes (or non-passive). In effect, a redox electrode 
has two roles: first, it acts as a reagent; and, secondly, it measures 
the energy of the redox couple of which it forms one part when 
connected to a voltmeter. 

Some metals, such as aluminium or magnesium, cannot function 
as redox electrodes because of a coating of passivating oxide. Oth- 
ers, such as calcium or lithium, are simply too reactive, and would 
dissolve if immersed in solution. 

But it is also extremely common for both redox states of a redox 
couple to be non-conductive. Simple examples might include dis- 
solving bromine in an aqueous solution of bromide ions, or the 
oxidation of hydrogen gas to form protons, at the heart of a hydro- 
gen fuel cell; see Equation (7.13). In such cases, the energy of 
the couple must be determined through a different sort of elec- 
trode, which we call an inert electrode. Typical examples of inert 
electrodes include platinum, gold, glassy carbon or (at negative 
potentials) mercury. The metal of an inert electrode itself does not 
react in any chemical sense: such electrodes function merely as a 
probe of the electrode potential for measurements at zero current, 
and as a source, or sink, of electrons for electrolysis processes if 
current is to flow. 

The final class of electrodes we encounter are amalgam elec- 
trodes, formed by 'dissolving' a metal in elemental (liquid) mer- 
cury, generally to yield a solid. We denote an amalgam with brack- 
ets, so the amalgam of sodium in mercury is written as Na(Hg). The 
properties of such amalgams can be surprisingly different from their 

Metallic mercury is a 
poor choice of inert 
electrode at positive 
potentials because it 
oxidizes to form Hg(II) 

We require an inert 
electrode when both 
parts of a redox couple 
reside in solution, or 
do not conduct: the 
electrode measures the 
energy of the couple. 

We denote an amal- 
gam by writing the 'Hg' 
in brackets after the 
symbol of the dissolved 
element; cobalt amal- 
gam is symbolized as 



constituent parts, so Na(Hg) is solid and, when prepared with certain concentrations 
of sodium, does not even react with water. 

Currents conduct 
through an electrolyte 
by means of ions. 

Why is electricity more dangerous in wet weather? 

Electrolytes for cells, and introducing ions 

Most electrical apparatus is safe when operated in a dry environment, but everyone 
should know that water and electricity represent a lethal combination. Only a minimal 
amount of charge conducts through air, so cutting dry grass with an electric mower 
is safe. Cutting the same grass during a heavy downpour risks electrocution, because 
water is a good conductor of electricity. 

But water does not conduct electrons, so the charges must move 
through water by a wholly different mechanism than through a 
metallic electrode. In fact, the charge carriers through solutions - 
aqueous or otherwise - are solvated ions. The 'mobility' /x of an 
ion in water is sufficiently high that charge conducts rapidly from 
a wet electrical appliance toward the person holding it: it behaves 
as an electrolyte. 

All cells comprise half-cells, electrodes and a conductive elec- 
trolyte; the latter component separates the electrodes and conducts 
ions. It is usually, although not always, a liquid and normally has 
an ionic substance dissolved within it, the solid dissociating in 
solution to form ions. Aqueous electrolytes are a favourite choice 
because the high 'dielectric constant' e of water imparts a high 
'ionic conductivity' k to the solution. 

Sometimes electrochemists are forced to construct electrochem- 
ical cells without water, e.g. if the analyte is water sensitive or 
merely insoluble. In these cases, we construct the cell with an 
organic solvent, the usual choice being the liquids acetonitrile, 
propylene carbonate (I), A^.A^-dimethylformamide (DMF) or di- 
methylsulphoxide (DMSO), each of which is quite polar because 
of its high dielectric constant e. 

Ultimately, the word 
'ion' derives from the 
Greek eimi 'to go', 
implying the arrival of 
someone or something. 
We get the English 
word 'aim' from this 

Ionic conductivity is 
often given the Greek 
symbol kappa (k) 
whereas electrical con- 
ductivity is given the 
different Greek symbol 
sigma (a). 








In some experiments, we need to enhance the ionic conductivity of a solution, 
so we add an additional ionic compound to it. Rather confusingly, we call both the 
compound and the resultant solution 'an electrolyte'. 



The preferred electrolytes if the solvent is water are KC1 and NaNC>3. If the solvent 
is a non-aqueous organic liquid, then we prefer salts of tetra-alkyl ammonium, such 
as tetra-n-butylammonium tetrafluoroborate, "Bu4N + BF4-. 

7.2 Introducing half-cells and electrode 

Why are the voltages of watch and car batteries 

Relationships between emf and electrode potentials 

Being a cell, a battery contains two half-cells separated by an electrolyte. The elec- 
trodes are needed to connect the half-cells to an external circuit. Each electrode may 
act as part of a redox couple, but neither has to be. 

The market for batteries is huge, with new types and applications 
being developed all the time. For example, a watch battery is a type 
of 'silver oxide' cell: silver in contact with silver oxide forms one 
half-cell while the other is zinc metal and dications. Conversely, 
a car battery is constructed with the two couples lead(IV)|lead 
and lead(IV)|lead(II). The electrolyte is sulphuric acid, hence this 
battery's popular name of 'lead-acid' cell (see further discussion 
on p. 347). 

The first difference between these two batteries is the voltage 
they produce: a watch battery produces about 3 V and a lead-acid 
cell about 2 V. The obvious cause of the difference in emf are 
the different half-cells. The 'electrode potential' E is the energy, 
expressed as a voltage, when a redox couple is at equilibrium. 
As a cell comprises two half-cells, we can now define the emf 
according to 

An "electrode poten- 
tial' E is the energy 
(expressed as a volt- 
age) when a redox 
couple is at equilibrium. 
The value of E cannot 
be measured directly 
and must be calculated 
from an experimental 

Two redox states of the 
same material form a 
redox couple. 


J (positive half-cell) 

' (negative half-cell) 


This definition is absolutely crucial. It does not matter if the 
values of E for both half-cells are negative or both are positive: 
is (positive) is defined as being the more positive of the two half-cells, 
and ^(negative) is the more negative. 

We now consider the emf in more detail, and start by saying 
that it represents the separation in potential between the two half- 
cell potentials; See Equation (7.22). In order for AG( ce ii) to remain 
positive for all fhermodynamically spontaneous cell discharges, the 
emf is defined as always being positive. 

It is impossible to 
determine the potential 
of a single electrode: 
only its potential rel- 
ative to another elec- 
trode can be measured. 



Another convention dictates that we write the more positive electrode on the right- 
hand side of a cell, so we often see Equation (7.22) written in a slightly different 

emf = £ ( rhs) - £(lhs) (7.23) 

Being a potential, the electrode potential has the symbol E. We must exercise care 
in the way we cite it. E is the energy of a redox couple, since it relates to two redox 
species, both an oxidized and reduced form, 'O' and 'R' respectively. We supplement 
the symbol E with appropriate subscripts, as £"o,R- 

Worked Example 7.9 Consider the electrode potentials for metallic lead within a lead- 
acid battery. The lead has three common redox states, Pb 4+ , Pb 2+ and Pb°, so there are 
three possible equilibria to consider: 

Pb 4 

Pb 4 


+ 2e~ 

(a q) + 4e- 


Pb 2+ ran , + 2e 

Pb 2 



Pb° (s 

for which E 


for which E 


for which E 


w+.pb 2 

We now see why it is so misleading to say merely E P \, or, worse, 'the electrode potential 
of lead'. 

We cite the oxidized 
form first, as E 0R . 

We usue 

Ily omit the 




on unch 




When choosing 


ween two ionic 



the name 

of the 

higher (more oxidized) 


ends with -ic 


the lower 


oxidized) form 




We conventionally cite the oxidized form first within each sym- 
bol, which is why the general form is Eo^, so £ Pb 4+ Pb 2+ is correct, 
but E Ph 2+ pb 4+ is not. Some people experience difficulty in decid- 
ing which redox state is oxidized and which is the reduced. A 
simple way to differentiate between them is to write the balanced 
redox reaction as a reduction. For example, consider the oxidation 
reaction in Equation (7.1). On rewriting this as a reduction, i.e. 
Al 3+ ( aq ) + 3e~ = Al( s ), the oxidized redox form will automatically 
precede the reduced form as we read the equation from left to 
right, i.e. are written in the correct order. For example, £o,r f° r 
the couple in Equation (7.1) is E Al i+ A1 . 

We usually cite an uncharged participant without a superscript. 
Considering the reaction Pb + + 2e~ = Pb, the expression E Pb 2+ Pb 
is correct but the '0' in E Pb 2+ Pb o is superfluous. 

SAQ 7.10 Consider the cobaltous ion | cobalt redox cou- 
ple. Write an expression for its electrode potential. 

With more complicated redox reactions, such as 2H + (aq ) + 2e~ = 
H2( g ), we would not normally write the stoichiometric number, 
so we prefer £h+,h 2 to £ , 2 h + ,h 2 ; tne additional '2' before H + is 
superfluous here. 



SAQ 7.11 Write down an expression for the electrode potential of the 
bromine | bromide couple. [Hint: it might help to write the balanced redox 
reaction first.] 

How do 'electrochromic' car mirrors work? 

Introducing an orbital approach to dynamic electrochemistry 

It's quite common when driving at night to be dazzled by the 
lights of the vehicle behind as they reflect from the driver's new- 
view or door mirror. We can prevent the dazzle by forming a 
layer of coloured material over the reflecting surface within an 
electrochromic mirror. Such mirrors are sometimes called 'smart 
mirrors' or electronic 'anti-dazzle mirrors'. 

These mirrors are electrochromic if they contain a substance that changes colour 
according to its redox state. For example, methylene blue, MB + (II), is a chromophore 
because it has an intense blue colour. II is a popular choice of electrochromic material 
for such mirrors: it is blue when fully oxidized, but it becomes colourless when 
reduced according to 

Electrochromic mirrors 
are now a common 
feature in expensive 

MB U - 

MB + 

+ e" 


(H 3 C) 2 N 




We can now explain how an electrochromic car mirror operates. The mirror is 
constructed with II in its colourless form, so the mirror functions in a normal way. 
The driver 'activates' the mirror when the 'anti-dazzle' state of the mirror is required, 
and the coloured form of methylene blue (MB + ) is generated oxidatively according 
to Equation (7.24). Coloured MB + blocks out the dazzling reflection at the mirror by 
absorbing about 70 per cent of the light. After our vehicle has been overtaken and 
we require the mirror to function normally again, we reduce MB + back to colourless 
MB via the reverse of Equation (7.24), and return the mirror to its 
colourless state. These two situations are depicted in Figure 7.6. 

We discuss 'colour' in Chapter 9, so we restrict ourselves here to 
saying the colour of a substance depends on the way its electrons 
interact with light; crucially, absorption of a photon causes an elec- 
tron to promote between the two frontier orbitals. The separation 
in energy between these two orbitals is E, the magnitude of which 
relates to the wavelength of the light absorbed X according to the 
Planck-Einstein equation, E = hc/X, where h is the Planck constant and c is the 

An electron is donated 
to an orbital during 
reduction. The electron 
removed during oxida- 
tion is taken from an 





Incident beam 

Emerging beam 



Incident beam 

Emerging beam 

Reflective Electrochromic 
surface layers 


Figure 7.6 Mirrors: (a) an ordinary car driver's mirror reflects the lights of a following car, which 
can dazzle the driver; (b) in an electrochromic mirror, a layer of optically absorbing chemical is 
electro-generated in front of the reflector layer, thereby decreasing the scope for dazzle. The width 
of the arrows indicates the relative light intensity 

speed of light in a vacuum. The value of E for MB corresponds to an absorption 
in the UV region, so MB appears colourless. Oxidation of MB to MB + causes a 
previously occupied orbital to become empty, itself changing the energy separation 
E between the two frontier orbitals. And if E changes, then the Planck-Einstein 
equation tells us the wavelength A of the light absorbed must also change. E for 
MB + corresponds to A of about 600 nm, so the ion is blue. 

The reasoning above helps explain why MB and MB + have different colours. To 
summarize, we say that the colours in an electrochromic mirror change following 
oxidation or reduction because different orbitals are occupied before and after the 
electrode reaction. 

Why does a potential form at an electrode? 

Formation of charged electrodes 

For convenience, we will discuss here the formation of charges with the example of 
copper metal immersed in a solution of copper sulphate (comprising Cu 2+ ions). We 
consider first the situation when the positive pole of a cell is, say, bromine in contact 
with bromide ions, causing the copper to be the negative electrode. 

Let's look at the little strip cartoon in Figure 7.7, which shows the surface of a 
copper electrode. For clarity, we have drawn only one of the trillion or so atoms 
on its surface. When the cell of which it is a part is permitted to discharge spon- 
taneously, the copper electrode acquires a negative charge in consequence of an 
oxidative electron-transfer reaction (the reverse of Equation (7.7)). During the oxi- 
dation, the surface-bound atom loses the two electrons needed to bond the atom to 
the electrode surface, becomes a cation and diffuses into the bulk of the solution. 




Surface atom 


Positive ion 

Electrode Excess surface charge 

Figure 7.7 Schematic drawing to illustrate how an electrode acquires its negative charge 


Positive ion 

Surface atom 


Electrode Excess surface charge 

Figure 7.8 Schematic drawing to illustrate how an electrode acquires its positive charge 

The two electrons previously 'locked' into the bond remain on the electrode surface, 
imparting a negative potential. 

We now consider a slightly different cell in which the copper half-cell is the positive 
pole. Perhaps the negative electrode is zinc metal in contact with Zn 2+ ions. If the cell 
discharges spontaneously, then the electron-transfer reaction is the reduction reaction 
in Equation (7.7) as depicted in the strip cartoon in Figure 7.8. A bond forms between 
the surface of the copper electrode and a Cu + cation in the solution The electrons 
needed to reduce the cation come from the electrode, imparting a net positive charge 
to its surface. 

Finally, we should note that the extent of oxidation or reduction needed to cause a 
surface charge of this type need not be large; and the acquisition of charge, whether 
positive or negative, is fast and requires no more than a millisecond after immersing 
the electrodes in their respective half-cells. 



7.3 Activity 

Why does the smell of brandy decrease after 
dissolving table salt in it? 

We mention the volatile 
alcohol here because it 
is responsible for the 

Real and 'perceived' concentrations 

At the risk of spoiling a good glass of brandy, try adding a little table salt to it and 
notice how the intensity of the smell is not so strong after the salt dissolves. 

We recall from Chapter 5 how the intensity of a smell we detect 
with our nose is proportional to the vapour pressure of the substance 
causing it. The vapour pressure of ethanol is Methanol) > its magni- 
tude being proportional to the mole fraction of ethanol in the brandy; 
brandy typically contains about 40 per cent (by volume) of alcohol. 
Although adding table salt does not decrease the proportion of 
the alcohol in the brandy, it does decrease the apparent amount. And because the 
perceived proportion is lowered, so the vapour pressure drops, and we discern the 
intensity of the smell has decreased. We are entering the world of 'perceived' con- 

Although the actual concentrations of the volatile components 
in solution remain unchanged after adding the salt, the system per- 
ceives a decrease in the concentration of the volatile components. 
This phenomenon - that the perceived concentration differs from 
the real concentration - is quite common in the thermodynamics 
of solution-phase electrochemistry. We say that the concentration persists, but the 
'activity' a has decreased by adding the salt. 

As a working definition, the activity may be said to be 'the perceived concentra- 
tion' and is therefore somewhat of a 'fudge factor'. More formally, the activity a is 
defined by 

The 'activity' a is the 
thermodynamically per- 
ceived concentration. 



where c is the real concentration. The concluding term y, termed the activity coeffi- 
cient, is best visualized as the ratio of a solute's 'perceived' and 'real' concentrations. 
The activity a and the activity coefficient y are both dimension- 
less quantities, which explains why we must include the additional 
'c & ' term, thereby ensuring that a also has no units. We say the 

We only add the term 
c e in order to render 
the activity dimension- 

value of c s is 1 moldm - when c is expressed in the usual units 
of mol dm -3 , and 1 molm -3 if c is expressed in the SI units of 
mol m~ 3 , and so on. 

Why does the smell of gravy become less intense after 
adding salt to it? 

The effect of composition on activity 

Gravy is a complicated mixture of organic chemicals derived from soluble meat 
extracts. Its sheer complexity forces us to simplify our arguments, so we will 



approximate and say it contains just one component in a water-based solution. Any 
incursions into reality, achieved by extending our thoughts to encompass a multi- 
component system, will not change the nature of these arguments at all. 

Adding table salt to gravy causes its lovely smell to become less intense. This is a 
general result in cooking: adding a solute (particularly if the solute is ionic) decreases 
the smell, in just the same way as adding table salt decreased the smell of brandy in 
the example directly above. 

The ability to smell a solute relies on it having a vapour pres- 
sure above the solution. Analysing the vapour above a gravy dish 
shows that it contains molecules of both solvent (water) and solute 
(gravy), hence its damp aroma. The vapour pressures above the 
gravy dish do not alter, provided that we keep the temperature con- 
stant and maintain the equilibrium between solution and vapour. 
The proportion of the solute in the vapour is always small because 
most of it remains in solution, within the heavier liquid phase. 

As a good approximation, the vapour pressure of each solute in the vapour above 
the dish is dictated by the respective mole fractions in the gravy beneath. As an 
example, adding water to the gravy solution dilutes it and, there- 
fore, decreases the gravy smell, because the mole fraction of the 
gravy has decreased. 

Putting ionic NaCl in the gravy increases the number of ions in 
solution, each of which can then interact with the water and the solute, 
which decreases the 'perceived concentration' of solute. In fact, we 
can now go further and say the thermodynamic activity a represents 
the concentration of a solute in the presence of interactions. 

The pressure above 
a solution relates to 
the composition of 
solution, according 
to Henry's law; see 
Section 5.6. 

An electrochemist asses- 
ses the number of ions 
and their relative influ- 
ence by means of the 
'ionic strength' I (as 
defined below). 

Why add alcohol to eau de Cologne? 

Changing the perceived concentration 

Fragrant eau de Cologne is a dilute perfume introduced in Col- 
ogne (Germany) in 1709 by Jean Marie Farina. It was probably 
a modification of a popular formula made before 1700 by Paul 
Feminis, an Italian in Cologne, and was based on bergamot and 
other citrus oils. The water of Cologne was believed to have the 
power to ward off bubonic plague. 

Eau de Cologne perfume is made from about 80-85 per cent 
water and 12-15 per cent ethanol. Volatile esters make up the 
remainder, and provide both the smell and colour. 

The vapour pressure of alcohol is higher than that of water, so 
adding alcohol to an aqueous perfume increases the pressure of the 
gases above the liquid. In this way, the activity a of the organic 
components imparting the smell will increase and thereby increase 
the perceived concentration of the esters. And increasing fl( es ter) 

The word 'perfume' 
comes from the Latin 
per fumem, meaning 
'through smoke'. 

These esters are sta- 
ble in the dark, but 
degrade in strong sun- 
light, which explains 
why so many perfumes 
are sold in bottles of 
darkened or frosted 



has the effect of making the eau de Cologne more pungent. Stated another way, the 
product requires less ester because the alcohol increases its perceived concentration. 
Incidentally, the manufacturer also saves money this way. 

Thermodynamic activity a 

Every day, electrochemists perform measurements that require a knowledge of the 
activity a. Measurements can be made in terms of straightforward concentrations if 
solutions are very dilute, but 'very dilute' in this context implies c « 10~ 4 moldm -3 , 
or less. Since most solutions are far more concentrated than millimoles per litre, from 
now on we will write all equations in terms of activities a instead of concentration c. 
The values of activity a and concentration c are the same for 
very dilute solutions, so the ratio of a and c is one because the 
real and perceived concentrations are the same. If a = c, then 
Equation (7.25) shows how the activity coefficient y has a value 
of unity at low concentration. 

By contrast, the perceived concentration is usually less than the 
real concentration whenever the solution is more concentrated, so 
y < 1. To illustrate this point, Figure 7.9 shows the relationship 
between the activity coefficient y (as 'v') and concentration (as 'x') for a few sim- 
ple solutes in water. The graph shows clearly how the value of y can drop quite 
dramatically as the concentration increases. 

The concept of activity 
was introduced in the 
early 20th century by 
one of the giants of 
American chemistry, G. 
N. Lewis. 



HCI [1 : 1] 
KCI [1 : 1] 
KOH [1 : 1] 

NaBr[1 : 1] 

H 2 S0 4 [1 :2] 
CuS0 4 [2 : 2] 
ln 2 (SQ 4 ) 3 [2 : 3] 


0.4 0.6 0.8 

Concentration / mol dm 


Figure 7.9 The dependence of the mean ionic activity coefficient y± on concentration for a few 
simple solutes 



Unit and unity here 
both mean 'one', so 
unit activity means 
a = 1. 

The activity of a solid The activity of a pure solid in its standard 
state is unity, so the activity of pure copper or of zinc metal elec- 
trodes is one. We write this as a^cu) or a (Zn) = 1- 

The activity of an impure solid is more complicated. Such an 
'impure' system might be represented by a solid metal with a dirty 
surface, or it might represent a mixture of two metals, either as an alloy or an amalgam 
with a metal 'dissolved' in mercury. 

For example, consider the bi-metallic alloy known as bronze, 
which contains tin (30 mol%) and copper (70 mol%). There are 
two activities in this alloy system, one each for tin and copper. The 
activity of each metal is obtained as its respective mole fraction x, 
so x ( sn) = a (Sn ) = 0.3, and a (Cu) = 0.7. 

An 'alloy' is a mixture 
of metals, and is not a 

Worked Example 7.10 A tooth filling is made of a silver amalgam that comprises 
37 mol% silver. What is the activity of the mercury, a 


The activity of the mercury tf(Hg) is the same as its mole fraction, 
X(Hg). By definition 

*(Hg) + *(Ag) = 1 


X (Hg) 

1 -X 



The sum of the mole 
fractions x must always 
add up to one because 
'the sum of the con- 
stituents adds up to 
the whole'. 

The activity is therefore 

*(H g ) = fl(Hg) = 0.63 

Reminder: the value of 
p & is 10 5 Pa. 

The activity of a gas The activity of a pure gas is its pressure 
(in multiples of p^), so a(H 2 ) = P(h 2 ) ^ P & - The activity of pure 
hydrogen gas a(H 2 ) at p e is therefore unity. 

In fact, for safety reasons it is not particularly common to employ 
pure gases during electrochemical procedures, so mixtures are pre- 
ferred. As an example, the hydrogen gas at the heart of the standard 
hydrogen electrode (SHE) is generally mixed with elemental nitro- 
gen, with no more than 10 per cent of H2 by pressure. We call the 
other gas a base or bath gas. Conversely, we might also say that hydrogen dilutes 
the nitrogen, and so is a diluent. 

In such cases, we can again approximate the activity to the mole fraction x. 

In a mixture of gases, 
we call the inert gas a 
base or bath gas. 

Worked Example 7.11 Hydrogen gas is mixed with a nitrogen 'bath gas'. The overall 
pressure is p® . If the mole fraction of the hydrogen is expressed as 10 per cent, what is 
its activity? 

By definition, x (X ) = partial pressure, Pqq, so 

a (H 2 ) — P(H 2 ) + P(total) =0.1 



Amalgams are liquid 
when very dilute, but 
are solid if the mole 
fraction of mercury 
drops below about 70 
per cent. 

The activity of a solution It is unwise to speak in broad terms of 'the activity of 
a solution' because so many different situations may be considered. For example, 
consider the following two examples. 

(1) The activity of a mixture of liquids. It is rarely a good idea to 
suggest that the activity of a liquid in a mixture is equal to its mole 
fraction x because of complications borne of intermolecular inter- 
actions (e.g. see Chapter 2 and Section 5.6 concerning Raoult's 
law). Thankfully, it is generally rare that an electrochemist wants 
to study liquid mixtures of this sort (except amalgams diluted to a 
maximum mole fraction of about 1 per cent metal in Hg), so we 
will not consider such a situation any further. 
(2) The activity of a solute in a liquid solvent. The activity a and concentration c 
may be considered to be wholly identical if the concentration is tiny (to a maximum 
of about 10~ 3 moldm -3 ), provided the solution contains no other solutes. Such a 
concentration is so tiny, however, as to imply slightly polluted distilled water, and is 
not particularly useful. 

For all other situations, we employ the Debye-Huckel laws (as below) to calculate 
the activity coefficient y. And, knowing the value of y, we then say that a = (c 4- 
c & ) x y (Equation (7.25)), remembering to remove the concentration units because 
a is dimensionless. 

Why does the cell em f alter after adding LiCI? 

In fact, a similar result 
is obtained when adding 
most ionic electrolytes. 

Ionic 'screening' 

Consider the Daniell cell Zn|Zn 2+ ||Cu 2+ |Cu. The cell emf is about 1.1 V when pre- 
pared with clean, pure electrodes and both solutions at unit activity. The emf decreases 
to about 1.05 V after adding lithium chloride to the copper half-cell. Adding more 
LiCI, but this time to the zinc solution, increases the emf slightly, to about 1.08 V. 
No redox chemistry occurs, so no copper ions are reduced to 
copper metal nor is zinc metal oxidized to form Zn 2+ . No com- 
plexes form in solution, so the changes in emf may be attributed 
entirely to changing the composition of the solutions. 

Lithium and chloride ions are not wholly passive, but interact 
with the ions originally in solution. Let us look at the copper ions, 
each of which can associate electrostatically with chloride ions, 
causing it to resemble a dandelion 'clock' with the central copper 
ion looking as though it radiates chloride ions. All the ions are 
solvated with water. These interactions are coulombic in nature, 
so negatively charged chloride anions interact attractively with the 
positive charges of the copper cations. Copper and lithium cations 
repel. Conversely, the additional Li + ions attract the negatively 
charged sulphates from the original solution; again, Cl~ and S0 4 ~ 
anions repel. 

We need a slightly 
different form of y 
when working with 
electrolyte solutions: 
we call it the mean 
ionic activity coefficient 
Y±, as below. 



'Associated ions' in 
this context means 
an association species 
held together (albeit 
transiently) via electro- 
static interactions. 

The ionic atmosphere moves continually, so we consider its com- 
position statistically. Crystallization of solutions would occur if the 
ionic charges were static, but association and subsequent dissocia- 
tion occur all the time in a dynamic process, so even the ions in a 
dilute solution form a three-dimensional structure similar to that in 
a solid's repeat lattice. Thermal vibrations free the ions by shaking 
apart the momentary interactions. 

The ions surrounding each copper cation are termed the ionic atmosphere. In the 
neighbourhood of any positively charged ion (such as a copper cation), there are likely 
to be more negative charges than positive (and vice versa). We say the cations are 
surrounded with a shell of anions, and each anion is surrounded by a shell of cations. 
The ionic atmosphere can, therefore, be thought to look much like an onion, or a 
Russian doll, with successive layers of alternate charges, with the result that charges 
effectively 'cancel' each other out when viewed from afar. 

Having associated with other ions, we say the copper ion is screened from anything 
else having a charge (including the electrode), so the full extent of its charge cannot be 
'experienced'. In consequence, the magnitude of the electrostatic interactions between 
widely separated ions will decrease. 

The electrode potential measured at an electrode relates to the 
'Coulomb potential energy' V 'seen' by the electrode due to the 
ions in solution. V relates to two charges z.\ and z.i ( one being the 
electrode here) separated by a distance r, according to 



4jre e r r 


The 'Coulomb potential 
energy' V is equal to 
the work that must be 
done to bring a charge 

z+ from infinity to a 
distance of r from the 
charge z~. 

where e is the permittivity of free space and e r is the relative 
permittivity of the solvent. In water at 25 °C, e r has a value of 78.54. 

The magnitude of V relates to interactions between the electrode and nearby ions 
nestling within the interface separating the electrode and the ionic solution. Since 
the 'effective' (visible) charge on the ions decreases, so the electrode perceives there 
to be fewer of them. In other words, it perceives the concentration to have dipped 
below the actual concentration. This perceived decrease in the number of charges 
then causes the voltmeter to read a different, smaller value of E Cu 2+ Cu . 

