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PROFESSOR: OK, so kinetics
we're continuing with today.

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We talked on Wednesday about
first order kinetics and

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we'll do a brief review
of some of that.

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And we're going to talk about
second order kinetics.

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Today we're going to come back
up and talk about chemical

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equilibrium, which is something
I love to do to review things

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that we've talked about before.
 

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And we're going to start in
on reaction mechanisms.

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So, I thought I would mention,
some of you have may have seen

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the activity in the Infinite
Corridor, but today is world

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AIDS day, and today a lot of
the compounds that are being

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used to treat HIV were actually
designed based on making

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inhibitors to enzymes.
 

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And so, to design those
pharmaceuticals, people had

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to understand the reaction
mechanism of the enzyme, and

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enzymes, of course, are
catalysts in the body.

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So, knowledge of what medical
individuals needed to know to

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design these inhibitors to
treat HIV are actually a lot

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from this unit that we're
going to be talking about.

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So, we'll be talking about
reaction mechanisms, and we're

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also going to be talking about
enzyme catalysis, which were

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key points in being able to
come up with some of the

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current treatments against HIV.
 

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All right, so just a little
review from last Wednesday.

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We talked about first
order half life.

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We talked about first order
kinetics, we came up with an

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integrated first order rate
law, and we also talked

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about half life.
 

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And you told me last time that
an example of first order half

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life is radioactivity,
which we're going to be

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talking about today.
 

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So, just a little review from
last time, you have your first

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order integrated rate law, and
the half life is defined as the

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time it takes for half of the
original material to go away,

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and half life is abbreviated t
1/2, that's the symbol for half

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life, so the time for half of
the original material

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to go away.
 

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If you plug original material
divided by 2 in there, then the

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00:02:34 --> 00:02:40
original material a to the o
for original, drops out and you

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come up with this equation of
the natural log of 1/2 equals

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minus k t 1/2, and k is, of
course, our rate constant.

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And so, then we can take
the natural log of a

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1/2 and we get a value.
 

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Rearranging that, you get
half life equals 0 .

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6 9 3 1 over k.
 

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And so you told me last time
for this plot for first order

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half life, each half life,
half of the original

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material goes away.
 

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So, one example of a first
order half life process

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is radioactive decay.
 

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And the reason why this is a
first order process is because

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the decay of the nucleus is
independent of the number of

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surrounding nuclei that
has been decayed.

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So it's independent of
the original starting

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concentration, and since it's
independent, notice that's a

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blank in your notes, since it's
independent, then that makes it

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a first order process.
 

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So, we can apply first
order integrated rate laws

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to radioactive decay.
 

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So here were some of the
equations we had last time.

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We had, this is a different
expression of the first order

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rate law where the material,
concentration material of

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material a, at some particular
time equals the original

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concentration, e to the minus k
t, where you have your rate

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constant and the time that has
elapsed, and we also just

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talked about first order half
lifes with this equation here.

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So, we can use those same
equations, but often you don't

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see it in terms of
concentration of a, you usually

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see these expressions in terms
of either the number of nuclei

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or a different a, which
is a for activity.

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So instead of concentration,
we're talking about the

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number of nuclei that
have decay or capital N.

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So, we can write this same
expression, but now using N

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instead of concentration of a,
same thing, number of nuclei

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equal the original number of
nuclei, e, to the minus k t,

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where k is our rate constant,
or in this case decay

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constant, and t is time.
 

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So, with chemical kinetics,
we're usually talking about the

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change in concentration of
things over time, but with

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nuclear kinetics, we're talking
about the number of decay

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events, the number of
nuclei that have decayed.

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And so, here with nuclear
kinetics, we measure these

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events using a Geiger counter.
 

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00:05:29 --> 00:05:34
So this can measure radiation,
and I'm going to come around

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and just check the room.
 

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And so, the gasket's
ionized and then you

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hear different clicks.
 

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See if you can hear the
clicks as I come around.

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00:05:47 --> 00:05:59
So let's just see if we have
any problems over here.

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Oh, maybe a little bit.
 

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No, this is fine.
 

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There's always a little bit of
radioactivity, it's all fine.

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So, this is a Geiger
counter, which will measure

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00:06:24 --> 00:06:29
nuclear kinetics, it will
measure radioactivity.

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And my lab has this particular
one, because we use x-rays

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00:06:33 --> 00:06:36
in our experiments.
 

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OK, we'll leave this on
low just to check things

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out as we go along.
 

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All right, so we do have
a term, A, that we talk

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about in this unit.
 

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Instead of concentration
of A, it's activity.

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And so, activity here, sort of
the decay rate, is also called

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activity, capital A, and so
this is equal to the change in

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00:07:04 --> 00:07:08
the number of nuclei or
our decay constant times

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00:07:08 --> 00:07:12
the number of nuclei.
 

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00:07:12 --> 00:07:14
And people will often talk
about the activity of

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particular radioactive
compounds.

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So, because activity is
proportional to the number of

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nuclei, you can also take
this expression and write

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00:07:26 --> 00:07:28
it as this expression.
 

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So you can have either the
number of nuclei equal the

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original number of nuclei, e to
the k t, or you can do it in

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terms of activity -- that the
activity at some time is equal

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to the original activity,
e to the minus k t.

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00:07:44 --> 00:07:51
So, all of these equations can
be re-written in this way.

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So, let's talk a
minute about units.

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All right, so the activity for
units, the new activity is Bq,

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Becquerel. and that, actually
is named after a French person.

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Henry was his first name,
and my French pronunciation

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is not very good.
 

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This is the current unit.
 

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It's equal to one radioactive
disintegration per second.

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The older unit, which you may
be familiar with, is called

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the Curie, and that is 3 .
 

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7 times 10 to the 10
disintegrations per second.

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Does anyone want to guess,
the Curie unit, who

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that was named after?
 

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STUDENT: Marie Curie?
 

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PROFESSOR: No, it was
named after her husband

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actually, Pierre Curie.
 

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And I actually always assume,
because Marie is actually more

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famous than her husband.
 

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But she, Marie Curie won two
Nobel prizes, so we shouldn't

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feel too sorry for her.
 

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Her husband shared the first
Nobel Prize with her in, I

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think it was 1903, but then in
1906 he was killed in a road

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accident, run over by something
that was crossing the street.

