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PROFESSOR: So last time we were
starting more complicated
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00:00:24 --> 00:00:37
mechanisms and we looked
at series reactions.
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00:00:37 --> 00:00:40
A goes B, which is
an intermediate.
12
00:00:40 --> 00:00:43
Goes to C with rates k1 and k2.
13
00:00:43 --> 00:00:49
And we solved the
problem exactly.
14
00:00:49 --> 00:00:52
Which turned out to be a
little bit complicated.
15
00:00:52 --> 00:00:58
And then we were looking
at special cases.
16
00:00:58 --> 00:01:05
And I left the special case
of k1 equals k2 to do
17
00:01:05 --> 00:01:08
as a homework problem.
18
00:01:08 --> 00:01:14
And then on the board we did
k1 much greater than k2.
19
00:01:14 --> 00:01:22
Which looks like, if k1 is much
greater than k2, the rate into
20
00:01:22 --> 00:01:28
k1 is much greater than k2, so
using the analogy of buckets
21
00:01:28 --> 00:01:37
and pipes, it's a big pipe
between k1 and between A and B.
22
00:01:37 --> 00:01:39
And a little pipe between B
and C and you started with
23
00:01:39 --> 00:01:41
a bunch of liquid on top.
24
00:01:41 --> 00:01:45
And you basically
dump it into here.
25
00:01:45 --> 00:01:49
And then you slowly
drip into C here.
26
00:01:49 --> 00:01:55
So the first thing that happens
is that A gets dumped, it
27
00:01:55 --> 00:01:56
becomes B very quickly.
28
00:01:56 --> 00:01:58
With a rate constant k1.
29
00:01:58 --> 00:02:02
In a first order fashion.
30
00:02:02 --> 00:02:07
So A is A0 e to the
minus k1 times time.
31
00:02:07 --> 00:02:10
Then B comes up sharply.
32
00:02:10 --> 00:02:18
And then drops down with
rate constant k2 slowly.
33
00:02:18 --> 00:02:24
So B is approximately A0 e
to the minus k2 times time.
34
00:02:24 --> 00:02:30
And then C has a very small
quadratic rise, followed by
35
00:02:30 --> 00:02:33
the exponential rise to
saturation that you'd expect.
36
00:02:33 --> 00:02:37
So C is approximately a0
times one minus e to the
37
00:02:37 --> 00:02:40
minus k2 times time.
38
00:02:40 --> 00:02:43
The rate constant k2
dominates the long times.
39
00:02:43 --> 00:02:45
And everything it's looking
like it's pseudo first
40
00:02:45 --> 00:02:47
order at long times.
41
00:02:47 --> 00:02:51
So you can do this either by
thinking about it, solve the
42
00:02:51 --> 00:02:55
problem by thinking about it,
which is perfectly fine.
43
00:02:55 --> 00:02:58
Or you can do it by
solving it exactly, as
44
00:02:58 --> 00:03:00
we have done last time.
45
00:03:00 --> 00:03:02
And then putting in the
right approximations.
46
00:03:02 --> 00:03:05
Which in this case was
Taylor's approximation,
47
00:03:05 --> 00:03:08
if I remember right.
48
00:03:08 --> 00:03:12
And know just by looking at the
equation and putting the right
49
00:03:12 --> 00:03:14
approximation, you get it out.
50
00:03:14 --> 00:03:22
And a third approximation,
which is the opposite one, k2
51
00:03:22 --> 00:03:28
much less than k1, so k1 much
less than k2, the way that
52
00:03:28 --> 00:03:39
that's going to look, we
can do it by inspection.
53
00:03:39 --> 00:03:43
So basically if we think of
buckets and pipes, that's a
54
00:03:43 --> 00:03:48
very thin pipe connecting the
bucket for A and B, so we drip
55
00:03:48 --> 00:03:51
from A to B, then we
have this very fat pipe
56
00:03:51 --> 00:03:53
connecting B and C.
57
00:03:53 --> 00:03:56
So as soon as a little
B gets in there, it
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00:03:56 --> 00:03:58
immediately becomes C.
59
00:03:58 --> 00:04:03
So, the rate determining step
is the first one, A goes to B.
60
00:04:03 --> 00:04:06
And so k1, we expect that
to be the rate that
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00:04:06 --> 00:04:09
dominates the long time.
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00:04:09 --> 00:04:17
And we never expect to get any
significant amount of B to pile
63
00:04:17 --> 00:04:19
up in the middle bucket,
because this pipe is much
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00:04:19 --> 00:04:22
bigger than that one here.
65
00:04:22 --> 00:04:28
Therefore, what we expect to
see is a slow decay in A, with
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00:04:28 --> 00:04:33
rate constant k1 dominating the
long times, A is approximately
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00:04:33 --> 00:04:40
A0 e to the minus
k1 times the time.
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00:04:40 --> 00:04:46
We don't expect B to amount
to very much at all.
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00:04:46 --> 00:04:49
It'll rise up a little bit at
the very early times, and
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00:04:49 --> 00:04:52
it'll pretty much stay
very, very small.
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00:04:52 --> 00:04:53
Not quite constant.