The zinc ions in the other half of the Daniell cell can similarly interact with ions 
added to solution, causing the zinc electrode to 'see' fewer Zn 2+ species, and the 
voltmeter again reads a different, smaller value of E Zn 2+ Zn . Since the emf represents 
the separation between the electrode potentials of the two half-cells, any changes in 
the emf illustrate the changes in the constituent electrode potentials. 

Background to the Debye-Huckel theory 

The interactions between the ions originally in solution and any added LiCl are 
best treated within the context of the Debye-Hiickel theory, which derives from a 
knowledge of electrostatic considerations. 


Firstly, we assume the ions have an energy distribution as defined by the Boltzmann 
distribution law (see p. 35). Secondly, we say that electrostatic forces affect the 
behaviour and the mean positions of all ions in solution. It should be intuitively clear 
that ions having a larger charge are more likely to associate strongly than ions having 
a smaller charge. This explains why copper ions are more likely to associate than 
are sodium ions. The magnitude of the force exerted by an ion with a charge z.\ on 
another charge zi separated by an inter-ion distance of r in a medium of relative 
permittivity e r is the 'electrostatic interaction' 0, as defined by 


4jte e r r 2 

Positive values of <t> 
imply repulsion, and a 
negative value attrac- 

Note how this equation states that the force is inversely proportional to the square 
of the distance between the two charges r, so the value of decreases rapidly as r 

Since cations and anions have opposite charges, is negative. 
The force between two anions will yield a positive value of 0. We 
see how a positive value of implies an inter-ionic repulsion and 
a negative value implies an inter-ionic attraction. 

The Debye-Hiickel theory suggests that the probability of find- 
ing ions of the opposite charge within the ionic atmosphere increa- 
ses with increasing attractive force. 

Why does adding NaCI to a cell alter the emf, but 
adding tonic water doesn't? 

The effects of ion association and concentration on y 

Sodium chloride - table salt - is a 'strong' ionic electrolyte because it dissociates 
fully when dissolved in water (see the discussion of weak and strong acids in 
Section 6.2). The only electrolytes in tonic water are sugar (which is not ionic) and 
sodium carbonate, which is a weak electrolyte, so very few ions are formed by adding 
the tonic water to a cell. 

The ratio of perceived to real concentrations is called the activity coefficient y 
(because, from Equation (7.25), y = a -4- c). Furthermore, from the definition of activ- 
ity in Equation (7.20), y will have a value in the range zero to one. The diagram 
in Figure 7.9 shows the relationship between y and concentration c for a few ionic 

Adding NaCI to solution causes y to decrease greatly because the number of ions 
in solution increases. Adding tonic water does not decrease the activity coefficient 
much because the concentration of the ions remains largely unchanged. The change 
in y varies more with ionic electrolytes because the interactions are far stronger. And 
if the value of y does not change, then the real and perceived concentrations will 
remain essentially the same. 



The extent of ionic screening depends on the extent of associa- 
tion. The only time that association is absent, and we can treat ions 
as though free and visible ('unscreened'), is at infinite dilution. 

Why does MgCI 2 cause a greater decrease 
in perceived concentration than KCI? 

The mean ionic activity coefficient y± 

Infinite dilution (extra- 
polation to zero con- 
centration) means so 
small a concentration 
that the possibility of 
two ions meeting, and 
thence associating, is 
tiny to non-existent. 

The value of y depends 
on the solute employed. 

The extent of ionic association depends on the ions we add to 
the solution. And the extent of association will effect the extent of 
screening, itself dictating how extreme the difference is between 
perceived and real concentration. For these reasons, the value of 
y(= a H- c) depends on the choice of solute as well as its concentration, so we ought 
to cite the solute whenever we cite an activity coefficient. 

The value of y is even more difficult to predict because solutes 
contain both anions and cations. In fact, it is impossible to dif- 
ferentiate between the effects of each, so we measure a weighted 
average. Consider a simple electrolyte such as KCI, which has 
one anion per cation. (We call it a '1:1 electrolyte'.) In KCI, the 
activity coefficient of the anions is called Y(cr) an d the activity 
coefficient of the cations is 7(k+)- We cannot know either y + or 
Y-\ we can only know the value of y±. Accordingly, we modify 
Equation (7.25) slightly by writing 

We cannot know either 
y + or y_; we can only 
know the value of their 
geometric mean y ± . 



We call KCI a 1:1 elec- 
trolyte, since the ratio 
of anions to cations is 

where the only change is the incorporation of the mean ionic activity coefficient y± . 
The mean ionic activity coefficient is obtained as a geometric mean via 

Y± = VT(K+) x y (cr) 


By analogy, the expression for the mean ionic activity coefficient y± for a 2:1 
electrolyte such as K2SO4 is given by 


yI x Y- 


where the cube root results from the stoichiometry, since K2SO4 contains three ions 
(we could have written the root term alternatively as ZJy + x y + x y_, with one y 
term per ion). Again, a 1:3 electrolyte such as FeCl3 dissolves to form four ions, so 
an expression for its mean ionic activity coefficient y± will include a fourth root, etc. 



SAQ 7.12 Write an expression similar to Equation (7.29) for the 2:3 
electrolyte Fe2(S0 4 )3- 

Why is calcium better than table salt at stopping soap 

Ionic strength I and the Debye-Huckel laws 

People whose houses are built on chalky ground find that their kettles and boilers 
become lined with a hard 'scale'. We say that the water in the area is 'hard', meaning 
that minute amounts of chalk are dissolved in it. The hard layer of 'scale' is chalk 
that precipitated onto the inside surface of the kettle or boiler during heating. 

It is difficult to get a good soapy froth when washing the hands 

We look at the actions 
of soaps in Chapter 10. 

in 'hard water' because the ions from chalk in the water associate 
with the long-chain fatty acids in soap, preventing it from ionizing 
properly. Conversely, if the water contains table salt - for example, 
when washing the dishes after cooking salted meat - there is less of a problem with 
forming a good froth. Although the concentrations of sodium and calcium ions may be 
similar, the larger charges on the calcium and carbonate ions impart a disproportionate 
effect, and strongly inhibit the formation of frothy soap bubbles. 

Having discussed ionic screening and its effects on the value 
of y±, we now consider the ionic charge z. When assessing the 
influence of z, we first define the extent to which a solute pro- 
motes association, and thus screening. The preferred parameter is 
the 'ionic strength' /, as defined by 

In 'dynamic' elec- 
trochemistry (when 
currents flow) we need 
to be careful not to 
mistake ionic strength 
and current, since both 
have the symbol I. 

1 '=' 



where z, is the charge on the ion i in units of electronic charge, and c, is its concen- 
tration. We will consider three simple examples to demonstrate how ionic strengths 
/ are calculated. 

Worked Example 7.12 Calculate the ionic strength of a simple 1 : 1 electrolyte, such as 
NaCl, that has a concentration of c = 0.01 moldm -3 . 

Inserting values into Equation (7.31) we obtain 

/= \ J]{ |t Na+] x (+ 1 ) 2 ] + ( [C1 " ] x (- 1 ) 2 



terms for the 

terms for the 

sodium ions 

chloride ions 



We next insert concentration terms, noting that one sodium ion and 
one chloride are formed per formula unit of sodium chloride (which 
is why we call it a 1:1 electrolyte). Accordingly, the concentrations 
of the two ions, [Na + ] and [Cl~], are the same as [NaCl], so 

/ = i{([NaCl] x 1) + ([NaCl] x 1)} 

so we obtain the result for a 1:1 electrolyte that /(NaCl) = [NaCl]. 

Note that / has the same units as concentration: inserting 
the NaCl concentration [NaCl] = 0.01 mol dm -3 , we obtain I — 
0.01 mol dm" 3 . 

NaCl is called a '1:1 
electrolyte' because the 
formula unit contains 
one anion and one 

We obtain the result 
I = c only for 1:1 (uni- 
valent) electrolytes. 

Worked Example 7.13 Calculate the ionic strength of the 2:2 electrolyte FeSCU at a 
concentration c — 0.01 moldm - . 

Inserting charges in Equation (7.31): 

I = ^{[Fe 2+ ] x (+2) 2 + [SO 2 "] x (-2) 2 } 

We next insert concentrations, again noting that one ferrous ion and one sulphate ion are 
formed per formula unit: 

/ = -{([FeS0 4 ] x 4) + ([FeS0 4 ] x 4)} 

so we obtain the result I — 4 x c for this, a 2:2 electrolyte. 

Inserting the concentration c of [FeSO/j] = 0.01 moldm -3 , we obtain / = 0.04 mol 
dm~ , which explains why hard water containing FeSC»4 has a greater influence than 
table salt of the same concentration. 

Worked Example 7.14 Calculate the ionic strength of the 1:2 electrolyte CuCl2, again 
of concentration 0.01 moldm -3 . 

Inserting charges into Equation (7.31): 

/ = ^{[Cu 2+ ] x (+2) 2 + [CI"] x (-1) 2 } 

We next insert concentrations. In this case, there are two chloride 
ions formed per formula unit of salt, so [CI"] — 2 x [CUCI2], but only 
one copper, so [Cu 2+ ] = [CuCl 2 ]. 

/ = -{([CuCl 2 ] x 4) + (2[CuCl 2 ] x 1)} 

Note how the calcu- 
lation requires the 
charge per anion, 
rather than the total 
anionic charge. 

so we obtain the result I = 3 x c for this, a 1:2 electrolyte. And, / = 0.03 moldm 
because [CuCl 2 ] = 0.01 moldm" 3 . 




SAQ 7.13 Calculate the relationship between concentration and ionic 
strength for the 1:3 electrolyte C0CI3. 

Ionic strength I is an 
integral multiple of 
concentration c, where 
integer means whole 
number. A calculation 
of/ not yielding a whole 
number is wrong. 

Table 7.5 summarizes all the relationships between concentra- 
tion and ionic strength / for salts of the type M x+ X y ~, listed as 
a function of electrolyte concentration. Notice that the figures in 
the table are all integers. A calculation of / not yielding a whole 
number is wrong. 

Ions with large charges generally yield weak electrolytes, so the 
numbers of ions in solution are often smaller than predicted. For 
this reason, values of / calculated for salts represented by the bot- 
tom right-hand corner of Table 7.5 might be too high. 

Why does the solubility ofAgCI change after adding 
MgS0 4 ? 

Calculating values of y± 

Silver chloride is fairly insoluble (see p. 332), with a solubility 
product K sv of 1.74 x 10~ 10 mol 2 dm~ 6 . Its concentration in pure 
distilled water will, therefore, be 1.3 x 10~ 5 moldm -3 , but adding 
magnesium sulphate to the solution increases it solubility appre- 
ciably; see Figure 7.10. 

This increase in solubility is not an example of the common ion 
effect, because there are no ions in common. Also impossible is 

the idea that the equilibrium constant has changed, because it is a constant. 

Strictly, we should formulate all equilibrium constants in terms of activities rather 

than concentrations, so Equation (7.32) describes K sp for dissolving partially soluble 

AgCl in water: 

We obtain the 



[Ag + ] = 

■- 1.3 x 

10- 5 as 

the square root 

of 1.74 

x 10- 10 

mol 2 

dnrr 6 . 





[Ag+][C1 ] x Y(A g +)Y(cr) 


Table 7.5 Summary of the relationship between ionic 
strength / and concentration c. As an example, sodium 
sulfate (a 1:2 electrolyte) has an ionic strength that is 
three times larger than c 


x 2 - 

x 3 - 

x 4 - 






M 2 + 





M 3 + 





M 4 + 







0.4 0.6 0.8 

[MgSO 4 ]/0.01 mol drrr 3 


Figure 7.10 The solubility s of AgCl (as 'y') in aqueous solutions of MgS0 4 against its concen- 
tration, [MgS0 4 ] (as 'x'). T = 298.15 K 

The exact structure of the equilibrium constant on the right-hand side of Equation 
(7.32) follows from the definition of activity a in Equation (7.25). The product of the 
two y terms is y±> 

The values of the activity coefficients decrease with increasing ionic strength / (as 
below). The only way for ^" sp to remain constant at the same time as the activity 
coefficient y± decreasing is for the concentrations c to increase. And this is exactly 
what happens: the concentration of AgCl has increased by about 50 per cent when 
the concentration of MgS0 4 is 1.2 mol dm -3 . 

Changes in solubility product are one means of experimentally determining a value 
of activity coefficient, because we can independently determine the concentrations 
(e.g. via a titration) and the values of all y± will be 'one' at zero 
ionic strength. 

Alternatively, we can calculate a value of y± with the Debye- 
Huckel laws. There are two such laws: the limiting and the simpli- 
fied laws. Calculations with the limiting law are only valid at very 

low ionic strengths (i.e. < / < 10 mol dm ), which is very 
dilute. The limiting law is given by 

An ionic strength of 
10 3 mol dm 3 could 
imply a concentra- 
tion as low as 10~ 4 
moldnrr 3 , because 
I >c. 

logioK± = -Mz + z, |V7 


where A is the so-called Debye-Hiickel 'A' constant (or factor), 
which has a value of 0.509 mol -1 ^ 2 dm 3 ^ 2 at 25 °C. z + and z.~ are 
the charges per cation and per anion respectively. The vertical mod- 
ulus lines '|' signify that the charges on the ions have magnitude, 
but we need to ignore their signs (in practice, we call them both 

The quantities bet- 
ween the two vertical 
modulus lines T have 
magnitude alone, so 
we ignore the signs 
on the charges z+ 
and z~ 



From Equation (7.28), we expect a plot of log 10 y± (as 'y') against v7 (as 'x') 
to be linear. It generally is linear, although it deviates appreciably at higher ionic 

Worked Example 7.15 What is the activity coefficient of copper in a solution of copper 
sulphate of concentration 10~ 4 moldm - ? 

Note how we ignore 
the sign of the negative 
charge here. 

When calculating with 
Equation (7.33), be 
sure to use 'log' (in 
base 10) rather than 
'In' (log in base e). 

At extremely low ionic 
strengths, the simpli- 
fied law becomes the 
limiting law. This fol- 
lows since the denomi- 
nator 'I + bv7' tends to 
one as ionic strength 
I tends to zero, caus- 
ing the numerator to 
become one. 

Copper sulphate is a 2:2 electrolyte so, from Table 7.5, the ionic 
strength / is four times its concentration. We say / = 4 x 10 -4 

moldm 3 . 

Inserting values into Equation (7.33): 

log 10 Y ± = -0.509 | + 2 x -2|(4 x 10 -4 ) 1/2 

log 10 y ± = -2.04 x (2 x 10 -2 ) 

logio Y± = -4-07 x 10 -2 

Taking the anti-log: 

Y± = 10 -00407 

Y± = 0.911 

We calculate that the perceived concentration is 91 percent of the real 

For solutions that are more concentrated (i.e. for ionic strengths 
in the range 10 -3 < I < 10 -1 ), we need the Debye-Hiickel sim- 
plified law: 

where all other terms have the same meaning as above, and b 
is a constant having an approximate value of one. We include b 
because its units are mor~ 1/2 dm 3/2 . It is usual practice to say b = 
1 mol - ' dm ' , thereby making the denominator dimensionless. 

SAQ 7.14 Prove that the simplified law becomes the limiting law at very 
low J. 

Worked Example 7.16 What is the activity coefficient of a solution of CUSO4 of con- 
centration 10~ 2 moldm - ? 

Again, we start by saying that I — 4 x c, so I — 4 x 10 -2 moldm -3 . Inserting values 
into Equation (7.34): 

log 10 y± = 

0.509 |2 x -2|V4 x 10~ 2 
1 + V4 x 10~ 2 



Table 7.6 Typical activity coefficients y± for ionic electrolytes as a function of concentration c 
in water 



mol drrT c - 

= 10- 3 /mol 

dm 3 c = 

= 10" 2 /mol 


- 3 c~- 

= 10" '/mol 

dm 3 c = 

= 1.0/moldm -3 

HC1 (1:1) 





KOH (1:1) 





CaCl 2 (1:2) 





CuS0 4 (2:2) 





In 2 (S0 4 ) 3 (2:3) 





logio Y± = ~ 

1 + 0.2 

logio Y± = -0.3393 

Y± = 10 



Y± = 0.458 

Table 7.6 cites a few sample values of y± as a function of concentration. Note how 
multi-valent anions and cations cause y± to vary more greatly than do mono-valent 
ions. The implications are vast: if an indium electrode were to be immersed in a 
solution of In2(S04)3 of concentration 0.1 mol dm -3 , for example, then a value of 
y± = 0.035 means that the activity (the perceived concentration) would be about 30 
times smaller! 

SAQ 7.15 From Worked Example 7.15, the mean ionic activity coefficient 
y± is 0.911 for CUSO4 at a concentration of 10 -4 mol drrT 3 . Show that 
adding MgS0 4 (of concentration 0.5 mol dm" 3 ) causes y± of the CUSO4 to 
drop to 0.06. [Hint: first calculate the ionic strength. [MgS0 4 ] is high, so 
ignore the CUSO4 when calculating the ionic strength /.] 

7.4 Half-cells and the Nernst equation 

Why does sodium react with water yet copper doesn't? 

Standard electrode potentials and the E e scale 

Sodium reacts with in water almost explosively to effect the reaction 

Na (s) + IT 1 


Na" 1 

(aq) + 2 H 2(g) 



The protons on the left-hand side come from the water. Being spontaneous, the value 
of AG r for Equation (7.25) is negative. The value of AG r comprises two components: 

(1) AG for the oxidation reaction Na ► Na + + e~; and 

(2) AG for the reduction reaction H + + 2e~ ► jH-2- 

Since these two equations represent redox reactions, we have effectively separated a 
cell into its constituent half-cells, each of which is a single redox couple. 

By contrast, copper metal does not react with water to liberate hydrogen in a 
reaction like Equation (7.35); on the contrary, black copper(II) oxide reacts with 
hydrogen gas to form copper metal: 

CuO (s) + H 2(g) ► Cu (s) + 2H+ + O 2- (7.36) 

The protons and oxide ions combine to form water. Again, the value of AG r for 
Equation (7.36) is negative, because the reaction is spontaneous. AG would be pos- 
itive if we wrote Equation (7.36) in reverse. The change in sign follows because the 
Gibbs function is a function of state (see p. 83). 

The reaction in Equation (7.36) can be split into its two constituent half-cells: 

(1) AG for the reaction Cu 2+ + 2e" ► Cu; and 

(2) AG for the reaction H 2 ► 2H+ + 2e". 

Let us look at Equation (7.36) more closely. The value of AG r comprises two 
components, according to Equation (7.36): 

AG r = (AG Cu 2 + ^ Cu ) + (AG H2 ^2H+) (7.37) 

If we wished to be wholly consistent, we could write both reac- 
tions as reduction processes. Reversing the direction of reaction 
(2) means that we need to change the sign of its contribution toward 
the overall value of AG r , so 

This change of sign 
follows from the 
change in direction of 
the second reaction. 

AG r = (AG Cu 2 + ^ Cu ) - (AG 2H +^h 2 ) (7.38) 

We remember from Equation (7.15) how AG( ce ii) = —nF x emf. We will now 
invent a similar equation, Equation (7.39), which relates AG for a half-cell and 
its respective electrode potential £o,r, saying: 

AGo.r = -nFE ^ (7.39) 

Substituting for AGo,r in Equation (7.38) with the invented expression in Equation 
(7.39) gives 

AG r = (-nFE Cu 2 +Cu ) - (-nFE H+ , H2 ) (7.40) 



The 'standard electrode 
potential' E^ R is the 
value of E ,R obtained 
at standard conditions. 

This expression does not relate to a true cell because the two electrode potentials 
are not measured with electrodes, nor can we relate AG r to the emf, because elec- 
trons do not flow from one half-cell via an external circuit to the other. Nevertheless, 
Equation (7.40) is a kind of proof that the overall value of AG r relates to the con- 
stituent half-cells. 

If we write a similar expression to that in Equation (7.40) for the reaction between 
sodium metal and water in Equation (7.35), then we would have to write the term 
for the hydrogen couple first rather than second, because the direction of change 
within the couple is reversed. In fact, any couple that caused hydrogen gas to form 
protons would be written with the hydrogen couple first, and any couple that formed 
hydrogen gas from protons (the reverse reaction) would be written with the hydrogen 
term second. 

This observation led the pioneers of electrochemical thermodynamics to construct 
a series of cells, each with the H + |H2 couple as one half-cell. The emf of each 
was measured. Unfortunately, there were always more couples than measurements, 
so they could never determine values for either E Cu 2+ Cu or £h+,h 2 
(nor, indeed, for any electrode potential), so they commented on 
their relative magnitudes, and compiled a form of ranking order. 

These scientists then suggested that the value of £h+,h 2 should 
be defined, saying that at a temperature of 298 K, pumping the 
hydrogen gas at a pressure of hydrogen of 1 atm through a solution 
of protons at unit activity generates a value of £h+,h 2 that is always 
zero. They called the half-cell 'H2( g )(/> = 1 atm)|H + (a = 1)' the 
standard hydrogen electrode (SHE), and gave it the symbol E^ + H . 
The '°' symbol indicates standard conditions. 

Then, knowing E^ + H , it was relatively easy to determine values 
of electrode potentials for any other couple. With this methodology, 
they devised the 'standard electrode potentials' E & scale (often 
called the 'E nought scale', or the 'hydrogen scale'). 

Table 7.7 contains a few such values of E & , each of which was 
determined with the same standard conditions as for the hydrogen 
couple, i.e. at T = 298 K, all activities being unity and p = 1 atm 
(the pressure is not, therefore, p^). 

Negative values of E & (such as £ Na + Na = —2.71 V) indicate 
that the reduced form of the couple will react with protons to form 
hydrogen gas, as in Equation (7.35). The more negative the value 
of E & , the more potent the reducing power of the redox state, so 
E e for the magnesium couple is —2.36 V, and E K+ K = —2.93. 
Zinc is a less powerful reducing agent, so E 2+ = —0.76 V, 

and a feeble reducing agent like iron yields a value of £.? 2 + F of 
only -0.44 V. 

And positive values of E e indicate that the oxidized form of 
the redox couple will oxidize hydrogen gas to form protons, again 

A pressure of p = 1 atm 
is not the same as 
p e , but its use is a 
permissible deviation 
within the SI scheme. 

Negative values of 
Eq r indicate that the 
reduced form of the 
couple will react with 
protons to form hydro- 
gen gas. 

Positive values of E^ R 
indicate that the oxi- 
dized form of the redox 
couple will oxidize 
hydrogen gas to form 



Table 7.7 The electrode potential series (against the SHE). The electrode potential series is an 
arrangement of reduction systems in ascending order of their standard electrode potential £" e 



E & /V Couple a 


Sm + + 2e = Sm 

Li+ + e~ = Li 

K+ + e" = K 

Rb+ + e" = Rb 

Cs+ + e" = Cs 

Ra 2+ + 2e" = Ra 

Ba 2+ + 2e" = Ba 

Sr 2+ + 2e~ = Sr 

Ca 2+ + 2e" = Ca 

Na+ + e~ = Na 

Ce 3+ + 3e- = Ce 

Mg 2+ + 2e" = Mg 

Be 2+ + 2e" = Be 

U 3+ + 3e" = U 

Al 3+ + 3e - = Al 

Ti 2+ + 2e" = Ti 

V 2+ + 2e" = V 

Mn 2+ + 2e" = Mn 

Cr 2+ + 2e - = Cr 

2H 2 + 2e - = H 2 + 20H~ 

Cd(OH) 2 + 2e" = Cd + 20H~ 

Zn 2+ + 2e" = Zn 

Cr 3+ + 3e" = Cr 

2 + e" = OJ 

In 3+ + e~ = In 2+ 

S + 2e" = S 2 ~ 

In 3+ + 2e" = In+ 

Fe 2+ + 2e" = Fe 

Cr 3+ + e~ = Cr 2+ 

Cd 2+ + 2e- = Cd 

In 2+ + e" = In+ 

Ti 3+ + e" = Ti 2+ 

PbS0 4 + 2e- = Pb + SO 2 " 

In 3+ + 3e" = In 

Co 2+ + 2e- = Co 

Ni 2+ + 2e" = Ni 

Agl + e~ = Ag + 1 - 

Sn 2+ + 2e _ = Sn 

In+ + e" = In 


Pb 2+ + 2e- = Pb 



Fe 3+ + 3e _ = Fe 



Ti 4 + + e - = Ti 3+ 



2 H + + 2e~ = H2 {by definition) 



AgBr + e~ = Ag + Br 



Sn 4+ + 2e - = Sn 2+ 



Cu 2+ + e" = Cu+ 



Bi 3+ + 3e" = Bi 



AgCl + e" = Ag + Cr 



Hg 2 Cl 2 + 2e - = 2Hg + 2C1" 



Cu 2+ + 2e" = Cu 



2 + 2H 2 + 4e" = 40H~ 



NiOOH + H 2 + e~ = 

: Ni(OH) 2 + OH" 



Cu + + e~ = Cu 



I 3 - + 2e - = 31" 



I 2 + 2e - = 21- 



MnO^ + 3e" = Mn0 2 



Hg 2 S0 4 + 2e - = 2Hg 

+ so 2 - 



Fe 3+ + e" = Fe 2+ 



AgF + e" = Ag + F 



Hg 2+ + 2e" = 2Hg 



Ag+ + e" = Ag 



2Hg 2+ + 2e" = Hg 2+ 



Pu 4+ + e" = Pu 3+ 



Br 2 + 2e" = 2Br" 



Pr 2+ + 2e" = Pr 



Mn0 2 + 4H+ + 2e" = 

Mn 2+ + 2H 2 



2 + 4H+ + 4e" = 2H 2 



Cl 2 + 2e- = 2Cr 



Au 3+ + 3e" = Au 



Mn 3+ + e" = Mn 2+ 



MnO^ + 8H+ + 5e" = 

= Mn 2+ + 4H 2 



Ce 4+ + e- = Ce 3+ 



Pb 4+ + 2e- = Pb 2+ 



Au + + e _ = Au 



Co 3+ + e" = Co 2+ 



Ag 2+ + e" = Ag+ 



S 2 2 " + 2e- = 2S0 2 " 



F 2 + 2e _ = 2F" 


a The more positive the value of E & , the more readily the half- reaction occurs in the direction left to right; the 

more negative the value, the more readily the reaction occurs in the direction right to left. 

b Elemental fluorine is the strongest oxidizing agent and Sm 2+ is the weakest. Oxidizing power increases from 

Sm 2+ to F 2 . 

c Samarium is the strongest reducing agent and F~ is the weakest. Reducing power increases from F~ to Sm. 



with the magnitude of E e indicating the oxidizing power: E^ 2+ 
a powerful oxidizing agent such as bromine has a value of E^ _ = 

= +0.34 V, but 

+ 1.09 V. 
In summary, sodium reacts with water and copper does not in consequence of their 
relative electrode potentials. 

y does a torch battery eventually *go fla 

The Nernst equation 

A new torch battery has a voltage of about 1.5 V, but the emf decreases with usage 
until it becomes too small to operate the torch for which we bought it. We say the 
battery has 'gone flat', and throw it away. 

We need to realize from Faraday's laws that chemicals within a battery are con- 
sumed every time the torch is switched on, and others are generated, causing the 
composition within the torch to change with use. Specifically, we alter the relative 
amounts of oxidized and reduced forms within each half-cell, causing the electrode 
potential to change. 

The relationship between composition and electrode potential is 
given by the Nernst equation 

Eo , R = E^ + ^m( a -p) 

nF \a {K) J 


Though it is relatively 
easy to formulate rela- 
tions like the Nernst 
equation here for a 
cell, Equation (7.41) 
properly relates to a 

where E Q R is the standard electrode potential determined at s.t.p. 
and is a constant, £o,r is the electrode potential determined at non 
s.t.p. conditions. R, T, n and F have their usual definitions. 

The bracket on the right of Equation (7.41) describes the relative activities of 
oxidized and reduced forms of the redox couple within a half-cell. The battery goes 
flat because the ratio a(o)/a(R) alters with battery usage, so the value of £o,r changes 
until the emf is too low for the battery to be useful. 

