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PROFESSOR: OK.
 

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So last time you started
kinetics which is a completely

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different topic from
thermodynamics.

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They're related and we'll see
a relationship at some point.

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And you did first
order kinetics.

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And today we're going to move
on and go ahead and do more

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complicated kinetics and
hopefully get to some

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interesting stuff in
a couple lectures.

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Right now we just have to do
the review of stuff that

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you've probably seen before.
 

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And so we're going to
go reasonably fast.

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So you saw first
order reactions.

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Today we're going to do
second order reactions.

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At least begin with
second order reactions.

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Second order kinetics.
 

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Of the form, and then
there are two kinds.

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There's first order, rather,
second order, in one reactant

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of the form A goes to products.
 

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And then you have first order
in two reactants first

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order in two reactants.
 

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So I'll have the form A
plus B goes of products.

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Where that's first order in
A and first order in B, and

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this is second order in A.
 

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So you can think of this
as A plus A goes to

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products if you want.
 

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And there's some rate
constant k, associated

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with this reaction.
 

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And we do the rate analysis.
 

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We write the rate
of this process.

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Minus dA/dt.
 

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00:02:03 --> 00:02:07
 


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And for the purpose of writing
on the board, and you might

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want to do this also in your
homework, when you're

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tired of writing.
 

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I'm going to skip the brackets.
 

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The little brackets that we
usually put for concentration.

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I'm going to skip those,
because it's just too

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much work to write them.
 

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And you can understand that
A is a concentration of A.

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So this is equal
to k A squared.

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Second order on A.
 

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And the units for k, it's
important to keep track of your

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units, at the end of the
calculations, often you want to

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make sure your units work out.
 

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So the units for k are
going to be of this, A

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is in moles per liter.
 

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The units for k, you're going
to have to be able to match

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the units on this side here.
 

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So the units for k are going
to be liters squared per

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mole squared per second.
 

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To make the units match.
 

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Then you integrate this,
on both sides, from zero

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to t, or from A0, the
initial rate, to A.

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Or from zero to t and this
doesn't go like this.

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So integrate with A0 to A.
 

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You've got dA over A squared,
you put all the A's on one

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side, all the t's on the other
side, from zero to t dt with a

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minus k on this side here.
 

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And then you get your rate
equation, integrated

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rate equation for this.
 

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Which gives you one over A is
equal to kt plus one over A0.

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So this gives you A as
a function of time.

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And this is a convenient
way to write it, because

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it's linear in time.
 

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So you always try to get
things to be linear in time.

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Because then you can plot
them as a straight line.

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Plot on this axis here you
plot one over A. on this

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axis here you plot t
as a function of time.

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And then you get a straight
line where the slope is k.

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Gives you the rate, and the
intercept is one over A0.

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Yes.
 

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STUDENT: [INAUDIBLE]
 

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PROFESSOR: k is moles per
liter squared per second.

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You're right, liters per
mole per second, yes.

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That is correct. because
it has to work.

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Otherwise it doesn't work.
 

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So I need to turn on my brain.
 

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OK, think.
 

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Liters per mole per second.
 

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Thank you.
 

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And the other thing that you
want to know is the half-life.

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What is the half-life.
 

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So you set 2 A0.
 

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So A0 over two, you look for
the time where you get to A0

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over two, so you put
A0 over two in here.

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00:05:43 --> 00:05:48
And that's going to be equal to
k times t 1/2 plus one over A0.

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So you solve for the half-life
and you get a half-life

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of one over k A0.
 

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So the half-life is inversely
proportional to the amount of

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stuff you started out with.
 

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00:06:02 --> 00:06:06
Unlike the first order
reaction, where the half-life

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00:06:06 --> 00:06:14
was independent of the amount
that you started out with.

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So this was the easy one.
 

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The next one is a little
bit more complicated.

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Which is when your first order
in each of two reactants.

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Then your rate, your
differential rate equation,

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looks like this.
 

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Because the k times A times B,
and now you don't know what

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to do with B, a priori.
 

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00:06:38 --> 00:06:43
So you want to rewrite this
equation a little bit

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differently in terms of the
amount of A that's used up.

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00:06:47 --> 00:06:51
So you define A, x is equal
to A0 minus A, this is the

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amount of A that's used up.
 

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This is what you started
out with, this is what

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00:06:56 --> 00:06:56
you're left with.
 

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And so the difference is
what's being used up.

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00:07:01 --> 00:07:09
And dx/dt is minus dA/dt.
 

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00:07:09 --> 00:07:15
And by stoichiometry, what
you've used up, of A, is also

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what you've used up of B.
 

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00:07:18 --> 00:07:22
Because for every A that
reacts, you have to

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00:07:22 --> 00:07:24
use up one mole of B.
 

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00:07:24 --> 00:07:26
For every mole of A
that reacts, you use

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00:07:26 --> 00:07:27
up one mole of B.
 

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And so you also have
x to B0 minus B.

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So, you can plug this in here.
 

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00:07:34 --> 00:07:37
And get a differential equation
which is purely in terms of

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one variable, which is x.
 

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Right here, it looks like it's
in terms of two variables,

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which makes it complicated.
 

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By doing this change of
variables, you see that A

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and B actually related.
 

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00:07:47 --> 00:07:51
Because of the reaction
stoichiometry.

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So you can rewrite that as
dx/dt is equal to k times A0

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minus x times B0 minus x.
 

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00:08:02 --> 00:08:06
And now you have a differential
equation in one variable, which

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00:08:06 --> 00:08:09
with some tricks you can solve.
 

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00:08:09 --> 00:08:13
So we want to have the
integrated equation.

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00:08:13 --> 00:08:15
So we take an integral
of both sides.

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00:08:15 --> 00:08:17
We put in all the
x's on one side.

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00:08:17 --> 00:08:19
All the times on
the other side.

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00:08:19 --> 00:08:28
We'll go from x equals zero to
x, dx, A0 minus x times B0

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00:08:28 --> 00:08:34
minus x is equal to k
from zero to t dt.

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00:08:34 --> 00:08:37
And now you have to dig back
into the last time you took

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00:08:37 --> 00:08:44
integration calculus, which for
me was about two centuries ago.

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00:08:44 --> 00:08:49
And figure out how to
do this integral here.

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00:08:49 --> 00:08:51
And the trick for doing
this integral is to

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00:08:51 --> 00:09:01
use partial fractions.
 

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00:09:01 --> 00:09:04
So you use partial fractions
to do this integral here.