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So he did not share the second
Nobel Prize, because by the

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time that came around about
1911, he had passed away.

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So, at first, we had the Curie,
but then that turned out

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to be a really big number.
 

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And so, when you were talking
about sort of safe units for

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workers to be exposed to, if
they were being exposed to

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things 10 to the 10, that
really isn't very healthy.

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So they wanted to have sort
of a much smaller unit.

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And so I guess that Marie Curie
at that point talked about how

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her husband would feel about
having the Curie not being the

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standard unit, but I guess she
was OK with it, because if we

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00:10:03 --> 00:10:07
had kept that same unit, then
people would have been using it

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00:10:07 --> 00:10:10
and it would of had to of been
a really, really small number,

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because it was sort of picked
to be set up to something that

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was too large, and she didn't
want her husband's name

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apparently associated with a
sort of an infinitesimally

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small quantity of something.
 

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So, the Curie was sort
of done away with.

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And Henry Becquerel, who
was one of the people who

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discovered radioactivity and
shared that first Nobel Prize,

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had the unit named after him.
 

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And I always ask the freshman
chemistry class that as they

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00:10:38 --> 00:10:43
go through MIT, if they ever
discover a unit that is named

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00:10:43 --> 00:10:46
after a female scientist, to
please come back

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and let me know.
 

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This was the one I thought was
named after a female scientist,

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but as it turns out, it
was actually Pierre Curie.

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So, if you hear of any,
please let me know

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for future reference.
 

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So, the current unit you'll
be using is Bq here

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for radioactivity.
 

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So, you're not responsible for
knowing all the different

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types of radioactivity.
 

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When you're working problems,
you can always get

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this information.
 

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00:11:17 --> 00:11:19
I'll just mention that a
number of different kinds of

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radioactivity, some involve a
mass change, some do not

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00:11:23 --> 00:11:25
involve a mass change.
 

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So, alpha decay, this isn't
actually in your notes,

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there's a reference to
where the table is.

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You're not responsible for
memorizing it, so I didn't

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00:11:32 --> 00:11:33
put it in the notes.
 

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An alpha decay is equivalent to
a helium 4 nucleus, so you lose

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two protons, two neutrons, so
that's a big mass change.

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Whereas say a beta decay
involves a loss of electrons,

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so there's no mass change
associated with that.

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00:11:49 --> 00:11:52
So just to be aware that there
are these differences in

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different types of radiation.
 

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There's also really big
differences in terms of half

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lives of radioactive isotopes,
and again, this information

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00:12:03 --> 00:12:06
would be given to you on a test
or a problem-set, so you

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00:12:06 --> 00:12:09
don't have to memorize it.
 

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00:12:09 --> 00:12:13
So, this table is similar to
one in your book, and the point

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00:12:13 --> 00:12:17
here is just how different
half lifes can be.

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00:12:17 --> 00:12:23
So the abbreviation a
here is year, d is day.

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So you see some of these half
lifes are in multiple years,

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some of them are days, so there
are big differences in terms

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00:12:32 --> 00:12:36
of the half life of some of
these radioactive isotopes.

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00:12:36 --> 00:12:39
Some of them stay around for
a really, really, really,

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00:12:39 --> 00:12:42
really long time.
 

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00:12:42 --> 00:12:46
So, I thought I would share
with you a poem about

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00:12:46 --> 00:12:49
half lifes today.
 

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00:12:49 --> 00:12:52
And this was written by a
former graduate student at MIT,

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00:12:52 --> 00:12:56
Mala Radhakrishnan, and she is
now a professor at Wellesley

210
00:12:56 --> 00:13:01
college right here in
Wellesley, Massachusetts.

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00:13:01 --> 00:13:05
So, her poem entitled "Days of
our half lives," is from her

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00:13:05 --> 00:13:09
collection of chemistry poetry,
"Chemistry for the Couch

213
00:13:09 --> 00:13:17
Potato." And this particular
poem involves the uranium

214
00:13:17 --> 00:13:21
238 decay series.
 

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00:13:21 --> 00:13:23
So, here we go.
 

216
00:13:23 --> 00:13:29
"Days of our half lives.
 

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00:13:29 --> 00:13:33
My dearest love, I writing
you to tell you all

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00:13:33 --> 00:13:34
that I've been through.
 

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00:13:34 --> 00:13:38
I've changed my whole
identity, but loved, I

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00:13:38 --> 00:13:40
can not pretend to be.
 

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00:13:40 --> 00:13:45
When I was uranium 238,
you were on my case to

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00:13:45 --> 00:13:46
start losing weight.
 

223
00:13:46 --> 00:13:52
For 5 billion years I'd hoped
and I prayed, and finally

224
00:13:52 --> 00:13:55
I had an alpha decay.
 

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00:13:55 --> 00:13:59
Two protons, two neutrons
went right out the door,

226
00:13:59 --> 00:14:03
and now I was thorium 234.
 

227
00:14:03 --> 00:14:07
But my nucleus was still unfit
for your eyes, not positive

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00:14:07 --> 00:14:10
enough for it's large size.
 

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00:14:10 --> 00:14:14
But this time my half life was
really not very long, because

230
00:14:14 --> 00:14:17
my will to change
was quite strong.

231
00:14:17 --> 00:14:20
It took just a month, not
even a millennium, to beta

232
00:14:20 --> 00:14:24
decay into protactinium.
 

233
00:14:24 --> 00:14:27
But you still rejected me right
off the bat, protactinium,

234
00:14:27 --> 00:14:30
who's heard of that?
 

235
00:14:30 --> 00:14:35
So, beta decay, I did much
more to become uranium 234.

236
00:14:35 --> 00:14:39
Myself again, but a new
isotope, you still

237
00:14:39 --> 00:14:42
weren't satisfied,
but I still had hope.

238
00:14:42 --> 00:14:45
Three alpha decays, it was
hard, but I stayed on

239
00:14:45 --> 00:14:51
through thorium, through
radium, and then radon.

240
00:14:51 --> 00:14:54
I thought I would finally
please you, my mass was a

241
00:14:54 --> 00:14:59
healthy 222, but you said,
although I like your mass,

242
00:14:59 --> 00:15:04
I do not want to be
with a noble gas.