Worked Example 7.17 A silver electrode is immersed into a dilute solution of silver 
nitrate, [AgN0 3 ] = 10~ 3 mol -3 . What is the electrode potential E Ag+Ag at 298 K? Take 


0.799 V. 

The Nernst equation, Equation (7.41), for the silver couple is 


E Ag \A g = £ Ag +, A g + ~y ln 

a (Ag + ) 

a {Ag) 

For simplicity, we assume that the concentration and activity of 
silver nitrate are the same, i.e. <3(A g +) = 10 -3 . We also assume that the 
silver is pure, so its activity is unity. 

We use the approxima- 
tion 'concentration 
= activity' because the 
solution is very dilute. 



The value of RT/F is 
0.0257 Vat 298 K. 

Inserting values into Equation (7.41): 

^Ag+.Ag = 0.799 V + 0.0257 Vln 


Note how, as a conse- 
quence of the laws of 
arithmetic, we multiply 
the RT/F term with the 
logarithm term before 
adding the value of 




Ea s \a s = 0.799 V + (0.0257 V x -6.91) 


0.799 V- 0.178 V 

Note how the difference 
between E and E e is 
normally quite small. 

£ Ag+ , Ag = 0.621 V 

SAQ 7.16 A wire of pure copper is immersed into a solu- 
tion of copper nitrate. If E 

0.24 V, what is the concentration of Cu' 
a (Cu 2 +) is the same as [Cu 2+ ]. 

= 0.34 V and E r ? 

,Cu Cu 


? Assume that 

The Nernst equation cannot adequately describe the relationship between an elec- 
trode potential £o,r and the concentration c of the redox couple it represents, unless 
we substitute for the activity, saying from Equation (7.28), a = c x y±. 

The form of Equation (7.41) will remind us of the equation of a straight line, so a 
plot of £o,r as the observed variable (as 'y'), against ln(fl(Q) 4- fl(R)) (as 'x') should 
be linear with a gradient of RT -r- nF and with E Q R as the intercept on the y-axis. 

Worked Example 7.18 Determine a value for the standard electrode potential Ef + . 
with the data below. Assume that y± — 1 throughout. 

[AgN0 3 ]/moldm- 3 0.001 0.002 0.005 0.01 0.02 0.05 0.1 

£ Ag+, Ag /V 0.563 0.640 0.664 0.682 0.699 0.723 0.741 

Figure 7.11 shows a Nernst graph drawn with the data in the table. The intercept of the 
graph is clearly 0.8 V. 

Why does E AgC \ iAg change after immersing an SSCE in a 
solution of salt? 

Further calculations with the Nernst equation 

Take a rod of silver, and immerse it in a solution of potassium 
chloride. A thin layer of silver chloride forms on its surface when 
the rod is made positive, generating a redox couple of AgCl|Ag. 
We have made a silver-silver chloride electrode (SSCE). 

Now take this electrode together with a second redox couple (i.e. 
half-cell) of constant composition, and dip them together in a series 

Care: In some books, 
SSCE is taken to mean 
a sodium chloride 
saturated calomel elec- 


-6 -4 

ln([Ag + ]/mol dm 3 ) 

Figure 7.11 Nernst graph of the electrode potential E Ag + Ag as 'y' against ln([Ag + ]/moldirT 3 ) 
as 'x'. A value of E + = 0.8 V is obtained as the intercept on the y-axis 

of salt solutions, and measure the emf. The magnitude of the emf will depend on the 
concentration of the salt. The silver rod and its outer layer of silver chloride do not 
alter, so why does the emf change? 

The electrode potential i?AgCi,Ag relates to the following redox reaction: 

AgCl (s) + e- ► Ag° (s) + cr 



This redox couple is more complicated than any we have encountered yet, so the 
Nernst equation will appear to be a little more involved than those above: 



£ A g ci,Ag + ~y 



*(Ag°) a (cn , 


Silver chloride is the oxidized form, so we write it on top of the bracketed fraction, 
and silver metal is the reduced form, so we write it beneath. But we must also write a 
term for the chloride ion, because Cl~( aq ) appears in the balanced reduction reaction 
in Equation (7.42). 

If we immerse a silver electrode bearing layer of AgCl in a concentrated solution 
of table salt, then the activity fl(cr) W11 l be high; if the solution of salt is dilute, then 
^AgCi.Ag will change in the opposite direction, according to Equation (7.43). 

SAQ 7.17 An SSCE electrode is immersed in a solution of [NaCI] = 
0.1 moldrrr 3 . What is the value of f Ag ci,Ag? Take E* gOAg = 0.222 V. Take 
all y± = 1. [Hint: the activities a ( A g ci) and a (Ag0) are both unity because both 
are pure solids.] 



Why 'earth' a plug? 

Reference electrodes 

An electrical plug has three connections (or 'pins'): 'live', 'neutral' and 'earth'. The 
earth pin is necessary for safety considerations. The potential of the earth pin is the 
same as that of the ground, so there is no potential difference if we stand on the 
ground and accidentally touch the earth pin in a plug or electrical appliance. We will 
not be electrocuted. Conversely, the potentials of the other pins are different from 
that of the earth - in fact, we sometimes cite their potentials with respect to the earth 
pin, effectively defining the potential of the ground as being zero. 

The incorporation of an earth pin is not only desirable for safety, it also enables 
us to know the potential of the other pins, because we cite them with respect to the 
earth pin. 

A reaction in an electrochemical cell comprises two half-cell reactions. Even when 
we want to focus on a single half-cell, we must construct a whole cell and determine 
its cell emf, which is defined as '^(positive electrode) - ^(negative electrode)'- Only when we 
know both the emf and the value of one of the two electrode potentials can we 
calculate the unknown electrode potential. 

A device or instrument having a known, predetermined electrode 
potential is called a reference electrode. A reference electrode is 
always necessary when working with a redox couple of unknown 
£o,r- A reference electrode acts in a similar manner to the earth 
pin in a plug, allowing us to know the potential of any electrode 
with respect to it. And having defined the potential of the refer- 
ence electrode - like saying the potential of earth is zero - we then 
know the potential of our second electrode. 

A reference electrode 
is a constant-potential 
device. We need such 
a reference to deter- 
mine an unknown 
electrode potential. 


At the heart of any reference electrode lies a redox couple of known composition: any 
passage of current through the reference electrode will change its composition (we argue 
this in terms of Faraday's laws in Table 7.1). This explains why we must never allow a 
current to flow through a reference electrode, because a current will alter its potential. 

The standard hydrogen electrode - the primary reference 

The internationally accepted primary reference is the standard 
hydrogen electrode (SHE). The potential of the SHE half-cell is 
defined as 0.000 V at all temperatures. We say the schematic for 
the half-cell is 

We define the value 
of F(she) as zero at al 

Pt|H 2 ( fl = l)|H+(a=l,aq)| 



The SHE is depicted in Figure 7.12, and shows the electrode 
immersed in a solution of hydrogen ions at unit activity (corre- 
sponding to 1.228 moldm -3 HC1 at 20 °C). Pure hydrogen gas at 
a pressure of 1 atm is passed over the electrode. The electrode 
itself consists of platinum covered with a thin layer of 'platinum 
black', i.e. finely divided platinum, electrodeposited onto the plat- 
inum metal. This additional layer thereby catalyses the electrode 
reaction by promoting cleavage of the H-H bonds. 

Table 7.8 lists the advantages and disadvantages of the SHE. 

We employ hydrochlo- 
ric acid of concentra- 
tion 1.228 moldm 3 
at 20 °C because the 
activity of H + is less 
than its concentration, 
i.e. y± < 1. 

Gaseous hydrogen 
(at a pressure of 1 atm) 




Aqueous acid 

with protons at 

unit activity 

Figure 7.12 Schematic depiction of the standard hydrogen electrode (SHE). The half-cell sche- 
matic is therefore Pt|H 2 (a = l)|H+(a = 1) 

Table 7.8 Advantages and disadvantages of using the standard 
hydrogen electrode (SHE) 

Advantage of the SHE 

The SHE is the international standard 

Disadvantages of the SHE 

Safety The SHE is intrinsically dangerous because H 2 

gas is involved 
Size The SHE requires cumbersome apparatus, 

including a heavy cylinder of hydrogen 
Cost The SHE can be expensive because of using 

H 2 gas 
Accuracy With the SHE it is difficult to ensure that the 

activity of the protons is exactly unity 
Precision The SHE is prone to systematic errors, e.g. 

cyclic fluctuations in the H 2 pressure 



Measurement with the hydrogen electrode The SHE is the primary reference 
electrode, so other half-cell potentials are measured relative to its potential. In practice, 

if we wish to determine the value of E 


then we construct a cell of the type 

Remember that the 
proton is always sol- 
vated to form a hydro- 
xonium ion, so a (H3 o+) 

= 3(H+). 

The words unit activ- 
ity here mean that 
the activity of silver 
ions is one (so the 
system perceives the 
concentration to be 
1 moldrrr 3 ). 

Just because a half- 
reaction appears in a 
table is no guarantee 
that it will actually 
work; such potentials 
are often calculated. 

The SHE is sometimes 
erroneously called a 
normal hydrogen elec- 
trode (NHE). 

Pt|H 2 (a= l)|H + (a= l)||M"+(a= 1)|M 

Worked Example 7.19 We immerse a piece of silver metal into a 
solution of silver ions at unit activity and at s.t.p. The potential across 
the cell is 0.799 V when the SHE is the negative pole. What is the 
standard electrode potential £ & of the Ag + , Ag couple? 

By definition 

Cmj — lL (positive electrode) -^ (negative electrode) 

Inserting values gives 

0.799 V = £* + Ag - £ (SHE ) 
0.799 V= Ef + . 

Ag + ,Ag 

because £(she) = 0. 

The value of i?Ag+,Ag i n this example is the standard electrode 
potential because a (Ag +^ — 1, and s.t.p. conditions apply. We say that 

£■*„+ a — E . + . = 0.799 V versus the SHE. Most of the values of 

E in Table 7.7 were obtained in a similar way, although some were 

We should be aware from Table 7.8 that the SHE is an ideal device, 
and the electrode potential will not be exactly V with non-standard 

Secondary reference electrodes 

The SHE is experimentally inconvenient, so potentials are often measured and quoted 
with respect to reference electrodes other than the SHE. By far the most common 
reference is the saturated calomel electrode (SCE). We will usually make our choice 
of reference on the basis of experimental convenience. 



the 'S 

' of 









The SCE 

By far the most common secondary reference electrode is the SCE: 
Hg|Hg 2 Cl 2 |KCl(sat'd)| 



■ Lead to voltmeter 

■ Glass tube 

■ Platinum contact 

- Saturated KCI solution 

■ Paste of Hg and Hg 2 CI 2 


- Crystals of KCI 
Porous sinter 
Figure 7.13 Schematic representation of the saturated calomel electrode (SCE) 

The potential of the SCE is 0.242 V at 298 K relative to the SHE. 

At the 'heart' of the SCE is a paste of liquid mercury and 
mercurous chloride (Hg 2 Cl2), which has the old-fashioned name 
'calomel'. Figure 7.13 depicts a simple representation of the SCE. 

The half-cell reaction in the SCE is 

Hg 2 Cl 2 + 2e" 

■* 2CP + 2Hg 


So i?(scE) = £Hg 2 ci 2 ,Hg- From this redox reaction, the Nernst equa- 
tion for the SCE is 

'Hg 2 Cl 2 ,Hg 

= E 

Hg 2 Cl 2 ,Hg 

Calomel is the old- 
fashioned name for 
mercurous chloride, 
Hg 2 CI 2 . Calomel was a 
vital commodity in the 
Middle Ages because 
it yields elemental 
mercury ('quick sil- 
ver') when roasted; the 
mercury was required 
by alchemists. 


the square terms for mercury and the chloride ion are needed in 
response to the stoichiometric numbers in Equation (7.44). Both 
mercury and calomel are pure substances, so their activities are 
unity. If the activity of the chloride ion is maintained at a constant 
level, then £(sce) will have a constant value, which explains why 
the couple forms the basis of a reference electrode. 

Changing fl(cr) must alter £'Hg 2 ci 2 ,Hg» since these two variables 
are interconnected. In practice, we maintain the activity of the chlo- 
ride ions by placing surplus KCI crystals at the foot of the tube. 
The KCI solution is saturated - hence the 'S' in SCE. For this reason, we should 
avoid any SCE not showing a crust of crystals at its bottom, because its potential will 
be unknown. Also, currents must never be allowed to pass through an SCE, because 
charge will cause a redox change in E^CE)- 

Table 7.9 lists the advantages and disadvantages of the SCE reference. Despite 
these flaws, the SCE is the favourite secondary reference in most laboratories. 

Oxidative currents 
reverse the reaction 
in Equation (7.44). 
cr and Hg are con- 
sumed and Hg 2 CI 2 
forms. The denomina- 
tor of Equation (7.45) 
decreases, and F(sce) 



Table 7.9 The advantages and disadvantages of using the saturated calomel 
electrode (SCE) 

Advantages of the SCE 




SCEs are easy to make, and hence they are cheap 
SCEs can be made quite small (say, 2 cm long 

and 0.5 cm in diameter) 
Unlike the SHE, the SCE is non-flammable 

Disadvantages of the SCE 

Contamination Chloride ions can leach out through the SCE sinter 

Temperature effects The value of dE/dT is quite large at 0.7 mVKT 1 

Solvent The SCE should not be used with non-aqueous 


The silver-silver chloride electrode 

We make the best films 
of AgCI by anodizing a 
silver wire in aqueous 
KCI, not in HCI. The 
reasons for the differ- 
ences in morphology 
are not clear. 

Photolytic breakdown 
is a big problem when 
we light the laboratory 
with fluorescent strips. 

The silver-silver chloride electrode (SSCE) is another secondary 
reference electrode. A schematic of its half-cell is: 

Ag|AgCl|KCl(aq, sat'd) 

The value of E Agci A = 0.222 V. 

The AgCI layer has a pale beige colour immediately it is made, 
but soon afterwards it assumes a pale mauve and then a dark purple 
aspect. The colour changes reflect chemical changes within the film, 
caused as a result of photolytic breakdown: 

AgCI + hv + electron donor ► Ag + CI" 


exposure to light. 

introduce colloids 

Table 7.10 lists 

The purple colour is caused by colloidal silver, formed in a sim- 
ilar manner to the image on a black-and-white photograph after 
For this reason, an SSCE should be remade fairly frequently. (We 
in Chapter 10.) 
the advantages and disadvantages of SSCEs. 

Table 7.10 The advantages and disadvantages of using the silver-silver chloride electrode (SSCE) 

Advantages of the SSCE 




Disadvantages of the SSCE 
Photochemical stability 

The SSCE is easy and extremely cheap to make 

The SSCE has the smallest temperature voltage coefficient 

of any common reference electrode 
An SSCE can be as large or small as desired; it can even be 

microscopic if the silver is thin enough 

The layer of AgCI must be remade often 

The solid AgCI loses Cl~ ions during photolytic breakdown 



7.5 Concentration cells 

Why does steel rust fast while iron is more passive? 

Concentration cells 

Steel is an impure form of iron, the most common contaminants 
being carbon (from the coke that fuels the smelting process) and 
sulphur from the iron oxide ore. 

Pure iron is relatively reactive, so, given time and suitable con- 
ditions of water and oxygen, it forms a layer of red hydrated iron 
oxide ('rust'): 

Many iron ores also 
contain iron sulphide, 
which is commonly 
called fool's gold. 

4Fe (s) + 30 2 + nH 2 ► 2Fe 2 3 • (H 2 0)„ (s 



By contrast, steel is considerably more reactive, and rusts faster and to a greater 

The mole fraction x of Fe in pure iron is unity, so the activity of 
the metallic iron is also unity. The mole fraction x of iron in steel 
will be less than unity because it is impure. The carbon is evenly 
distributed throughout the steel, so its mole fraction x ( q is con- 
stant, itself ensuring that the activity is also constant. Conversely, 
the sulphur in steel is not evenly distributed, but resides in small 
(microscopic) 'pockets'. In consequence, the mole fraction of the 
iron host X(p e ) fluctuates, with x being higher where the steel is more pure, and lower 
in those pockets having a high sulphur content. To summarize, there are differences 
in the activity of the iron, so a concentration cell forms. 

The emf of a concentration cell (in this case, on the surface of 
the steel where the rusting reaction actually occurs) is given by 

We define a concen- 
tration cell as a cell in 
which the two half-cells 
are identical except for 
their relative concen- 






The electrolyte on the 
surface of the iron com- 
prises water containing 
dissolved oxygen (e.g. 
rain water). 

Notice how this emf has no standard electrode potential E & terms 
(unlike the Nernst equation from which it derives; see Justification 
Box 7.2). 

A voltage forms between regions of higher irons activity a\ and 
regions of lower iron activity a 2 (i.e. between regions of high purity 
and low iron purity); see Figure 7.14. We can write a schematic 
for a microscopic portion of the iron surface as: 

Fe(a 2 ),S,C|0 2 ,H 2 0|Fe(ai),S,C 

The commas in this 
schematic indicate that 
the carbon and sulphur 
impurities reside within 
the same phase as the 



Humid air 

Sulphur cluster: 

area of steel having 

a low mole fraction of iron 

steel having a 

relatively high 

mole fraction of iron 

Figure 7.14 Concentration cells: a voltage forms between regions of higher iron activity ci\ and 
regions of lower iron activity a 2 (i.e. between regions of high purity and low iron purity). The 
reaction at the positive 'anode' is 4Fe + 30 2 -*■ 2Fe 2 03, and the reaction at the negative 'cathode' 
is S + 2e - -» S 2 ~ 

There is no salt bridge or any other means of stopping current flow in the micro- 
scopic 'circuit' on the iron surface, so electrochemical reduction occurs at the right- 
hand side of the cell, and oxidation occurs at the left: 

at the LHS, the oxidation reaction is formation of rust (Equation (7.47)); 

at the RHS, the reduction reaction is usually formation of sulphide, via S + 

2e~ -► S 2 ". 

We can draw several important conclusions from the example of rusting steel. Firstly, 
if the impurities of carbon and sulphur are evenly distributed throughout the steel then, 
whatever their concentrations, the extent of rusting will be less than if the impurities 
cluster, because the emf of a concentration cell is zero when the ratio of activities is 

Secondly, it is worth emphasizing that while oxide formation would have occurred 
on the surface of the iron whether it was pure or not, the steel containing impurities 
rusts faster as a consequence of the emf, and also more extensively than pure iron 

If the two half-cells 
were shorted then 
reduction would occur 
at the right-hand half- 
cell, Cu 2+ (aq) + 2e- -> 
CU( S ), and oxidation 
would proceed at the 
left-hand side, Cu (s) -> 

Cu 2+ (a q) + 2e". 

Thermodynamics of concentration cells 

A concentration cell contains the same electroactive material in 
both half-cells, but in different concentration (strictly, with different 
activities). The emf forms in response to differences in chemical 
potential [i between the two half-cells. Note that such a concen- 
tration cell does not usually involve different electrode reactions 
(other than, of course, that shorting causes one half-cell to undergo 
reduction while the other undergoes oxidation). 

Worked Example 7.20 Consider the simple cell Cu|Cu 2+ (a = 
0.002)||Cu 2+ (a = 0.02)|Cu. What is its emf! 


Inserting values into Equation (7.48): 

0.0257 V / 0.02 

emf = In 

2 V 0.002 

emf = 0.0129 Vln(10) 
emf = 0.0129 V x 2.303 
emf = 29.7 mV 

so the emf for this two-electron concentration cell is about 30 mV. For an analogous 
one-electron concentration cell, the emf would be 59 mV. 

SAQ 7.18 The concentration cell Zn|[Zn 2+ ](c = 0.0112 moldnrr 3 )|[Zn 2+ ] 
(c = 0.2 moldnrr 3 )|Zn is made. Calculate its emf, assuming all activity 
coefficients are unity. 

Justification Box 7.2 

Let the redox couple in the two half-cells be O + «e~ = R. An expression for the emf 
of the cell is 

emf = £(rhs) £(lhs) 
The Nernst equation for the 0,R couple on the RHS of the cell is: 

£o,R = < R+ ^ ln(^ 
nF \a {R)R 

and the Nernst equation for the same 0,R couple on the LHS of the cell is: 

£o,R = < R + ^ln(^w) 

nF W)lhs/ 

Substituting for the two electrode potentials yields an emf of the cell of 

emf = E * R +^L ln (IB**) -E% R -— In ^ (0) - 
°' R nF \a mm J °' R nF 

It will be seen straightaway that the two £ & terms cancel to leave 

nF W) RHS 7 nF V«(R)l 

A good example of a concentration cell would be the iron system in the worked example 
above, in which a(R) = <Z(Cu) = U accordingly, for simplicity here, we will assume that 
the reduced form of the couple is a pure solid. 



The emf of the concentration cell, therefore, becomes 


emf = — ■ ln[fl (0 )RHs] ln[fl (0 )LHs] 

nF nF 

which, through the laws of logarithms, simplifies readily to yield Equation (7.48). 

If we assume that the activity coefficients in the left- and right-hand half-cells are the 
same (which would certainly be a very reasonable assumption if a swamping electrolyte 
was also in solution), then the activity coefficients would cancel to yield 


RT ln f [Q]RHS 

nF \[0]lhs 

How do pH electrodes work? 

The pH -glass electrode 

A pH electrode is sometimes also called a 'membrane' electrode. Figure 7.15 shows 
how its structure consists of a glass tube culminating with a bulb of glass. This bulb 
is filled with a solution of chloride ions, buffered to about pH 7. A slim silver wire 
runs down the tube centre and is immersed in the chloride solution. It bears a thin 
layer of silver chloride, so the solution in the bulb is saturated with AgCl. 

The bulb is usually fabricated with common soda glass, i.e. glass containing a high 
concentration of sodium ions. Finally, a small reference electrode, such as an SCE, 
is positioned beside the bulb. For this reason, the pH electrode ought properly to be 
called a pH combination electrode, because it is combined with a reference electrode. 
If the pH electrode does not have an SCE, it is termed a glass electrode (GE). The 
operation of a glass electrode is identical to that of a combina- 
tion pH electrode, except that an external reference electrode is 

To determine a pH with a pH electrode, the bulb is fully immer- 
sed in a solution of unknown acidity. The electrode has fast res- 
ponse because a potential develops rapidly across the layer of glass 

Empirical means found 
from experiment, rather 
than from theory. 

Buffer solution 
(containing chloride ions) 

Thin-walled glass bulb 
Silver wire 

Deposit of silver chloride 
Figure 7.15 Schematic representation of a pH electrode (also called a 'glass electrode') 


between the inner chloride solution and the outer, unknown acid. Empirically, we find 
the best response when the glass is extremely thin: the optimum seems to be 50 (xm or 
so (50 |xm = 0.05 mm = 50 x 10 -6 m). Unfortunately, such thin glass is particularly 
fragile. The glass is not so thin that it is porous, so we do not need to worry about 
junction potentials £j (see Section 7.6). The non-porous nature of the glass does 
imply, however, that the cell resistance is extremely large, so the circuitry of a pH 
meter has to operate with minute currents. 

The magnitude of the potential developing across the glass de- 

A pH meter is essen- 
tially a precalibrated 

pends on the difference between the concentration of acid inside 

the bulb (which we know) and the concentration of the acid outside 

the bulb (the analyte, whose pH is to be determined). In fact, the 

emf generated across the glass depends in a linear fashion on the 

pH of the analyte solution provided that the internal pH does not alter, which is why 

we buffer it. This pH dependence shows why a pH meter is really just a pre-calibrated 

voltmeter, which converts the measured emf into a pH. It uses the following formula: 

2.303 RT 

emf = K+ pH (7.49) 


SAQ 7.19 An emf of 0.2532 V was obtained by immersing a glass elec- 
trode in a solution of pH 4 at 25 °C. Taking E (S ce) = 0.242 V, calculate the 
'electrode constant' K. 

SAQ 7.20 Following from SAQ 7.19, the same electrode 
was then immersed in a solution of anilinium hydrochloride 
of pH = 2.3. What will be the new emf? 

In practice, we do not 
know the electrode 
constant of a pH elec- 

Electrode 'slope' 

We can readily calculate from Equation (7.49) that the emf of a pH electrode should 
change by 59 mV per pH unit. It is common to see this stated as 'the electrode 
has a slope of 59 mV per decade' . A moment's pause shows how this is a simple 
statement of the obvious: a graph of emf (as 'y') against [H + ] (as 'x') will have 
a gradient of 59 mV (hence 'slope'). The words 'per decade' point to the way that 
each pH unit represents a concentration change of 10 times, so a pH of 3 means 
that [H+] = 10" 3 moldm" 3 , a pH of 4 means [H+] = 10" 4 moldm" 3 and a pH of 
5 means [H + ] = 10~ 5 moldm -3 , and so on. If the glass electrode does have a slope 
of 59 mV, its response is said to be Nernstian, i.e. it obeys the Nernst equation. The 
discussion of pH in Chapter 6 makes this same point in terms of Figure 6.1. 

Table 7.11 lists the principle advantages and disadvantages encountered with the 
pH electrode. 

SAQ 7.21 Effectively, it says above: 'From this equation, it can be readily 
calculated that the emf changes by 59 mV per pH unit'. Starting with the 
Nernst equation (Equation (7.41)), show this statement to be true. 


Table 7.11 Advantages and disadvantages of the pH electrode 


1 . If recently calibrated, the GE and pH electrodes give an accurate response 

2. The response is rapid (possibly millisecond) 

3. The electrodes are relatively cheap 

4. Junction potentials are absent or minimal, depending on the choice of reference electrode 

5. The electrode draws a minimal current 

6. The glass is chemically robust, so the GE can be used in oxidizing or reducing 
conditions; and the internal acid solution cannot contaminate the analyte 

7. The pH electrode has a very high selectivity - perhaps as high as 10 5 : 1 at room 
temperature, so only one foreign ion is detected per 100000 protons (although see 
disadvantage 6 below). The selectivity does decrease a lot above ca 35 °C 


Both the glass and pH electrodes alike have many disadvantages 

1. To some extent, the constant K is a function of the area of glass in contact with the acid 
analyte. For this reason, no two glass electrodes will have the same value of K 

2. Also, for the same reason, K contains contributions from the strains and stresses 
experienced at the glass. 

3. (Following from 2): the electrode should be recalibrated often 

4. In fact, the value of K may itself be slightly pH dependent, since the strains and stresses 
themselves depend on the amount of charge incorporated into the surfaces of the glass 

5. The glass is very fragile and, if possible, should not be rested against the hard walls or 
floor of a beaker or container 

6. Finally, the measured emf contains a response from ions other than the proton. Of these 
other ions, the only one that is commonly present is sodium. This error is magnified at 
very high pH (>11) when very few protons are in solution, and is known as the 
'alkaline error' 

Justification Box 7.3 

At heart, the pH 

electrode operates as a simple 

concentration cell. Consider the schematic 

H+(a 2 )||H+(o 1 ), 

then the Nernst equation can 


emf = In 

■> p 

be written as Equation 



which, if written in terms of logarithms in base 10, becomes 


emf = 


logl " (?) 


Subsequent splitting of the logarithm terms gives 


emf = log 10 fl2 


F logio 0i 




If we say that a\ is the analyte of known concentra- 
tion (i.e. on the inside of the bulb), then the last term in 
the equation is a constant. If we call the term associated 
with 02 'K', then we obtain 


emf = K -\ log 10 a 2 

If ci2 relates to the acidic solution of unknown con- 
centration then we can substitute for 'log 10 fl2\ by say- 
ing that pH = — log 10 [H + ], so: 

emf = K -\ x — pH 

which is the same as Equation (7.49) 


Care: we have assu- 
med here that the 
activities and concen- 
trations of the solvated 
protons are the same. 

This derivation is based 
on the Nernst equation 
written in terms of 
ionic activities, but pH 
is usually discussed in 
terms of concentration. 

7.6 Transport phenomena 

How do nerve cells work? 