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00:09:04 --> 00:09:15
Which means that if you take
this ratio, one over A0 minus x

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00:09:15 --> 00:09:20
times B0 minus x and rewrite it
as some number, n1 divided by

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00:09:20 --> 00:09:30
A0 minus x, plus some number n2
divided by B0 minus x, you

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00:09:30 --> 00:09:35
solve for n1 and n2, and you
find that n1 here is equal to

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00:09:35 --> 00:09:45
one over B0 minus A0 and n2 is
minus one over B0 minus A0.

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00:09:45 --> 00:09:48
And so you plug,
now, this in here.

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00:09:48 --> 00:09:52
And instead of having this
complicated denominator, you

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00:09:52 --> 00:09:55
have a sum of two integrals
that you know how to do.

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00:09:55 --> 00:10:00
Because they're basically
of the form one over x.

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00:10:00 --> 00:10:03
And in doing this, you also
realize that you have to be

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00:10:03 --> 00:10:06
careful because when A0 is
equal to B0, when you have the

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00:10:06 --> 00:10:10
same amount of A and B, then
things blow up and

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00:10:10 --> 00:10:10
you're in trouble.
 

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00:10:10 --> 00:10:13
So that's going to
be a special case.

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00:10:13 --> 00:10:16
So always look out for special
cases for these things.

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00:10:16 --> 00:10:23
So you assume, then, that
B0 is not the same as A0.

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00:10:23 --> 00:10:25
And then you can go
forward with solving it.

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00:10:25 --> 00:10:29
So you integrate, and at the
end of the process, I'm not

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00:10:29 --> 00:10:33
going to go through it, it's
really complicated, you get

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00:10:33 --> 00:10:34
something that looks like this.
 

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00:10:34 --> 00:10:43
A0 minus B0 log of
A B0 over A0 B.

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00:10:43 --> 00:10:46
And you have your
equation here.

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00:10:46 --> 00:10:49
Now, we don't really have
a good way to plot it

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00:10:49 --> 00:10:51
against one variable.
 

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00:10:51 --> 00:10:54
And the usual thing is
look at specific cases.

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00:10:54 --> 00:10:56
And limiting cases.
 

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00:10:56 --> 00:11:00
And there's one limiting
case we already brought up.

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00:11:00 --> 00:11:02
Which is when you started with
the same amount of material.

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What does it look like if
you have the same amount

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00:11:04 --> 00:11:06
of material to begin with?
 

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00:11:06 --> 00:11:11
So if you have A0 equal to B0,
you start out with the same

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00:11:11 --> 00:11:15
amount of stuff, but if you
start out with the same amount

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00:11:15 --> 00:11:18
of stuff, then this doesn't
look so different from, at

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00:11:18 --> 00:11:21
least mathematically, from
this one right here.

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00:11:21 --> 00:11:24
If you start out with A0 is
equal to B0, and for every mole

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00:11:24 --> 00:11:27
of A that you use up you use a
mole of B, then throughout

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00:11:27 --> 00:11:30
the whole reaction, the
concentration of A and the

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00:11:30 --> 00:11:33
concentration of B are
going to be the same.

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00:11:33 --> 00:11:38
So for the whole reaction, if
you started with this, A is

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00:11:38 --> 00:11:41
equal to B for all times.
 

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00:11:41 --> 00:11:42
That makes it easy.
 

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00:11:42 --> 00:11:45
Because now you can go back and
instead of writing A times B,

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00:11:45 --> 00:11:47
you can write A times A,
which is A squared, which

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00:11:47 --> 00:11:47
is what we had here.
 

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00:11:47 --> 00:11:51
And then you have the
whole thing solved.

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00:11:51 --> 00:11:57
So in that case here you just
have minus dA/dt is equal to k

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00:11:57 --> 00:12:06
A squared, and one over A is
equal to kt plus one over A0.

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00:12:06 --> 00:12:09
You don't even have to do any
math, you just look at it.

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00:12:09 --> 00:12:10
So that's one case.
 

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00:12:10 --> 00:12:13
Another case is if one of
the reactants is in much

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00:12:13 --> 00:12:16
higher concentration
than the other reactant.

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00:12:16 --> 00:12:18
And that's called flooding.
 

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00:12:18 --> 00:12:21
You basically flood the
system with one reactant.

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00:12:21 --> 00:12:27
And that's something that
we'll use again, hopefully

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00:12:27 --> 00:12:29
by the end of class today.
 

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00:12:29 --> 00:12:33
Let's say that we take, so this
is another limiting case.

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00:12:33 --> 00:12:35
Let's say we take A0 to
be much bigger than B0.

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00:12:35 --> 00:12:39
So we flood the system with A0.
 

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00:12:39 --> 00:12:45
As a result, the concentration
of A doesn't change

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00:12:45 --> 00:12:47
very much in my pot.
 

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00:12:47 --> 00:12:50
It's hugely concentrated
in A, there's a little

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00:12:50 --> 00:12:53
bit of B around.
 

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00:12:53 --> 00:12:54
Again, with the process.
 

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00:12:54 --> 00:12:57
If I use all of the B up,
the difference in A is

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00:12:57 --> 00:12:58
going to be very small.
 

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00:12:58 --> 00:13:00
So at the end of the process,
I'm basically still going

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00:13:00 --> 00:13:04
to have A0 left in the pot.
 

214
00:13:04 --> 00:13:07
So during the whole process,
during the whole time period,

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00:13:07 --> 00:13:13
I might as well assume
that A is equal to A0.

216
00:13:13 --> 00:13:14
And that makes my
life much easier.

217
00:13:14 --> 00:13:17
Because now if I write my
differential equation in terms

218
00:13:17 --> 00:13:24
of B instead of A, so the rate
of destruction of B, k A times

219
00:13:24 --> 00:13:30
B, instead of writing A here,
it's pretty much constant

220
00:13:30 --> 00:13:30
for the whole time.
 

221
00:13:30 --> 00:13:34
I'm just going to write A0.
 

222
00:13:34 --> 00:13:37
So now, if k times A0 is a
constant, and this looks

223
00:13:37 --> 00:13:42
awfully like a first
order reaction.

224
00:13:42 --> 00:13:44
So I can solve for it.
 

225
00:13:44 --> 00:13:48
And I get that, so I can just
write the answer because

226
00:13:48 --> 00:13:49
I've done this already.
 

227
00:13:49 --> 00:13:50
I don't have to do it again.
 

228
00:13:50 --> 00:13:56
The concentration of B then
goes like B0 e to the minus

229
00:13:56 --> 00:13:59
k prime t, that's the
first order reaction.

230
00:13:59 --> 00:14:03
Where k prime here is this
new rate constant, this new

231
00:14:03 --> 00:14:10
number, which is k, times A0.
 

232
00:14:10 --> 00:14:14
So that's easy to solve also.
 