243
00:15:04 --> 00:15:08
You had a point, I wasn't
reactive, so in order to please

244
00:15:08 --> 00:15:10
you, I stayed proactive.
 

245
00:15:10 --> 00:15:13
A few days later I found you
and said, two more alpha

246
00:15:13 --> 00:15:18
decays and now I am lead.
 

247
00:15:18 --> 00:15:22
You shook your head, you
were not too keen on

248
00:15:22 --> 00:15:24
my mass number of 214.
 

249
00:15:24 --> 00:15:28
I had a bad experience
with that mass before, an

250
00:15:28 --> 00:15:32
unstable acitone walked
right out the door.

251
00:15:32 --> 00:15:35
So in order to change, I
went away, but all I could

252
00:15:35 --> 00:15:38
do was just beta decay.
 

253
00:15:38 --> 00:15:41
My hopes and my dreams started
to go under, because beta

254
00:15:41 --> 00:15:45
decays don't change
a mass number.

255
00:15:45 --> 00:15:50
To bismuth, then polonium, I
hoped and I beckoned, my half

256
00:15:50 --> 00:15:53
life was 164 micro seconds.
 

257
00:15:53 --> 00:15:56
And then finally I alpha
decayed and then I was

258
00:15:56 --> 00:16:02
lead with the prize
worthy mass of 210.

259
00:16:02 --> 00:16:05
I've got to admit I was getting
quite tired, my patience

260
00:16:05 --> 00:16:08
with you had nearly expired.
 

261
00:16:08 --> 00:16:12
You were more demanding than
any I dated, and much of my

262
00:16:12 --> 00:16:17
energy had already
been liberated.

263
00:16:17 --> 00:16:20
But you still weren't happy,
but you had a fix, I really

264
00:16:20 --> 00:16:26
like the number of 206, So I
waited for years until the day,

265
00:16:26 --> 00:16:30
which began with another beta
decay, and then one more, and

266
00:16:30 --> 00:16:37
finally in the end I alpha-ed
to lead 206, my friend.

267
00:16:37 --> 00:16:40
To change any further I
wouldn't be able, no longer

268
00:16:40 --> 00:16:42
active, but happily stable.
 

269
00:16:42 --> 00:16:46
It took me billions of years to
do, and look how I've changed

270
00:16:46 --> 00:16:49
and all just for you.
 

271
00:16:49 --> 00:16:53
And wait, what did you say?
 

272
00:16:53 --> 00:16:57
You've gotten so old that
I'd rather be with a

273
00:16:57 --> 00:17:02
young lass of gold?
 

274
00:17:02 --> 00:17:04
Well, I give up, we're
through, my pumpkin.

275
00:17:04 --> 00:17:07
Shouldn't all my effort be
counting for something?

276
00:17:07 --> 00:17:10
Well, you won't be able to rule
me any more, because I'm

277
00:17:10 --> 00:17:16
leaving you, not for
one atom, but four.

278
00:17:16 --> 00:17:20
That's right, when you
were away defusing, I met

279
00:17:20 --> 00:17:24
some chlorines that I
found quite amusing.

280
00:17:24 --> 00:17:30
So we're going to form lead
c l 4, and you won't be

281
00:17:30 --> 00:17:33
hearing from me any more.
 

282
00:17:33 --> 00:17:37
See, over the years I've grown
quite wise, I've learned that

283
00:17:37 --> 00:17:39
love is about compromise.
 

284
00:17:39 --> 00:17:43
You still have half of your
half lives to live, so now

285
00:17:43 --> 00:17:48
you go out there, it's
your turn to give."

286
00:17:48 --> 00:17:58
And that is "The days of our
half lives." So, Mala takes

287
00:17:58 --> 00:18:02
great effort to make sure that
all her poetry not only rhymes,

288
00:18:02 --> 00:18:04
but it is chemically correct.
 

289
00:18:04 --> 00:18:08
So, it's a good way to review
material to read "Chemistry

290
00:18:08 --> 00:18:12
from the Couch Potato."
 

291
00:18:12 --> 00:18:17
All right, so let's do an
example now and think about how

292
00:18:17 --> 00:18:19
things will change over time.
 

293
00:18:19 --> 00:18:23
So we have an example, we want
to know the original activity,

294
00:18:23 --> 00:18:29
and the activity after 17 years
of a sample of plutonium.

295
00:18:29 --> 00:18:33
So let's take a look at how
we'll do this problem.

296
00:18:33 --> 00:18:37
So first, given the information
up there, the first thing

297
00:18:37 --> 00:18:44
we want to do is find the
original number of nuclei.

298
00:18:44 --> 00:18:53
So first, capital N o, the
original number of nuclei.

299
00:18:53 --> 00:18:57
So, we're given information
about grams, so we have 0 .

300
00:18:57 --> 00:19:00
5 grams.
 

301
00:19:00 --> 00:19:04
And now if we want to know the
number of nuclei, what's the

302
00:19:04 --> 00:19:08
first thing I have to do?
 

303
00:19:08 --> 00:19:10
Convert from grams to what?
 

304
00:19:10 --> 00:19:11
STUDENT: Moles.
 

305
00:19:11 --> 00:19:13
PROFESSOR: Moles, right.
 

306
00:19:13 --> 00:19:19
And here we want to use the
molecular mass that's given to

307
00:19:19 --> 00:19:20
us in the form of that isotope.
 

308
00:19:20 --> 00:19:29
So here, we are given
information about 239, and so

309
00:19:29 --> 00:19:34
that's the number we want to
use in our conversions.

310
00:19:34 --> 00:19:37
So we can convert that over,
but that's going to give

311
00:19:37 --> 00:19:42
us moles, so how do we go
from moles to molecules?

312
00:19:42 --> 00:19:50
Avagadro's number, 6.022
times 10 to the 23.

313
00:19:50 --> 00:19:53
This time we're going to talk
about it in terms of nuclei per

314
00:19:53 --> 00:20:00
mole, and so that's going to
give us 1.3 times 10

315
00:20:00 --> 00:20:03
to the 21 nuclei.
 

316
00:20:03 --> 00:20:13
OK, so now we know the
original number of nuclei.

317
00:20:13 --> 00:20:16
The next thing we're going
to want to do is find k.

318
00:20:16 --> 00:20:23
And k is our rate constant for
decay or our decay constant.