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00:04:53 --> 00:05:01
And in fact if you look at the
full result, and you put in
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00:05:01 --> 00:05:05
your approximations, and I'll
let you do that, you would find
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00:05:05 --> 00:05:11
that B is essentially related
to A through a constant term
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00:05:11 --> 00:05:15
where k1 is very small
compared to k2.
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00:05:15 --> 00:05:17
So this is a very
small number here.
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00:05:17 --> 00:05:19
B is equal to a very
small number times A.
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00:05:19 --> 00:05:24
So it basically follows A, but
at a very very, low level.
79
00:05:24 --> 00:05:33
And then you would find that C
rises up in approximately e to
80
00:05:33 --> 00:05:37
the minus k1 times t, so
with rate constant k1.
81
00:05:37 --> 00:05:45
One minus e to the minus
k1 times the time.
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00:05:45 --> 00:05:49
So the dominant rate is k1, and
in this case here this is an
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00:05:49 --> 00:05:52
approximation that we're
going to see again.
84
00:05:52 --> 00:05:56
We're going to revisit this
approximation here when we get
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00:05:56 --> 00:05:58
to more complicated mechanisms,
it's going to become this
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00:05:58 --> 00:06:01
steady state approximation.
87
00:06:01 --> 00:06:03
Where the amount of the
intermediate, or the
88
00:06:03 --> 00:06:07
concentration of intermediate,
is low at all times.
89
00:06:07 --> 00:06:09
And doesn't change very fast.
90
00:06:09 --> 00:06:11
The slope here is very slow.
91
00:06:11 --> 00:06:14
Because k1 is so slow coming
down, and B is related to A,
92
00:06:14 --> 00:06:19
which means it's related to
this slow decrease in A, the
93
00:06:19 --> 00:06:21
amount of B is small and the
rate of change in B
94
00:06:21 --> 00:06:24
is also very small.
95
00:06:24 --> 00:06:28
And hopefully we'll get to that
very soon in this lecture.
96
00:06:28 --> 00:06:31
OK, so that's a summary of
what we did last time.
97
00:06:31 --> 00:06:36
Any questions?
98
00:06:36 --> 00:06:50
So the next thing now is to
go to the next mechanism.
99
00:06:50 --> 00:06:51
Which is, so far we've
looked at reactions that
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00:06:51 --> 00:06:53
proceed all the way.
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00:06:53 --> 00:06:55
We haven't looked at
equilibrium yet.
102
00:06:55 --> 00:06:58
But we know thermodynamics
is all about equilibrium.
103
00:06:58 --> 00:07:01
So now we're going to connect
kinetics and thermodynamics.
104
00:07:01 --> 00:07:05
We're going to look at a
mechanism which is an
105
00:07:05 --> 00:07:09
equilibrium mechanism.
106
00:07:09 --> 00:07:16
Equilibrium or
reversible reactions.
107
00:07:16 --> 00:07:21
Where you have A goes to B,
with summary constant k1, but
108
00:07:21 --> 00:07:24
you can have the reverse
process B goes back to A with
109
00:07:24 --> 00:07:27
the rate constant k minus one.
110
00:07:27 --> 00:07:35
Or you can write is as A k1, k
minus one to B, where k1 could
111
00:07:35 --> 00:07:39
be k forward and k minus
one could be k backwards.
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00:07:39 --> 00:07:42
Also written as k sub
f, and this is also
113
00:07:42 --> 00:07:46
written as k sub b.
114
00:07:46 --> 00:07:49
So now we have more information
than we had before.
115
00:07:49 --> 00:07:51
Because now we know this is
an equilibrium problem.
116
00:07:51 --> 00:07:54
And we already know
about equilibrium
117
00:07:54 --> 00:07:55
from thermodynamics.
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00:07:55 --> 00:08:01
And we know that at
equilibrium, in addition to all
119
00:08:01 --> 00:08:03
the rate laws that we're going
to write down to solve the
120
00:08:03 --> 00:08:07
problem, we also know that at
equilibrium, the equilibrium
121
00:08:07 --> 00:08:11
constant is related to the
equilibrium concentrations.
122
00:08:11 --> 00:08:14
To the ratio of the
equilibrium concentrations.
123
00:08:14 --> 00:08:19
This is something
additional that we have.
124
00:08:19 --> 00:08:20
So let's write down
everything we know.
125
00:08:20 --> 00:08:24
And our goal is to find out
the time, the dynamics
126
00:08:24 --> 00:08:25
of this problem.
127
00:08:25 --> 00:08:28
Suppose you start at some time
t equals zero, at a certain
128
00:08:28 --> 00:08:31
amount of A0 in your pot, a
certain amount of B0, how
129
00:08:31 --> 00:08:33
do you get to equilibrium.
130
00:08:33 --> 00:08:37
What's the rate, what
does it look like?
131
00:08:37 --> 00:08:41
So the first thing is to
write down the rates.
132
00:08:41 --> 00:08:47
So the forward rate minus
dA/dt, which is the forward
133
00:08:47 --> 00:08:52
rate, R sub forward is
equal to k1 times A.
134
00:08:52 --> 00:08:57
And the backwards rate, minus
dB/dt, which we can write as R
135
00:08:57 --> 00:09:11
sub backwards, that's
k minus one times B.
136
00:09:11 --> 00:09:16
At equilibrium, we
have no dynamics.
137
00:09:16 --> 00:09:19
Or no ensemble dynamics.