233
00:14:14 --> 00:14:15
So always go to the
limiting cases, because

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00:14:15 --> 00:14:18
they tend to easy.
 

235
00:14:18 --> 00:14:25
And if you were to go to the
full solution and put in this

236
00:14:25 --> 00:14:28
limiting case, then you'd
find that you get the right

237
00:14:28 --> 00:14:30
answer this way as well.
 

238
00:14:30 --> 00:14:33
You can directly go to the easy
way of doing it, or you can go

239
00:14:33 --> 00:14:37
through the whole process of
solving it and putting the

240
00:14:37 --> 00:14:39
approximation in there.
 

241
00:14:39 --> 00:14:41
And do the cancellations
and get this.

242
00:14:41 --> 00:14:43
But this is much easier.
 

243
00:14:43 --> 00:14:46
Just writing the answer down
is always much easier.

244
00:14:46 --> 00:14:48
So it's a pseudo first
order reaction.

245
00:14:48 --> 00:14:53
We call this a pseudo
first order reaction.

246
00:14:53 --> 00:14:56
So we're done with the
simple stuff now.

247
00:14:56 --> 00:15:04
Any questions on first order
and second order reactions?

248
00:15:04 --> 00:15:07
So the next step is,
you've got a reaction.

249
00:15:07 --> 00:15:11
It could be a gas phase
reaction, it could be a

250
00:15:11 --> 00:15:13
solution phase reaction.
 

251
00:15:13 --> 00:15:17
There's some quantity, some
property, of the solution

252
00:15:17 --> 00:15:17
that's going to change.
 

253
00:15:17 --> 00:15:20
That's going to allow
you to follow it as

254
00:15:20 --> 00:15:21
a function of time.
 

255
00:15:21 --> 00:15:22
And that property
could be many things.

256
00:15:22 --> 00:15:24
It could be spectroscopic.
 

257
00:15:24 --> 00:15:27
It could be that there's an
absorption in the visible that

258
00:15:27 --> 00:15:33
changes as the concentration of
one of your reactants changes.

259
00:15:33 --> 00:15:37
Or one of the products could
have an absorption band that

260
00:15:37 --> 00:15:38
you could follow in time.
 

261
00:15:38 --> 00:15:39
Or you could using
infrared spectroscopy

262
00:15:39 --> 00:15:41
to follow it in time.
 

263
00:15:41 --> 00:15:44
Or if you have a reaction in
the gas phase, and you have

264
00:15:44 --> 00:15:46
more products or less products
than the reactants, and the

265
00:15:46 --> 00:15:48
pressure is going to change in
time if you have

266
00:15:48 --> 00:15:49
a finite volume.
 

267
00:15:49 --> 00:15:52
So there's usually some
quantity that you can use

268
00:15:52 --> 00:15:59
to follow the reaction in
time to extract out data.

269
00:15:59 --> 00:16:03
And then from that data, that
you want to know what are the

270
00:16:03 --> 00:16:04
kinetics of this reaction.
 

271
00:16:04 --> 00:16:07
Because eventually you're going
to try to find a mechanism.

272
00:16:07 --> 00:16:09
You're going to try to
find a mechanism that's

273
00:16:09 --> 00:16:12
consistent with the data.
 

274
00:16:12 --> 00:16:13
So you get data.
 

275
00:16:13 --> 00:16:18
Then you want extract out of
the rate constants and orders.

276
00:16:18 --> 00:16:21
So let's assume that you've
found a way to get data.

277
00:16:21 --> 00:16:24
And now you've got to
analyze your data.

278
00:16:24 --> 00:16:30
And suppose that you've found a
way to measure the reactant

279
00:16:30 --> 00:16:33
concentration as a
function of time.

280
00:16:33 --> 00:16:38
And let's say that in the first
case, the simplest case is

281
00:16:38 --> 00:16:43
that you have one reactant.
 

282
00:16:43 --> 00:16:47
So you have one reactant, A.
 

283
00:16:47 --> 00:16:50
So A goes to products.
 

284
00:16:50 --> 00:16:53
And you've managed to extract
A as a function of time.

285
00:16:53 --> 00:16:57
Well, the obvious thing to
do is to plot A versus time

286
00:16:57 --> 00:16:59
and see what it fits like.
 

287
00:16:59 --> 00:17:03
So you take A versus
time, and you plot it.

288
00:17:03 --> 00:17:07
And you know that if you plot
log A versus time and it's a

289
00:17:07 --> 00:17:10
straight line, well, that's
going to be a first

290
00:17:10 --> 00:17:12
order process.
 

291
00:17:12 --> 00:17:16
Plot log A versus time in a
straight line, you know that's

292
00:17:16 --> 00:17:18
going to be first order.
 

293
00:17:18 --> 00:17:20
If it doesn't go to a
straight line you know

294
00:17:20 --> 00:17:21
it's not first order.
 

295
00:17:21 --> 00:17:26
So then you go ahead and plot
one over A versus time.

296
00:17:26 --> 00:17:32
And if it's a straight line,
you know it's second order.

297
00:17:32 --> 00:17:35
And if it's not second order,
it's not a straight line.

298
00:17:35 --> 00:17:38
It's not a straight line, you
look for some other order.

299
00:17:38 --> 00:17:39
So that's one way to do it.
 

300
00:17:39 --> 00:17:44
And you've got to have enough
points on your graph, because

301
00:17:44 --> 00:17:48
if I were to plot a, let's say
this is my data point, if I

302
00:17:48 --> 00:17:52
have a second order process, at
the beginning it's going to

303
00:17:52 --> 00:17:54
look an awful lot like
a first order process.

304
00:17:54 --> 00:17:56
It's not until after a
while that it's going

305
00:17:56 --> 00:17:59
to start to deviate.
 

306
00:17:59 --> 00:18:03
So you've got to have enough
points down in time to make

307
00:18:03 --> 00:18:07
sure that you can differentiate
between a straight line and

308
00:18:07 --> 00:18:10
a line that's not straight.
 

309
00:18:10 --> 00:18:15
And usually, that's
often a mistake that

310
00:18:15 --> 00:18:16
experimentalists make.
 

311
00:18:16 --> 00:18:17
They look at the beginning.
 

312
00:18:17 --> 00:18:21
They say, oh, it's a straight
line, work is done.

313
00:18:21 --> 00:18:22
Go home.
 

314
00:18:22 --> 00:18:30
But usually you need to have a
good amount of the reactant

315
00:18:30 --> 00:18:32
consumed before you can tell
the difference between

316
00:18:32 --> 00:18:37
first and second order.
 

317
00:18:37 --> 00:18:41
So you'll have an opportunity
to do this on the homework.