319
00:20:23 --> 00:20:29
So what do we know about k
for a first order process?

320
00:20:29 --> 00:20:34
We know the equation for what?
 

321
00:20:34 --> 00:20:36
For first order
half life, right.

322
00:20:36 --> 00:20:38
So that's o .
 

323
00:20:38 --> 00:20:44
6 9 3 1 over t 1/2.
 

324
00:20:44 --> 00:20:48
And in this problem we were
given, the half life, and often

325
00:20:48 --> 00:20:52
you will be given the half life
or you can look it up, and so

326
00:20:52 --> 00:20:55
we can put that in,
so we have 0 .

327
00:20:55 --> 00:20:59
6 9 3 1 over 7 .
 

328
00:20:59 --> 00:21:04
6 times 10 to the 11 seconds.
 

329
00:21:04 --> 00:21:09
And we can calculate our
constant, which is 9 .

330
00:21:09 --> 00:21:18
1 times 10 to the
minus 13 per second.

331
00:21:18 --> 00:21:21
So now we were asked to find
the original activity and the

332
00:21:21 --> 00:21:27
activity after 17 years.
 

333
00:21:27 --> 00:21:34
So, first we'll find the
original activity, and the

334
00:21:34 --> 00:21:39
original activity is going to
be equal to our rate constant

335
00:21:39 --> 00:21:42
times our original
number of nuclei.

336
00:21:42 --> 00:21:45
So we've just solved for both
of these, so we can plug

337
00:21:45 --> 00:21:48
these in, so we had 9 .
 

338
00:21:48 --> 00:21:55
1 times 10 the minus 13
per second times the

339
00:21:55 --> 00:21:57
number of nuclei, 1 .
 

340
00:21:57 --> 00:22:05
3 times 10 to the 21
nuclei, equals 1 .

341
00:22:05 --> 00:22:12
2 times 10 to the 9, and
what are the units here?

342
00:22:12 --> 00:22:22
It's just like a hum,
it's hard to understand.

343
00:22:22 --> 00:22:25
STUDENT: Nuclei per second.
 

344
00:22:25 --> 00:22:30
PROFESSOR: Nuclei per second,
which is the same as what?

345
00:22:30 --> 00:22:34
That's equal to something else.
 

346
00:22:34 --> 00:22:39
Yup, so that's the same as the
Becquerel or the Bq, so it's

347
00:22:39 --> 00:22:43
defined as nuclei per second,
or number of disintegrations

348
00:22:43 --> 00:22:47
per second.
 

349
00:22:47 --> 00:22:59
All right, so let's
do the last one.

350
00:22:59 --> 00:23:10
OK, so now after 17 years, so
now we can say that the

351
00:23:10 --> 00:23:14
activity at some time is equal
to the original activity, e to

352
00:23:14 --> 00:23:20
the minus k t, and we can put
in the activity that we

353
00:23:20 --> 00:23:21
just found, which is 1 .
 

354
00:23:21 --> 00:23:31
2 times 10 to the 9 Bq, times e
to the minus k, which is 9 .

355
00:23:31 --> 00:23:37
1 times 10 to the minus 13
per second times 17 years,

356
00:23:37 --> 00:23:40
which in seconds is 5 .
 

357
00:23:40 --> 00:23:44
4 times 10 to the 8 second.
 

358
00:23:44 --> 00:23:47
So here, we want to make sure
that our units are going to

359
00:23:47 --> 00:23:50
cancel, and this is where
people often run into problems.

360
00:23:50 --> 00:23:55
They'll plug in 17 years, and
then a rate constant, which was

361
00:23:55 --> 00:23:58
calculated in seconds, and
things will not cancel

362
00:23:58 --> 00:23:59
appropriately.
 

363
00:23:59 --> 00:24:02
So make sure that you get your
units consistent so that your

364
00:24:02 --> 00:24:07
seconds are going to cancel.
 

365
00:24:07 --> 00:24:11
And so this term, if we do the
math out here with the number

366
00:24:11 --> 00:24:15
significant figures, we find
that that equals 1.2

367
00:24:15 --> 00:24:18
times 10 to 9 Bq.
 

368
00:24:18 --> 00:24:22
That term is insignificant
in our problem.

369
00:24:22 --> 00:24:28
So, the original radioactivity
and the activity after 17 years

370
00:24:28 --> 00:24:32
are the same in terms of
the significant figures.

371
00:24:32 --> 00:24:37
And so, I choose this problem
to emphasize a problem that

372
00:24:37 --> 00:24:40
we have, and that is
radioactive waste.

373
00:24:40 --> 00:24:46
It takes a very long time for
some compounds to decay.

374
00:24:46 --> 00:24:50
And so you have to think about
storing radioactive waste, and

375
00:24:50 --> 00:24:54
think about a container
that will outlast that

376
00:24:54 --> 00:24:55
radioactive waste.
 

377
00:24:55 --> 00:24:58
And how do you know that the
container is going to outlast

378
00:24:58 --> 00:25:00
the radioactive waste.
 

379
00:25:00 --> 00:25:02
You can't really do an
experiment because the time

380
00:25:02 --> 00:25:05
involved in doing the
experiment, anyone who designs

381
00:25:05 --> 00:25:08
the container won't be alive by
the time you're concerned about

382
00:25:08 --> 00:25:12
whether the container is
going to be stable or not.

383
00:25:12 --> 00:25:16
So taking radioactivity
is an issue.

384
00:25:16 --> 00:25:19
You heard some in the
presidential campaign about

385
00:25:19 --> 00:25:24
whether both candidates believe
in nuclear energy or not, and I

386
00:25:24 --> 00:25:28
think that both of them said,
it needs to be considered, we

387
00:25:28 --> 00:25:31
need to have everything
on the table.

388
00:25:31 --> 00:25:34
If we're going to have a real
uniform energy policy, we need

389
00:25:34 --> 00:25:35
to think about everything.
 

390
00:25:35 --> 00:25:40
So, issues of radioactive waste
and how to handle radioactivity

391
00:25:40 --> 00:25:43
safely are going to come back
as being current,

392
00:25:43 --> 00:25:44
important topics.
 