138
00:09:19 --> 00:09:21
The pot looks constant.
139
00:09:21 --> 00:09:24
It doesn't look like
anything's going on in there.
140
00:09:24 --> 00:09:26
The concentration of
A doesn't change.
141
00:09:26 --> 00:09:27
The concentration of
B doesn't change.
142
00:09:27 --> 00:09:32
They're constant at the
equilibrium concentration.
143
00:09:32 --> 00:09:36
So that implies that the rate
of formation of B must be
144
00:09:36 --> 00:09:39
equal to the rate of
destruction of B.
145
00:09:39 --> 00:09:43
Or, in other words, that the
rate forward at equilibrium has
146
00:09:43 --> 00:09:46
to be equal the rate backwards.
147
00:09:46 --> 00:09:50
Otherwise, there'd be
changes in concentrations.
148
00:09:50 --> 00:09:54
So if that's true, then that
means that at equilibrium, k1
149
00:09:54 --> 00:10:02
A equilibrium has to be equal
to k minus one B equilibrium.
150
00:10:02 --> 00:10:10
Which means that we can
get this ratio of B to A.
151
00:10:10 --> 00:10:16
And see that this is equal
to k1 over k minus one.
152
00:10:16 --> 00:10:19
So we've just, this is
pretty major here.
153
00:10:19 --> 00:10:23
Looks pretty simple, but
we've taken an equilibrium
154
00:10:23 --> 00:10:24
property here.
155
00:10:24 --> 00:10:27
And related to kinetics.
156
00:10:27 --> 00:10:34
To a rate property.
157
00:10:34 --> 00:10:37
So here we have kinetics.
158
00:10:37 --> 00:10:46
Here we have thermo.
159
00:10:46 --> 00:10:49
And this is going to turn out
to be, this kind of ratio
160
00:10:49 --> 00:10:52
between forward rates and
backwards rates, related to
161
00:10:52 --> 00:10:55
equilibrium constants is going
to be valid, not just for this
162
00:10:55 --> 00:10:58
very simple mechanism we have
here with just two species.
163
00:10:58 --> 00:11:01
But it's going to be valid
for more complicated.
164
00:11:01 --> 00:11:03
And we have two species
reacting to form another
165
00:11:03 --> 00:11:09
species, or bimolecular
stuff, et cetera.
166
00:11:09 --> 00:11:11
OK, so now we've laid
the groundwork, let's
167
00:11:11 --> 00:11:12
look at the dynamics.
168
00:11:12 --> 00:11:14
Let's look at the
time evolution.
169
00:11:14 --> 00:11:19
So that means that we
need to take, well this
170
00:11:19 --> 00:11:19
is not quite right.
171
00:11:19 --> 00:11:22
I shouldn't write minus dA/dt.
172
00:11:22 --> 00:11:26
This is minus dA/dt rate
forward in the absence
173
00:11:26 --> 00:11:38
of reverse process.
174
00:11:38 --> 00:11:42
It says just the destruction of
A and not the creation of A.
175
00:11:42 --> 00:11:48
So if you look at the full
kinetics, the full mechanism,
176
00:11:48 --> 00:11:51
this also, the rate backwards
is minus dB/dt, in the absence
177
00:11:51 --> 00:11:55
of forming B back through
the reverse process.
178
00:11:55 --> 00:11:58
So both of these are in
the absence of their
179
00:11:58 --> 00:11:59
reverse process.
180
00:11:59 --> 00:12:06
The full kinetics, for
the destruction of A.
181
00:12:06 --> 00:12:10
A is the rate backwards, or the
rate forwards, which is k1
182
00:12:10 --> 00:12:12
times A, and then the rate
backwards, which is
183
00:12:12 --> 00:12:14
the creation of A.
184
00:12:14 --> 00:12:15
That's the two rates together.
185
00:12:15 --> 00:12:18
Minus k minus one times B.
186
00:12:18 --> 00:12:20
There's the formation
of A here, this is
187
00:12:20 --> 00:12:21
the destruction of A.
188
00:12:21 --> 00:12:24
This is the full differential
equation for the
189
00:12:24 --> 00:12:24
full mechanism.
190
00:12:24 --> 00:12:26
Not just one part of it.
191
00:12:26 --> 00:12:28
This one here is just for the
first step, and that one here
192
00:12:28 --> 00:12:31
is for the second step.
193
00:12:31 --> 00:12:34
This is what we need to solve.
194
00:12:34 --> 00:12:40
But now we have a lot of help
by what we've already done.
195
00:12:40 --> 00:12:44
So, what we can do is well, the
problem here is that's it looks
196
00:12:44 --> 00:12:47
hard because there's
B included in here.
197
00:12:47 --> 00:12:48
This is not just A,
there's B in here.
198
00:12:48 --> 00:12:50
But there's only two species.
199
00:12:50 --> 00:12:52
And stoichiometry is
going to help us out.
200
00:12:52 --> 00:12:55
Remember last time, or a few
times ago, we said the last
201
00:12:55 --> 00:12:57
species is the easy one.
202
00:12:57 --> 00:13:00
Because it's always related to
the concentration of all the
203
00:13:00 --> 00:13:02
other species by stoichiometry.
204
00:13:02 --> 00:13:04
And the last one here is B.