Ionic transport across membranes 

The brain relays information around the body by means of nerves, allowing us to 
register pain, to think, or to instruct the legs to walk and hands to grip. Although the 
way nerves operate is far from straightforward, it is nevertheless clear that the nerve 
pathways conduct charge around the body, with the charged particles (electrons and 
ions) acting as the brain's principal messengers between the brain and body. 

The brain does not send a continuous current through the nerve, but short 'spurts'. 
We call them impulses, which transfer between nerve fibres within the synapses of 
cells (see Figure 7.16). The cell floats within an ionic solution called plasma. The 
membrane separating the synapse from the solution with which the nerve fibre is in 
contact surrounding the cell is the axon, and is essential to the nerve's operation. 

The charge on the inside of a cell is negative with respect to 
the surrounding solution. A potential difference of about —70 mV 
forms across the axon (cell membrane) when the cell is 'at rest', i.e. 
before passing an impulse - we sometimes call it a rest potential, 
which is caused ultimately by differences in concentration either 
side of the axon (membrane). 

Movement of charge across the membrane causes the potential to change. A huge 
difference in concentration is seen in composition between the inside of the axon 
and the remainder of the nerve structure. For example, consider the compositional 

No potential difference 
forms along the mem- 
brane surface, only 
across it. 


Cell exterior 

Cell membrane ('axon') 
to separate the cell interior and exterior 

Cell interior 

Ions leaving 
cell interior 

Ions entering cell 
interior from exterior 

Figure 7.16 Schematic diagram showing a portion of a cell, the membrane ('axon') and the way 
ions diffuse across the axon 

Table 7.12 

Concentrations of ions 

inside and outside nerve 


[Na + ]/moldirT 3 

[K + ]/moldrrr 3 

[Cn/moldirT 3 

Inside the axon 
Outside the axon 




Source: J. Koryta, Ions, Electrodes and Membranes, Wiley, Chichester, 1991, p. 172. 

In this context, per- 
meable indicates that 
ions or molecules can 
pass through the mem 
brane. The mode of 
movement is probably 
diffusion or migration. 

differences in Table 7.12. The data in Table 7.12 refer to the nerves 
of a squid (a member of the cephalopod family) data for other 
species show a similar trend, with massive differences in ionic 
concentrations between the inside and outside of the axon. These 
differences, together with the exact extent to which the axon mem- 
brane is selectively permeable to ions, determines the magnitude 
of the potential at the cell surface. 

The membrane encapsulating the axon is semi-permeable, there- 
by allowing the transfer of ionic material into and out from the 
axon. Since the cell encapsulates fluid and also floats in a fluid, 
we say the membrane represents a 'liquid junction'. A potential 
forms across the membrane in response to this movement of ions 
across the membrane, which we call a 'junction potential' E y If 
left unchecked, ionic movement across the membrane would occur 
until mixing was intimate and the two solutions were identical. 
For a nerve to transmit a 'message' along a nerve fibre, ions traverse the axons and 
transiently changing the sign of the potential across the membrane, as represented 
schematically in Figure 7.17. We call this new voltage an action potential, to differ- 
entiate it from the rest potential. To effect this change in potential, potassium cations 

Some texts give the 
name 'diffusion poten 
tial'to E } . 








Time f/ms 

/ \ ' l ' 

J Action \ 

/ potential \ Rest potentia | 

Onset of impulse 

Figure 7.17 The potential across the axon-cell membrane changes in response to a stimulus, 
causing the potential to increase from its rest potential to its action potential 

move from inside the axon concurrently with sodium ions moving 
in from outside. With a smaller difference in composition either 
side of the membrane, the junction potential decreases. 

A nerve consists of an immense chain of these axons. Impulses 
'conduct' along their length as each in turn registers an action 
potential, with the net result that messages transmit to and from 
the brain. 

To achieve this other- 
wise difficult process, 
chemical 'triggers' pro- 
mote the transfer of 

Liquid junction potentials 

A liquid junction potential £j forms when the two half-cells of a cell contain different 
electrolyte solutions. The magnitude of £j depends on the concentrations (strictly, the 
activities) of the constituent ions in the cell, the charges of each moving ion, and 
on the relative rates of ionic movement across the membrane. We record a constant 
value of £j because equilibrium forms within a few milliseconds of the two half-cells 
adjoining across the membrane. 

Liquid junction potentials are rarely large, so a value of £j as 
large as 0.1 V should be regarded as exceptional. Nevertheless, 
junction potentials of 30 mV are common and a major cause of 
experimental error, in part because they are difficult to quantify, 
but also because they can be quite irreproducible. 

We have already encountered expressions that describe the emf 
of a cell in terms of the potentials of its constituent half-cells, e.g. 
Equation (7.23). When a junction potential is also involved - and 
it usually is - the emf increases according to 

emf = .Expositive half-cell) ~ ^negative half-cell) + Ej (7.53) 

which explains why we occasionally describe £j as 'an additional source of potential'. 

In most texts, the liquid 
junction potential is 
given the symbol E r In 
some books it is written 
as Efin or even E, 





While it is easy to measure a value of emf we do not know the magnitude of E y 
SAQ 7.21 illustrates why we need to minimize E y 

SAQ 7.22 The emf of the cell SHE |Ag+|Ag, is 0.621 V. Use the Nernst 
equation to show that a (Ag + } = 10~ 3 if fj = V, but only 4.6 x 10 -4 if 

fj = 20 mV. E^ g+ Ag = 0.799 V. [Hint: to compensate for Ej in the sec- 
ond calculation, say that only 0.601 V of the emf derives from the Ag + |Ag 
half-cell, i.e. E Ag+ Ag = 0.601 V.] 

What is a 'salt bridge'? 

Minimizing junction potentials 

In normal electrochemical usage, the best defence against a junction 
potential £j is a salt bridge. In practice, the salt bridge is typically a 
thin strip of filter paper soaked in electrolyte, or a U-tube containing 
an electrolyte. The electrolyte is usually KC1 or KNO3 in relatively 
high concentration; the U-tube contains the salt, perhaps dissolved 
in a gelling agent such as agar or gelatine. 
We connect the two half-cells by dipping either end of the salt 
bridge in a half-cell solution. A typical cell might be written in schematic form as: 

It's called a bridge 
because it connects the 
two half-cells, and salt 
because we saturate 
it with a strong ionic 

Zn (s) |Zn 2+ (aq) |S|Cu 2+ (aq) |Cu 


We write the salt bridge as '|S|\ where the S is the electrolyte within the salt bridge. 

But how does the salt bridge minimize Ej? We recognize first how the electrolyte 
in the bridge is viscous and gel -like, so ionic motion through the bridge is slow. 
Secondly, the ionic diffusional processes of interest involve only the two ends of the 
salt bridge. Thirdly, and more importantly, the concentration of the salt in the bridge 
should greatly exceed the concentrations of electrolyte within either half-cell (exceed, 
if possible, by a factor of between 10-100 times). 

The experimental use of a salt bridge is depicted in Figure 7.18. The extent of 
diffusion from the bridge, as represented by the large arrows in the diagram, is seen 
to be much greater than diffusion into the bridge, as represented by the smaller of 
the two arrows. A liquid junction forms at both ends of the bridge, each generating 
its own value of E y If the electrolyte in the bridge is concentrated, then the diffusion 
of ions moving from the bridge will dominate both of these two E y Furthermore, 
these £j will be almost equal and opposite in magnitude, causing them to cancel each 
other out. 

Table 7.13 shows how the concentration of the salt in the bridge has a large effect 
on E-y it is seen that we achieve a lower value of £j when the bridge is constructed 
with larger concentrations of salt. A junction potential £j of as little as 1-2 mV can 
be achieved with a salt bridge if the electrolyte is concentrated. 


Salt bridge 


Figure 7.18 The two half-cells in a cell are joined with a salt bridge. Inset: more ions leave the 
bridge ends than enter it; the relative sizes of the arrows indicate the relative extents of diffusion 

Table 7.13 

Values of junc- 

tion potential 

in aqueous cells 

as a function of the concentra- 

tion of inert KC1 within a salt 


[KCl]/c s 








4.2 (sat'd) 


Minimizing junction potentials with a swamping electrolyte 

The second method of minimizing the junction potential is to employ a 'swamping 
electrolyte' S. We saw in Section 4.1 how diffusion occurs in response to entropy 
effects, themselves due to differences in activity. Diffusion may be minimized by 
decreasing the differences in activity, achieved by adding a high concentration of 
ionic electrolyte to both half-cells. Such an addition increases their ionic strengths /, 
and decreases all activity coefficients y± to quite a small value. 

If all values of y± are small, then the differences between activities also decrease. 
Accordingly, after adding a swamping electrolyte, fewer ions diffuse and a smaller 
junction potential forms. 

7.7 Batteries 

How does an electric eel produce a current? 

Introduction to batteries 

The electric eel (Electrophorus electricus) is a thin fish of length 3-5 feet; see 
Figure 7.19. It is capable of delivering an electric shock of about 600 V as a means 


Figure 7.19 The electric eel (Electrophorus electricus) is a long, thin fish (3-5 feet) capable of 
delivering an electric shock of about 600 V. (Figure reprinted from Ions, Electrodes and Membranes 
by Jiri Koryta. Reproduced by permission of John Wiley and Sons Ltd) 

of self-protection or for hunting. The eel either stuns a possible aggressor, or becomes 
an aggressor itself by stunning its prey, prior to eating it. 

Fundamentally, the eel is simply a living battery. The tips of its head and tail 
represent the poles of the eel's 'battery'. As much as 80 per cent of its body is an 
electric organ, made up of many thousands of small platelets, which are alternately 
super-abundant in potassium or sodium ions, in a similar manner to the potentials 
formed across axon membranes in nerve cells (see p. 339). In effect, the voltage 
comprises thousands of concentration cells, each cell contributing a potential of about 
160 mV. It is probable that the overall eel potential is augmented with junction 
potentials between the mini-cells. 

The eel produces its electric shock when frightened, hungry or when it encounters 
its prey. The shock is formed when the eel causes the ionic charges on the surfaces 
of its voltage cells to redistribute (thereby reversing their cell polarities), and has the 
effect of summing the emfs of the mini-cells, in just the same way as we sum the 
voltages of small batteries incorporated within a series circuit. The ionic strength of 
seawater is very high, so conduction of the current from the eel to its prey is both 
swift and efficient. 

Battery terminology 

A battery is defined as a device for converting chemical energy into electrical energy. 
A battery is therefore an electrochemical cell that spontaneously produces a current 
when the two electrodes are connected externally by a conductor. The conductor will 
be the sea in the example of the eel above, or will more typically be a conductive 



A battery is sometimes 
called a galvanic cell. 

metal such as a piece of copper wire, e.g. in a bicycle headlamp. The battery pro- 
duces electrons as a by-product of the redox reaction occurring at the cathode. These 
electrons pass through the load (a bulb, motor, etc.) and do work, before re-entering 
the battery where the anode consumes them. Electrochemical reduction occurs at the 
positive pole (the anode) of the battery simultaneously with electrochemical oxidation 
at the negative pole (the cathode). 

There are several types of battery we can envisage. A majority of 
the batteries we meet are classed as primary batteries, i.e. a chem- 
ical reaction occurs in both compartments to produce current, but 
when all the chemicals have been consumed, the battery becomes 
useless, so we throw it away. In other words, the electrochemical reactions inside 
the battery are not reversible. The most common primary batteries are the Leclanche 
cell, as described below, and the silver-oxide battery, found inside most watches and 
slim-line calculators. 

By contrast, secondary batteries may be reused after regenerat- 
ing their original redox chemicals. This is achieved by passing a 
current through the battery in the opposite direction to that during 
normal battery usage. The most common examples of secondary 
batteries are the lead-acid cell (there is one inside most cars) and 
nickel-cadmium batteries (commonly called 'NiCad' batteries). 

In the shops, sec- 
ondary batteries are 
usually called rechar- 
geable batteries. 

What is the earliest known battery? 

Battery types 

We have evidence that batteries were not unknown in the ancient world. The Parthians 
were a race living in the Mediterranean about 2000 years ago, from ca 300 bc until 
ad 224, when they were wiped out by the Romans. They are mentioned in the Bible, 
e.g. see The Acts of the Apostles, Chapter 2. 

A device was found in 1936 near what is now modern Baghdad, the capital of 
Iraq. It consisted of a copper cylinder housing a central iron rod. The identity of the 
ionic electrolyte is now wholly unknown. The device was held together with asphalt 
as glue. 

If the copper was tarnished and the iron was rusty (i.e. each was covered with 
a layer of oxide), then an approximate emf for this 2000-year-old battery would 
probably be in the range 0.6-0.7 V. We do not know what the battery was used for. 

The Daniell cell 

One of the first batteries in recent times was the Daniell cell, 

Zn|Zn 2+ :Cu 2+ |Cu. This battery comprised two concentric terra- 
cotta pots, the outer pot containing a zinc solution and the inner 
pot containing a copper solution. Metallic rods of copper and 

The vertical dotted 
line in this schematic 
indicates a porous 



zinc were then immersed in their respective solutions. The electrode reaction at the 
zinc anode is Zn — »■ Zn 2+ + 2e~ , while reduction occurs at the positive electrode, 
Cu 2+ + 2e" -+ Cu. 

Although this battery was efficient, it was never popular because it required aqueous 
solutions, which can be a danger if they slopped about. Its market share also suffered 
when better batteries were introduced to the market. 

The Leclanche 'dry-cell' battery 

The Leclanche cell was first sold in 1880, and is still probably the most popular 
battery in the world today, being needed for everyday applications such as torches, 
radios, etc. It delivers an emf of ca 1 .6 V. 

Figure 7.20 depicts the Leclanche cell in schematic form. The zinc can is generally 
coated with plastic for encapsulation (i.e. to prevent it from splitting) and to stop the 
intrusion of moisture. Plastic is an insulator, and so we place a conductive cap of 
stainless steel at the base of the cell to conduct away the electrons originating from 
the dissolution of the zinc from the inside of the can. A carbon rod then acts as an 
inert electrode to conduct electrons away from the reduction of Mn02 at the cathode. 

The reaction at the cathode is given by 

2Mn0 2(s) + 2H 2 + 2e~ 

2MnO(OH) (s) +20H 



and the reaction at the zinc anode is: Zn — > Zn 2+ + 2e~. 

We incorporate an ammonium salt to immobilize the Zn 2+ ions: NH4CI is prepared 
as a paste, and forms a partially soluble complex with zinc cations produced at the 

Positive terminal 

Mixture of zinc chloride 
ammonium chloride, carbon powder 
and manganese dioxide 

Carbon rod 

Zinc casing (anode) 

Negative terminal 
(in electrical contact 
with the zinc case) 

Figure 7.20 Schematic depiction of the Leclanche cell 


Table 7.14 Advantages and disadvantages of the Leclan- 
che cell 


It is cheap to make 

It has a high energy density 

It is not toxic 

It contains no liquid electrolytes 


Its emf decreases during use as the material is consumed 
It cannot readily deliver a high current 

anode. We sometimes add starch to the paste to provide additional stiffness. The 
juxtaposition of the zinc ions with the zinc of the casing forms a redox buffer, 
thereby decreasing the extent to which the potential of the zinc half-cell wanders 
while drawing current. 

Table 7.14 lists the advantages and disadvantages of the Leclan- 
che cell. 

The lead -acid battery 

The lead-acid cell was invented by Plante in 1859, and has remai- 
ned more-or-less unchanged since Faure updated it in 1881. The 
lead-acid cell is the world's most popular choice of secondary 
battery, meaning it is rechargeable. It delivers an emf of about 
2.0 V. Six lead-acid batteries in series produce an emf of 12 V. 

Table 7.15 Advantages and disadvantages of the lead-acid battery 


It is relatively easy to make, and so can be quite cheap 

It has a high energy density, producing much electrical energy per unit mass 

It can readily deliver a very high current 


It contains toxic lead 

Also, since it contains lead, its power density is low 

The acid is corrosive 

Furthermore, the acid is a liquid electrolyte 

Given time, lead sulphate (which is non-conductive) covers the electrode. Having 

'sulphated up', the energy density of the battery is greatly impaired. To avoid 

sulphating up, it ought to be recharged often 

Alkaline manganese 
cells are broadly sim- 
ilar in design to the 
Leclanche cell, but they 
contain concentrated 
KOH as the electrolyte 
instead of NH 4 CI. 


Plates of lead, each coated with lead dioxide, are immersed in fairly concentrated 
sulphuric acid. Lead is oxidized at the lead anode during discharge: 

Pb (s) + HS0 4 " (aq) ► PbS0 4(s) + H+ (aq) + 2e" (7.55) 

The reaction at the cathode during discharge is 

Pb0 2(s) + 2H+ (aq) + H 2 S04(a q) + 2e" ► PbS0 4(s) + 2H 2 (7.56) 

Both half-cell reactions are fully electro-reversible. In practice, 

Spongy lead has a 
higher surface area 
than normal lead. 

there are two types of lead: the 'collector' electrode is made of lead 
alloy 'mesh' in order to give it greater structural strength, and is 
made with about 5 per cent antimony. 'Spongy lead' (Pb + Pb0 2 ) 
is introduced into the holes of the mesh. 
Table 7.15 lists the advantages and disadvantages of the lead-acid battery. 


Chemical kinetics 


In previous chapters, we considered questions like: 'How much energy does a reaction 
liberate or consume?' and 'In which direction will a reaction proceed?' We then asked 
questions like: 'To what extent will a reaction proceed in that direction, before it 
stops?' and even 'Why do reactions occur at all?' In this chapter, we look at a different 
question: 'How fast does a reaction proceed?' Straightaway, we make assumptions. 
Firstly, we need to know whether the reaction under study can occur: there is no 
point in looking at how fast it is not going if a reaction is not thermodynamically 
feasible! So we first assume the reaction can and does occur. 

Secondly, we assume that reactions can be treated according to their type, so 'reac- 
tion order' is introduced and discussed in terms of the way in which concentrations 
vary with time in a manner that characterizes that order. 

Finally, the associated energy changes of reaction are discussed in terms of the 
thermodynamic laws learnt from previous chapters. Catalysis is discussed briefly 
from within this latter context. 

8.1 Kinetic definitions 

Why does a 'strong' bleach clean faster 
than a weaker one does? 

Introduction to kinetics: rate laws 

We often clean away the grime and dirt in a kitchen with bleach, 
the active ingredient of which is the hypochlorite ion C10~ . The 
cleaning process we see by eye ('the bleaching reaction') occurs 
between an aqueous solution of C10~ ion and coloured species 
stuck to the kitchen surfaces, which explains why the dirt or grease, 
etc., appears to vanish during the reaction. The reaction proceeds 
concurrently with colour loss in this example. 

Care: a supermarket 
uses the word 'strong' 
in a different way from 
chemists: remember 
from Chapter 6 that the 
everyday word 'strong' 
has the specific chem- 
ical meaning 'a large 
equilibrium constant of 













is in eq 


with chlorine 

This reaction could be 
one of the steps in 
a more complicated 
series of reactions, 
in a so-called multi- 
step reaction. If this 
reaction is the rate- 
determining step of 
the overall compli- 
cated series, then this 
rate law still holds; 
see p. 357. 

We define the rate of 
reaction as the speed at 
which a chemical con- 
version proceeds from 
start to its position 
of equilibrium, which 
explains why the rate 
is sometimes written 
as d£/dt, where ? is the 
extent of reaction. 

We formulate the rate 
of a reaction by multi- 
plying the rate constant 
of the reaction by the 
concentration of each 
reactant, i.e. by each 
species appearing at 
the tail end of the 
arrow. We can only do 
this if the reaction is 
elementary (proceeds 
in a single step) 

We soon discover that a 'strong' bleach cleans the surfaces faster 
than a more dilute bleach. The reason is that 'strong' bleaches are 
in fact more concentrated, since they contain more C10~ ions per 
unit volume than do 'weaker' bleaches. 

We will consider the chemical reaction between the hypochlo- 
rite ion and coloured grease to form a colourless product P (the 
'bleaching' reaction) as having the following stoichiometry: 

CIO + grease 


We wish to know the rate at which this reaction occurs. The rate 
is defined as the number of moles of product formed per unit time. 
We define this rate according to 

rate = 



As far as equations like Equation (8.2) are concerned, we tend to 
think of a chemical reaction occurring in a forward direction, so the 
product in Equation (8.2) is the chemical at the head of the arrow 
in Equation (8.1). Consequently, the concentration of product will 
always increase with time until the reaction reaches its position of 
equilibrium (when the rate will equal zero). This explains why the 
rate of reaction always has a positive value. The rate is generally 
cited with the units of mol dm~ s _1 , i.e. concentration change 
per second. 

The numerical value of the rate of reaction is obtained from a 
rate equation, which is obtained by first multiplying together the 
concentrations of each reactant involved in the reaction. (Before 
we do this, we must be sure of the identities of each reactant - in 
a complicated multi-step reaction, the reacting species might differ 
from those mentioned in the stoichiometric equation.) The follow- 
ing simple equation defines exactly the rate at which the reaction 
in Equation (8.1) occurs: 

rate = k [CIO ] [grease] 


where the constant of proportionality k is termed the rate constant. 
The value of k is generally constant provided that the reaction 
is performed at constant temperature T. Values of rate constant 
are always positive, although they may appear to be negative in 
some of the more complicated mathematical expressions. Table 8.1 
contains a few representative values of k. 

We see from Equation (8.3) that the reaction proceeds faster (has 
a faster rate) when performed with a more concentrated ('strong') 



Table 8.1 Selection of rate constants k 


Phase Temperature/ °C kl units 3 

First-order reactions 

SO2CI2 -*■ S0 2 + Cl 2 

Cyclopropane — >• propene 

C2H6 — > 2CH3 

CIH2C-CH2CI -► CH 2 =CHC1 + HC1 


Second-order reactions 

CIO" + Br" -► BrO + CI" Aqueous 

CH3COOC2H5 + NaOH -> CH 3 C0 2 Na + C 2 H 5 OH Aqueous 


H 2 + I 2 -*■ 2HI Gas 

2N0 2 -► 2NO + 2 Gas 

H 2 + 2NO -► N 2 + 2H 2 Gas 

21 -► I 2 Gas 

H+ + OH" -*■ H 2 Aqueous 






2 x 10" 5 

6.71 x 10" 5 

5.36 x 10" 4 

4.4 x 10" 3 

4.2 x 10" 7 
1.07 x 10" 2 

2.42 x 10" 2 


7 x 10 9 
1.35 x 10 11 

a For first-order reactions, k has the units of s . For second-order reactions, k has units of dm mol s 

bleach because the concentration term '[CIO ]' in Equation (8.3) 
has increased. 

SAQ 8.1 Consider the reaction between ethanoic acid 
and ethanol to form the pungent ester ethyl ethanoate 
and water: 



Care: A 'rate constant' 
is written as a lower 
case k in contrast to the 
more familiar 'equilib- 
rium constant', which 
is written as an upper 
case K. 

Write an expression for the rate of this reaction in a similar form to that in 
Equation (8.3), assuming the reaction proceeds in a single step as written. 

SAQ 8.2 Write an expression for the rate of the reaction Cu 2+ (aq ) + 
4NH 3(a q) ->■ [Cu(NH 3 )4] 2+ (a q), assuming that the reaction proceeds in a 
single step as written. 

Why does the bleaching reaction eventually stop? 

Calculating rates and rate constants 

When cleaning in the kitchen with a pool of bleach on tables and surfaces, there 
comes a time when the bleaching action seems to stop. We might say that the bleach 
is 'exhausted', and so pour out some more bleach from the bottle. 

When thinking about reaction kinetics, we need to appreciate that reactions involve 
chemical changes, with reactants being consumed during a reaction, and products 



being formed. After a time of reacting, the concentration of one or more of the 
chemicals will have decreased to zero. We generally say the chemical is 'used up'. 
Now look again at Equation (8.3). The rate of reaction depends on the concentrations 
of both the grease and the C10~ ion from the bleach. If the concentration [C10~] 
has decreased to zero, then the rate will also be zero, whatever the value of k or the 
concentration of grease. And if the rate is zero, then the reaction stops. 

Although we appreciate from Equation (8.3) that the reaction 
will stop when one or both of the concentration terms reaches zero, 
we should also appreciate that the concentration terms reach zero 
faster if the value of k is large, and the concentrations deplete more 
slowly if k is smaller. We see how the value of the rate constant 
is important, because it tells us how fast a reaction occurs. 

The value of the rate 
constant is important 
because it tells us how 
fast a reaction occurs. 

Worked Example 8.1 In solution, the cerium(IV) ion reacts with aqueous hydrogen 
peroxide with a 1:1 stoichiometry. The reaction has a rate constant of 1.09 x 
10 6 dm mol~ s _1 . How fast is the reaction that occurs between Ce IV and H2O2, if 
[Ce IV ] = 1(T 4 moldm" 3 and [H 2 2 ] = 1(T 3 moldnT 3 ? 

By 'how fast', we are in effect asking 'What is the value of the rate of this reaction?' 
The reaction has a 1:1 stoichiometry; so, following the model in Equation (8.3), the 
rate equation of reaction is 



Placing the concen- 
tration brackets 
together - without 
a mathematical 
sign between 
them - implies that 
the concentrations are 
to be multiplied. 

rate = £[Ce lv ][H 2 2 ] 

where k is the rate constant. Inserting values for k and for the two 

rate = 1.09 x 10 6 dm 3 mol -1 s _1 x 10" 4 moldm -3 

x 10 -3 moldm -3 

rate = 0.109 moldm 3 s" 

or, stated another way, the concentration of product changes (increases) by the amount 
0.109 moldm -3 per second, or just over 1 mol is formed during 10 s. This is quite 
a fast reaction. 

SAQ 8.3 Show that the rate of reaction in Worked Example 8.1 quadru- 


pies if both [Ce lv ] and [H 2 2 ] are doubled. 

The rate constant k is 
truly constant and only 
varies with tempera- 
ture and (sometimes) 
with ionic strength I. 

Worked Example 8.1 shows a calculation of a reaction rate from 
a rate constant k of known value, but it is much more common to 
know the reaction rate but be ignorant of the rate constant. A rate 
equation such as Equation (8.3) allows us to obtain a numerical 
value for k. And if we know the value of k, we can calculate 
from the rate equation exactly what length of time is required 
for the reaction to proceed when performed under specific reac- 
tion conditions. 



Worked Example 8.2 Ethyl ethanoate (0.02 moldirT ) hydrolyses during reaction with 
aqueous sodium hydroxide (0.1 moldm" ). If the rate of reaction is 3 x 10 2 moldm" s _1 , 
calculate the rate constant k. 

The rate equation of reaction is given by 

rate = /fc[CH 3 COOCH 2 CH 3 ][NaOH] 
Rearranging Equation (8.5) to make k the subject gives 



[CH 3 COOCH 2 CH 3 ] [NaOH] 

Inserting values yields 

3 x 10 2 moldm 3 s 

3 c-1 


0.02 moldm x 0.1 moldm 

i=1.5x 10 5 mor'dnrV 1 


This is quite a large value of k. 

Worked Example 8.2 yields a value for the rate constant k, but an alternative and 
usually more accurate way of obtaining k is to prepare a series of solutions, and to 
measure the rate of each reaction. A graph is then plotted of 'reaction rate' (as 'v') 
against 'concentration(s) of reactants' (as 'x') to yield a linear graph of gradient equal 
to k. 

Worked Example 8.3 We continue with the reaction between Ce IV and H2O2 from 
Worked Example 8.1. Consider the following data: 

[Ce IV ]/moldnT 3 
[H 2 2 ]/moldm- 3 

rate/mol dm~ 3 s _1 

3 x 10~ 5 
8 x 10~ 6 

5 x 10" 5 
2 x 10~ 5 

8 x 10" 5 
3 x 10~ 5 

1 x 10" 4 
5 x 10- 5 

Equation (8.4) above can be seen to have the form 'y = mx', in which 
'rate' is 'y' and the mathematical product '[Ce IV ] x [H 2 2 ]' is 'x'. 
The gradient m will be equal to k. 

Figure 8.1 shows such a graph. The gradient of the graph is 1.09 x 
10 6 dm 3 s" 1 (which is the same value as that cited in Worked Example 
8.1). Notice how the intercept is zero, which confirms the obvious 
result that the rate of reaction is zero (i.e. no reaction can occur) when 
no reactants are present. 