393
00:25:44 --> 00:25:47
And so these may be topics that
you will, in your lifetime,

394
00:25:47 --> 00:25:50
have to deal with, either as a
scientist trying to come up

395
00:25:50 --> 00:25:55
with new technologies, or as a
citizen deciding whether having

396
00:25:55 --> 00:25:59
a radioactive plant in your
hometown is a good idea or not.

397
00:25:59 --> 00:26:03
A lot of people are happy about
nuclear energy, as long as the

398
00:26:03 --> 00:26:06
power plants are nowhere
located near them.

399
00:26:06 --> 00:26:09
But, these are things that
you'll have to face, and you

400
00:26:09 --> 00:26:12
probably will be voting on
this in the future, if not

401
00:26:12 --> 00:26:13
dealing with it directly.
 

402
00:26:13 --> 00:26:19
So that's how you do
a problem in this.

403
00:26:19 --> 00:26:24
All right, so let's talk about
a medical use of radioactivity.

404
00:26:24 --> 00:26:27
Radioactivity can definitely
be our friend, as well

405
00:26:27 --> 00:26:30
as something to be
concerned about.

406
00:26:30 --> 00:26:34
And I think I mentioned this in
the first day of class, one of

407
00:26:34 --> 00:26:37
the ways that the Chemistry
Department has moved to being

408
00:26:37 --> 00:26:41
the number one ranked Chemistry
Department of U.S. News and

409
00:26:41 --> 00:26:45
World Report over the years, is
a little extra money that came

410
00:26:45 --> 00:26:50
in from the work of -- a patent
from Professor Alan Davidson

411
00:26:50 --> 00:26:53
that we were able to do some
pretty exciting things with

412
00:26:53 --> 00:26:55
that money over the years.
 

413
00:26:55 --> 00:26:59
So I always like to mention all
the great money-making

414
00:26:59 --> 00:27:03
discoveries that occurred using
511-1 material, and this

415
00:27:03 --> 00:27:05
is another example.
 

416
00:27:05 --> 00:27:10
So, he used an isotope of
technetium, and it's being used

417
00:27:10 --> 00:27:17
organ scanning, bone scans,
it's one of the leading

418
00:27:17 --> 00:27:19
ones for heart imaging.
 

419
00:27:19 --> 00:27:22
It's also been used
recently in breast cancer.

420
00:27:22 --> 00:27:27
It's estimated 7 million uses
annually in the U.S. And

421
00:27:27 --> 00:27:33
so, this was patented as
cardiolite, and it's really

422
00:27:33 --> 00:27:35
just very simple chemistry.
 

423
00:27:35 --> 00:27:38
So you're using a d block
metal, an isotope of a d

424
00:27:38 --> 00:27:42
block metal, which has
your exciting d orbitals.

425
00:27:42 --> 00:27:47
And what did he do, he made a
coordination complex with that

426
00:27:47 --> 00:27:51
metal, an isotope of it, and he
found ligands, cyanide ligands,

427
00:27:51 --> 00:27:53
those are pretty
common ligands.

428
00:27:53 --> 00:27:56
You've seen a lot of
coordination complexes with

429
00:27:56 --> 00:27:59
cyanide ligands, and he tried
different ligands to get the

430
00:27:59 --> 00:28:03
desired properties of
stability and solubility,

431
00:28:03 --> 00:28:05
and that's all it was.
 

432
00:28:05 --> 00:28:08
So he used some knowledge of
radioactivity, knowledge of

433
00:28:08 --> 00:28:11
inorganic chemistry -- he was
an inorganic chemist,

434
00:28:11 --> 00:28:13
he's retired now.
 

435
00:28:13 --> 00:28:17
And simple coordination
chemistry, and made an enormous

436
00:28:17 --> 00:28:21
amount of money for MIT, and
particularly, the Chemistry

437
00:28:21 --> 00:28:26
Department, and also, this
has saved a lot of lives.

438
00:28:26 --> 00:28:30
So, imaging is something that
chemists do a lot of, actually.

439
00:28:30 --> 00:28:35
Not just imaging for cancer or
imaging of organs, but also

440
00:28:35 --> 00:28:38
imaging of live cells to try
to understand how the cell

441
00:28:38 --> 00:28:40
works when it's healthy.
 

442
00:28:40 --> 00:28:43
And so recently, Professor
Alice Ting, in the Chemistry

443
00:28:43 --> 00:28:49
Department, received an NIH
pioneer award, NIH is National

444
00:28:49 --> 00:28:52
Institutes of Health, and
started giving these pioneer

445
00:28:52 --> 00:28:57
awards for people coming up
with very innovative ideas, the

446
00:28:57 --> 00:29:01
kind of innovative ideas that
most people would not want to

447
00:29:01 --> 00:29:03
fund, because there's a good
chance it might not work, but

448
00:29:03 --> 00:29:05
if it did, it would
be spectacular.

449
00:29:05 --> 00:29:09
So she received one of these
awards for trying to develop

450
00:29:09 --> 00:29:13
technology to image
protein-protein interactions

451
00:29:13 --> 00:29:16
in living cells, which is
something that people would

452
00:29:16 --> 00:29:17
really, really love
to be able to do.

453
00:29:17 --> 00:29:20
And so she is involved in
developing technology.

454
00:29:20 --> 00:29:23
So developing of imaging tools
is something that a lot of

455
00:29:23 --> 00:29:26
chemists do, it's a very
popular area in chemistry.

456
00:29:26 --> 00:29:28
And if it's something that
you're interested in, there's

457
00:29:28 --> 00:29:30
definitely a lot of people
around that you could

458
00:29:30 --> 00:29:33
think about working with
for a UROP position.

459
00:29:33 --> 00:29:37
OK, so that is first order.
 

460
00:29:37 --> 00:29:41
And now let's go on and
talk about second order

461
00:29:41 --> 00:29:43
integrated rate laws.
 

462
00:29:43 --> 00:29:45
And we're going to have a
little derivation for you, I

463
00:29:45 --> 00:29:48
always like to warn people that
it's coming, because all of a

464
00:29:48 --> 00:29:52
sudden equations are coming in
and out, and you just want to

465
00:29:52 --> 00:29:55
know where these equations
are coming from.

466
00:29:55 --> 00:29:58
So, as we talked about
last time, this is an

467
00:29:58 --> 00:30:00
expression for rate law.
 