205
00:13:04 --> 00:13:07
So we know that B
is related to A.
206
00:13:07 --> 00:13:11
B is whatever you
started out with.
207
00:13:11 --> 00:13:14
Plus whatever you
used up for A.
208
00:13:14 --> 00:13:17
That's one way of writing it.
209
00:13:17 --> 00:13:23
So you start with this amount,
and you create this amount.
210
00:13:23 --> 00:13:24
By having A react.
211
00:13:24 --> 00:13:27
This is what you started with,
A, and this is what's left.
212
00:13:27 --> 00:13:30
The difference has
to go into B.
213
00:13:30 --> 00:13:34
So stroke stoichiometry give
you a relationship, which you
214
00:13:34 --> 00:13:41
can use to plug into your
differential equation here.
215
00:13:41 --> 00:13:44
And now, after rearranging your
differential equation, putting
216
00:13:44 --> 00:13:50
B in there, you can rewrite
this as k1 plus k minus one
217
00:13:50 --> 00:14:01
times A minus k minus
one times B0 plus A0.
218
00:14:01 --> 00:14:03
Something which depends
on A, something which
219
00:14:03 --> 00:14:07
is constant here.
220
00:14:07 --> 00:14:08
OK, and what else do we know?
221
00:14:08 --> 00:14:10
We know something
about equilibrium.
222
00:14:10 --> 00:14:14
We know something about when we
reach a state where the rate
223
00:14:14 --> 00:14:16
forward is equal to
the rate backward.
224
00:14:16 --> 00:14:21
When the rate forward is equal
to the rate backwards, the
225
00:14:21 --> 00:14:24
rate of change of A is zero.
226
00:14:24 --> 00:14:25
A is not changing.
227
00:14:25 --> 00:14:27
The concentration of
A is not changing.
228
00:14:27 --> 00:14:38
So at equilibrium, dA/dt
is equal to zero.
229
00:14:38 --> 00:14:43
And just like in this example
here, this is going to turn out
230
00:14:43 --> 00:14:45
to be related to an
approximation called the steady
231
00:14:45 --> 00:14:47
state approximation,
which we're going to see
232
00:14:47 --> 00:14:51
later this morning.
233
00:14:51 --> 00:14:55
Writing at equilibrium that the
change in the concentration of
234
00:14:55 --> 00:14:58
one of the species is equal to
zero, we're going to use that
235
00:14:58 --> 00:15:00
as an approximation also
later, when we get to more
236
00:15:00 --> 00:15:02
complicated mechanisms.
237
00:15:02 --> 00:15:04
This is going to be the
equilibrium approximation.
238
00:15:04 --> 00:15:08
And this is going to be the
steady state approximation.
239
00:15:08 --> 00:15:12
Let's keep going with our,
we're solving this guy here.
240
00:15:12 --> 00:15:14
So at equilibrium, dA/dt
is equal to zero.
241
00:15:14 --> 00:15:27
And then we can replace our
concentrations here with, and
242
00:15:27 --> 00:15:30
so if you write equal to zero
here, then we can write it
243
00:15:30 --> 00:15:32
as equilibrium right here.
244
00:15:32 --> 00:15:34
And we'll be able to solve for
A equilibrium in terms of
245
00:15:34 --> 00:15:40
k1, k minus one, B0, and A0.
246
00:15:40 --> 00:15:44
And then if you do that, you
get A equilibrium is equal to
247
00:15:44 --> 00:15:53
k minus one over k1 plus k
minus one times B0 plus A0.
248
00:15:53 --> 00:16:00
And we're going to need that.
249
00:16:00 --> 00:16:05
Because now we're going to be
able to plug this, so there's
250
00:16:05 --> 00:16:07
A0 plus B0 sitting here.
251
00:16:07 --> 00:16:09
And we're going to replace
A0 plus B0 in terms
252
00:16:09 --> 00:16:12
of A equilibrium.
253
00:16:12 --> 00:16:15
There's a minus sign here.
254
00:16:15 --> 00:16:21
And between this A and this A
equilibrium here, so by
255
00:16:21 --> 00:16:27
rewriting, by plugging in A B0
A equilibrium instead of B0
256
00:16:27 --> 00:16:36
plus A0, we can rewrite that
equation as dA/dt is equal to
257
00:16:36 --> 00:16:45
k1 plus k minus one times
A minus A equilibrium.
258
00:16:45 --> 00:16:48
This is nice because, this
quantity, A minus A
259
00:16:48 --> 00:16:52
equilibrium, describes the
difference between where
260
00:16:52 --> 00:16:55
are you are and where
you want to be.
261
00:16:55 --> 00:16:56
It's a nice variable to have.
262
00:16:56 --> 00:16:59
It's going to change in time
and at infinite time, this
263
00:16:59 --> 00:17:01
is going to go to zero.
264
00:17:01 --> 00:17:03
A is going to go
to equilibrium.
265
00:17:03 --> 00:17:05
A equilibrium is a constant.
266
00:17:05 --> 00:17:07
Now we have a differential
equation that relates
267
00:17:07 --> 00:17:10
A to this difference.
268
00:17:10 --> 00:17:12
But A equilibrium
is a constant.