We are saying here that 
the rate is of the form 
y = mx (straight line). 
The intercept is zero. 




"i i i i r 

0.00E+00 1.00E-09 2.00E-09 3.00E-09 4.00E-09 5.00E-09 6.00E-09 
[H 2 2 ]x[Ce lv ]/mol 2 drrr 6 

Figure 8.1 Graph of reaction rate against the product '[Ce IV ] x [H2O2]'. The numerical value of 
the gradient of the graph is the rate constant k 

Why does bleach work faster on some greases than on 

Rates expressions, reaction stoichiometry and reaction order 

Each reaction has a unique value of rate constant k. For example, the value of k 
in Worked Example 8.2 would have been different if we had chosen ethyl formate, 
or ethyl propanoate, or ethyl butanoate, etc., rather than ethyl ethanoate. The value 
of k depends ultimately on the Gibbs function of forming reaction intermediates, as 
discussed below. 

Most forms of grease in the kitchen derive from organic materials in the 
home - some derive from meat, some from vegetable oils and some from pets in 
the home, or even human tissue such as oily fingerprints. (Kitchen grease is, in fact, 
a complicated mixture of chemicals, each of which reacts with bleach at a different 
rate and, therefore, with a different value of k.) 

But the reaction conditions are still more complicated because the stoichiometry 
of reaction might alter. Many greases and oils comprise the esters of long-chain 
fatty acids. The hydrolysing reaction between NaOH and an ester such as ethyl 
ethanoate proceeds with a stoichiometry of 1:1, but a tri-ester, such as most natural 
oils (e.g. olive oil or sunflower oil), occurs with a 1:3 stoichiom- 
etry, consuming one hydroxide ion per ester bond. Clearly, the 
hydroxide will be consumed more quickly when hydrolysing a tri- 
ester than a mono-ester. The rate depends on the stoichiometry 
of reaction. 

The rate of reaction 
depends on the stoi- 


Worked Example 8.4 The active ingredient within many weedkillers is methyl violo- 
gen, MV+* (I). 

H 3 C-N X > ( 7 N-CH 3 

\=/ \=/ 



Being a weedkiller, MV + " is said to be phytotoxic. As a complication, two radicals of 
MV + " will dimerize in solution to form the non-toxic dimer species (MV)2 2+ : 

2 MV+' ► (MV) 2 2+ (8.6) 

Why does the rate of the reaction in Equation (8.6) quadruple when we double the con- 
centration of the radical cation, MV + "? 

The rate of reaction is written as the rate constant of reaction multiplied by the concen- 
tration of each reacting species on the tail end of the arrow. Accordingly, we write 

rate = k[MY + '] 2 (8.7) 

Why does the rate quadruple if we double [MV + *]? We start by saying that 
[MV + *] (ne w) — 2 x [MV + *] (old) . By doubling the concentration of MV + " and (from 
Equation (8.7)) squaring its new value, we see that the new rate = k x {2 x [MV + *]( id)} 2 . 
This new rate is the same as 4 x £[MV +, ] 2 old) , which explains why the rate quadruples 
when the concentration of MV + * doubles. 

The reason why the rate equation includes the square term [MV + *] 2 rather than just 
[MV +# ] should not surprise us. Notice that we could have written the equation for the 
chemical reaction in a slightly different way, as 

MV+' + MV+* ► (MV) 2 2+ (8.8) 

which is clearly the same equation as Equation (8.6). If we derive a rate equation for this 
alternative way of writing the reaction, with a concentration term for each participating 
reactant, then the rate equation of the reaction is 

rate = fc[MV + '][MV +# ] (8.9) 

and multiplying the two MV + * terms together yields [MV + *] 2 . We see that Equation (8.9) 
is the same rate equation as that obtained in Equation (8.7). 

SAQ 8.4 Write the rate equations for the following reactions. In each 
case, assume the reaction proceeds according to its stoichiometric 



(1) 2N0 2( g) "* N 2 4 (g) 

(2) 2S0 2(g) 

(3) Ag^ 





2S0 3( g) 

> AgCI (s) 

(4) HCI (aq) + NaOH (aq) -► NaCI (aq) 

H 2 

Reaction order 

The order of a reac- 
tion is the same as 
the number of concen 
tration terms in the 
rate expression. 

The order of a reaction is the same as the number of concentration 
terms in the rate expression. Consider the general rate equation: 

rate = fc[A] fl [B]* 


The sum of the powers is equal to the order of the reaction. 

We see that Equation (8.9) involves two concentration terms, so 
it is said to be a second-order reaction. The rate expression for 

SAQ 8.4 (1) involves two concentration terms, so it is also a second -order reaction. 

Although each of the two concentrations in Equation (8.7) is the same, there are 

nevertheless two concentrations, and Equation (8.7) also represents a second-order 

reaction. In fact, the majority of reactions are second order. 

The rate expression for SAQ 8.4 (2) involves three concentra- 
tions, so it is a third-order reaction. Third-order reactions are very 
rare, and there are no fourth-order or higher reactions. 

Third-order reactions 
are very rare. 

SAQ 8.5 Analyse the following rate equations, and determine the orders 
of reaction: 

(1) rate = /c[Cu 2+ ][NH 3 ] 

(2) rate = /c[OI-r] 

(3) rate = /c[NO] 2 [0 2 ] 

It is increasingly common to see the rate constant given a subscripted descriptor 
indicating the order. The rate equation in SAQ 8.5 (1) would therefore be written 

as k 2 . 

Why do copper ions amminate so slowly? 

The kinetic treatment of multi-step reactions, and the 
rate-determining step 

Addition of concentrated ammonia to a solution of copper(II) yields a deep-blue 
solution of [Cu(NH 3 )4] 2+ . The balanced reaction is given by 

Cu 2+ (aq) + 4NH 


-> [Cu(NH 3 ) 4 ] 2+ 





A quick look at the reaction suggests that the rate of this ammination reaction should 
be &[Cu 2+ ( aq )][NH3( aq )] 4 , where the power of '4' derives from the stoichiometry 
(provided that the reaction as written was the rate-determining step). It would be 
a fifth-order reaction, and we would expect that doubling the concentration of ammo- 
nia would cause the rate to increase 16-fold (because 2 4 = 16). But the increase in 
rate is not 16-fold; and, as we have just seen, a fifth-order reaction is not likely. 
In fact, the ammination reaction forming [Cu(NH 3 )4] 2+ occurs 

stepwise, with first one ammonia ligand bonding to the copper ion, 

then a second, and so forth until the tetra-amminated complex is 

formed. And if there are four separate reaction steps, then there 

are four separate kinetic steps - one for each ammination step, 

with each reaction having its own rate constant k - we call them 

£(i) > k(2) , &(3) and k^. This observation helps explain why the increase in reaction 

rate is not 16-fold when we double the concentration of ammonia. 

Each step in a multi- 
step reaction sequence 
proceeds at a different 


When we write k(i) 

with the 

subscripted '1' in brackets, 


mean the rate constant of 

the first 

step in a multi-step 

reaction. When we write k\ 

with a subscripted '1' but no 


we mean 

the rate constant of a first-order reaction. 

We adopt this convention 

to avoid confusion. 

Most organic reactions occur in multi-step reactions, with only 
a small minority of organic reactions proceeding with a single 
step. We find, experimentally, that it is extremely unlikely for any 
two steps to proceed with the same rate constant, which means 
that we can only follow one reaction at a time. And the reaction 
that can be followed is always the slowest reaction step, which 
we call the rate-determining step - a term we often abbreviate 
to RDS. 

A simple analogy from everyday life illustrates the reasonable- 
ness of this assumption. Imagine driving north from Italy to Nor- 
way in a journey involving travel along a fast motorway, along 
a moderately fast main road, and crawling through a 'contraflow' 
system, e.g. caused by road works. No matter how fast we travel 
along the main roads or the motorway, the overall time required 
for the journey depends crucially on the slowest bit, the tedious 
stop-start journey through the 'contraflow'. In a similar way, the 
only step in a multi-step reaction that we are able to follow experi- 
mentally is the slowest. We call it the rate-determining step because 
it 'determines' the rate. 

It is not possible to 
follow the rates in 
a multi-step reac- 
tion sequence: only 
the slowest step can 
be followed. 

We call the slow- 
est step in a multi- 
step process its rate- 
determining step, often 
abbreviated to 'RDS'. 
The overall (observed) 
rate of a multi-step 
reaction is equal to 
the rate of the rate- 
determining reaction. 



The separate reac- 
tion steps will proceed 
with the same rates in 
the unlikely event that 
all steps are diffusion 
controlled; see p. 416. 

The pair of reactions 
in SAQ 8.6 explains 
how old wine 'goes 
off', i.e. becomes too 
acidic to drink when 
left too long. 

We will discuss multiple-step reactions in much greater detail in 
Section 8.4. 

SAQ 8.6 Consider the oxidation of aqueous ethanol (e.g. 
in wine) to form vinegar: 

first reaction 

ethanol + 2 > ethanal + H 2 (8.12) 

second reaction 

2 ethanal + 2 > 2 ethanoic acid (8.13) 


Decide which step is rate determining, and then write 
an expression for its rate (assuming that each reac- 
tion proceeds according to the stoichiometric equation 

How fast is the reaction that depletes the ozone layer? 

Use of pressures rather than concentrations in a rate equation 

Many substances react in the gas phase rather than in solution. An important example 
is the process thought to deplete the ozone layer: the reaction between gaseous ozone, 
O3, and chlorine radicals, high up in the stratosphere. Ultimately, the chlorine derives 
from volatile halocarbon compounds, such as the refrigerant Freon-12 or the methyl 
chloroform thinner in correction fluid. 

The ozone-depleting reaction involves a rather complicated series of reactions, all 
of which occur in the gas phase. Equation (8.14) describes the rate-determining step: 

The CI* radical is retrie- 
ved quantitatively, and 
is therefore a catalyst. 

CI* + 3 

-* CIO* + 2 


The CIO* radical product of Equation (8.14) then reacts further, 
yielding the overall reaction: 

20 3(el + CI* 

(as a catalyst) 

30 2( g) +C1* 

(retrieved catalyst) 


The Earth is constantly irradiated with UV light, much of which is harmful to our 
skin. Ozone absorbs the harmful UV frequencies and thereby filters the light before 
it reaches the Earth's surface. Normal diatomic oxygen, O2, does not absorb UV in 
this way, so any reaction that removes ozone has the effect of allowing more harmful 
UV light to reach us. The implications for skin health are outlined in Chapter 9. 

The rate equation for Equation (8.14) is expressed in terms of 'pressures' p rather 
than concentrations, such as [CI*]. We write the rate as 

rate = kp(CV)p(0 3 ) 




where k here is a gas-phase rate constant. This rate expression has a similar form to 
the rate expressions above, except that the now-familiar concentration terms are each 
replaced with a pressure term, causing the units of k to differ. 

The kinetics of reactions such as those leading to ozone depletion are treated in 
greater depth in subsequent sections. 

Why is it more difficult to breathe when up a mountain 
than at ground level? 

The dependence of rate on reactant pressure 

We all breathe oxygen from the air to maintain life. Although we require oxygen, 
the air contains other gases. Normal air contains about 21 per cent of oxygen, the 
remainder being about 1 per cent argon and 78 per cent nitrogen. 

We term the component of the total air pressure due to oxygen its 'partial pressure', 
p{02)- From Dalton's law (see Section 5.6), the partial pressure of oxygen p(C>2) is 
obtained by multiplying the mole fraction of the oxygen by the overall pressure of 
the gases in air. 

SAQ 8.7 The air we breathe at sea level has a pres- 
sure of 100 kPa. Show that the partial pressure of oxygen 
is 21 kPa. 

'Standard pressure' 
p e equals 10 5 Pa, and 
is sometimes called 
1 bar. 

We start to feel a bit breathless if the partial pressure of oxygen 
decreases below about 15 kPa; and we will feel quite ill (light- 
headed, breathless and maybe nauseous) if p(02) drops below 
10 kPa. 

Air is taken into the lungs when we breathe. There, it is trans- 
ported through the maze of progressively smaller bronchial tubes 
until it reaches the tiny sacs of delicate tissue called alveoli. Each 
sac look like bunches of grapes. The alveoli are the sites where oxy- 
gen from the air enters the blood, and the carbon dioxide from the 
body passes into the air (Figure 8.2). Oxygenated blood then flows 
around the body. Each alveolus is tiny, but there are 300000000 
in each lung. 

We can think of the oxygen transfer from the lung to the blood as a simple chemical 
reaction: molecules of gas strike the alveoli. By analogy with simple solution-phase 
reactions, the rate equation describing the rate at which oxygen enters the blood is 
formulated according to 

As the molar masses of 
oxygen, nitrogen and 
argon are so similar, 
we can approximate 
the mole fractions of 
the gases to their per- 
centage compositions. 

rate = k x /(O2) x /(alveoli) 


where each / simply means 'function of. The thermodynamic function chosen to 
represent the oxygen is its partial pressure, p(02)- The alveoli are solid, so we omit 




Oxygenated air 
in the lung 

Capillaries within 
the lung, containing 



4t J Principal bronchus 

Figure 8.2 Oxygen from the air is taken into the lungs and is transported across the delicate 
alveoli tissue covering the inside of the lung and into the blood 

them from the rate equation, Equation (8.17). Accordingly, Equation (8.17) becomes 

rate = k x p{02) 


which is seen to be very similar to the rate equations we formulated earlier for a 
reaction between solution-phase chemicals, with partial pressures replacing concen- 

The overall pressure of the air decreases as we journey upwards, away from sea 
level, but the proportions of the gases in the air remains constant. At a height of 
2 miles above sea level, the air pressure has dropped to about two-thirds of p & . 

Worked Example 8.5 If the overall air pressure at a height of about 2 miles has dropped 
to 67 kPa, what is the partial pressure of oxygen? 

Strategy, we calculate the partial pressure of oxygen p(02) from Dalton's law (see 
p. 221), saying /KO2) = total pressure of the air x mole fraction of oxygen in the air. 
Substituting values into the above equation: 

p(0 2 ) = 67 kPa x 0.21 

So the partial pressure of oxygen is 14 kPa. 

This simple calculation shows why it is more difficult to breathe when up a mountain 
than at ground level: the pressure term in Equation (8.18) decreases, so the rate at which 
oxygen enters the blood decreases in proportion to the decrease in the oxygen partial 
pressure. And the partial pressure is smaller at high altitudes than at sea level. 


By corollary, more oxygen can enter the blood if the oxygen partial pressure is increased. 
Two simple methods are available to increase piOx)'- 

(1) Breathe air of normal composition, but at a greater overall 
pressure. An example of this approach is the diver who 
breathes underwater while fitted out with SCUBA gear. 

'SCUBA' is an acronym 
for 'self-contained 
underwater breathing 

(2) In cases where the lungs are damaged, a doctor will place a 
patient in an 'oxygen tent'. The overall pressure of gas is the 
same as air pressure (otherwise the tent would explode) but the percentage of 
oxygen in the air is much greater than in normal air. For example, if the overall 
air pressure is the same as p , but the air comprises 63% oxygen rather than 
the 21% in normal air, then the partial pressure of oxygen will treble. 

SAQ 8.8 Consider the equilibrium reaction between hydrogen and chlo- 
rine to form HCI: 


H2(g) + Cl2(g) — ~ > 2HCI(g) 

Write two separate rate laws, one for the forward reaction and one for 
back reactions. 

8.2 Qualitative discussion of concentration 

Why does a full tank of petrol allow a car to travel 
over a constant distance? 

Qualitative reaction kinetics: molecularity 


Cars and buses are fuelled by a volatile mixture of hydrocarbons. The mixture is 
called 'petrol' in the UK, and 'gas' (short for gasoline) in the USA. One of the 
main chemicals in petrol is octane, albeit in several isomeric forms. In the internal 
combustion engine, the carburettor first vaporizes the petrol to form an aerosol (see 
Section 10.2) comprising tiny droplets of petrol suspended in air (Figure 8.3). This 
vaporization process is similar to that which converts liquid perfume into a fine spray. 


Spray (aerosol) of petrol 
droplets in air 

Figure 8.3 A carburettor in a car engine vaporizes the petrol to form an aerosol comprising tiny 
droplets of petrol suspended in air 



These droplets burn in a controlled manner inside the car engine to release their 
chemical energy while the petrol combines chemically with oxygen: 

C8Hi 8 ( both g and i) + 12.502(g) ► 8C02(g) + 9H 2 



This equation relates the overall ratios of reactants and products. 
It may be termed a fully balanced equation, or is more commonly 
termed a stoichiometric equation. 

We employ octane as the fuel rather than, say, paper because 
burning octane releases such a large amount of heat energy (which 
is converted into kinetic energy). Inside the cylinders of the engine, 
this energy is released quickly, causing the gaseous products of 
combustion to heat up rapidly, which causes the pressure inside 
the cylinder to increase greatly. An additional cause of increasing 
pressure is the change in the number of moles of gas formed during 
combustion, since 13.5 mol of reactant form 17 mol of gaseous 
product. The ultimate cause of the car's motion is the large amount 
of pressure work performed during burning. 
A constant amount of energy is released per mole of petrol. Furthermore, the size 

of the car's petrol tank predetermines the car's capacity to contain the alkane fuel; 

so we see how the overall amount of energy available to the car between refuelling 

stops cannot alter much. As the amount of energy is constant, the amount of work 

that the car can perform is also constant. 

A well-tuned car will, therefore, travel essentially the same number of miles per 

tank of petrol because the amount of energy released for work is simply a function 

of the tank's capacity. 

The word 'stoichiom- 
etry' comes from the 
Greek word stoic, 
meaning 'indifference'. 
A stoichiometric reac- 
tion is, therefore, indif- 
ferent to all external 
conditions, and pro- 
ceeds with a predeter- 
mined ratio of products 
and reactants. 

Why do we add a drop of bromine water to a solution 
of an a I ken e? 

Reaction stoichiometry and molecularity 

Bromine readily adds across an alkenic double bond by electrophilic addition 
(Figure 8.4). The brominated compound is usually colourless, but bromine in solution 
('bromine water') has a red colour. Addition of bromine water to an alkene is 
accompanied by a loss of the red colour as reaction proceeds. The stoichiometry of 
reaction is almost always 1:1, with one molecule of bromine reacting per double bond. 
Elemental analysis can be employed to show that the reaction between Br 2 and 
a C=C double bond always occurs with this stoichiometry of 1:1. It is a law of 

Figure 8.4 The red colour of elemental bromine is lost during addi- 
tion across an alkenic double bond; the brominated compound is 
usually colourless 



But sometimes we find that the kinetic data obtained experimentally bear little 
resemblance to the fully balanced reaction. In other words, the reaction proceeds by 
a mechanism that is different from the fully balanced equation. 

The argument (above) concerning petrol centres on the way that chemicals always 
react in fixed proportions. The reaction above (Equation (8.19)) is the stoichiometric 
reaction because it cites the overall amounts of reaction occurring. But we should 
appreciate that the reaction as written will not occur in a single step: it is impos- 
sible even to imagine one molecule of octane colliding simultaneously with 12.5 
molecules of oxygen. The probability is simply too vast. Even if there is a prob- 
ability of getting the molecules together, how do we conceive of one and a half 
molecules of oxygen? How can we have half a molecule? Even if we could, is 
it possible to arrange these 12.5 molecules of oxygen physically around a single 
molecule of octane? 

Similarly, we have also seen already how the copper(II) tetrakis 
(amine) complex forms in a step manner with four separate stages, 
rather than in a single step, forming the mono-ammine complex, 
then the bis-ammine, the tris-ammine and finally the tetrakis -ammine, 
complex. So we start to appreciate that the actual reaction occur- 
ring during the burning of octane is more complicated than it first 
appears to be: the ratios in the stoichiometric equation are not 
useful in determining the reaction mechanism. 

In the case of burning a hydrocarbon, such as octane, the first step of the reac- 
tion usually occurs between a peroxide radical 02* generated by the spark of the 
sparking plug. A radical inserts into a C-H bond of a hydrocarbon molecule with the 
likely mechanism: 

~ C-H + 2 ' >~ C-O'-O-H (8.20) 

The stoichiometric 
ratios in a fully bal- 
anced equation are 
usually not useful for 
determining the reac- 
tion mechanism. 

The peroxide bond in the product is weak and readily cleaves to form additional radicals. 
Because more radicals are formed, any further reaction proceeds by a chain reaction, 
termed radical propagation, until all the petrol has been consumed. 

The rate-limiting process in Equation (8.20) involves the two 
species (peroxide and octane) colliding within the car cylinder 
and combining chemically. Because two species react in the rate- 
limiting reaction step, we say that the reaction step represents 
a bi molecular reaction. In alternative phraseology, we say 'the 
molecularity of the reaction is two'. 

The overwhelming majority of reactions are bimolecular. Some 
reactions are unimolecular and a mere handful of processes proceed 
as a trimolecular reactions. No quadrimolecular (or higher order) 
reactions are known. 

We must appreciate the essential truth that the molecularity of 
a reaction and the stoichiometric equation are two separate things, 
and do not necessarily coincide. Luckily, we find that reactions are 
quite often 'simple' (or 'elementary'), by which we mean that they involve a single 
reaction step. The molecularity and the reaction order are the same if the reaction 

We describe the mole- 

cularity with the famil- 

iar Latin-based des- 

criptors 'uni'= l/bi' = 

2 and Mxi" = 3. 

Care: the molecularity 
and the order of a 
reaction need not be 
the same. 



Kinetically, a reaction 
is simple if the molecu- 
larity and the reaction 
order are the same, 
usually implying that 
the reaction proceeds 
with a single step. 

involves a single step, so we say that many inorganic reactions are 
simple because they are both second order and involve a bimolec- 
ular reaction mechanism. 

In SAQ 8.2, we considered the case of forming [Cu(NH3)4] 2+ 
from copper(II) and ammonia. We have already seen that a reaction 
cannot be quintimolecular - five species colliding simultaneously. 
In fact, involves a sequence of a bimolecular reactions. 

When magnesium dissolves in aqueous acid, why does 
the amount of fizzing decrease with time? 

Reaction profiles 

Magnesium ribbon reacts with sulphuric acid to cause a vigorous reaction, as demon- 
strated by the large volume of hydrogen gas evolved, according to 

Mg (s) + H 2 SO 


MgS0 4(aq) + H 2(g) f 


We know the reaction is complete when no more grey metal remains and the solution 

is clear. In fact, we often say the magnesium dissolves, although such dissolution is 

in fact a redox reaction (see Chapter 7). 

But observant chemists will notice that the rate at which the gas is formed decreases 

even before the reaction has stopped. Stated another way, the rate at which H2 is 

formed will decrease during the course of the reaction. 
We can explain this in terms of a rate expression, as follows. First, consider the 

case where 1 mol of magnesium is reacted with 1 mol of sulphuric acid (we will 
also say that the overall volume of the solution is 1 dm 3 ). Initially 
there is no product, but at the end of reaction there will be 1 mol of 
MgS0 4 and mol of magnesium metal or sulphuric acid. There- 
fore, the amount of product in creases while the amounts of the two 
reactants will both decrease as the reaction proceeds. The amounts 
of material change and, as the volume of solution is constant, the 
concentrations change. 
We have already seen that the concentrations of reactants dictate the rate of reaction. 

For the consumption of magnesium in acid, the rate of reaction is given by 

The concentrations of 
all reactants and prod- 
ucts change during the 
course of a reaction. 

rate = £[Mg][H 2 S0 4 ] 


This result was obtained by recalling how the rate of a reaction is equal to the rate 
constant of the process, multiplied by the concentration of each species at the tail 
end of the arrow in Equation (8.21). 

There is 1 mol of magnesium at the start of reaction, and the concentration of 
sulphuric acid was 1 mol dm -3 (because there was 1 mol of sulphuric acid in 1 dm 3 ). 



We next consider the situation after a period of time has elapsed such that half of the 
magnesium has been consumed. By taking proportions, 0.5 mol of the acid has also 
reacted, so its new concentration is 0.5 mol dm -3 . Accordingly, the value of 'rate' 
from Equation (8.22) is smaller, meaning that the rate has slowed down. And so the 
rate at which hydrogen gas is produced will also decrease. 

Further reflection on Equation (8.22) shows that the concentrations of the two 
reactants will always alter with time, since, by the very nature of a chemical reaction, 
reactants are consumed. Accordingly, the rate of reaction will decrease continually 
throughout the reaction. The rate will reach zero (i.e. the reaction will stop) when one 
or both of the concentrations reaches zero, i.e. when one or all of the reactants have 
been consumed completely. The rate at which hydrogen gas is formed will reach zero 
when there is no more magnesium to react. 

Figure 8.5 is a graph of the amount of sulphuric acid remaining as a function of 
time. The trace commences with 1 mol of H2SO4 and none is left at the end of the 
reaction, so [H2SO4] = 0. The rate of reaction is zero at all times 
after the sulphuric acid is consumed because the rate equation, 
Equation (8.22), involves multiplying the k and [Mg] terms by 
[H2SO4], which is zero. Note that the abscissa (y-axis) could have 
been written as 'concentration', because the volume of the solution 
does not alter during the course of reaction, in which case the trace 
is called a concentration profile. 

Also depicted on the graph in Figure 8.5 is the number of moles of magnesium 
sulphate produced. It should be apparent that the two concentration profiles (for 
reactant and product) are symmetrical, with one being the mirror image of the other. 
This symmetry is a by-product of the reaction stoichiometry, with 1 mol of sulphuric 
acid forming 1 mol of magnesium sulphate product. 

We call a graph of 
concentration of reac- 
tant or product (as V) 
against time (as V) a 
concentration profile. 


1 •* ' 
1 / 
1 / 
1 / 








1 1 


Time f/min 



Figure 8.5 Concentration profiles for the reaction between sulphuric acid and magnesium to 
form magnesium sulphate, (a) Profile for the consumption of sulphuric acid, and (b) profile for the 
formation of magnesium sulphate. The initial concentration of H2SO4 was 1.0 moldm~ 




Figure 8.6 Concentration profile for the reaction between sulphuric 
acid and magnesium to form magnesium sulphate: reaction with 
->• three concentrations of [H2S04], = o. The same amount of magnesium 
reacted in each case, and cj, > C2 > C\ 

Figure 8.6 is a similar graph to Figure 8.5, showing the concentration profiles of 
sulphuric acid as a function of its initial concentration. The gradients of each trace 
are different, with the more concentrated solutions generating the steepest traces. 
In fact, we have merely rediscovered the concept of rate, because the gradient of 
the concentration profile is reaction rate, being the rate at which a compound or 
chemical reacts as a function of time. We say the rate of reaction is the gradient, after 
Equation (8.22). 

Graphs such as those in Figures 8.5 and 8.6 are an ideal means 
of determining the rates of reaction. To obtain the rate, we plot 
the concentration of a reactant or product as a function of time, 
and measure the slope. (Strictly, since the slopes are negative for 
reactants, so the rate is 'slope x — 1'.) 

We obtain the rate 
of reaction as the 
gradient of a concen 
tration profile. 


Because product forms at the expense of reactant, the magnitudes of the rates of forming 
product and consuming reactant are the same. Although the magnitudes of the rates 
are the same, their signs are not: the rate of forming product is positive because the 
concentration increases with time. 

For example, consider the reaction, a A + bB = cC + dD: 

d[Product] d[reactant] 




For the individual chemical components, we say, 

1 d[A] 

reactants rate = — 1 x — x 

a At 

At t [ d[C] 

products rate = - x 

F c dr 

1 d[B] 

rate = — 1 x - x 

b dr 
1 d[D] 

rate = — x 

d dr 

In each case, the minus sign indicates a decreasing concentration with time. 


The rate of loss of reactant is negative because the concentration decreases with 
time The gradient at the start of the reaction is called the initial rate. Analysing 
the initial rates method is an extremely powerful way of determining the order of 
a reaction. 

Worked Example 8.6 The following kinetic data were obtained for the reaction between 
nitric oxide and hydrogen at 700 °C. Determine: (1) the order of the reaction of the reaction 
at this temperature; (2) the rate constant of reaction. 