468
00:30:00 --> 00:30:03
You have your rate constant,
your concentration of

469
00:30:03 --> 00:30:07
something, a, and it's raised
to a coefficient, and here that

470
00:30:07 --> 00:30:12
coefficient is 2, indicating
it's a second order process.

471
00:30:12 --> 00:30:17
So if there's nothing
up there, that's 1.

472
00:30:17 --> 00:30:21
And then 2, and again, the
order of the reaction can be

473
00:30:21 --> 00:30:25
positive, negative, it can be
integers, it can be fractions.

474
00:30:25 --> 00:30:29
But this is second
order, so we have 2.

475
00:30:29 --> 00:30:32
Now, as we did with the first
order expression, we're going

476
00:30:32 --> 00:30:36
to separate our concentration
terms and our time terms.

477
00:30:36 --> 00:30:39
So we're going to bring our
concentration term over to one

478
00:30:39 --> 00:30:42
side, another concentration
term here, and we're going to

479
00:30:42 --> 00:30:47
have our rate constant and our
time term on the other side.

480
00:30:47 --> 00:30:49
And now we're going to
integrate, because it is

481
00:30:49 --> 00:30:51
an integrated rate law.
 

482
00:30:51 --> 00:30:55
So we can integrate from the
original concentration of a to

483
00:30:55 --> 00:30:59
the concentration of a at some
time, t, and then we'll also

484
00:30:59 --> 00:31:02
integrate from zero time to
that time, t, on

485
00:31:02 --> 00:31:04
the other side.
 

486
00:31:04 --> 00:31:07
Now, I'm going to take this
expression and just bring it up

487
00:31:07 --> 00:31:09
to the top of the page, so
that's the exact same

488
00:31:09 --> 00:31:11
expression, nothing
has happened.

489
00:31:11 --> 00:31:14
And now we're going to
solve that integral.

490
00:31:14 --> 00:31:17
So we can solve that integral,
and if you want to look at

491
00:31:17 --> 00:31:20
these -- the back of your
textbook has all of these

492
00:31:20 --> 00:31:23
conversions, if you
want to look at them.

493
00:31:23 --> 00:31:26
So, we're going to solve that
integral, now we have minus

494
00:31:26 --> 00:31:31
parentheses 1 over the
concentration of a at some

495
00:31:31 --> 00:31:34
time, t, minus 1 over the
original concentration

496
00:31:34 --> 00:31:36
equals minus k t.
 

497
00:31:36 --> 00:31:40
We can get rid of some
of these minus signs.

498
00:31:40 --> 00:31:46
So we're going to bring the
concentration of time, t, over

499
00:31:46 --> 00:31:49
on this side, we have our k t,
and now we have this other

500
00:31:49 --> 00:31:51
term, the original
concentration term is on the

501
00:31:51 --> 00:31:58
other side, and this is
expressed in a certain way that

502
00:31:58 --> 00:32:01
gives you the equation
for a straight line.

503
00:32:01 --> 00:32:06
And again, kinetics, you need
experimental data for kinetics,

504
00:32:06 --> 00:32:09
and so when you measure your
data, you plot your data, and

505
00:32:09 --> 00:32:12
so there's a lot of equations
for straight lines that you

506
00:32:12 --> 00:32:15
have, because it's all about
trying to plot data, and figure

507
00:32:15 --> 00:32:19
out what the order
is experimentally.

508
00:32:19 --> 00:32:24
So, here's an equation for a
straight line, and we can plot

509
00:32:24 --> 00:32:59
this, and you can tell me what
the intercept of this line is.

510
00:32:59 --> 00:33:13
OK, let's just take
10 more seconds.

511
00:33:13 --> 00:33:17
Yup, so all of you know how
to analyze the equation

512
00:33:17 --> 00:33:20
for a straight line.
 

513
00:33:20 --> 00:33:26
So, here we have 1 over the
original concentration, and

514
00:33:26 --> 00:33:32
then our slope is equal
to what here? k.

515
00:33:32 --> 00:33:36
So, you can plot your data
as your concentration

516
00:33:36 --> 00:33:37
of a changes with time.
 

517
00:33:37 --> 00:33:41
You can plot the data, and if
the data, if it's plotted as 1

518
00:33:41 --> 00:33:44
over the concentration of a
versus time and it gives you a

519
00:33:44 --> 00:33:47
straight line, that's
consistent with it being a

520
00:33:47 --> 00:33:50
second order process.
 

521
00:33:50 --> 00:33:54
So, in terms of second order
half life, we talked about

522
00:33:54 --> 00:33:59
first order half life, and for
any half life, it's just the

523
00:33:59 --> 00:34:04
time it takes for half of the
original material to go away.

524
00:34:04 --> 00:34:09
So we can rewrite this and take
a to the t, and substitute in

525
00:34:09 --> 00:34:14
the original concentration
divided by 2, and then we can

526
00:34:14 --> 00:34:19
have t have a special name t
1/2, so that's the half life.

527
00:34:19 --> 00:34:23
And now we can just
simplify this expression.

528
00:34:23 --> 00:34:26
We can bring the 2 up here,
and now we can combine

529
00:34:26 --> 00:34:28
our concentration
terms on the side.

530
00:34:28 --> 00:34:32
So we take 2, we bring
this over, minus 1.

531
00:34:32 --> 00:34:36
And that simplifies 1 over
the concentration of the

532
00:34:36 --> 00:34:40
original material here.
 

533
00:34:40 --> 00:34:44
And now we can solve for it
in terms of the half life.

534
00:34:44 --> 00:34:48
So the half life for a second
order process equals 1 over

535
00:34:48 --> 00:34:54
k times the original
concentration of the material.

536
00:34:54 --> 00:35:01
So, a second order half life
depends on the starting

537
00:35:01 --> 00:35:02
concentration.
 

538
00:35:02 --> 00:35:06
So that's very different from a
first order half life process

539
00:35:06 --> 00:35:10
where concentration term
cancels out entirely.

540
00:35:10 --> 00:35:14
So for a first order process,
the concentration of the

541
00:35:14 --> 00:35:19
original material does not
affect the half life, or for

542
00:35:19 --> 00:35:23
radioactive decay, the original
number of nuclei -- it's

543
00:35:23 --> 00:35:27
independent of how many nuclei
were around at the time.