318
00:18:41 --> 00:18:45
This kind of exercise of
extracting the order

319
00:18:45 --> 00:18:48
of a simple reaction.
 

320
00:18:48 --> 00:18:51
Another way to do it if you
have a simple reaction is

321
00:18:51 --> 00:18:53
to look at half-lives.
 

322
00:18:53 --> 00:19:00
That would be the
half-life method.

323
00:19:00 --> 00:19:03
If you can measure A as a
function of time, then you know

324
00:19:03 --> 00:19:07
when you've gotten A over two.
 

325
00:19:07 --> 00:19:13
So, you know that if I look at
the half-life versus the

326
00:19:13 --> 00:19:16
concentration, the initial
concentration, of my reactant,

327
00:19:16 --> 00:19:18
that tells me something
about the order.

328
00:19:18 --> 00:19:22
Because we saw that for our
first order, t 1/2 was

329
00:19:22 --> 00:19:30
independent of the
initial concentration.

330
00:19:30 --> 00:19:34
And that for a second order, t
1/2 was proportional to one

331
00:19:34 --> 00:19:39
over the inverse of the
initial concentration.

332
00:19:39 --> 00:19:45
So if you plot t 1/2 versus A0,
or have a few A0 versus t 1/2s,

333
00:19:45 --> 00:19:47
then you can tell the
difference between first order

334
00:19:47 --> 00:19:51
and a second order reaction,
and see which one fits.

335
00:19:51 --> 00:19:59
Sometimes to get even more
solid numbers, because from

336
00:19:59 --> 00:20:01
here you can also extract k.
 

337
00:20:01 --> 00:20:04
If you have a bunch of points,
of t 1/2 versus A0, you can

338
00:20:04 --> 00:20:06
extract k, the rate constant.
 

339
00:20:06 --> 00:20:08
Not just the order, but
also the rate constant

340
00:20:08 --> 00:20:10
out of this data.
 

341
00:20:10 --> 00:20:14
You can use multiple lifetimes
if you have enough data.

342
00:20:14 --> 00:20:15
Multiple lifetimes.
 

343
00:20:15 --> 00:20:21
So you can define, you
can define a t 3/4.

344
00:20:21 --> 00:20:26
Which is the amount of time it
takes for the concentration

345
00:20:26 --> 00:20:30
of A to be 1/4 of what
you started out with.

346
00:20:30 --> 00:20:34
So 3/4 is gone.
 

347
00:20:34 --> 00:20:39
And then you can put that into
your first order rate law.

348
00:20:39 --> 00:20:46
So when you have log of A over
A0 is equal to minus kt, you

349
00:20:46 --> 00:20:51
get that t 3/4, we can solve
for t 3/4, and you get that

350
00:20:51 --> 00:20:57
that's equal to two
log two, over k.

351
00:20:57 --> 00:20:58
And then you can do
the same thing for a

352
00:20:58 --> 00:21:01
second order process.
 

353
00:21:01 --> 00:21:05
So this is first order.
 

354
00:21:05 --> 00:21:09
You plug in t 3/4 and A is
equal to 1/4 A0, and you

355
00:21:09 --> 00:21:16
solve for t 3/4 for a
second order process.

356
00:21:16 --> 00:21:21
And you get that this is equal
to three over A0 times k.

357
00:21:21 --> 00:21:27
So it's the same functional
form as these two.

358
00:21:27 --> 00:21:28
But there's a pre-factor
that's different here.

359
00:21:28 --> 00:21:31
So here there's a two
that comes in there.

360
00:21:31 --> 00:21:34
And here there's there's a
three that comes in there.

361
00:21:34 --> 00:21:37
And so there's an obvious way
to tell, then, if you have the

362
00:21:37 --> 00:21:39
t 1/2 and the t 3/4 signs.
 

363
00:21:39 --> 00:21:43
Basically, you follow, instead
of having many reactions,

364
00:21:43 --> 00:21:46
instead of having many
reactions to do, with different

365
00:21:46 --> 00:21:49
A0's here you can
do one reaction.

366
00:21:49 --> 00:21:51
If you do one reaction and
you watch the reactant go

367
00:21:51 --> 00:21:53
away, and you time it.
 

368
00:21:53 --> 00:21:57
When 1/2 of it is gone, that's
one time, then you keep

369
00:21:57 --> 00:22:00
going, like 3/4 is gone,
that's another time.

370
00:22:00 --> 00:22:03
Then you can take the
ratio of those two times.

371
00:22:03 --> 00:22:10
Of t 3/4 versus t 1/2.
t 3/4 versus t 1/2.

372
00:22:10 --> 00:22:13
And if it's a first order
process, the ratio

373
00:22:13 --> 00:22:19
here is just two.
 

374
00:22:19 --> 00:22:24
And if you take t 3/4 over t
1/2 and it's a second order

375
00:22:24 --> 00:22:32
process, the ratio is three.
 

376
00:22:32 --> 00:22:38
So with one experiment, then,
you can extract out the order.

377
00:22:38 --> 00:22:39
You can't extract out,
well, you can extract

378
00:22:39 --> 00:22:41
out the rate constant.
 

379
00:22:41 --> 00:22:44
If you know the order then you
know which equation fits, and

380
00:22:44 --> 00:22:46
you can extract out the the
rate constant, with

381
00:22:46 --> 00:22:46
a big error bar.
 

382
00:22:46 --> 00:22:50
You're always better off doing
many multiple lifetimes of

383
00:22:50 --> 00:22:55
different A0's or many of
these just to get more

384
00:22:55 --> 00:22:58
statistics in the result.
 

385
00:22:58 --> 00:23:00
So this is the simple process.
 

386
00:23:00 --> 00:23:02
And you always try, if you have
something complicated, you

387
00:23:02 --> 00:23:06
always try to bring it back to
a one component process by

388
00:23:06 --> 00:23:08
doing something like flooding.
 

389
00:23:08 --> 00:23:14
So if you have five different
reactants, if you make four of

390
00:23:14 --> 00:23:19
them in very large quantities
and keep one of them in very

391
00:23:19 --> 00:23:22
small quantities, then all
the four that are in large

392
00:23:22 --> 00:23:25
quantities are basically
constant over the process.

393
00:23:25 --> 00:23:29
And you basically are looking
at only one reactant

394
00:23:29 --> 00:23:31
going away.
 

395
00:23:31 --> 00:23:34
And then you can use these
methods to figure out what the

396
00:23:34 --> 00:23:39
order is for that one reactant.
 

397
00:23:39 --> 00:23:45
So, questions about simple one
reactant sort of processes.

398
00:23:45 --> 00:23:48
It's pretty straightforward.
 