269
00:17:12 --> 00:17:16
So there's nothing that forbids
us from just writing minus A
270
00:17:16 --> 00:17:20
equilibrium here. d of
a constant dt is zero.
271
00:17:20 --> 00:17:24
So I'm just basically
subtracting zero here.
272
00:17:24 --> 00:17:26
So and I have something of
the form dx/dt is equal
273
00:17:26 --> 00:17:28
to constant times x.
274
00:17:28 --> 00:17:30
And I know how to write that.
275
00:17:30 --> 00:17:32
That looks just like a
first order process.
276
00:17:32 --> 00:17:39
I know the solution is A minus
A equilibrium is equal to my
277
00:17:39 --> 00:17:42
initial A, my initial state.
278
00:17:42 --> 00:17:52
A0 times e to the minus k1
plus k minus one times time.
279
00:17:52 --> 00:17:55
And I've solved the problem.
280
00:17:55 --> 00:18:00
I've described how, as a
function of time, how A,
281
00:18:00 --> 00:18:04
the concentration of
A, gets equilibrium.
282
00:18:04 --> 00:18:06
And if you want to know B,
it's just the same thing.
283
00:18:06 --> 00:18:09
You replace B here and B0 here.
284
00:18:09 --> 00:18:11
You put k minus one here, k1
here, but it's the same thing.
285
00:18:11 --> 00:18:17
286
00:18:17 --> 00:18:34
So if you were to sketch it out
as a function of time, there's,
287
00:18:34 --> 00:18:37
eventually you'll get to the
concentration A equilibrium.
288
00:18:37 --> 00:18:40
If you're above it, you're
going to decay in a first
289
00:18:40 --> 00:18:44
order fashion, with
a rate constant k1.
290
00:18:44 --> 00:18:51
A minus A equilibrium is A0
minus A equilibrium e to the
291
00:18:51 --> 00:18:58
minus k prime times t, where k
prime is k1 plus k minus one.
292
00:18:58 --> 00:19:02
And this is when A is greater
than A0, A equilibrium.
293
00:19:02 --> 00:19:03
A0 is greater than equilibrium.
294
00:19:03 --> 00:19:08
And if you start below, you're
going to come up like this.
295
00:19:08 --> 00:19:11
A0 is less than A equilibrium.
296
00:19:11 --> 00:19:14
And both rates come into
here, in a very simple way.
297
00:19:14 --> 00:19:17
Just the sum of the two.
298
00:19:17 --> 00:19:22
So experimental, if you want to
find out these rates, you can
299
00:19:22 --> 00:19:27
measure their equilibrium
concentrations.
300
00:19:27 --> 00:19:34
Measure k equilibrium by
obtaining the equilibrium
301
00:19:34 --> 00:19:36
concentrations of B and A.
302
00:19:36 --> 00:19:40
And that gives you the ratio
of k1 plus k minus one.
303
00:19:40 --> 00:19:43
And then you can just start
the process with some
304
00:19:43 --> 00:19:44
concentration of A.
305
00:19:44 --> 00:19:46
Which is different
in equilibrium.
306
00:19:46 --> 00:19:55
Then watching it in time,
extract out, observe and
307
00:19:55 --> 00:19:58
measure k1 plus k minus one.
308
00:19:58 --> 00:19:59
In this kinetic equation.
309
00:19:59 --> 00:20:01
At kinetic relation.
310
00:20:01 --> 00:20:06
And then you have two
results, two pieces of data.
311
00:20:06 --> 00:20:09
You've measured k prime, which
gives you the sum of the two.
312
00:20:09 --> 00:20:11
And you've measured K
equilibrium, which gives
313
00:20:11 --> 00:20:12
you the ratio of the two.
314
00:20:12 --> 00:20:16
And you've got your two
rate constants out.
315
00:20:16 --> 00:20:21
OK, any questions for the
equilibrium problem?
316
00:20:21 --> 00:20:22
We're slowly -- yes.
317
00:20:22 --> 00:20:27
STUDENT: [INAUDIBLE]
318
00:20:27 --> 00:20:29
PROFESSOR: There is not yet
a relationship between
319
00:20:29 --> 00:20:31
k1 and k minus one.
320
00:20:31 --> 00:20:38
It depends on the problem.
321
00:20:38 --> 00:20:45
And we're going to study the,
when we get to potential
322
00:20:45 --> 00:20:52
barriers, we look at,
we'll look at this issue.
323
00:20:52 --> 00:20:54
OK, it's a good question.
324
00:20:54 --> 00:20:59
We're going to do that
probably next time.
325
00:20:59 --> 00:21:01
So let's get going, then,
with reversible reactions.
326
00:21:01 --> 00:21:03
So now we have, let's make it a
little bit more complicated.
327
00:21:03 --> 00:21:07
Let's, instead of two species,
let's make it three species.
328
00:21:07 --> 00:21:14
So we have A plus B goes to C,
where the rate constant k1, and
329
00:21:14 --> 00:21:23
then C goes to A plus to B with
a rate constant k minus one
330
00:21:23 --> 00:21:34
that you can rewrite as A plus
B goes to C, k1, k minus one,
331
00:21:34 --> 00:21:36
and then you write down your,
you want to solve this.