544
00:35:27 --> 00:35:30
But for second order process,
the starting concentration

545
00:35:30 --> 00:35:36
does matter.
 

546
00:35:36 --> 00:35:38
So again, chemistry
is experimental.

547
00:35:38 --> 00:35:41
And so what you would be doing
in a lab, you would be trying

548
00:35:41 --> 00:35:45
to figure out what the order of
the reaction is, and so you

549
00:35:45 --> 00:35:47
could try out your data.
 

550
00:35:47 --> 00:35:50
You say I don't know if it's
first or second order, so for a

551
00:35:50 --> 00:35:54
first order plot, you're going
to be plotting the natural log

552
00:35:54 --> 00:35:57
of your concentrations versus
time, and if second order,

553
00:35:57 --> 00:36:00
you plot 1 over the
concentration versus time.

554
00:36:00 --> 00:36:04
And so you could plot your data
and see that oh, look at this,

555
00:36:04 --> 00:36:08
it fits a straight line really
well if I plot it as 1

556
00:36:08 --> 00:36:09
over the concentration.
 

557
00:36:09 --> 00:36:13
If I plot it as the natural log
versus time, the data doesn't

558
00:36:13 --> 00:36:14
fit a straight line at all.
 

559
00:36:14 --> 00:36:18
So this is not a first order
process, this is much

560
00:36:18 --> 00:36:21
more consistent with a
second order process.

561
00:36:21 --> 00:36:24
So again, figuring out where
something is first order or

562
00:36:24 --> 00:36:31
second order is done
experimentally.

563
00:36:31 --> 00:36:32
All right.
 

564
00:36:32 --> 00:36:35
Now we're going to talk
about kinetics and

565
00:36:35 --> 00:36:37
equilibrium constants.
 

566
00:36:37 --> 00:36:40
So, I always get very excited,
as you know, when we come back

567
00:36:40 --> 00:36:45
to equilibrium constants, so am
always very happy at this

568
00:36:45 --> 00:36:48
time in the course when we
can relate kinetics and

569
00:36:48 --> 00:36:50
equilibrium constants.
 

570
00:36:50 --> 00:36:54
So, at equilibrium, another way
to think about what's happening

571
00:36:54 --> 00:36:57
at equilibrium, is that the
rate of the forward reaction

572
00:36:57 --> 00:37:01
and the rate of the reverse
reaction are equal

573
00:37:01 --> 00:37:04
to each other.
 

574
00:37:04 --> 00:37:08
So, we can now talk about
big letter K, which is our

575
00:37:08 --> 00:37:11
equilibrium constant again,
and our little letter k's,

576
00:37:11 --> 00:37:13
which our rate constant.
 

577
00:37:13 --> 00:37:17
So the equilibrium constant for
a chemical reaction a plus b

578
00:37:17 --> 00:37:21
equals c plus d is going to
be equal to what, what

579
00:37:21 --> 00:37:25
do I put on the top?
 

580
00:37:25 --> 00:37:31
Concentration of? c, and
concentration of d, right.

581
00:37:31 --> 00:37:35
So our products, and at the
bottom we put our concentration

582
00:37:35 --> 00:37:41
of our reactants, or a and b.
 

583
00:37:41 --> 00:37:47
Now, we can also think about
this reaction in terms of

584
00:37:47 --> 00:37:50
little rate constants.
 

585
00:37:50 --> 00:37:56
So we have small letter k 1 on
top, and small letter k to

586
00:37:56 --> 00:37:59
the minus 1 on the bottom.
 

587
00:37:59 --> 00:38:03
So, the forward reaction, the
rate of forward reaction is

588
00:38:03 --> 00:38:08
going to be equal to k 1 times
the concentration of a and

589
00:38:08 --> 00:38:11
the concentration of b.
 

590
00:38:11 --> 00:38:17
And on the bottom, our rate is
going to be equal to the little

591
00:38:17 --> 00:38:20
rate constant, so for the
reverse reaction it's the

592
00:38:20 --> 00:38:23
reverse rate constant, k minus
1, and in the reverse

593
00:38:23 --> 00:38:27
direction, our reactants are
the products for the forward

594
00:38:27 --> 00:38:32
direction, or c and d.
 

595
00:38:32 --> 00:38:37
So, here we have these rates,
and at equilibrium, those

596
00:38:37 --> 00:38:39
rates are going to be equal.
 

597
00:38:39 --> 00:38:43
So at equilibrium, little k 1,
a times b is going to be equal

598
00:38:43 --> 00:38:47
to little k minus 1
times c times d.

599
00:38:47 --> 00:38:59
And at equilibrium we had c d
over a b is equal to then k 1

600
00:38:59 --> 00:39:04
over k minus 1, so if we just
rearrange this expression and

601
00:39:04 --> 00:39:08
move the rate constants to one
side and concentration terms to

602
00:39:08 --> 00:39:12
the other side, this expression
is the same as this expression,

603
00:39:12 --> 00:39:15
and we also know what this
expression is equal to,

604
00:39:15 --> 00:39:18
which is our big K.
 

605
00:39:18 --> 00:39:22
So, therefore, our equilibrium
constant equals the rate

606
00:39:22 --> 00:39:26
constant for the forward
reaction over the rate constant

607
00:39:26 --> 00:39:29
for the reverse direction.
 

608
00:39:29 --> 00:39:32
And so here is an expression
that compares equilibrium

609
00:39:32 --> 00:39:37
constants with rate constants.
 

610
00:39:37 --> 00:39:41
So now, let's think about
what is true about this.

611
00:39:41 --> 00:39:44
So our equilibrium constant,
then, is the ratio or the

612
00:39:44 --> 00:39:48
forward rate over the
reverse rate for these

613
00:39:48 --> 00:39:51
elementary reactions.
 

614
00:39:51 --> 00:39:56
And if we think about rate
constants in kinetics terms, if

615
00:39:56 --> 00:39:59
k is greater than 1, if there
are more products than

616
00:39:59 --> 00:40:03
reactants at equilibrium,
what's true about k

617
00:40:03 --> 00:40:06
1 and k minus 1?
 

618
00:40:06 --> 00:40:12
Is k 1 greater than or
less than k minus 1?

619
00:40:12 --> 00:40:13
Right.
 