399
00:23:48 --> 00:24:02
So now, let's say we have
more complicated reactions.

400
00:24:02 --> 00:24:05
There are two ways that
we can deal with that.

401
00:24:05 --> 00:24:07
The first, I already mentioned,
which is the flooding

402
00:24:07 --> 00:24:09
which we'll get back to.
 

403
00:24:09 --> 00:24:14
And another way is
called, so we have some

404
00:24:14 --> 00:24:18
complicated reactions.
 

405
00:24:18 --> 00:24:20
Complex reactions, with
multiple reactants, that's A

406
00:24:20 --> 00:24:24
plus B plus C goes to products.
 

407
00:24:24 --> 00:24:25
And there could be
some stoichiometry

408
00:24:25 --> 00:24:28
in front of there.
 

409
00:24:28 --> 00:24:29
So one of the ways to deal
with that is called the

410
00:24:29 --> 00:24:33
initial rate method.
 

411
00:24:33 --> 00:24:38
We want to find out orders
and rate constants.

412
00:24:38 --> 00:24:46
Initial rate method.
 

413
00:24:46 --> 00:24:53
So if I look at minus dA/dt,
one of the reactants or the

414
00:24:53 --> 00:24:57
rate of the reaction near
time t equals zero.

415
00:24:57 --> 00:24:59
That's the initial rate.
 

416
00:24:59 --> 00:25:00
Right as the reaction starts.
 

417
00:25:00 --> 00:25:03
I mix everything together and
I, just as after I mix it

418
00:25:03 --> 00:25:06
together, I watch the process
of A disappearing or B

419
00:25:06 --> 00:25:08
disappearing or C disappearing.
 

420
00:25:08 --> 00:25:13
And so in reality what I'm
doing is minus delta A / delta

421
00:25:13 --> 00:25:18
t near t equals zero, where
delta t is a small interval.

422
00:25:18 --> 00:25:20
So there's not much
change in delta A.

423
00:25:20 --> 00:25:21
Experimentally
that's what I do.

424
00:25:21 --> 00:25:24
And that's pretty much,
essentially getting

425
00:25:24 --> 00:25:26
this number out.
 

426
00:25:26 --> 00:25:28
And we're going to call
that the initial rate.

427
00:25:28 --> 00:25:32
And the initial rate is k,
and at the beginning I

428
00:25:32 --> 00:25:33
haven't used up anything.
 

429
00:25:33 --> 00:25:36
So all the initial
concentrations are there A0

430
00:25:36 --> 00:25:41
to the alpha, B0 to the
beta, C0 to the gamma.

431
00:25:41 --> 00:25:46
Et cetera if you have
more reactants.

432
00:25:46 --> 00:25:47
So you measure this.
 

433
00:25:47 --> 00:25:55
You measure this R0, and then
you repeat the same process

434
00:25:55 --> 00:25:59
with a new concentration of
one of those three reactants.

435
00:25:59 --> 00:26:03
So with A0 prime, let's say.
 

436
00:26:03 --> 00:26:09
And then you get a
new R0, R0 prime.

437
00:26:09 --> 00:26:13
Then you take the ratios of
these R0 and R0 primes,

438
00:26:13 --> 00:26:16
R0 divided by R0 prime.
 

439
00:26:16 --> 00:26:23
So R0 is k A0 to the alpha,
B0 to the beta, C0 to the

440
00:26:23 --> 00:26:29
gamma, then you have k
A0 prime to the alpha.

441
00:26:29 --> 00:26:38
B0 to the beta, C0 to
the gamma, and the

442
00:26:38 --> 00:26:40
k's you disappear.
 

443
00:26:40 --> 00:26:42
The B0's disappear.
 

444
00:26:42 --> 00:26:43
The C0's disappear.
 

445
00:26:43 --> 00:26:45
The only thing that we've
changed is the concentration

446
00:26:45 --> 00:26:46
of A0 to begin with.
 

447
00:26:46 --> 00:26:50
So the ratios of these initial
rates is the ratios of the

448
00:26:50 --> 00:26:56
A0's to the alpha power, to
the order in terms of A0.

449
00:26:56 --> 00:27:00
So if you're clever about your
choice of ratios, then you

450
00:27:00 --> 00:27:02
can get alpha pretty easily.
 

451
00:27:02 --> 00:27:16
So if you choose, so if you now
choose A0 prime to be equal to

452
00:27:16 --> 00:27:27
1/2 A0, then you measure
R0 over R0 prime.

453
00:27:27 --> 00:27:35
And if you get one, then you
know that that alpha's zero.

454
00:27:35 --> 00:27:38
Alpha has to be zero here.
 

455
00:27:38 --> 00:27:39
Then you know alpha is zero,
that gives you the order.

456
00:27:39 --> 00:27:41
It's a zero order reaction.
 

457
00:27:41 --> 00:27:45
If you get that R0 over R0
prime is square root of two, or

458
00:27:45 --> 00:27:51
rather 1/2, then a square root
of two, square root of 2, then

459
00:27:51 --> 00:27:54
you know that alpha is 1/2.
 

460
00:27:54 --> 00:27:56
And that's a half order.
 

461
00:27:56 --> 00:28:00
We haven't seen any half order
reactions yet, but we will.

462
00:28:00 --> 00:28:04
Those are indicative of a
complicated mechanisms.

463
00:28:04 --> 00:28:07
But that's what you would
get out of this experiment.

464
00:28:07 --> 00:28:15
If you get that it's equal to
two, if you get that this ratio

465
00:28:15 --> 00:28:16
is equal to two, then you know
that alpha is equal

466
00:28:16 --> 00:28:19
to one, et cetera.
 

467
00:28:19 --> 00:28:24
So it's a pretty easy
way to get the order.

468
00:28:24 --> 00:28:26
Then you repeat the experiment.
 

469
00:28:26 --> 00:28:30
Now instead of changing
A0, you keep A0 constant

470
00:28:30 --> 00:28:31
and you change B0.
 

471
00:28:31 --> 00:28:38
Or, you can use flooding or
isolation, which is the

472
00:28:38 --> 00:28:40
next process to get
the other orders.

473
00:28:40 --> 00:28:43
And eventually you get
the rate constant.

474
00:28:43 --> 00:28:46
Once you have all the orders,
and you have R0, then you

475
00:28:46 --> 00:28:50
have the rate constant.
 

476
00:28:50 --> 00:28:52
OK, so that's one
way of doing it.

477
00:28:52 --> 00:28:57
And a second way of doing it is
the way we already mentioned.

478
00:28:57 --> 00:29:02
To solve the second order
reaction in two components.

479
00:29:02 --> 00:29:08
Which is to flood the reaction
with everything except for one.