332
00:21:36 --> 00:21:38
So the first thing you do is
your write down your
333
00:21:38 --> 00:21:43
differential equations. dA/dt
is equal to, put all the rates
334
00:21:43 --> 00:21:49
in there. k1 A B minus
k minus one times C.
335
00:21:49 --> 00:21:57
And at equilibrium, we
set that equal to zero.
336
00:21:57 --> 00:22:01
So you can write equilibrium
here, equilibrium here,
337
00:22:01 --> 00:22:04
equilibrium here.
338
00:22:04 --> 00:22:11
And when you bring this term
over to the other side, you
339
00:22:11 --> 00:22:17
find, then that at equilibrium
the ratio of the forward rate
340
00:22:17 --> 00:22:21
to the backward rates is equal
to C equilibrium divided
341
00:22:21 --> 00:22:23
by A equilibrium.
342
00:22:23 --> 00:22:24
The equilibrium, which,
you know is the
343
00:22:24 --> 00:22:26
equilibrium constant.
344
00:22:26 --> 00:22:30
So again, for this slightly
more complicated problem, the
345
00:22:30 --> 00:22:32
ratio of the forward and
backward rates are related to
346
00:22:32 --> 00:22:39
the equilibrium constant.
347
00:22:39 --> 00:22:41
So now you want to solve this.
348
00:22:41 --> 00:22:42
Well, we're not
even going to try.
349
00:22:42 --> 00:22:44
Because it's going to
be too complicated.
350
00:22:44 --> 00:22:48
So as soon as you get away from
first order kinetics and go to
351
00:22:48 --> 00:22:55
second order kinetics with
multiple steps, it's a mess.
352
00:22:55 --> 00:22:59
So instead you try to
find approximations.
353
00:22:59 --> 00:23:00
In this case here
there's one we can use.
354
00:23:00 --> 00:23:02
Or a limiting case, at least.
355
00:23:02 --> 00:23:09
One we can use, which is an
obvious one, which is flooding.
356
00:23:09 --> 00:23:11
As a limiting case.
357
00:23:11 --> 00:23:16
If we want to isolate a
small part of the problem,
358
00:23:16 --> 00:23:17
we can use flooding.
359
00:23:17 --> 00:23:22
Because we can overwhelm the
system with either A or B.
360
00:23:22 --> 00:23:27
Take B0 much greater
than A0 and C0.
361
00:23:27 --> 00:23:29
So over the course of the
reaction of the process,
362
00:23:29 --> 00:23:33
the concentration of
B hardly changes.
363
00:23:33 --> 00:23:39
And so when we write our
kinetic equation, our rate law,
364
00:23:39 --> 00:23:46
k1 A B minus k minus one C, we
could put a little naught here,
365
00:23:46 --> 00:23:47
because we know the
concentration of B
366
00:23:47 --> 00:23:49
is not changing.
367
00:23:49 --> 00:23:53
And so now we're left with
the problem we had before.
368
00:23:53 --> 00:23:56
Which was a reversible
first order process which
369
00:23:56 --> 00:24:00
we have just solved.
370
00:24:00 --> 00:24:04
And you go through the
experiment and you extract out
371
00:24:04 --> 00:24:08
k1 times B0 and k minus one.
372
00:24:08 --> 00:24:10
You do the experiment again
with a different concentration
373
00:24:10 --> 00:24:12
of B0 to start out with.
374
00:24:12 --> 00:24:14
And you can extract out k1.
375
00:24:14 --> 00:24:18
So this is a process of
getting the rates by using a
376
00:24:18 --> 00:24:21
simple approximation here.
377
00:24:21 --> 00:24:26
Simple limiting case.
378
00:24:26 --> 00:24:30
Any questions?
379
00:24:30 --> 00:24:32
So we've just finished
putting all the building
380
00:24:32 --> 00:24:35
blocks together now.
381
00:24:35 --> 00:24:43
Let me remind you what
these building blocks are.
382
00:24:43 --> 00:24:45
We have a bunch of
approximations under our belt.
383
00:24:45 --> 00:24:55
Flooding, we've looked at rates
being much faster than others.
384
00:24:55 --> 00:25:01
And we have three simple
mechanisms, we've looked
385
00:25:01 --> 00:25:03
at parallel reactions.
386
00:25:03 --> 00:25:06
We looked at series reactions.
387
00:25:06 --> 00:25:13
And we've looked at
reversible reactions.
388
00:25:13 --> 00:25:15
So a complicated mechanism
will basically put these
389
00:25:15 --> 00:25:19
three building blocks
together in a series.
390
00:25:19 --> 00:25:22
And they'll all be happening
at the same time somehow.
391
00:25:22 --> 00:25:27
And obviously, since it was,
since we threw up our hands,
392
00:25:27 --> 00:25:31
just with a simple reversible
process like here, it's clear
393
00:25:31 --> 00:25:35
that we're going to throw
up our hands a lot.
394
00:25:35 --> 00:25:37
When we write down
these mechanisms.
395
00:25:37 --> 00:25:38
So we're going to
need approximations.
396
00:25:38 --> 00:25:40
We're going to need something
that we'll automatically
397
00:25:40 --> 00:25:43
go to, and say this is
going to be too hard.
398
00:25:43 --> 00:25:45
I'm not even going to try.
399
00:25:45 --> 00:25:50
Let's do an approximation.