Experiment A 

Experiment B 

Experiment C 

Initial concentration of 




NO/mol dm" 3 

Initial concentration of 




H2/mol dm~ 3 

Initial rate/mol dnT 3 s _1 

2.4 x 10~ 6 

1.2 x 10~ 6 

0.6 x 10~ 6 

The rates listed in the table were obtained as the gradients of graphs like those in 
Figure 8.6. 

Answer Strategy 

(1) To determine the order of reaction. It is always good research strategy to change 
only one variable at a time. That way, the measured response (if any) can be attributed 
unambiguously to the change in that variable. And the variable of choice in this example 
will be one or other of the two initial concentrations. 

The rate equation for the reaction will have the following form: 

rate=fc[NOr[H 2 ]- y (8.23) 

We determine values for the exponents x and y varying [NO] and [H2]. First, consider 
experiments A and B. In going from experiment B to experiment A, the concentration 
[NO] remains constant but we double [H2], as a consequence of which the rate dou- 
bles. There is, therefore, a linear relationship between [H2] and rate, so the value of y 
is '1'. 

Second, consider experiments A and C. In going from C to A, we double the con- 
centration [NO] and the rate increases by a factor of four. Accordingly, the value of x 
cannot be '1' because the increase in rate is not linear. In fact, as 2 2 = 4, we see how 
the exponent x has a value of '2'. 

The reaction is first order in H2, second order in NO and, therefore, third order overall. 
Inserting values into Equation (8.23), we say, rate = A:[NO] 2 [H2]. 

(2) To determine the rate constant of reaction. We know the rate and concentrations for 
several sets of experimental conditions, so we rearrange Equation (8.23) to make k the 
subject and insert the concentrations. 

k = 

[NO] 2 [H 2 ] 


The question cites data from three separate experiments, each of which will give the same 
answer. We will insert data from 'experiment A': 

2.4 x 1(T 6 moldings" 1 

[0.25 moldm~'] 2 [0.01 moldm _j ] 


k = 3.84 x 10 3 dm 6 mor 2 s _1 

SAQ 8.9 Iodide reacts with thiosulphate to form elemental iodine. If the 
reaction solution contains a tiny amount of starch solution, then this I2 is 
seen by eye as a blue complex. The data below were obtained at 298 K. 
Determine the order of reaction, and hence its rate constant k. 

Experiment A 

Experiment B 

Experiment C 

Initial concentration of 




I~/mol dm" 3 

Initial concentration of 




S 2 8 2 "/mol dm 3 

Initial rate/mol 

1.5 x 10- 4 

7.5 x 10- 5 

3.0 x 10- 4 

dm~ 3 s * 

8.3 Quantitative concentration changes: 
integrated rate equations 

Why do some photographs develop so slowly? 

Extent of reaction and integrated rate equations 

A common problem for amateur photographers who develop their own photographs 
is gauging the speed necessary for development. When the solution of thiosulphate 
is first prepared, the photographs develop very fast, but this speed decreases quite 
rapidly as the solution 'ages', i.e. the concentration of S20s 2 ~ decreases because 
the thiosulphate is consumed. So most inexperienced photographers have, at some 
time, ruined a film by developing with an 'old' solution: they wait what seems like 
forever without realizing that the concentration of thiosulphate is simply too low, 
meaning that development will never occur. They ruin the partially processed film 
by illuminating it after removal from the developing bath. These amateurs need to 
answer the question, 'How much thiosulphate remains in solution as a function of 
timeT We need a mathematical equation to relate concentrations and time. 



The concentrations of each reactant and product will vary during the course of 
a chemical reaction. The so-called integrated rate equation relates the amounts of 
reactant remaining in solution during a reaction with the time elapsing since the 
reaction started. The integrated rate equation has a different form according to the 
order of reaction. 

Let us start by considering a first-order reaction. Because the reactant concentration 
depends on time t, we write such concentrations with a subscript, as [A] ( . The initial 
reactant concentration (i.e. at time t = 0) is then written as [A]o. The constant of 
proportionality in these equations will be the now-familiar rate constant k\ (where 
the subscripted '1' indicates the order). 

The relationship between the two concentrations [A]o, [A] r and t is given by 





Equation (8.24) is the integrated first-order rate equation. Being 
a logarithm, the left-hand side of Equation (8.24) is dimensionless, 
so the right-hand side must also be dimensionless. Accordingly, 
the rate constant k will have the units of s _1 when the time is 
expressed in terms of the SI unit of time, the second. 

The first-order rate 
constant k will have 
the units of s _1 . 

Worked Example 8.7 Methyl ethanoate is hydrolysed when dissolved in excess hydro- 
chloric acid at 298 K. The ester's concentration was 0.01 moldm -3 at the start of the 
reaction, but 8.09 x 10~ 2 after 21 min. What is the value of the first-order rate con- 
stant kyl 

Answer Strategy. (1) we convert the time into SI units of seconds; (2) we insert values 
into Equation (8.24). 

(1) Convert the time to SI units: 21 min = 21 min x 60 smin = 1260 s 

(2) Next, inserting values into Equation (8.24) 


0.01 moldm~ J 
0.008 09 moldm" 

= 1260 s x k\ 

Notice how the units 
of concentration will 
cancel here. 


ln(1.236) = ki x 1260 s 

Taking the logarithm yields 

0.211 = Jfc, x 1260 s 



and rearranging to make k\ the subject gives 



1.68 x 10~ 4 sT l 

1260 s 
This value of k\ is relatively small, indicating that the reaction is rather slow. 

SAQ 8.10 (Continuing with the same chemical example): what is the 
concentration of the methyl ethanoate after a time of 30 min? Keep the 
same value of ki - it's a constant. 

We can calculate a 
rate constant k without 
knowing an abso- 
lute value for [A] 
by following the frac- 
tional changes in the 
time-dependent con- 
centration [A] t . 

We note, when looking at the form of Equation (8.24), how the 
bracket on the left-hand side contains a ratio of concentrations. 
This ratio implies that we do not need to know the actual concen- 
trations of the reagent [A]o when the reaction starts and [A], after 
a time t has elapsed since the reaction commenced; all we need 
to know is the fractional decrease in concentration. Incidentally, 
this aspect of the equation also explains why we could perform the 
calculation in terms of a percentage (i.e. a form of ratio) rather 
than a 'proper' concentration. 
In fact, because the integrated first-order rate equation (Equation (8.24)) is written 
in terms of a ratio of concentrations, we do not need actual concentrations in moles 
per litre, but can employ any physicochemical parameter that is proportional to con- 
centration. Obvious parameters include conductance, optical absorbance, the angle 
through which a beam of plane-polarized light is rotated (polarimetry), titre from a 
titration and even mass, e.g. if a gas is evolved. 

Worked Example 8.8 Consider the simple reaction 'A — > product' . After 3 min, 20 per 
cent of A has been consumed when the reaction occurs at 298 K. What is the rate constant 
of reaction k\ ? 

We start by inserting values into the integrated Equation (8.24), noting 
that if 20 per cent has been consumed then 80 per cent remains, so: 

If we know the amount 
of reactant consumed, 
then we will need to 
calculate how much 

Remember that we 
should always cite a 
rate constant at the 
temperature of mea- 
surement, because k 
itself depends on tem- 

/100% of [A] 

In I 

V 80% 

k x x (3 x 60 s) 

of [A] 
The logarithmic term on the left-hand side is ln(1.25) = 0.223, so 


k, = 

180 s 

where the term '180 s' comes from the seconds within the 3 min of 
observation time. We see that k\ — 1.24 x 10~ 3 s" 1 at 298 K. 

SAQ 8.11 If the molecule A reacts by a first-order mech- 
anism such that 15% is consumed after 1276 s, what is 
the rate constant k? 



Justification Box 8.1 

Integrated rate equations for a first-order reaction 

The rate law of a first-order reaction has the form 
'rate = &i[A]'. And, by 'rate' we mean the rate of 
change of the concentration of reactant A, so 

rate = 


= *i[A] 


The minus sign in 
Equation (8.25) is 
essential to show that 
the concentration of A 
decreases with time. 

We will start at t = with a concentration [A]o, with the concentration decreasing with 
time t as [A] t . The inclusion of a minus sign is crucial, and shows that species A is a 
reactant and thus the amount of it decreases with time. 

Separating the variables (i.e. rearranging the equation) and indicating the limits, 
we obtain 

/ — d[A] = *! / 

•/fAln L A Jt Jo 


Note how we can place the rate constant outside the integral, because it does not change 
with time. Integration then yields 


And, after inserting the limits, we obtain 

- (ln[A], - ln[A] ) = *if 

Using the laws of logarithms, the equation may be tidied further to yield Equation (8.24): 


which is the integrated first-order rate equation. 

Note that if a multiple-step reaction is occurring, then this equation relates only to the 
case where the slowest (i.e. rate-limiting) step is kinetically first order. We will return 
to this idea when we consider pseudo reactions in Section 8.4. 

Graphical forms of the rate equations 

Similar to the integrated first-order rate equation is the linear first-order rate equation : 




t + 





We obtain the first- 
order rate constant k 
by drawing a graph 
with the integrated 
first-order rate equa- 
tion, and multiplying its 
slope by -1. 

which we recognize as having the form of the equation for a straight 
line: plotting ln[A] ( (as 'y') against time (as 'x') will be linear for 
reactions that are first order. 

The rate constant k\ is obtained from such a first-order rate graph 
as (—1 x gradient) if the time axis is given with units of seconds. 
Accordingly, the units of the first-order rate constant are s _1 . 

Worked Example 8.9 Consider the following reaction: hydrogen 
peroxide decomposes in the presence of excess cerous ion Ce m (which 
reacts to form eerie ion Ce IV ) according to a first-order rate law. The 
following data were obtained at 298 K: 

This reaction appears 
to be first order 
because the cerium 
ions are in excess, so 
the concentration does 
not really change. We 
look at pseudo-order 
reactions on p. 387. 

Time f/s 
[H 2 2 ],/moldm _3 







Time f/s 
[H 2 2 L/moldm- 3 







Figure 8.7 shows the way the concentration of hydrogen perox- 
ide decreases with time. The trace is clearly curved, and Figure 8.8 
shows a graph constructed with the linear form of the first-order 
integrated rate equation, Equation (8.26). This latter graph is clearly 

The rate constant is obtained from the figure as (— 1 x 'gradient'), 
so A: = 0.11 s" 1 at 298 K. 







10 15 

Time t/ms 

Figure 8.7 Plot of [H2O2L against time. Notice the pronounced plot curvature 




Figure 8.8 Graph constructed by drawing ln[H 2 02L (as 'y') against time t (as V), i.e. with the 
axes of a linear first-order rate law. Notice the linearity of the trace 

Justification Box 8.2 

The linear form of the integrated first-order rate equation 

If we start with the now-familiar integrated rate equation of Equation (8.24): 




Using the laws of logarithms, we can split the left-hand side: 

ln[A] - ln[A], = kit 

Next, we multiply by — 1 : 

ln[AT - ln[A] = -M 

Then, by adding the [A]o term to the right-hand side we obtain Equation (8.26): 

ln[A] t = -kit + ln[A] 

which is the linear form of the equation, as desired. We note that the intercept in 
Equation (8.26) is ln[A]o- 

We could have obtained Equation (8.26) alternatively by integrating without limits 
during the derivation in Justification Box 8.1. 



The SI unit of time is 
the second (see p. 15), 
but it is sometimes 
more convenient to cite 
k in terms of non-SI 

units, such as min 
even year -1 . 


Worked Example 8.10 What is the relationship between the values 
of rate constant expressed in units of s~ : , and expressed in units of 
min -1 and year 


By Analogy. Let the rate constant be k. Conversion into SI is easy: 
k in min -1 = 60 x (k in s _1 ) because there are 60 s per minute, 
so 60 times as much reaction can occur during a minute. 

Via Dimensional Analysis. Again, let the rate constant be k. In 
this example, imagine the value of k is 3.2 year -1 . 
The number of seconds in a year is (60 x 60 x 24 x 365.25) = 3.16 x 10 7 s. We 
can write this result as 3.16 x 10 7 syr -1 , which (by taking reciprocals) means that 
2.37 x 10" 8 yrs" 1 . 

To obtain k with units of s _1 , we say 

To convert k from SI 
units to the time unit 
of choice, just mul- 
tiply the value of k 
by the fraction of the 
time interval occurring 
during a single second. 


(3.2 yr" 1 ) x (2.37 x 10" s yr s" 1 ) 

original value of k conversion factor 

sok= 1.01 x 10" 7 s _1 . 

SAQ 8.12 Show that the rate constants 1.244 x 10 4 yr" 1 and 3.94 x 
10" 4 s _1 are the same. 

Integrated rate equations: second-order reactions 

For a second-order reaction, the form of the integrated rate equation is different: 

Notice that th 

e units 

of the second 


rate constant 

k 2 are 

dm 3 mor 1 s -1 


are, in effect, 


tration) -1 s _1 . 

We do not need to 
know the temperature 
in order to answer this 
question; but we do 
need to know that T 
remained constant, i.e. 
that the reaction was 



[A], [A] 

= k 2 t 


where the subscripted '2' on k reminds us that it represents a 
second-order rate constant. The other subscripts and terms retain 
their previous meanings. 

SAQ 8.13 Show that the units of the second-order rate 
constant are dm 3 mor 1 s~ 1 . [Hint: you will need to per- 
form a simple dimensional analysis of Equation 8.27.] 

Worked Example 8.11 We encountered the dimerization of methyl 
viologen radical cation MV + " in Equation (8.6) and Worked 
Example 8.4. Calculate the value of the second-order rate constant 
£2 if the initial concentration of MV + " was 0.001 moldm -3 and the 
concentration dropped to 4 x 10~ 4 mol dirT 3 after 0.02 s. (The tem- 
perature was 298 K.) 



Inserting values into Equation (8.27): 




4 x 10" 4 moldm" 

2500 (moldm -3 ) - 

10 3 moldm 

1000 (moldm -3 )" 

= k 2 x 0.02 s 

k 7 x 0.02 s 

1500 (moldm -3 )" 1 

k 7 x 0.02 s 

Rearranging, we say 

1500 (moldm -3 )" 


k 2 = 7.5 x 10 4 dm 3 mol" 1 s" 1 at 298 K 
which is relatively fast. 

SAQ 8.14 Remaining with the same system from Worked Example 8.11, 
having calculated k 2 (i.e. having 'calibrated' the reaction), how much MV + " 
remains after 40 ms (0.04 s)? 

SAQ 8.15 Consider a second-order reaction which consumes 15 per cent 
of the initial material after 12 min and 23 s. If [A]o was 1 x 10 3 mol dm 3 , 
calculate k 2 . [Hint: first calculate how much material remains.] 

An alternative form of the integrated rate equation is the so-called linear form 




= h 

t + 





which we recognize as relating to the equation of a straight line, 
so plotting a graph of l/[A] r (as 'y') against time (as 'x') will be 
linear for a reaction that is second order. The rate constant k 2 is 
obtained directly as the gradient of the graph. 

We obtain the second- 
order rate constant k 2 
as the slope of a graph 
drawn according to 
the integrated second- 
order rate equation. 

Worked Example 8.12 Consider the data below, which relate to the second-order racem- 
ization of a glucose in aqueous hydrochloric acid at 17 °C. The concentrations of glucose 
and hydrochloric acid are the same, '[A]'. 

Time f/s 
[A] /mol dm 









Care: The gradient 
is only truly k if the 
time axis is given with 
the SI units of time 
(the second). 

A graph of the concentration [A] (as 'y') against time (as 'x') is 
clearly not linear; see Figure 8.9(a). Conversely, a different, linear, 
graph is obtained by plotting 1/[A] ( (as 'y') against time (as 'x'); see 
Figure 8.9(b). This follows the integrated second-order rate equation. 
The gradient of Figure 8.9(b) is the second-order rate constant ki, and 
has a value of 6.00 x 10 -4 dm 3 mol~' s" 1 . 

1000 1500 

Time // s 



1000 1500 2000 2500 

Time f/s 

Figure 8.9 Kinetics of a second-order reaction: the racemization of glucose in aqueous mineral 
acid at 17 °C: (a) graph of concentration (as 'y') against time (as 'jc'); (b) graph drawn according 
to the linear form of the integrated second-order rate equation, obtained by plotting 1/[A], (as 'y') 
against time (as '*'). The gradient of trace (b) equals the second-order rate constant k 2 , and has a 
value of 6.00 x 10~ 4 dnVmor's" 1 



SAQ 8.16 Consider the following data concerning the 
reaction between triethylamine and methyl iodide at 20 °C 
in an inert solvent of CCI 4 . The initial concentrations of 
[CH 3 I] and [N(CH 3 ) 3 ] are the same. Draw a suitable 
graph to demonstrate that the reaction is second order, 
and hence determine the value of the second-order rate 
constant k 2 . 


ber that 





the concentration at 

t= is 

also a 


data po 


Time r/s 
[CH 3 I]o = 

[N(CH 3 ) 3 ] /mol dm 







Warning: if a chemical process comprises several reaction steps, only the progress 
of the slowest step can be followed kinetically. These graphical methods of deter- 
mining k are only useful for obtaining the rate-determining step (RDS) of such 
reactions. Although the reaction may appear kinetically simple, it is wisest to assume 

Justification Box 8.3 

Integrated rate equations for a second-order reaction 

We will only consider the derivation of the simplest case of a second-order reaction, 
where the concentrations of the two reacting species are the same. Being second order, 
the rate law has the form rate = ^[A] 2 . The subscript 
'2' on k indicates a second-order process. Again, by 
'rate' we mean the rate of change of the concentration 
of reactant A. 

rate ; 


k 2 [A] z 


The minus sign in 
Equation (8.29) is ess- 
ential to show that 
the concentration of A 
decreases with time. 

As before, we shall start at t = with a concentration [A]o. The value of [A], 
decreases with time t, hence a minus sign is inserted. 

Rearranging the equation, and indicating the limits yields 

/•[A], J ft 

/ —2 d[A] = k 2 / 

.'[Alo |AJ, Jo 

Integrating gives 

-I [A], 


= k 2 [t]' 

-I [Alo 


The two minus 

signs on 

the left will cancel. 



and rearranging 


gives Equation 


1 1 

= k 2 t 

This equation is known 

as the 

integrated second-order rate equation. 

Second-order reactions of unequal concentration 

We will start with the reactants A and B having the concentrations [A]o and [B]o 
respectively. If the rate constant of reaction is k^, and if the concentrations at time t 
are [A], and [B] ( respectively, then it is readily shown that 


[B] - [A], 

, , [A]„x[B]. 

x In = K2I 

[B] x [A], ' - 


We will need to look further at this equation when thinking about kinetic situations in 
which one of the reactants is in great excess (the so-called 'pseudo order' reactions 
described in Section 8.4). 

SAQ 8.17 A 1:1 reaction occurs between A and B, and is second order. 
The initial concentrations of A and B are [A]o = 0.1 moldrrT 3 and [B]n = 

0.2 moldrrr 3 . What is k 2 if [A] t = 0.05 moldm -3 
to work out a value for [B] f as well. 

after 0.5 h? Remember 

Why do we often refer to a 'half-life' when speaking 
about radioactivity? 


A radioactive substance is one in which the atomic nuclei are unstable and sponta- 
neously decay to form other elements. Because the nuclei decay, the amount of the 
radioactive material decreases with time. Such decreases follow the straightforward 
kinetic rate laws we discussed above. 

But many people talk emotionally of radioactivity 'because 
radioactive materials are so poisonous', and one of the clinching 
arguments given to explain why radioactivity is undesirable is that 
radioactive materials have long 'half-lives'. What is a half-life? 
And why is this facet of their behaviour important? And, for that 
matter, is it true that radioactive materials are poisonous? 

We shall look at why 
radioactive materials 
are toxic in Section 8.3, 



Table 8.2 Half-lives of radioactive isotopes (listed in order of increasing atomic number) 



Source of radioactive isotope 

l2 B 

14 C 

40 K 
60 Co 


239p u 




1.3 x 

10 9 




1.6 x 

10 7 


4.5 x 

10 s 


2.4 x 

10 4 


Unnatural (manmade) 


Natural: 0.011% of all natural potassium 

Unnatural: made for medicinal uses 

Unnatural: fallout from nuclear weapons 

Natural: 99.27% of all uranium 

Unnatural: by-product of nuclear energy 

A 'half-life' t 1/2 is the 
time required for the 
amount of material 
to halve. 

The half-life of radioactive decay or of a chemical reaction is 
the length of time required for exactly half the material under 
study to be consumed, e.g. by chemical reaction or radioactive 
decay. We often give the half-life the symbol t\/2, and call it 
'tee half. 

The only difference between a chemical and a radioactive half-life is that the 
former reflects the rate of a chemical reaction and the latter reflects the rate of 
radioactive (i.e. nuclear) decay. Some values of radioactive half-lives are given in 
the Table 8.2 to demonstrate the huge range of values ti/2 can take. The difference 
between chemical and radioactive toxicity is mentioned in the Aside box on p. 382. 
A chemical half-life is the time required for half the material to have been consumed 
chemically, and a radioactive half-life is the time required for half 
of a radioactive substance to disappear by nuclear disintegration. 
Since most chemicals react while dissolved in a constant volume 
of solvent, the half-life of a chemical reaction equates to the time 
required for the concentration to halve. 

Worked Example 8.13 The half-life of bU Co is 10.5 min. If we start 
with 100 g of 60 Co, how much remains after 42 min? 

60 Co is a favourite 
radionuclide within the 
medical profession, 
because its half life 
is conveniently short. 

Answer Strategy 

1 . We determine how many of the half-lives have occurred during the 
time interval. 

2. We then successively halve the amount of 60 Co, once per half-life. 

(1) The number of half-lives is obtained by dividing 10.5 min into 42 min; so four 
half-lives elapse during 42 min. 

(2) If four half-lives have elapsed, then the original amount of 60 Co has halved, 
then halved again, then halved once more and then halved a fourth time: 


1st 2nd 3rd 4th 

half-life _ half-life „, half-life ,_ half-life ^ „, 
► 50 g ► 25 g ► 12.5 g ► 6.25 g 


We see that one-sixteenth of the 60 Co remains after four half-lives, because ( j) 4 = i. 
In fact, a general way of looking at the amount remaining after a few half-lives is to 
say that 

fraction remaining = (|)" (8.31) 

where n is the number of half-lives. 

SAQ 8.18 The half-life of radioactive 14 C is 5570 years. If we start with 
10 g of 14 C, show that the amount of 14 C remaining after 11140 years 
is 2.5 g. 

Often, though, we don't know the half-life. One of the easier ways to determine a 
value of ti/2 is to draw a graph of amount of substance (as ' v') - or, if in solution, 
of concentration - against time (as 'x'). 

Worked Example 8.14 The table below shows the amount of a biological metabolite 
T-IDA as a function of time. 

1. What is the half-life of T-IDA? 

2. Show that the data follow the integrated first-order rate equation. 

Time f/min 

















dm" 3 

Figure 8.10 shows a plot of the amount of material (as 'v') as a function of time / (as 
'x'), which is exponential. This shape should not surprise us, because Equation (8.31) is 
also exponential in form. 

To obtain the half-life, we first choose a concentration - any concentration will do, 
but we will choose [T-IDA] = 50 pmoldm -3 . We then draw a horizontal arrow from this 
concentration on the y-axis, note the time where this arrow strikes the curve, and then read 
off the time on the x-axis, and call it t\ . Next, we repeat the process, drawing an arrow 
from half this original concentration, in this case from [T-IDA] = 
25 |xmoldm~ . We note this new time, and call it t2. The half-life is 
simply the difference in time between t\ and ti- It should be clear that 
the half-life in this example is 10 min. 

But then we notice that the time needed to decrease from 60 to 
30 |imoldm~ , or from 2 to 1 p,moldm~ will also be 10 min each. 
In fact, we deduce the important conclusion that the half-life of a 
first-order reaction is independent of the initial concentration of 

As long as a reaction 
is first order, the dura- 
tion of a half-life will 
be the same length 
of time regardless of 
the initial amount of 
material present. 



Figure 8.10 Kinetic trace concerning the change in concentration as a function of time: graph 
of [T-IDA] (as 'y') against time (as 'x') to show the way half-life is independent of the initial 

Figure 8.11 Graph plotted with data from Figure 8.10, plotted with the axes of the linear form 
of the integrated first-order rate equation, with ln[A] as 'y' against time t as 'x' 

To show that this reaction is kinetically first order, we take the logarithm of the con- 
centration, and plot ln[A]f (as 'y') against time t (as '*'); see Figure 8.11. That the graph 
in Figure 8.11 is linear with this set of axes demonstrates its first-order character. 

The half-life of second- or third-order reactions is not independent of the initial con- 
centration in this way (see p. 387). 



The difference between chemical and radiochemical toxicity 

It is good that we should be concerned about the environmental impact of what we, 
as chemists, do to our planet. But many environmental campaigners too easily confuse 
radioactive toxicity and chemical toxicity. For example, the radon gas emanating from 
naturally occurring granite rocks is chemically inert, because it is a rare gas, but it is 
toxic to humans because of its radioactivity. Conversely, sodium cyanide contains no 
radioactive constituents yet is chemically toxic. 

The conceptual problems start when considering materials such as plutonium, which is 
a by-product of the nuclear electricity industry. Plutonium is one of the most chemically 
toxic materials known to humanity, and it is also radioactive. The half-life of 238 Pu is so 
long at 4.5 x 10 8 years (see Table 8.2) that we say with some certainty that effectively 
none of it will disappear from the environment by radioactive decay; and if none of it 
decays, then it cannot have emitted ionizing a and ft particles, etc. and, therefore, cannot 
really be said to be a radioactive hazard. Unfortunately, the long half-life also means that 
the 238 Pu remains more-or-less for ever to pollute the environment with its lethal chemistry. 

But if we accept that plutonium is chemically toxic, 
then we must also recognize that the extent of its toxic- 
ity will depend on how the plutonium is bonded chemi- 
cally, i.e. in what redox and chemical form it is present. 
As an example, note how soldiers were poisoned with 
chlorine gas during the First World War (when it was 
called Mustard Gas), but chloride in table salt is vital 
for life. Some plutonium compounds are more toxic 
than others. 

At the other extreme are materials with very short 
radioactive half-lives, such as 12 B (which has a relatively short t\/2 of 0.02 h). 12 B is 
less likely to cause chemical poisoning than 238 Pu simply because its residence time is so 
short that it will transmute to become a different element and, therefore, have little time to 
interact in a chemical sense with anything in the environment (such as us). On the other 
hand, its short half-life means that the speed of its radioactive decay will generate many 
subatomic particles (a, fi and y particles) responsible for radioactive poisoning per unit 
time, causing a larger dose of radioactive poisoning. 

This Aside is not intended to suggest that the threat of radioactivity is to be ignored 
or marginalized; but we should always aim to be well informed when confronting an 
environmental problem. 

If we accept that plu- 
tonium is chemically 
toxic, then we also 
need to recognize that 
the extent of its toxicity 
will depend on how the 
plutonium is bonded 
chemically (see p. 59). 

How was the Turin Shroud 'carbon dated'? 

Quantitative studies with half-lives 

The Turin Shroud is a long linen sheet housed in Italy's Turin Cathedral. Many 
people believe that the surface of the cloth bears the image of Jesus Christ (see 



Figure 8.12 Many people believe the Turin Shroud bears an image of Jesus Christ, imprinted soon 
after his crucifixion. Radiocarbon dating suggests that the flax of the shroud dates from 1345 AD 

Figure 8.12), imprinted soon after he died by crucifixion at the hands of the Roman 
authorities in about ad 33. There has been constant debate about the Turin Shroud and 
its authenticity since it first came to public notice in ad 1345: some devout people 
want it to be genuine, perhaps so that they can know what Jesus actually looked like, 
while others (many of whom are equally devout) believe it to be a fake dating from 
the Middle Ages. 

An accurate knowledge of the Turin Shroud's age would allow us to differentiate 
between these two simplistic extremes of ad 33 and ad 1345, effectively distinguishing 
between a certain fake and a possible relic of enormous value. 