480
00:29:08 --> 00:29:14
So this is called
flooding or isolation.

481
00:29:14 --> 00:29:20
Basically, you isolate one
reactant and watch it.

482
00:29:20 --> 00:29:22
You're trying to get
back to a system, which

483
00:29:22 --> 00:29:23
is a simple system.
 

484
00:29:23 --> 00:29:25
Which is a system
of one reaction.

485
00:29:25 --> 00:29:30
So let's say you take A0
to be much smaller than

486
00:29:30 --> 00:29:31
all the other species.
 

487
00:29:31 --> 00:29:35
You flood with B and C,
and you isolate A0.

488
00:29:35 --> 00:29:41
And then your rate minus
dA/dt, it's going to

489
00:29:41 --> 00:29:43
be A to the alpha.
 

490
00:29:43 --> 00:29:47
And instead of B, well, during
that process B's going to

491
00:29:47 --> 00:29:49
say pretty much constant.
 

492
00:29:49 --> 00:29:50
Because it's hugely
concentrated in B.

493
00:29:50 --> 00:29:51
You can replace B with B0.
 

494
00:29:51 --> 00:29:54
You can replace C with C0.
 

495
00:29:54 --> 00:29:58
So now you have an effective
constant, an effective rate

496
00:29:58 --> 00:30:06
constant, and then you have a
process which is effectively,

497
00:30:06 --> 00:30:09
or pseudo, one reactant.
 

498
00:30:09 --> 00:30:14
Then you can use these methods
here, you can plot A versus

499
00:30:14 --> 00:30:17
time, you can find path lines,
et cetera, to gather

500
00:30:17 --> 00:30:19
the order for it.
 

501
00:30:19 --> 00:30:28
Then you can get
alpha and k prime.

502
00:30:28 --> 00:30:33
And if you can change B0 and
C0, then you get k out of this.

503
00:30:33 --> 00:30:36
So this is basically all fairly
straightforward, just tedious

504
00:30:36 --> 00:30:42
experimentation to get
all these numbers out.

505
00:30:42 --> 00:30:44
OK, any questions on this?
 

506
00:30:44 --> 00:30:46
You'll get experience
on the homework.

507
00:30:46 --> 00:30:49
There's likely to be a question
on the final where you're given

508
00:30:49 --> 00:30:59
data and asked to extract out
orders and rate constants.

509
00:30:59 --> 00:31:00
So let's move on now.
 

510
00:31:00 --> 00:31:07
So, so far we've looked at
first and second order

511
00:31:07 --> 00:31:10
elementary processes, and we've
looked at taking data and

512
00:31:10 --> 00:31:13
extracting out rate
and rate constant.

513
00:31:13 --> 00:31:19
And the next step is
to build mechanisms.

514
00:31:19 --> 00:31:32
So a mechanism is when you take
a complicated reaction, like A

515
00:31:32 --> 00:31:36
plus B plus C goes to D plus E.
 

516
00:31:36 --> 00:31:39
And you break it up
into elementary steps.

517
00:31:39 --> 00:31:40
What's an elementary step?
 

518
00:31:40 --> 00:31:43
An elementary step is a
step which happens in

519
00:31:43 --> 00:31:45
a single reaction.
 

520
00:31:45 --> 00:31:51
So I could hypothesize that
this complicated reaction

521
00:31:51 --> 00:31:55
happens in three steps, where I
need to have a molecule of A

522
00:31:55 --> 00:31:59
and a molecule of B collide
with each other to first

523
00:31:59 --> 00:32:01
form an intermediate F.
 

524
00:32:01 --> 00:32:04
Then I want a molecule of F
plus a molecule of B to

525
00:32:04 --> 00:32:09
collide to form intermediate
G plus a product D.

526
00:32:09 --> 00:32:12
Then have the intermediate
G plus reactant C collide

527
00:32:12 --> 00:32:16
together to form the product E.
 

528
00:32:16 --> 00:32:19
So this set of elementary
steps, where at each step you

529
00:32:19 --> 00:32:24
have a collision of two or
three molecules together, three

530
00:32:24 --> 00:32:28
is not so common but two is
very common, those elementary

531
00:32:28 --> 00:32:33
steps are called the
steps of the mechanisms.

532
00:32:33 --> 00:32:44
And these elementary steps you
can define something called

533
00:32:44 --> 00:32:53
molecularity, which is the
number of species that you need

534
00:32:53 --> 00:32:57
to collide with each other in
one of these elementary steps.

535
00:32:57 --> 00:32:59
So the molecularity here
would be two, you need

536
00:32:59 --> 00:33:01
two molecules to react.
 

537
00:33:01 --> 00:33:03
Here it's two, here it's two.
 

538
00:33:03 --> 00:33:06
If I have an elementary step
which is a zero order in

539
00:33:06 --> 00:33:09
one reactant, then the
molecularity would be one.

540
00:33:09 --> 00:33:12
Or I could have A plus A,
the same molecules have to

541
00:33:12 --> 00:33:13
collide with each other.
 

542
00:33:13 --> 00:33:15
Molecularity would be two.
 

543
00:33:15 --> 00:33:18
And the molecularity
and the order of the

544
00:33:18 --> 00:33:20
reaction are connected.
 

545
00:33:20 --> 00:33:23
So if you have something which
is a molecularity of one,

546
00:33:23 --> 00:33:25
then it's going to be a
first order reaction.

547
00:33:25 --> 00:33:29
One reactant is just sitting
by itself, falls apart,

548
00:33:29 --> 00:33:31
like in radioactive decay.
 

549
00:33:31 --> 00:33:33
Molecularity of one, that's
a first order of process.

550
00:33:33 --> 00:33:36
If I need to have two molecules
come together, then it's

551
00:33:36 --> 00:33:37
a second order process.
 

552
00:33:37 --> 00:33:40
If I have to have three
molecules collide at the same

553
00:33:40 --> 00:33:44
time together, molecularity of
three, then it's going to

554
00:33:44 --> 00:33:49
depend on the concentration of
all three at the same time.

555
00:33:49 --> 00:33:52
That's called a ternary
reaction, and those are

556
00:33:52 --> 00:33:54
really quite rare.
 

557
00:33:54 --> 00:33:57
Termolecular reactions,
you need to have your

558
00:33:57 --> 00:34:01
concentrations very, very high
to statistically get an event

559
00:34:01 --> 00:34:07
happening where all three
molecules collide together.

560
00:34:07 --> 00:34:12
So three body reaction is hard
and anything higher than three

561
00:34:12 --> 00:34:15
body is essentially impossible.
 

562
00:34:15 --> 00:34:19
So that limits your
choices, which is nice.