400
00:25:50 --> 00:25:52
And two approximations that
we're going to talk about are
401
00:25:52 --> 00:25:55
ones I already mentioned
at the beginning.
402
00:25:55 --> 00:26:01
Which is the steady state
approximation, where one rate,
403
00:26:01 --> 00:26:07
the rate to go into the
intermediate is very slow, or
404
00:26:07 --> 00:26:10
the rate to get out of the
intermediate is very fast.
405
00:26:10 --> 00:26:14
And the equilibrium
approximation where the system
406
00:26:14 --> 00:26:19
sets up a very fast equilibrium
and you're allowed to use
407
00:26:19 --> 00:26:22
thermodynamics to help you out
in solving the problem.
408
00:26:22 --> 00:26:23
These are the two
approximations that we'll
409
00:26:23 --> 00:26:27
use when we put these
things together.
410
00:26:27 --> 00:26:30
OK, so let's look at the
first, simple, more
411
00:26:30 --> 00:26:32
complicated mechanism.
412
00:26:32 --> 00:26:34
To see where these
approximations come in.
413
00:26:34 --> 00:26:36
So the first one is
going to be a series.
414
00:26:36 --> 00:26:38
We're going to start putting
these things together.
415
00:26:38 --> 00:26:40
First one is a
series reversible.
416
00:26:40 --> 00:26:41
Those two together.
417
00:26:41 --> 00:26:46
Series reversible.
418
00:26:46 --> 00:26:50
So we have first, a
reversible process.
419
00:26:50 --> 00:26:54
Everything is first order
here. k minus one goes to B.
420
00:26:54 --> 00:26:57
And then B goes to C
with some rate constant
421
00:26:57 --> 00:27:01
k2, all first order.
422
00:27:01 --> 00:27:03
If we were to turn the crank,
we'd say, oh, I've got to write
423
00:27:03 --> 00:27:04
down all my rate laws here.
424
00:27:04 --> 00:27:09
Minus dA/dt is all the ways
that I destroy and create B, so
425
00:27:09 --> 00:27:18
there's a k1 times A minus k
minus one times B. dB/dt, all
426
00:27:18 --> 00:27:22
the ways that create and
destroy B is that you create it
427
00:27:22 --> 00:27:25
through destruction of A.
428
00:27:25 --> 00:27:29
I destroy it through the
backwards rate to make A.
429
00:27:29 --> 00:27:32
And I destroy it by
making the product, C.
430
00:27:32 --> 00:27:33
There's the intermediate.
431
00:27:33 --> 00:27:35
And then for C, that's
pretty simple.
432
00:27:35 --> 00:27:38
There's only one channel into
C, and that's the destruction
433
00:27:38 --> 00:27:41
of B to form C, k2 B.
434
00:27:41 --> 00:27:42
All the rate laws.
435
00:27:42 --> 00:27:45
I write everything I know here.
436
00:27:45 --> 00:27:48
Couple of differential
equations.
437
00:27:48 --> 00:27:52
We know it's going to
be hard to solve.
438
00:27:52 --> 00:27:58
So, let's remind ourselves
of what I just erased.
439
00:27:58 --> 00:28:08
Which was the case where the
intermediate concentration
440
00:28:08 --> 00:28:14
was always very small and
didn't change very fast.
441
00:28:14 --> 00:28:20
And the process was
dominated by k1 here.
442
00:28:20 --> 00:28:22
That was the rate
limiting step.
443
00:28:22 --> 00:28:29
Remember that our first
example this morning.
444
00:28:29 --> 00:28:34
So this is where the
concentration of B is
445
00:28:34 --> 00:28:36
roughly constant over time.
446
00:28:36 --> 00:28:41
It's small.
447
00:28:41 --> 00:28:45
And over any small time period,
it's roughly constant.
448
00:28:45 --> 00:28:52
Which means that dB/dt is
roughly equal to zero.
449
00:28:52 --> 00:28:53
If you look at a long term,
obviously it's going to
450
00:28:53 --> 00:28:56
change as you go from
this point to that point.
451
00:28:56 --> 00:28:58
In fact, there's a relationship
between A which is
452
00:28:58 --> 00:28:59
changing in time and B.
453
00:28:59 --> 00:29:02
But A is also changing
in time very slowly.
454
00:29:02 --> 00:29:05
Because the rate
k1 is very small.
455
00:29:05 --> 00:29:08
So dB/dt, which is related to
this rate k1, is going to be
456
00:29:08 --> 00:29:10
very slowly changing in time.
457
00:29:10 --> 00:29:16
We can approximate it at zero.
458
00:29:16 --> 00:29:19
What are the ways in terms of
matching these rate constants
459
00:29:19 --> 00:29:22
that we can get to
the approximation.
460
00:29:22 --> 00:29:28
To a diagram that
looks like that?
461
00:29:28 --> 00:29:34
But we don't want B to pile up.
462
00:29:34 --> 00:29:38
That means we have to have the
rates out of the intermediate
463
00:29:38 --> 00:29:41
to be much faster, at least one
of the rates out of the
464
00:29:41 --> 00:29:43
intermediate to be much faster
than the rate that creates
465
00:29:43 --> 00:29:45
the intermediate.