The age of the cloth was ascertained in 1988 when the Vatican (which has juris- 
diction over Turin Cathedral) allowed a small piece of the cloth to be analysed by 
radiocarbon dating. By this means, the shroud was found to date from ad 1320 ± 65. 
Even after taking account of the uncertainty of ±65 years, the age of the shroud is 
consistent with the idea of a medieval forgery. It cannot be genuine. 

But the discussion about the shroud continues, so many people now assert that 
the results of the test itself are part of a 'cover up', or that the moment of Jesus' s 
resurrection occurred with a burst of high-energy sub-atomic particles, which upset 
the delicate ratio of carbon isotopes. 



In 1946, Frank Libby of 
the Institute of Nuclear 
Sciences in Chicago 
initiated the dating of 
carbon-based artifacts 
by analysing the extent 
of radioactive decay. 

Radiocarbon dating 

The physicochemical basis behind the technique of radiocarbon 
dating is the isotopic abundances of carbon's three isotopes: 12 C 
is the 'normal' form and constitutes 98.9 per cent of all naturally 
occurring carbon. 13 C is the other naturally occurring isotope, with 
an abundance of about 1 per cent. 14 C does not occur naturally, 
but tiny amounts of it are formed when high-energy particles from 
space collide with gases in the upper atmosphere, thus causing 
radiochemical modification. 
All living matter is organic and, therefore, contains carbon; and since all living 

material must breathe CO2, all carbon-based life forms ingest 14 C. Additionally, living 

matter contains 14 C deriving from the food chain. 

But 14 C is radioactive, meaning that atoms of 14 C occasionally 

The beta particle emit- 
ted during radioactive 
decay is an ener- 
getic electron. 

self-destruct to form a beta particle, /3 , and an atom of 14 N 

14 N + /J 



Radiocarbon dating is 
also called 'radiometric 
dating' or 'radiochemi- 
cal dating'. 

The half-life of the process in Equation (8.32) is 5570 years. 
Following death, flora and fauna alike cease to breathe and eat, so the only 14 C in 
a dead body will be the 14 C it died with. And because the amounts of 14 C decrease 
owing to radioactive decay, the amount of the 14 C in a dead plant or 
person decreases whereas the amounts of the 12 C and 13 C isotopes 
do not. We see why the proportion of 14 C decreases steadily as a 
function of time following the instant of death. 

By corollary, if we could measure accurately the ratio of 12 C to 
14 C in a once-living sample, we could then determine roughly how 
long since it was last breathing. This explains why we can 'date' 
a sample by analysing the residual amounts of 14 C. 

In the carbon-dating experiment, a sample is burnt in pure oxy- 
gen and converted into water and carbon dioxide. Both gases are 
fed into a specially designed mass spectrometer, and the relative 
abundances of 12 CC>2 and 14 CC>2 determined. The proportion of 
14 C02 formed from burning an older sample will be smaller. 

Knowing this ratio, it is a simple matter to back calculate to 
ascertain the length of time since the sample was last alive. For 
example, we know that a time of one half-life tin has elapsed if 
a sample contains exactly half the expected amount of 14 C, so the 
sample died 5570 years ago. 

Great care is needed during the preparation of the sample, since 
dirt, adsorbed CO2 and other impurities can all contain additional 
sources of carbon. The dirt may come from the sample, or it could 
have been adsorbed during sample collection or even contamination 
during the dating procedure. The more recent the contamination, 
the higher the proportion of carbon that is radioactive 14 C that has 

Oil and oil-based prod- 
ucts contain no 14 C, 
because the crea- 
tures from which the 
oil was formed died 
so many millions of 
years ago. Accordingly, 
Equation (8.32) has 
proceeded to its 

Great care is needed 
during the preparation 
of the sample before 
dating to eliminate the 
possibility of contami- 
nation with additional 
sources of carbon. 



not yet decayed, causing the artifact to appear younger. It has been suggested, for 
example, that the Turin Shroud was covered with much 'modern' pollen and dust at 
the time of its radiocarbon dating, so the date of ad 1320 refers to the age of the 
modern pollen rather than that of the underlying cloth itself. 

How old is Otzi the iceman? 

Calculations with half-lives 

Approximately 5000 years ago, a man set out to climb the Tyrolean Alps on the 
Austrian-Italian border. At death, he was between 40 and 50 years old and suffered 
from several medical ailments. Some scientists believe he was caught in a heavy 
snowfall, fell asleep, and froze to death. Others suppose he was murdered during 
his journey. Either way, his body was covered with snow almost immediately and, 
due to the freezing weather, rapidly became a mummy - 'The Iceman'. In 1991, his 
body was re-exposed and discovered by climbers in the Otzal Alps, explaining why 
the 'Iceman', as he was called, was given the nickname 'Otzi' (or, 
more commonly, as just Otzi). 

His body (see Figure 8.13) was retrieved and taken to the Dep- 
artment of Forensic Medicine at the University of Innsbruck. Their 
analytical tests - principally radiocarbon dating - suggest that 
Otzi died between 3360 and 3100 bc. Additional radiocarbon dat- 
ing of wooden artifacts found near his body show how the site of 
his death was used as a mountain pass for millennia before and 
after his lifetime. 

But how, having defined the half-life tin as the time necessary 
for half of a substance to decay or disappear, can we quantita- 
tively determine the time elapsing since Otzi died? In Justification 
Box 8.4, we show how the half-life and the rate constant of decay 
k are related according to 

Scientific techniques, 
such as radiocarbon 
dating, applied to 
archaeology is some- 
times termed archaeo- 




Equation (8.33) sug- 
gests the half-life is 
independent of the 
amount of material 
initially present, so 
radioactive decay fol- 
lows the mathematics 
of first-order kinetics. 

Figure 8.13 The body of 'Otzi the Iceman' was preserved in the freezing depths of a glacier. 
Radiocarbon dating suggests that Ozti froze to death in about BC 3360-3100 



Worked Example 8.15 A small portion of Otzi's clothing was removed and burnt care- 
fully in pure oxygen. The amount of 14 C was found to be 50.93 per cent of the amount 
expected if the naturally occurring fabric precursors had been freshly picked. How long 
is it since the crop of flax was picked, i.e. what is its age? 

We start by inserting the known half-life t\/2 into Equation (8.33) to obtain a 'rate constant 
of radioactive decay'. 


By calculating k, we 
are in effect calibrating 
the experiment. We 
only need to do this 
calibration once. 

k = 

5570 yr 

By this means, we calculate the rate constant as k — 1.244 x 
10~ 4 yr -1 . (Alternatively, we could have calculated k in terms of 
the SI unit of time (the second), in which case k has the value 

3.94 x 10 



Using our calculators, 
we need to type 
'ln(0.5093)' as the 
numerator rather than 
a percentage. The 
minus sign comes from 
the laws of logarithms. 

To calculate t, the length of time since the radioactive decay 
commenced (i.e. since the fabric precursor died), we again insert 
values into the integrated form of the first-order rate equation, 
Equation (8.33). We then insert our previously calculated value of k: 


■1.244 x 10" 4 yr" 

5420 yr 

So the interval of time t since the flax was picked and thence woven 
into cloth is 5420 yr, so the cloth dates from about 3420 bc. 

SAQ 8.19 To return to the example of the Turin Shroud. Suppose a 
sloppy technique caused the precision of the 14 C measurement to decrease 
from 92.23 per cent to 90 ±2 percent. Calculate the range of ages for 
the shroud. [Hint: perform two calculations, one for either of the extreme 
values of percentage.] 

Justification Box 8.4 

Justification Box 8.1 shows how the integrated first-order rate equation is given by 
Equation (8.24): 





After a length of time equal to one half-life ti/2, the concentration of A will be [A] t , 
which, from the definition of half-life, has a value of jfAJo. 
Inserting these two concentrations into Equation (8.24) gives 



= kt 



The two [A]o terms cancel, causing the bracket on the left to simplify to just '2'. 
Accordingly, Equation (8.24) becomes 

ln(2) = kt l/2 

So, after rearranging to make t\/2 the subject, the half-life is given by Equation (8.33): 


fl/2 = — — 

SAQ 8.20 Show from Equation (8.27) that the half-life 
of a second-order equation is given by the expression: 


tl/2 = 

[A] /f2 

The half-life of a 
second-order reaction 
is not independent of 
the initial concentra- 
tions used. 

Why does the metabolism of a hormone not cause a 
large chemical change in the body? 

'Pseudo-order' reactions 

A hormone is a chemical that transfers information and instructions between cells in 
animals and plants. They are often described as the body's 'chemical messengers', but 
they also regulate growth and development, control the function of various tissues, 
support reproductive functions, and regulate metabolism (i.e. the process used to break 
down food to create energy). 

Hormones are generally quite small molecules, and are chemically uncomplicated. 
Examples include adrenaline (II) and /2-phenylethylamine (III). 



CH 2 CH 2 NH 2 



Most hormones are produced naturally in the body (e.g. adrenaline (II) is formed 
in the adrenal glands). From there, the hormone enters the bloodstream and is con- 
sumed chemically (a physiologist would say 'metabolized') at the relevant sites in 
the body - in fact, adrenaline accumulates and is then broken down chemically in 
the muscles and lungs. Adrenaline is generated in equal amounts in men and women, 



and promotes a stronger, faster heartbeat in times of crisis or panic. The body is 
thus made ready for aggression ('fight') or necessary feats of endurance to escape 
('flight'). Adrenaline is also administered artificially in medical emergencies, e.g. 
immediately following an anaphylactic shock, in order to give the heart a 'kick start' 
after a heart attack. 

/3-Phenylethylamine (III) is a different type of hormone, and 
is metabolized in women's bodies to a far higher extent than in 
a man's. The mechanism of metabolism is still a mystery, but it 
appears to cause feelings of excitement, alert feelings and in terms 
of mood, perhaps a bit of a 'high'. Unfortunately, the bodies of 
most men do not metabolize this hormone, so they do not feel the 
same 'high'. 

Hormones are potent and are produced in tiny concentrations 
(generally with a concentration of nanomoles per litre). By con- 
trast, the chemicals in the body with which the hormone reacts 
have a sizeable concentration. For example, reaction with oxy- 
gen in the blood is one of the first processes to occur during the 
metabolism of adrenaline. The approximate range of [02](biood) is 
0.02-0.05 moldm - , so the change in oxygen concentration is 
virtually imperceptible even if all the adrenaline in the blood is 
consumed. As a good corollary, then, the only concentration to 
change is that of the hormone, because it is consumed. 

Although such a reaction is clearly second order, it behaves like a 
^ reaction because only one of the concentrations changes. 
We say it is a pseudo first-order reaction. 

We say a reaction is 
pseudo first-order if 
it is second- or third- 
order, but behaves 
mathematically as 
though it were first 

A pseudo-order reac- 
tion proceeds with all 
but one of the reactants 
in excess. This ensures 
that the only con- 
centration to change 
appreciably is that of 
the minority reactant. 

Why do we not see radicals ft 

orming in 

the skin whili 

Pseudo-order rate constants 

One of the most common causes of skin cancer is excessive sunbathing. Radicals 
are generated in the skin during irradiation with high-intensity UV-light, e.g. while 
lying on a beach. These radicals react with other compounds in the skin, which may 
ultimately cause skin cancer. But we never see these radicals by eye, because their 
concentration is so minuscule. And the concentration is so small because the radicals 
react so fast. (Photo-ionization is discussed in Chapter 9.) 

Reaction intermediates are common in mechanistic studies of 
organic reactions. They are called 'intermediates' because they 
react as soon as they form. Such intermediates are sometimes 
described as 'reactive' because they react so fast they disappear 
more or less 'immediately'. Indeed, these intermediates are so reac- 
tive, they may react with solvent or even dissolved air in solution, 
i.e. with other chemicals in high concentration. 

The accumulated con- 
centration of such a 
fast-reacting inter- 
mediate will always 
be extremely low: 
perhaps as low as 
10- 10 moldm" 3 . 



Intermediates react 
fast because their 
activation energy is 
small - see 
Section 8.5. 

Since the reaction of intermediates is so fast, the concentration 
of the radical intermediate changes dramatically, yet the concentra- 
tions of the natural compounds in the skin with which the interme- 
diate reacts (via second-order processes) do not change perceptibly. 
How in practice, then, do we determine kinetic parameters for 
pseudo-order reactions such as these? 

We will call the intermediate 'A' and the other reagent, which is in excess, will be 
called 'B'. In the example here, B will be the skin, but is more generally the solvent 
in which the reaction is performed, or an additional chemical in excess. 

Because the concentration of B, [B], does not alter, this reaction will obey a first- 
order kinetic rate law, because only one of the concentrations changes with time. 
Because the reaction is first order (albeit in a pseudo sense), a plot of ln[A] ( (as ' v') 
against time (as 'x') will be linear for all but the longest times 
(e.g. see Figure 8.14). In an analogous manner to a straightforward 
first-order reaction, the gradient of such a plot has a value of 'rate 
constant x — 1' (see Worked Example 8.9). We generally call the 
rate constant k', where the prime symbol indicates that the rate 
constant is not a true rate constant, but is pseudo. 

The proper rate law for reaction for the second -order reaction between A and B is: 

Pseudo-order rate con- 
stants are generally 
indicated with a prime, 
e.g. k'. 

rate = fc 2 [A][B] 


as for any normal second-order reaction. By contrast, the perceived 
rate law we measure is first order, as given by 

rate = k'[A] 

where k! is the perceived rate constant. 


Writing a pseudo rate 
constant k' without an 
order implies that it is 
pseudo first order. 

Time t 

Figure 8.14 The reaction of A and B, with B greatly in excess is a second-order reaction, but 
it follows a kinetic rate law for a first-order reaction. We say it is pseudo first-order reaction. 
The deviation from linearity at longer times occurs because the concentration of B (which we 
assume is constant) does actually change during reaction, so the reaction no longer behaves as a 
first-order reaction 



Comparing Equations (8.34) and (8.35), we obtain 

The concentration [B] t 
barely changes, so 
we can write [B] in 
Equations (8.34) and 

*' = k 2 [B] 


This little relationship shows that a pseudo-order rate constant 
k! is not a genuine rate constant, because its value changes in 
proportion to the concentration of the reactant in excess (in this 
case, with [B]). 

Worked Example 8.16 The reaction of the ester ethyl methanoate and sodium hydrox- 
ide in water is performed with NaOH in great excess ([NaOH]o = 0.23 moldm~ 3 ). The 
reaction has a half-life that is independent of the initial concentration of ester present. 
13.2 per cent of the ester remains after 14 min and 12 s. What is the second-order rate 
constant of reaction ktl 

Strategy. (1) We ascertain the order of reaction, (2) we determine the pseudo rate constant 
k', (3) from k', we determine the value of the second-order rate constant k 2 . 

(1) Is the reaction a pseudo first-order process? The question says that the half-life t\/2 
of reaction is independent of initial concentration of ester, so the reaction must behave as 
though it was a. first-order reaction in terms of [ester]. In other words, NaOH is in excess 
and its concentration does not vary with time. 

(2) What is the value of the pseudo first-order rate constant k'l We calculate the 
pseudo first-order rate constant k' by assuming that the reaction obeys first-order kinetics. 
Accordingly, we write from Equation (8.24): 

/ [ester] o 
V [ester]. 






= k' x [(14 x 60) + 12] s 


ln(7.57) = k' x 852 s 

Because ln(7.57) = 2.02 we say 



852 s 

2.37 x 10~ 3 s" 

Note how manipulating 
the units in the fraction 
yields the correct units 
for k 2 . 

(3) What is the value of the second-order rate constant k 2 l The 
value of rate constant k 2 can be determined from Equation (8.36), as 
k' 4- [NaOH], so 

fe = 

2.37 x 10' 

0.23 moldm 

= 1.03 x 10~ 3 dm 3 mor 



SAQ 8.21 Potassium hexacyanoferrate(III) in excess oxidizes an alco- 
hol at a temperature of 298 K. The concentration of l<3[Fe(CN) 6 ] is 
0.05 moldrrT 3 . The concentration of the alcohol drops to 45 per cent 
of its t — value after 20 min. Calculate first the pseudo first-order rate 
constant k', and thence the second-order rate constant /C2- 

Justification Box 8.5 

When we first looked at the derivation of integrated rate equations, we looked briefly 
at the case where two species A and B were reacting but [A]o does not equal [B]o. The 
integrated rate equation for such a case is Equation (8.30): 

1 /[A] x[B]A 

x In = kit 

[B] -[A] t V[B] x[A]J 

Though we do not need to remember this fearsome-looking equation, we notice a few 
things about it. First, we assume that [B]o ^> [A]o, causing the first term on the left-hand 
side to behave as l/[B]o. 

Also, the major change in the logarithm bracket is 
the change in [A], since the difference between [B]o 
and [B] f will be negligible in comparison, causing a 
cancellation of the [B] terms on top and bottom. 

Accordingly, Equation (8.30) simplifies to 


x In 



Since ln([A]o/[A],) = k't for a pseudo first-order reac- 
tion (by analogy with Equation (8.24)), we say that 

— — x(kt) = k 2 t 

and cancelling the two t terms, and rearranging yields 

We argue this state- 
ment by saying that if 
[B] is, say, 100 times 
larger than [A], then 
a complete consump- 
tion of [A] (i.e. a 100 
per cent change in its 
concentration) will be 
associated with only a 
1 per cent change in 
[B] - which is tiny. 

k' = k 2 [B] 

so we re-obtain Equation (8.36). 

Alternatively, the value of the true second-order rate constant may be obtained by 
treating Equation (8.36) as the equation of a straight line, with the form y = mx: 


k 2 





Accordingly, we perform the kinetic experiment with a series of concentrations [B]o, 
the reactant in excess, and then plot a graph of k' (as 'y') against [B]o (as 'x'). The 
gradient will have a value of £2- 

A graphical method such as this is usually superior to a simplistic calculation of 
£2 = k! -r- [NaOH] (e.g. in the Worked Example 8.16), because scatter and/or chem- 
ical back or side reactions will not be detected by a single calculation. Also, the 
involvement of a back reaction (see next section) would be seen most straightfor- 
wardly as a non-zero intercept in a plot of k! (as 'v') against [reagent in excess] 
(as 'x'). 

Worked Example 8.17 The following kinetic data were obtained for the second-order 
reaction between osmium tetroxide and an alkene, to yield a 1,2-diol. Values of k' are 
pseudo-order rate constants because the OSO4 was always in a tiny minority. Determine 
the second-order rate constant k^ from the data in the following table: 

mol dm -3 

/fc'/KT 4 s" 1 

0.01 0.02 0.03 
3.2 6.4 9.6 

0.04 0.05 0.06 0.07 0.08 

12.8 16.0 19.2 22.4 25.6 




& 0.0015 



0.03 0.04 0.05 
[alkene] /mol drrr 3 


Figure 8.15 The rate constant of a pseudo-order reaction varies with the concentration of the 
reactant in excess: graph of k! (as 'y') against [alkene]o (as V). The data refer to the formation 
of a 1,2-diol by the dihydrolysis of an alkene with osmium tetroxide. The gradient of the graph 
yields k 2 , with a value of 3.2 x 10~ 2 dm 3 moP 1 s _1 



The data clearly show that k' is not a true rate constant, because 
its value varies as the concentration of the alkene increases. That k' 
increases linearly with increased [alkene] suggests a straightforward 
pseudo first-order reaction. Figure 8.15 shows a graph of k! (as 'y') 
against [alkene]o (as 'x'). The graph is linear, and has a gradient of 
3.2 x 10~ 2 dm moP 1 s _1 , which is also the value of ki. 

Graphs such as that 
in Figure 8.15 gen- 
erally pass through 
the origin. 

8.4 Kinetic treatment of complicated 

Why is arsenic poisonous? 

Care: it is unwise to call 
a complicated reaction 
such as these a 'com- 
plex reaction', since 
the word 'complex' 
implies an associa- 
tion compound. 

"Concurrent' or "competing' reactions 

Arsenic is one of the oldest and best known of poisons. It is so 
well known, in fact, that when the wonderful Frank Capra com- 
edy Arsenic and Old Lace was released, everyone knew that it was 
going to be a murder mystery in which someone would be poi- 
soned. In fact, it has even been rumoured that Napoleon died from 
arsenic poisoning, the arsenic coming from the green dye on his 
wallpaper. We deduce that even a small amount of arsenic will 
cause death, or at least an unpleasant and lingering illness. 

Arsenic exists in several different redox states. The characteristic 
energy at which one redox state converts to the other depends on 
its electrode potential E (see Chapter 7). The nervous system in a human body is 
'instructed' by the brain much like a microprocessor, and regulated by electron 'relay 
cycles' as the circuitry, which consume or eject electrons. Unfortunately, the electrons 
acquired or released by arsenic in the blood interfere with these naturally occurring 
electron relay cycles, so, following arsenic poisoning, the numbers of electrons in 
these relay cycles is wrong - drastically so, if a large amount of arsenic is ingested. 

Improper numbers of electrons in the relay cycles cause them not to work prop- 
erly, causing a breakdown of those bodily functions, which require exact amounts 
of charge to flow. If the nervous system fails, then the lungs are 
not 'instructed' how to work, the heart is not told to beat, etc., at 
which point death is not too far away. 

But arsenic is more subtle a poison than simply a reducing 
or oxidizing agent. Arsenic is a metalloid from Group 
V(B) of the periodic table, immediately below the elements 
nitrogen and phosphorus, both of which are vital for health. 
Unfortunately, arsenic is chemically similar to both nitrogen and 
phosphorus, and is readily incorporated into body tissues following 
ingestion. Arsenic effectively tricks the body into supposing 
that straightforward incorporation of nitrogen or phosphorus has 

Arsenic and nitrogen 
compete for the 
electrons participating 
in the natural electron- 
relay cycles in the 
body. The number of 
electrons transferred 
by nitrogen depends on 
the number of arsenic 
atoms competing 
for them. 




Product 1 

Product 2 

Figure 8.16 Arsenic and nitrogen compete for electrons both 
for and from the natural relay cycles in the body. The overall 
rate at which electrons are transferred by the nitrogen will alter 
when arsenic competes for them. The arsenic is poisonous, since 
these two pathways yield different products 

'Sequester' means to 
confiscate, seize or 
take control of some- 
thing to prevent its 
further use. The word 
comes from the Latin 
sequester, meaning 
a 'trustee' or 'agent' 
whose job was to 
seize property. 

The first time the body realizes that arsenic has been incorpo- 
rated is when the redox activity (as above) proceeds at potentials 
when nitrogen or phosphorus are inert. By the time we detect 
the arsenic poisoning (i.e. we feel unwell), it is generally too 
late, since atoms of arsenic are covalently bound within body tis- 
sue and cannot just be flushed out or treated with an antidote. 
The arsenic sequesters electrons that might otherwise be involved 
in other relay cycles, which is a concurrent kinetic process; see 
Figure 8.16. 

So arsenic is toxic because it has 'fooled' the body into thinking 
it is something else. 

Concurrent (or com- 
peting) reactions are 
so called because two 
reactions occur at 
the same time. We 
occasionally call them 
simultaneous reac- 
tions, although this 
terminology can be 

A beam of plane- 
polarized light is 
caused to rotate by 
an angle 9 as it passes 
through a solution of 
a chiral compound. 
The magnitude of 
6 depends on the 
concentration of the 
chiral compound. 

Why is the extent of Walden inversion 
smaller when a secondary alkyl halide 
reacts than with a primary halide? 

Reaction profiles for complicated reactions 

Alkyl halides react by a substitution reaction with hydroxide ions 
to yield an alcohol. A primary halide, such as 1-bromopentane, 
reacts by a simple bimolecular Sn2 mechanism, where the 'S' 
stands for substitution, the 'N' for nucleophilic (because the 
hydroxide ion is a nucleophile) and the '2' reminds us that 
the reaction is bimolecular. Being a single-step reaction, the 
substitution reaction is necessarily the rate-determining step. The 
reaction is accompanied by stereochemical inversion about the 
central tetrahedral carbon atom to which the halide is attached - we 
call it Walden inversion; see Figure 8.17(a). Each molecule of 
primary halide inverts during the Sn2 reaction. For this reason, we 
could monitor the rate of the Sn2 reaction by following changes in 
the angle of rotation 9 of plane polarized light. This angle 9 changes 

as a function of the extent of reaction £, so we know the reaction is complete when 

9 remains constant at a new value that we call 0(fi na i). 








,./ Br — " 





R R2 



-H R 

OH" I© "OH 

. R3 R2 . 


R % 


Figure 8.17 Reaction of an alkyl halide with hydroxide ion. (a) A primary halide reacts by an S N 2 
mechanism, causing Walden inversion about the central, chiral carbon, (b) A tertiary halide reacts 
by an Sn 1 mechanism (the rate-determining step of which is unimolecular dissociation, minimizing 
the extent of Walden inversion and maximizing the extent of racemization). Secondary alcohols 
often react with both SnI and Sn2 mechanistic pathways proceeding concurrently 

Polarimetry is the 
technique of following 
the rotation of plane- 
polarized light. 

By contrast, tertiary halides, such as 2-bromo-2,2-dimethylpro- 
pane, cannot participate in an Sn2 mechanism because it would be 
impossible to fit two methyl groups, one bromine and a hydroxide 
around a single carbon. The steric congestion would be too great. 
So the tertiary halide reacts by a different mechanism, which we 
call SnI- It's still a nucleophilic substitution reaction (hence the 
'S' and the 'N') but this time it is a unimolecular reaction, hence 
the '1'. The rate-determining step during reaction is the slow uni- 
molecular dissociation of the alkyl halide to form a bromide ion 
and a carbocation that is planar around the reacting carbon. 

Addition of hydroxide occurs as a rapid follow-up reaction. Even 
if the alkyl halide was chiral before the carbocation formed, racem- 
ization occurs about the central carbon atom because the hydrox- 
ide can bond to the planar central carbon from either side (see 
Figure 8.17(b)). Statistically, equal numbers of each racemate are 
formed, so the angle through which the plane polarized light rotated during reaction 
will, therefore, decrease toward 0°, when reaction is complete. 

In summary, primary halides react almost wholly by a bimolecular process and 
tertiary halides react by a unimolecular process. Secondary halides are structurally 
between these two extreme structural examples, since reaction occurs by both Sn2 and 
SnI routes. These two mechanisms proceed in competition, and occur concurrently. 

When following the (dual-route) reaction of a secondary halide with hydroxide ion, 
we find that the angle through which plane polarized light is rotated will decrease, 
as for primary and tertiary halides, but will not reach zero at completion. In fact, the 
final angle will have a value between 0° and #fi na i because of the mixtures of products, 
itself a function of the mixture of SnI and Sn2 reaction pathways. 

Genuinely first-order 
reactions are unusual. 
It is likely that the alkyl 
halide collides with 
another body (such 
as solvent) with suffi- 
cient energy to cause 
bond cleavage. 



We looked briefly at reaction profiles in Section 8.2. Before we look at the reaction 
profile for the concurrent reactions of hydrolysing a secondary alkyl halide, we will 
look briefly at the simpler reaction of a primary alkyl halide, which proceeds via a 
single reaction path. And for additional simplicity, we also assume that the reaction 
goes to completion. We will look not only at the rate of change of the reactants' 
concentration but also at the rate at which product forms. 

Consider the graph in Figure 8.18, which we construct with the 
data in Worked Example 8.12. We have seen the upper, lighter line 
before: it represents the rate of a second-order decay of molecule A 
with time as it reacts with the stoichiometry A + B — > product. The 
bold line in Figure 8.18 represents the concentration of the product. 
It is a 1 : 1 reaction, so each molecule of A consumed by the reaction 
will generate one molecule of product, with the consequence that 
the two traces are mirror images. Stated another way, the rate of 
forming product is the same as the rate of consuming A: 

We include the minus 
sign in Equation 
(8.37) to show how 
the product concentra- 
tion INcreases while 
the reactant concentra- 
tion DEcreases. 

rate = 

d[A] d [product] 
At At 


where the minus sign is introduced because one concentration increases while the 
other decreases. 