563
00:34:19 --> 00:34:21
So that's the mechanism.
 

564
00:34:21 --> 00:34:23
And so what we're going
to do next is go through

565
00:34:23 --> 00:34:24
some mechanisms.
 

566
00:34:24 --> 00:34:28
Some simple mechanisms and
build up the complexity.

567
00:34:28 --> 00:34:33
Any questions about
mechanisms here?

568
00:34:33 --> 00:34:37
What we're doing here is, we're
formulating a framework.

569
00:34:37 --> 00:34:39
Where we can go back and look
at things that are more

570
00:34:39 --> 00:34:42
complicated, like chain
reactions or explosions or

571
00:34:42 --> 00:34:47
enzymatic reactions, and know
when to apply approximations

572
00:34:47 --> 00:34:49
and et cetera.
 

573
00:34:49 --> 00:34:58
So we basically, here, are just
laying down the ground rules.

574
00:34:58 --> 00:35:03
So let's go to our first
example of a mechanism, a

575
00:35:03 --> 00:35:05
more complicated reaction.
 

576
00:35:05 --> 00:35:08
And what we're going to do is
we're going to extract out

577
00:35:08 --> 00:35:12
integrated rate laws out
of all these mechanisms.

578
00:35:12 --> 00:35:15
And see what it looks like,
as a function of time.

579
00:35:15 --> 00:35:20
So the first one we're
going to do is called

580
00:35:20 --> 00:35:22
parallel reactions.
 

581
00:35:22 --> 00:35:23
Simple mechanism.
 

582
00:35:23 --> 00:35:30
In this case here I have
one reactant, and that

583
00:35:30 --> 00:35:34
reactant has a choice.
 

584
00:35:34 --> 00:35:36
You can think of it as a
radioactive decay, an

585
00:35:36 --> 00:35:41
atom decaying in two
different channels.

586
00:35:41 --> 00:35:44
So it can decay into B,
or it can decay into C.

587
00:35:44 --> 00:35:50
There are two rate
constants, k1 and k2.

588
00:35:50 --> 00:35:52
So you can write it like
this, or you can write it

589
00:35:52 --> 00:35:56
as A goes to B plus C.
 

590
00:35:56 --> 00:35:58
This is how you would
write a reaction.

591
00:35:58 --> 00:36:01
And this is how you would
write your mechanism.

592
00:36:01 --> 00:36:08
A goes to B and A goes to C.
 

593
00:36:08 --> 00:36:11
Each elementary step, these are
the elementary steps and this

594
00:36:11 --> 00:36:13
is the complex reaction,
each elementary step

595
00:36:13 --> 00:36:15
is unimolecular.
 

596
00:36:15 --> 00:36:18
It's a first order process.
 

597
00:36:18 --> 00:36:20
So in all of these examples,
the first thing you do is

598
00:36:20 --> 00:36:24
you write your rate law.
 

599
00:36:24 --> 00:36:28
The rate at which A gets
created or destroyed.

600
00:36:28 --> 00:36:31
And there are two paths.
 

601
00:36:31 --> 00:36:34
It gets destroyed and into B,
with a rate which is

602
00:36:34 --> 00:36:37
proportional to the
concentration of A, and it gets

603
00:36:37 --> 00:36:47
destroyed into C, proportional
to the concentration of A.

604
00:36:47 --> 00:36:51
So you write down all the ways
that A can get destroyed.

605
00:36:51 --> 00:36:52
There are two ways here.
 

606
00:36:52 --> 00:36:54
Two channels.
 

607
00:36:54 --> 00:36:59
This one happens to be fairly
easy to solve. k1 plus k2 times

608
00:36:59 --> 00:37:00
A, and you've seen this before.
 

609
00:37:00 --> 00:37:03
It's minus dA/dt as
a constant times A.

610
00:37:03 --> 00:37:04
That's a first order process.
 

611
00:37:04 --> 00:37:06
So you can just write
down the answer.

612
00:37:06 --> 00:37:08
You don't need to
do any math here.

613
00:37:08 --> 00:37:11
You recognize that we just
call this one k prime.

614
00:37:11 --> 00:37:15
And that the rate is A
as a function of time.

615
00:37:15 --> 00:37:22
This is A0 e to the minus
k1 plus k2 times t.

616
00:37:22 --> 00:37:24
And everything you've learned
about plotting first order

617
00:37:24 --> 00:37:28
processes et cetera, is
applicable here, where the

618
00:37:28 --> 00:37:32
rate constant is the
sum of these two.

619
00:37:32 --> 00:37:34
So that's for the reactant.
 

620
00:37:34 --> 00:37:36
The products are also
interesting to plot as a

621
00:37:36 --> 00:37:39
function of time, to see
how they are related to

622
00:37:39 --> 00:37:41
each other in terms of
their concentrations.

623
00:37:41 --> 00:37:52
So let me go through this also.
 

624
00:37:52 --> 00:37:54
When things get more
complicated we'll quickly go

625
00:37:54 --> 00:37:56
and make approximations.
 

626
00:37:56 --> 00:37:59
But for now, we can still
do everything exactly.

627
00:37:59 --> 00:38:04
So you write down your rate
law for the product. dB/dt

628
00:38:04 --> 00:38:12
is equal to k1 A. dC/dt
is equal to k2 times A.

629
00:38:12 --> 00:38:16
The formation of B
depends linearly on A.

630
00:38:16 --> 00:38:18
The formation of C depends
linearly on A, because they're

631
00:38:18 --> 00:38:21
both first order processes.
 

632
00:38:21 --> 00:38:23
To make B and C.
 

633
00:38:23 --> 00:38:25
And you integrate.
 

634
00:38:25 --> 00:38:31
You integrate here
from zero to B, dB.

635
00:38:31 --> 00:38:38
Is equal from zero
to t, A dt, k1.

636
00:38:38 --> 00:38:39
A is a function of time.
 

637
00:38:39 --> 00:38:41
And we've already
solved for that.

638
00:38:41 --> 00:38:43
It's this exponential up there.
 

639
00:38:43 --> 00:38:51
So you plug in here A of time.
 

640
00:38:51 --> 00:38:53
And you turn the crank
and you integrate, and

641
00:38:53 --> 00:38:53
it's an exponential.
 

642
00:38:53 --> 00:38:56
So it's not so hard
to integrate.

643
00:38:56 --> 00:39:01
And you get that B is a
function of time is k1 times A0

644
00:39:01 --> 00:39:10
over k1 plus k2 times one minus
e to the minus k1 plus

645
00:39:10 --> 00:39:13
k2 times the time.
 