620
00:40:13 --> 00:40:16
So the forward rate constant
is greater than the

621
00:40:16 --> 00:40:19
reverse rate constant.
 

622
00:40:19 --> 00:40:23
And if K, big equilibrium
constant, K is less than 1, if

623
00:40:23 --> 00:40:28
at equilibrium there are more
reactants than products, what

624
00:40:28 --> 00:40:33
is true about this
relationship?

625
00:40:33 --> 00:40:35
It would be less than.
 

626
00:40:35 --> 00:40:39
So, you can think about
equilibrium constants now in

627
00:40:39 --> 00:40:41
terms of rate constants,
which we'll be doing a

628
00:40:41 --> 00:40:43
lot on Wednesday, too.
 

629
00:40:43 --> 00:40:44
All right.
 

630
00:40:44 --> 00:40:48
So let me introduce you to
a couple of more terms

631
00:40:48 --> 00:40:50
in the last few minutes.
 

632
00:40:50 --> 00:40:53
So reactions don't usually
occur in one step, but occur

633
00:40:53 --> 00:40:55
in a series of steps.
 

634
00:40:55 --> 00:41:02
Each step is called an
elementary reaction.

635
00:41:02 --> 00:41:07
So, the overall reaction, the
order and the rate law, can be

636
00:41:07 --> 00:41:13
derived from the stoichiometry
for an overall reaction, you

637
00:41:13 --> 00:41:16
can't use the stoichiometry,
but for an elementary

638
00:41:16 --> 00:41:18
reaction you can.
 

639
00:41:18 --> 00:41:21
So, for an elementary reaction,
say one step in the reaction

640
00:41:21 --> 00:41:25
mechanism, that step occurs
exactly as written so you

641
00:41:25 --> 00:41:27
can use stoichiometry.
 

642
00:41:27 --> 00:41:31
And that's going to be handy
in coming up with mechanisms.

643
00:41:31 --> 00:41:35
So, let's just look at one
example very briefly.

644
00:41:35 --> 00:41:38
So here we have the
decomposition of ozone, which

645
00:41:38 --> 00:41:42
is another environmental issue
that you will be faced

646
00:41:42 --> 00:41:45
with in your lifetime.
 

647
00:41:45 --> 00:41:50
So this is the overall
reaction, and you can't use the

648
00:41:50 --> 00:41:53
stoiciomery to figure out the
order of the reaction, but if

649
00:41:53 --> 00:41:57
you divide it up into
elementary reaction steps, then

650
00:41:57 --> 00:42:01
you can use the stoiciomery to
write the rate law for each

651
00:42:01 --> 00:42:04
step of that reaction.
 

652
00:42:04 --> 00:42:08
So, the first step here
is a unimolecular step.

653
00:42:08 --> 00:42:15
You have one thing going to two
things, and molecularity is the

654
00:42:15 --> 00:42:18
number of reactant molecules
the come together

655
00:42:18 --> 00:42:20
to form product.
 

656
00:42:20 --> 00:42:24
So, unimolecular you just have
one thing that's forming

657
00:42:24 --> 00:42:26
some kind of product.
 

658
00:42:26 --> 00:42:29
What do you think it's called
if you have two things

659
00:42:29 --> 00:42:32
forming a product?
 

660
00:42:32 --> 00:42:34
Bimolecular.
 

661
00:42:34 --> 00:42:37
These are good little one or
two point questions on a test,

662
00:42:37 --> 00:42:41
they're not very hard, they
should be pretty easy

663
00:42:41 --> 00:42:42
to think about.
 

664
00:42:42 --> 00:42:45
So, we have unimolecular,
examples would be some

665
00:42:45 --> 00:42:48
kind of decomposition
or radioactive decay.

666
00:42:48 --> 00:42:51
Bimolecular, two reactants
coming together

667
00:42:51 --> 00:42:52
to form products.
 

668
00:42:52 --> 00:42:56
And termolecular is three
reactants coming together

669
00:42:56 --> 00:43:00
to form a product,
and that's rare.

670
00:43:00 --> 00:43:04
And you can remember that it's
rare if you think about how you

671
00:43:04 --> 00:43:07
would hold three tennis balls
in your hand and have them all

672
00:43:07 --> 00:43:11
come together at the same time
to form product, that is

673
00:43:11 --> 00:43:12
a difficult thing to do.
 

674
00:43:12 --> 00:43:15
Two things coming together
is easy, bimolecular

675
00:43:15 --> 00:43:16
is very common.
 

676
00:43:16 --> 00:43:21
Termolecular not very common,
that they'd all come together

677
00:43:21 --> 00:43:25
at the same time to
form a product.

678
00:43:25 --> 00:43:32
All right, so we can write
rate laws for each step here.

679
00:43:32 --> 00:43:36
For the first step here, the
rate would be equal to a k -- I

680
00:43:36 --> 00:43:38
don't have the k written up
here, but there's always going

681
00:43:38 --> 00:43:40
to be a little k
over the arrow.

682
00:43:40 --> 00:43:44
So, k times the reactant
would be here.

683
00:43:44 --> 00:43:48
For bimolecular, again,
assume a k over there.

684
00:43:48 --> 00:44:13
What's that rate going
to be equal to?

685
00:44:13 --> 00:44:16
OK, let's just take 10
more seconds since

686
00:44:16 --> 00:44:27
class is almost over.
 

687
00:44:27 --> 00:44:34
Yup, so it's this one
right down here.

688
00:44:34 --> 00:44:39
So we have the rate is equal to
k times these two reactants.

689
00:44:39 --> 00:44:43
You can sum up the steps and
get the overall reactants.

690
00:44:43 --> 00:44:47
Notice that o is an
intermediate, it's formed

691
00:44:47 --> 00:44:48
here, decayed here.
 

692
00:44:48 --> 00:44:52
It goes away, and so o
doesn't appear in the

693
00:44:52 --> 00:44:54
overall expression.
 

694
00:44:54 --> 00:44:56
So we're going to talk
a lot about reaction

695
00:44:56 --> 00:44:58
intermediates next time.
 

696
00:44:58 --> 00:45:02
And also, remember that you
can't prove reaction mechanisms

697
00:45:02 --> 00:45:05
to be correct, they're just
consistent with the

698
00:45:05 --> 00:45:07
data that you have.