Now, to return to the hydrolysis of the secondary alkyl halides, we will call the 
reactions (1) and (2), where the '1' relates to the SnI reaction and the '2' relates to 
the Sn2 reactions. (And we write the numbers with brackets to avoid any confusion, 
i.e. to prevent us from thinking that the '1' and '2' indicate first- and second-order 
reactions respectively.) We next say that the rate constants of the two concurrent 
reactions are k(\) and k(2) respectively. As the two reactions proceed with the same 1:1 






1000 1500 

Time f/s 


Figure 8.18 Concentration profile for a simple reaction of a primary alkyl halide + OH -* 
alcohol. The bolder, lower line represents the concentration of product as a function of time, 
and the fainter, upper line represents the concentration of reactant 





| 0.025 

° 0.015 

Reaction (2) 

1000 1500 

Time f/s 


Figure 8.19 Concentration profiles for a concurrent reaction, e.g. of a secondary alkyl 
halide + OH~ — y alcohol: reaction (2) is twice as fast as reaction (1) in this example 

stoichiometry, the ratio of the products will relate to the respective 
rate constants in a very simple way, according to 

C(i) moles of product formed via reaction (1) 
C(2) moles of product formed via reaction (2) 


so we see that the amounts of product depend crucially on the 
relative magnitudes of the two rate constants. (We shall return to 
this theme when we look at the way the human body generates a 
high temperature to cure a fever in Section 8.5). 

We now look at the concentration profiles for reactions (1) and 
(2). For simplicity, we shall say that reaction (2) is twice as fast 
as reaction (1), which is likely - unimolecular reactions are often 
quite slow. We should note how, at the end of the reaction at the 
far right-hand side of the profile in Figure 8.19, the total sum of 
product will be the same as the initial concentration of reactant. 
Also note how, at all stages during the course of the reaction, the 
ratio of products will be 2:1, since that was the ratio of the two 
rate constants. 

Care: do not confuse 
kx and k (1) . ki is the 
rate constant of a first- 
order process, and k m 
is the rate constant of 
the first process in a 
multi-step reaction. 

We can inter-convert 
between 'number of 
moles' and 'concen- 
tration' here because 
the reactions are per- 
formed within a con- 
stant volume of sol- 

Why does 'standing' a bottle of wine cause it to smell 
and taste better? 

Consecutive reactions 

It is often said that a good wine, particularly if red, should 'stand' before serving. By 
'standing', we mean that the bottle should be uncorked some time before consumption, 



A wine is also said to 
age and breathe, which 
means the same thing. 

to allow air into the wine. After a period of about an hour or so, 
the wine should taste and smell better. 

Wines contain a complicated mixture of natural products, many 
of which are alcohols. Ethanol is the most abundant alcohol, at a 
concentration of 3-11 per cent by volume. The amounts of the 
other alcohols generally total no more than about 0.1-1 per cent. 

The majority of the smells and flavours found in nature comprise 
esters, which are often covalent liquids with low boiling points and 
high vapour pressures. For that reason, even a very small amount 
of an ester can be readily detected on the palate - after all, think 
how much ester is generated within a single rose and yet how 
overwhelmingly strong its lovely smell can be! 

Esters are the product of reaction between an alcohol and a car- 
boxylic acid. Although the reaction can be slow - particularly at 
lower temperatures - the equilibrium constant is sufficiently high 
for the eventual yield of ester to be significant. We see how even 
a small amount of carboxylic acid and alcohol can generate a suf- 
ficient amount of ester to be detected by smell or taste. 

This helps explain why a wine should be left to stand: some of 
the natural alcohols in the wine are oxidized by oxygen dissolved from the air to 
form carboxylic acids. These acids then react with the natural alcohols to generate a 
wide range of esters, which connoisseurs of wines will recognize as a superior taste 
and 'bouquet'. 

And the reason why the wine must 'stand' (rather than the 
reaction occurring 'immediately' the oxygen enters the bottle on 
opening) is that the reaction to form the ester is not a straight- 
forward one-step reaction: the first step (Equation (8.39)) is quite 
slow and occurs in low yield: 

The enzymes in wine 
are killed if the percent- 
age of alcohol exceeds 
about 13 per cent. Fer- 
mentation alone cannot 
make a stronger wine, 
so spirits are prepared 
by distilling a wine. 
Adding brandy to wine 
makes a fortified wine, 
such as sherry. 

The 'mash' that fer- 
ments to form wine 
sometimes includes 
grape skins; which 
are rich in enzymes. 
It is likely, there- 
fore, that the oxidation 
in Equation (8.39) is 
mediated (i.e. catal- 
ysed) enzymatically. 

alcohol (aq ) + O 

first reaction 


-> carboxylic acid, , + H2O 


where the rate constant of Equation (8.39) is that of a second- 
order reaction. Once formed, the acid reacts more rapidly to form 
the respective ester: 

carboxylic acid, ■, + alcohol 

second reaction 


-> ester 



where the rate constant for this second step is larger than the respective rate constant 
for the first step, implying that the second step is faster. 

In summary, we see that esterification is a two-step process. The first 
step - production of a carboxylic acid - is relatively slow because its rate is 
proportional to the concentration of dissolved oxygen; and the [02]( SO in) is l° w - Only 
after the wine bottle has been open for some time (it has had sufficient time to 



'breathe') will [02]( SO in) be higher, meaning that, after 'standing', the esters that lend 
additional flavour and aroma are formed in higher yield. 

The case of esterification is an example of a whole class of reac- 
tions, in which the product of an initial reaction will itself undergo 
a further reaction. We say that there is a sequence of reaction steps, 
so the reaction as a whole is sequential or consecutive. 

Although this example comprises two reactions in a sequence, 
many reactions involve a vast series of steps. Some radioactive 
decay routes, for example, have as many as a dozen species 
involved in a sequence before terminating with an eventual 

For simplicity, we will denote the reaction sequence by 

To remember this latter 
terminology, we note 
that for a consecutive 
reaction to occur, the 
second step proceeds 
as a consequence of 
the first. 

-> B 

-> C 


All the other reactants will be ignored here to make the analysis more straightforward, 
even if steps (1) and (2) are, in fact, bimolecular. We again write the reaction number 
within brackets to avoid confusion: we do not want to mistake the subscripted number 
for the order of reaction. We call the rate constant of the first reaction k(\) and the 
rate constant of the second will be k(2)- 

The material A is the initial reactant or precursor. It is usually 
stable and only reacts under the necessary conditions, e.g. when 
mixed and/or refluxed with other reactants. We know its concentra- 
tion because we made or bought it before the reaction commenced. 
Similarly, the material C (the product) is usually easy to handle, 
so we can weigh or analyse it when the reaction is complete, and 
examine or use it. By contrast, the material B is not stable - if it 
was, then it would not react further to form C as a second reaction step. Accordingly, 
it is rare that we can isolate B. We call B an intermediate, because it forms during 
the consumption of A but before the formation of C. 

The concentration profile of a simple 1 : 1 reaction is always easy to draw because 
product is formed at the expense of the reactant, so the rate at which reactant is 
consumed is the same as the rate of product formation. No such simple relation holds 
for a consecutive reaction, because two distinct rate constants are involved. Two 
extreme cases need to be considered when dealing with a consecutive reaction: when 

The maximum rate of 
reaction occurs when 
the concentration of 
the intermediate B 
is highest. 


C (2) 

and when k 


k(2). We shall treat each in turn. 

Why fit a catalytic converter to a car exhaust? 

Consecutive reactions in which the second reaction is slow 

First, we consider the case where the first reaction is very fast compared with the 
second, so k(i) > k(2)- This situation is the less common of the two extremes. When 



all the A has been consumed (to form B), the second reaction is only just starting to 
convert the B into the final product C. A graph of concentration against time will show 
a rapid decrease in [A] but a slow increase in the concentration of the eventual product, 
C. More specifically, the concentration of intermediate B will initially increase and 
only at later times will its concentration decrease once more as the slower reaction 
(the one with rate constant k^)) has time to occur to any significant 
extent. The concentration profile of B, therefore, has a maximum; 
see Figure 8.20. 

A good example of a consecutive process in which the second 
reaction is much slower than the first is the reaction occurring 
in a car exhaust. The engine forms carbon monoxide (CO) as its 
initial product, and only at later times will CO( g ) oxidize to form 
CC>2(g). In fact, the second reaction (CO( g ) + 502(g) -*■ C02( g )) is 
so slow that the concentration of CO( g ) is often high enough to 
cause serious damage to health. 

Most modern cars are 
fitted with a 'catalytic 
converter', one pur- 
pose of which is to 
speed up this second 
(slower) process in the 
reaction sequence. 

Why do some people not burn when sunbathing? 

Iff Sf 

Consecutive reactions in which the second reaction is fast 

If k ( i) < k ( 2), an inter- 
mediate is generally 
termed a reactive inter 
mediate to emphasize 
how soon it reacts. 

We now look at the second situation, i.e. when k 




first reaction produces the intermediate B very slowly. B is con- 
sumed immediately - as soon as each molecule of intermediate is 
formed - by the second reaction, which forms the final product C. 
Accordingly, we call it a reactive intermediate, to emphasize how 
rapidly it is consumed. 

The rate of the first reaction in the sequence is slow because k(i) 
is small, so the rate of decrease of [A] is not steep. Conversely, since k@) is fast, we 
would at first expect the rate d[C]/df to be quite high. A moment's thought shows 


Figure 8.20 Schematic graph of concentration against time ('a concentration profile') for a con- 
secutive reaction for which ktn > ken- Note the maximum in the concentration of the intermediate, 


B. This graph was computed with k@) being five times slower than k 




that this is not necessarily the case: 

rate of forming the final product, C 




We see from Equation (8.42) that the rate of forming C is quite slow because [B] is 
tiny no matter how fast the rate constant k(2)- 

We now consider the concentration profile of the reactive intermediate B. Because 
the second reaction is so much faster than the first, the concentration of B is never 
more than minimal. Its concentration profile is virtually horizontal, although it will 
show a very small maximum (it must, because [B] = at the start of the reaction at 
t = 0; and all the intermediate B has been consumed at the end of reaction, when 
[B] is again zero). At all times between these two extremes, the concentration of B 
at time t, [B] t , is not zero. The concentration profile is shown in Figure 8.21. 

Sunbathing to obtain a tan, or simply to soak up the heat, is an inadvertent means 
of studying photochemical reactions in the skin. It is also a good example of a 
consecutive reaction for which k(\) < k(2). 

Small amounts of organic radicals are formed continually in the 
skin during photolysis (in a process with rate constant &(i)). The 
radicals are consumed 'immediately' by natural substances in the 
skin, termed antioxidants (in a different process with rate constant 
k(2)). Vitamin C (L-(+)-ascorbic acid, IV) is one of the best natu- 
rally occurring antioxidants. Red wine and tea also contain efficient 

An antioxidant is a 
chemical that prevents 
oxidation reactions. 

CH 2 OH 


Figure 8.21 Schematic graph of concentration against time ('a concentration profile') for a con- 
secutive reaction for which k (2 ) > k(i). This graph was computed with k (2 ) being five times faster 

than k, 




The vast majority of organic radical reactions involve the radical as a reactive 
intermediate, since their values of k(2) are so high, although we need to note that the 
second reaction need not be particularly fast: it only has to be fast in relation to the 
first reaction. As a good generalization, the intermediate may be treated as a reactive 
intermediate if &(2)/&(i) > 10~ 3 . 

Integrated rate equations for consecutive reactions 

Our consecutive reaction here has the general form A — > B — >• C. Deriving an inte- 
grated rate equation for a consecutive reaction is performed in much the same way as 
for a simple one-step reaction (see Section 8.2), although its complexity will prevent 
us from attempting a full derivation for ourselves here. 
For the precursor A, the rate of change of [A] is given by 

rate = — — = k m [A] 


where the minus sign reflects the way that [A] is consumed, meaning its concentration 

decreases with time. 

The rate of change of C has been given already as Equation 
(8.42). Equations (8.42) and (8.43) show why the derivation of 
integrated rate equations can be difficult for consecutive reactions: 
while we can readily write an expression for the rate of forming 
C, the rate expression requires a knowledge of [B], which first 
increases, then decreases. The problem is that [B] is itself a function 
of time. 

The rate of change of [B] has two components. Firstly, we form 
B from A with a rate of '+&(i) x [A]'. The second part of the 
rate equation concerns the subsequent removal of B, which occurs 

with a rate of '—£(2) x [B]'. The minus sign here reminds us that [B] decreases in 

consequence of this latter process. 

Combining the two rate terms in Equations (8.42) and (8.43) 

Writing an integrated 
rate expression for 
a complicated reac- 
tion is difficult because 
we don't readily know 
the time-dependent 
concentration of the 
intermediate, [B] t . 

The rate expression of 
a complicated reaction 
comprises one term 
for each reaction step. 
In this case, species 
B is involved in two 
reactions, so the rate 
equation comprises 
two terms. 


k(i)[A] ~ k ( 2)[B] 


rate of the process 
forming the B 

rate of the process 
removing the B 

Equation (8.44) helps explain why the concentration profile in 
Figure 8.20 contains a maximum. Before the peak, and at short 
times, the second term on the right-hand side of Equation (8.44) is tiny because [B] 
is small. Therefore, the net rate d[B]/df is positive, meaning that the concentration of 
B increases. Later - after the peak maximum - much of the [A] has been consumed, 



meaning more [B] resides in solution. At this stage in the reaction, the first term on 
the right-hand side of Equation (8.44) is relatively small, because [A] is substantially 
depleted, yet the second term is now quite large in response to a higher value of [B]. 
The second term in Equation (8.44) is consequently larger than the first, causing the 
overall rate to be negative, which means that [B] decreases with time. 

Table 8.3 lists the concentrations [A] t , [B] t and [C] ( , which are exact, and will 
give the correct concentrations of A, B and C at any stages of a reaction. We will not 
derive them here, nor will we need to learn them. But it is a good idea to recognize 
the interdependence between [A]^, [B] ( and [C] t . 


ow do Reactolite sunglasses won 

The kinetics of reversible reactions 

Reactolite sunglasses are photochromic: they are colourless in the 
dark, but become dark grey-black when strongly illuminated, e.g. 
on a bright summer morning. The reaction is fully reversible (in the 
thermodynamic sense), so when energy is removed from the sys- 
tem, e.g. by allowing the lenses to cool in the dark, the photochem- 
ical reaction reverses, causing the lenses to become uncoloured and 
fully transparent again. 

The photochromic lenses contain a thin layer of a silver- 
containing glass, the silver being in its +1 oxidation state. Absorp- 
tion of a photon supplies the energy for an electron-transfer reaction 
in the glass, the product of which is atomic silver: 

The word 'photo- 
chromic' comes from 
the Greek words 
photos, meaning 
'light' and khromos 
meaning 'colour'. 
A photochromic 
substance acqui- 
res colour 
when illuminated. 

Ag + + e + hv 



where the electron comes from some other component within the 
glass. It is the silver that we perceive as colour. In effect, the colour 

Excited states are 
defined in Chapter 9. 

Table 8.3 Mathematical equations to describe the 
concentrations 3 of the three species A, B and C invol- 
ved in the consecutive reaction, A — > B — > C 

[A], = [A]„ exp(-* ( i)0 



H2) - K(l) 

-[A] {exp(fc (1) - exp(£ (2) f)} 

[C], = [A] 1 - 

fc(2) exp(-fc (1) + fc ( i)f 
k(2) - fc(i) 

a Rate constants are denoted as k, where the subscripts indi- 
cate either the first or the second reaction in the sequence. 
The subscript '0' indicates the concentration at the com- 
mencement of the reaction. Concentrations at other times 
are denoted with a subscript t. 



indicates a very long-lived excited state, the unusually long time is achieved because 
colouration occurs in the solid state, so the reaction medium is extremely viscous. 

Atomic silver is constantly being formed during illumination, but, at the same 
time, the reverse reaction also occurs during 'cooling' (also termed 'relaxation' - see 
Chapter 9) in a process called charge recombination. Such recombination is only 
seen when removing the bright light that caused the initial coloration reaction, so the 
reaction proceeds in the opposite direction. 

Reversible reactions 
are also termed equi- 
librium or opposing 

The reaction might 
be so slow that the 
reaction never actually 
reaches the RHS during 
a realistic time scale, 
but the direction of 
the reaction is still 
the same. 

For a thermodynami- 
cally reversible reac- 
tion, the rate constants 
of the forward and 
reverse reactions are 
k n and k- n respectively. 

Rate laws for reversible reactions 

All the reactions considered so far in this chapter have been irre- 
versible reactions, i.e. they only go in one direction, from fully 
reactants on the left-hand side to fully products on the right-hand 
side. (They might stop before the reaction is complete.) 
We now consider the case of a reversible reaction: 





where k\ is the rate constant for the forward reaction and k-\ is the 
rate constant for the reverse reaction. The minus sign is inserted to 
tell us that the reaction concerned is that of the reverse reaction. 

When writing a rate expression for such a reaction, we note that 
two arrows involve the species B, so, straightaway, we know that 
the rate expression for species B has two terms. This follows since 
the rate of change of the concentration of B involves two separate 
processes: one reaction forms the B (causing [B] to increase with 
time) while the other reaction is consuming the B (causing [B] to 
decrease). The rate is given by 

rate of change of [B] 




where the minus and (implicit) plus signs indicate that the concentrations decrease 
or increase with time respectively. Note that a new situation has arisen whereby the 
expression to describe the rate of change of [B] itself involves [B] - this is a general 
feature of rate expressions for simple reversible reactions. Similarly: 

rate of change of [A] = 

The concentrations 
stop changing when 
the reaction reaches 


= -*i[A] + *_i[B] 


Note that the rate of change of [A] is equal but opposite to 
the rate of change of [B], which is one way of saying that A is 
consumed at the expense of B; and B is formed at the expense 
of A. 



When the reaction has reached equilibrium, the rate of change of both species must 
be zero, since the concentrations do not alter any more - that is what we mean by 
true 'equilibrium'. From Equation (8.48): 

= *i[A] (eq) -*_i[B] (eq) 


fcl[A] (eq ) = *-l[B] 


which, after a little algebraic rearranging, 
ing result: 

kl [B](eq) 

k-l [A] ( eq) 

gives a rather surpris- 


We recognize the right-hand side of the equation as the equilibrium 
constant K. We give the term microscopic reversibility to the idea 
that the ratio of rate constants equals the equilibrium constant K. 

The principle of 
'microscopic rever- 
sibility' demonstrates 
how the ratio of rate 
constants (forward to 
back) for a reversible 
reaction equals the 
reaction's equilibrium 
constant K. 

Worked Example 8.18 Consider the reaction between pyridine and heptyl bromide, 
to make 1 -heptylpyridinium bromide. It is an equilibrium reaction with an equilibrium 
constant K — 40. What is the rate constant of back reaction k_y if the value of the forward 
rate constant k\ — 2.4 x 10 3 dm 3 mol _1 s _1 ? 

We start with Equation (8.49): 

K = 

and rearrange it to make k-\ the subject, to yield 

K_i = — 


We then insert values: 

so fc_ i = 60 dm 3 mol ' s ' . 

2.4 x 10 3 dm 3 moP's" 1 

SAQ 8.22 A simple first-order reaction has a forward rate constant of 
120 s _1 while the rate constant for the back reaction is 0.1 s -1 . Calculate 
the equilibrium constant K of this reversible reaction by 
invoking the principle of microscopic reversibility. 

Integrated rate equations for reversible reactions 

In kinetics, we often term the concentrations at equilibrium 'the 


infinity concentration'. The Reactolite glasses do not become 

In kinetics, the equi- 
librium concentration 
[A] ( eq) is often termed 
the infinity concentra- 
tion, and cited with an 
infinity sign as [A]^. 



We should never throw 
away a reaction solu- 
tion without measuring 
a value of [A] (eq) . 

progressively darker with time because the concentration of 
atomic silver reaches its infinity value [Ag]( eq ). 

We will consider the case of a first-order reaction, A T± product. 
Following integration of an expression similar to Equation (8.48), 
we arrive at 


[A]q - [A] (eg) 

[A], - [A] 


= (ki + k-i)t 


which is very similar to the equation we had earlier for a simple reaction (i.e. one 
proceeding in a single direction), Equation (8.24). There are two simple differences. 
Firstly, within the logarithmic bracket on the left-hand side, each term on top and 
bottom has the infinity reading subtracted from it. Secondly, the time t is not multi- 
plied by a single rate constant term, but by the sum of both rate constants, forward 
and back. 

This equation can be rearranged slightly, by separating the logarithm: 

ln([A], - [A] (eq) ) 



+ ln([A] - [A] (eq) ) 


Note that the final term on the right-hand side is a constant. Accordingly, a plot of 
In ([A]; — [A]( eq )) (as 'y') against time (as 'x') will yield a straight line of gradient 

Worked Example 8.19 The data below relate to the first-order isomerization of 2- 
hexene at 340 K, a reaction for which the equilibrium constant is known from other 
studies to be 10.0. What are the rate constants k\ and &_i? 

Time r/min 20 47 80 107 140 

([A], - [A] (eq) )/mol dm" 3 0.114 0.103 0.091 0.076 0.066 0.055 

Strategy. (1) we need to plot a graph of ln([A], — [A]( eq )) (as 'v') against time t (as 'x'); 
(2) determine its gradient; (3) then, knowing the equilibrium constant K, we will be able 
to determine the two rate constants algebraically. 

(1) Figure 8.22 shows a graph for a reversible first-order reaction with the axes for 
an integrated rate equation ln([A] t — [A] (eq) ) (as 'y') against time (as 'x'). The 
gradient is —5.26 x 10~ 3 min -1 . 

The microscopic- 
reversibility relation- 
ship K = k\ -4-/f_i can- 
not be applied unless 
we know we have a 
simple reversible reac- 

(2) The gradient is -(ki + k-i), so (k x + Jfc_i) = +5.26 x 10~ 3 
min - . 

(3) Next, we perform a little algebra, and start by saying that 
K — ky 4- k-\, i.e. we invoke the principle of microscopic 
reversibility . Multiplying the bracket (k t + k-i) by k_\ 4- k_\, 
i.e. by 1, yields 

gradient = k-i(K + 1) (8.52) 



1 1 








1 1 

50 100 

Time f/min 


Figure 8.22 Kinetic graph for a reversible first-order reaction with the axes for an integrated rate 
equation ln([A], — [A]( eq )) (as 'y') against time (as 'x'). The gradient is —5.26 x 10~ 3 min" 1 

Substituting values into Equation (8.52) gives 

5.26 x 10~ 3 min" 1 = fc_i(10.0 + 1) 


5.26 x lfr 3 min" 1 , , 

k-i = = 4.78 x 10" 4 min" 1 


Next, since we know both K and &_], we can calculate k\. We know from the gradient 
of the graph that 

(hi +Jfc_i) = 5.26x 10" 3 min" 1 

which, after a little rearranging, gives 

fci =5.26 x 10" 3 min -1 - Jfc_, 

and, after inserting values 

3 ™;„-l 

hi = (5.26 x 10"' - 4.78 x 10" 4 ) min -1 = 4.78 x 10" J min 

In summary, k\ — 4.78 x 10 3 min ' and fe_] = 4.78 x 10 4 

Notice how the ratio 
'/fi -e-/f_i' yields the 
value of K. 



It is wrong (but com- 
mon) to see a reversible 
reaction written with a 
double-headed arrow, 
as 'A +* B'. Such an 
arrow implies reso- 
nance, e.g. between 
the two extreme val- 
ence-bond structures of 
Kekule benzene. 

SAQ 8.23 Consider a reversible first-order reaction. Its 
integrated rate equation is given by Equation (8.50). People 
with poor mathematical skills often say (erroneously!) that 
taking away the infinity reading from both top and bottom is 
a waste of time because the two infinity concentration terms 
will cancel. Show that the infinity terms cannot be cancelled 
in this way; take [A] (eq ) = 0.4 moldrrT 3 , [A]o = 1 moldrrT 3 
and [A] f = 0.7 moldm -3 . 

8.5 Thermodynamic considerations: 

activation energy, absolute reaction rates 
and catalysis 

Why prepare a cup of tea with boiling water? 

The temperature dependence of reaction rate 

The instructions printed on the side of a tea packet say, 'To make a perfect cup of 
tea, add boiling water to the tea bag and leave for a minute'. The stipulation for 
'one minute' suggests the criterion for brewing tea in water at 100 °C is a kinetic 
requirement. In fact, it reflects the rate at which flavour is extracted from the tea bag 
and enters the water. 

Pure water boils at 100 °C (273.15 K). If the tea is prepared with 
cooler water, then the time taken to achieve a good cup of tea is 
longer (and by 'good', here, we mean a solution of tea having a 
sufficiently high concentration). If the water is merely tepid, then a 
duration as long as 10 min might be required to make a satisfactory 
cup of tea; and if the water is cold, then it is possible that the 
tea will never brew, and will always remain dilute. In summary, 
the rate of flavour extraction depends on the temperature because 
the rate constant of flavour extraction increases with increasing 

Remember that the 
temperature of boil- 
ing water 7~ (b oii> is 
itself a function of 
the external pres- 
sure, according to the 
equation (see 
Section 5.3). 

Why store food in a fridge? 

The temperature dependence of rate constants 

Food is stored in a fridge to prevent (or slow down) the rate at which it perishes. 
Foods such as milk or butter will remain fresh for longer if stored in a fridge, but they 
decompose or otherwise 'go off more quickly if stored in a warmer environment. 

The natural processes that cause food to go bad occur because of enzymes and 
microbes, which react with the natural constituents of the food, and multiply. When 



these biological materials have reached a certain concentration, the food smells and 
tastes bad, and is also likely to be toxic. 

The growth of each microbe and enzyme occurs with its own 
unique rate. A fridge acts by cooling the food in order to slow these 
rates to a more manageable level. At constant temperature, the rate 
of each reaction equals the respective rate constant k multiplied 
by the concentrations of all reacting species. For example, the rate 
of the reaction causing milk to 'go off occurs between lactic acid 
and an enzyme. The rate of the process is written formally as 

The rate constant k 2 is 
truly a constant at a 
fixed temperature, but 
can vary significantly. 

rate = £2 [lactose] [enzyme] 

where, as usual, the subscripted '2' indicates that the reaction is 
second order. Neither [lactose] nor [enzyme] will vary with temper- 
ature, so any variations in rate caused by cooling must, therefore, 
arise from changes in k 2 as the temperature alters. 

The rate constant ki is truly a constant at a fixed temperature, 
but can vary significantly: increasing as the temperature increases 
and decreasing as the temperature decreases. This result explains 
why rate of reaction depends so strongly on temperature. 


The rate constant 
increases as the tem- 
perature increases and 
decreases as the tem- 
perature decreases. 

Why do the chemical reactions involved in cooking 
require heating? 

Activation energy E a and the Arrhenius theory 


'Naturally occurring' 
was the old-fashioned 
definition of 'organic 
chemistry', and per- 
sisted until nearly the 
end of the 19th cen- 

Cooking is an applied form of organic chemistry, since the mole- 
cules in the food occur naturally. We heat the food because the 
reactions occurring in, say, a pie dish require energy; and an oven 
is simply an excellent means of supplying large amounts of energy 
over extended periods of time. 

The natural ingredients in food are all organic chemicals, and 
it is rare for organic reactions to proceed without an additional 
means of energy, which explains why we usually need to reflux a 
reaction mixture. 

It is easy to see why an endothermic reaction requires energy to react - the energy 
to replace the bonds, etc. must be supplied from the surroundings. But why does an 
exothermic reaction require additional energy? Why do we need to add any energy, 
since it surely seeks to lose energy? 

At the heart of this form of kinetic theory is the activated complex. In this context, 
the word 'activated' simply means a species brimming with energy, and which will 
react as soon as possible in order to decrease that energy content. 

A reaction can be thought of as a multi-step process: first the reactants approach and 
then they collide. Only after touching do they react. One of the more useful definitions 



The Franck- Condon 
principle states that 
atomic nuclei are sta- 
tionary during a reac- 
tion, with only elec- 
trons moving - 
see p. 451. 

of reac