646
00:39:13 --> 00:39:15
Things are already starting
to get a little bit

647
00:39:15 --> 00:39:18
more messy in the math.
 

648
00:39:18 --> 00:39:20
And then to get C, you actually
don't need to do anything.

649
00:39:20 --> 00:39:23
Because you notice that the
only difference between B and C

650
00:39:23 --> 00:39:26
here is replacing k2 with k1.
 

651
00:39:26 --> 00:39:28
So don't worry about
doing any math.

652
00:39:28 --> 00:39:33
Just write down the answer.
k2 A0, you interchange

653
00:39:33 --> 00:39:36
k1 and k2 at every step.
 

654
00:39:36 --> 00:39:42
One minus e to the minus k1
plus k2 times the time.

655
00:39:42 --> 00:39:45
The only difference is
up here in the k2 term.

656
00:39:45 --> 00:39:48
And those are your
equations for k1 and k2.

657
00:39:48 --> 00:39:53
And what you find, is the ratio
of B to C is a constant.

658
00:39:53 --> 00:39:56
If I divide B by C, everything
cancels out except for

659
00:39:56 --> 00:39:58
the k1 and the k2 here.
 

660
00:39:58 --> 00:40:02
Is equal to k1 over k2.
 

661
00:40:02 --> 00:40:11
And that is called
the branching ratio.

662
00:40:11 --> 00:40:13
The branching ratio, because
there are two branches

663
00:40:13 --> 00:40:15
out of the reactions.
 

664
00:40:15 --> 00:40:18
And this gives you the ratio of
which one is more likely to

665
00:40:18 --> 00:40:20
happen than the other one.
 

666
00:40:20 --> 00:40:22
And so if k1 is much
bigger than k2, the

667
00:40:22 --> 00:40:25
rate is per unit time.
 

668
00:40:25 --> 00:40:30
The units of k1 are per second
or per minute or per hour.

669
00:40:30 --> 00:40:33
So if this is big, if k1 is
big, then mostly you're going

670
00:40:33 --> 00:40:38
from A to B, and only a
little bit of C is formed.

671
00:40:38 --> 00:40:40
And the ratio of B and
C is always constant.

672
00:40:40 --> 00:40:47
And so you can plot, then,
you can plot the result.

673
00:40:47 --> 00:40:57
You can sketch out the result.
 

674
00:40:57 --> 00:41:03
So you know that A is going
to come down exponentially.

675
00:41:03 --> 00:41:06
Time on this axis here,
concentrations on

676
00:41:06 --> 00:41:10
this axis here.
 

677
00:41:10 --> 00:41:13
So this is A as a
function of time.

678
00:41:13 --> 00:41:15
That's that equation up
here, in exponential decay.

679
00:41:15 --> 00:41:17
And the quantity of A.
 

680
00:41:17 --> 00:41:24
And B and C are going to come
up in time, also, with this

681
00:41:24 --> 00:41:27
exponential format here.
 

682
00:41:27 --> 00:41:30
B is going to saturate at
this ratio right here, k1

683
00:41:30 --> 00:41:31
A0 divided by k1 plus k2.
 

684
00:41:31 --> 00:41:35
C is going to saturate
at this quantity here.

685
00:41:35 --> 00:41:38
So they're going to
start both at zero.

686
00:41:38 --> 00:41:52
And B is eventually going to
go to k1 A0 over k1, plus k2.

687
00:41:52 --> 00:42:08
And C, eventually, will go
to k2 A0 over k1 plus k2.

688
00:42:08 --> 00:42:13
And this, and the ratio of
these two lines at every

689
00:42:13 --> 00:42:22
point is k1 over k2.
 

690
00:42:22 --> 00:42:25
So, a simple question, for
instance, that you might be of

691
00:42:25 --> 00:42:32
the type that you might be
asked to look at is, suppose

692
00:42:32 --> 00:42:39
that k1 is 1/10 of k2.
 

693
00:42:39 --> 00:42:41
Which would you expect?
 

694
00:42:41 --> 00:42:48
Would you expect to
have, let's see.

695
00:42:48 --> 00:42:51
So C is in green here.
 

696
00:42:51 --> 00:42:55
C, B.
 

697
00:42:55 --> 00:43:05
Or do you expect,
so A comes down.

698
00:43:05 --> 00:43:10
B comes up.
 

699
00:43:10 --> 00:43:17
C comes up like this, or do you
expect the last choice, I'm

700
00:43:17 --> 00:43:39
going to put the last choice
here, so this is, let's call

701
00:43:39 --> 00:43:41
this choice number one.
 

702
00:43:41 --> 00:43:44
Choice number two.
 

703
00:43:44 --> 00:43:49
Choice number three.
 

704
00:43:49 --> 00:43:54
So k1, the rate k1 is
1/10 of the rate k2.

705
00:43:54 --> 00:43:57
That tells you something
about the branching ratio.

706
00:43:57 --> 00:44:00
So do you expect this one
here to be the right one?

707
00:44:00 --> 00:44:05
How many people think
this is the right one?

708
00:44:05 --> 00:44:06
What about this one here?
 

709
00:44:06 --> 00:44:09
How many people think
this is the right one?

710
00:44:09 --> 00:44:11
One person.
 

711
00:44:11 --> 00:44:13
What about this one here?
 

712
00:44:13 --> 00:44:15
So the branching ratio is
the ratio of the two.

713
00:44:15 --> 00:44:16
It's 1/10.
 

714
00:44:16 --> 00:44:21
So this is approximately, in my
sketch, poor sketch, granted,

715
00:44:21 --> 00:44:24
but this is approximately
10 times bigger than this.

716
00:44:24 --> 00:44:26
So that's the ratio
that you'd expect.

717
00:44:26 --> 00:44:30
And it's the right, here
k1 is the rate into B.

718
00:44:30 --> 00:44:33
It's slower than
the rate into C.

719
00:44:33 --> 00:44:38
So, you got it right.
 

720
00:44:38 --> 00:44:41
So for more complicated, we
have a more complicated

721
00:44:41 --> 00:44:43
process, we're going to ask
you the same sort of stuff.

722
00:44:43 --> 00:44:48
And it won't be as
straightforward.

723
00:44:48 --> 00:44:52
Any questions on this
beginning here?

724
00:44:52 --> 00:44:59
Next time we're going to finish
with the parallel first and

725
00:44:59 --> 00:45:01
second order processes.
 

726
00:45:01 --> 00:45:05
And hopefully we'll get done
with the complex reactions and

727
00:45:05 --> 00:45:11
mechanisms and move on to, I
forgot what's next on the list.

728
00:45:11 --> 00:45:17
But some explosions
or chain reactions.

729
00:45:17 --> 00:45:18