466
00:29:45 --> 00:29:50
So the different ways of
creating that approximation are
467
00:29:50 --> 00:29:53
if, and the length of the
arrows now is going to be
468
00:29:53 --> 00:29:55
proportional to the rate.
469
00:29:55 --> 00:29:58
We want it to be very
hard to make B.
470
00:29:58 --> 00:30:03
And as soon as we make B,
we want it to go away.
471
00:30:03 --> 00:30:05
So one of the ways to do
that is to have the reverse
472
00:30:05 --> 00:30:09
rate much faster than
the initial rate.
473
00:30:09 --> 00:30:12
And we could have a slow
rate into C, that's fine.
474
00:30:12 --> 00:30:16
We could have, again, a
very slow rate into B.
475
00:30:16 --> 00:30:20
We could have a slow reverse
rate and a fast rate into C.
476
00:30:20 --> 00:30:21
Perfectly fine.
477
00:30:21 --> 00:30:22
We could have both.
478
00:30:22 --> 00:30:28
Fast rate out of B through
A, or out of B through C.
479
00:30:28 --> 00:30:31
As long as this first rate into
the B is small compared to one
480
00:30:31 --> 00:30:33
of those two rates, we're
never going to pile up B.
481
00:30:33 --> 00:30:37
It's going to go away
as soon as we make it.
482
00:30:37 --> 00:30:42
So in all these cases,
k2 plus k minus one is
483
00:30:42 --> 00:30:46
much bigger than k1.
484
00:30:46 --> 00:30:49
When we have that situation,
then we can make the
485
00:30:49 --> 00:30:51
approximation up here.
486
00:30:51 --> 00:30:54
That the rate of change
in B is very small.
487
00:30:54 --> 00:30:56
Almost zero, which means
it's basically a constant.
488
00:30:56 --> 00:31:01
And instead of writing B in
this case here, we're going to
489
00:31:01 --> 00:31:04
write it as B steady state.
490
00:31:04 --> 00:31:08
Which is basically a constant
in terms of solving the
491
00:31:08 --> 00:31:10
differential equations.
492
00:31:10 --> 00:31:14
And that's going to make
our life much easier.
493
00:31:14 --> 00:31:19
Because instead of having
to solve these coupled
494
00:31:19 --> 00:31:22
differential equations, we're
just going to have to solve
495
00:31:22 --> 00:31:24
coupled algebraic equations.
496
00:31:24 --> 00:31:29
Which is really messy,
but less hard.
497
00:31:29 --> 00:31:34
If you're a bean-counter
than this is heaven.
498
00:31:34 --> 00:31:37
So now we're going to put
steady state, this is
499
00:31:37 --> 00:31:38
going to be a constant.
500
00:31:38 --> 00:31:40
And this is going to
be equal to zero.
501
00:31:40 --> 00:31:42
And then here this is going
to be steady state here.
502
00:31:42 --> 00:31:44
This is going to be
steady state here.
503
00:31:44 --> 00:31:47
And the first thing to do now
is that we've eliminated
504
00:31:47 --> 00:31:48
this differential equation.
505
00:31:48 --> 00:31:54
We now have an algebraic
equation where we can
506
00:31:54 --> 00:32:01
solve for B steady state.
507
00:32:01 --> 00:32:07
So if you recognize that you're
in this situation here,
508
00:32:07 --> 00:32:08
forget about trying to solve.
509
00:32:08 --> 00:32:13
Immediately go to the process
of putting in a constant for B.
510
00:32:13 --> 00:32:16
Setting dB/dt for the
intermediate equal to zero.
511
00:32:16 --> 00:32:22
And then starting to turn
the crank on the algebra.
512
00:32:22 --> 00:32:26
So let me go ahead and
turn the crank here.
513
00:32:26 --> 00:32:30
Go through the steps, which
is basically typical
514
00:32:30 --> 00:32:37
of these problems.
515
00:32:37 --> 00:32:38
So you turn the crank.
516
00:32:38 --> 00:32:42
You solve for the steady
state concentration.
517
00:32:42 --> 00:32:50
You get, in terms of A,
over k minus one, plus k2.
518
00:32:50 --> 00:32:55
Then you plug this
back in here.
519
00:32:55 --> 00:33:01
And you get a new differential
equation, minus dA/dt is equal
520
00:33:01 --> 00:33:06
to k1 times A minus k minus one
times B steady state, so
521
00:33:06 --> 00:33:09
we plug this in here.
522
00:33:09 --> 00:33:12
And you get, so there's
A sitting here.
523
00:33:12 --> 00:33:13
A sitting here.
524
00:33:13 --> 00:33:16
It's going to be of the form,
effective rate times A.
525
00:33:16 --> 00:33:18
It's going to be basically
a first order form.
526
00:33:18 --> 00:33:21
Which is what we'd expect
from sketching the
527
00:33:21 --> 00:33:24
diagram just like that.
528
00:33:24 --> 00:33:30
So you get minus dA/dt is an
effective rate. k1 times
529
00:33:30 --> 00:33:34
k2 over k1 plus k2.
530
00:33:34 --> 00:33:36
Times A.
531
00:33:36 --> 00:33:39
This is k prime.
532
00:33:39 --> 00:33:41
So if you were to do an
experiment under these
533
00:33:41 --> 00:33:47
conditions, you'd find that A
behaves, it's basically a
534
00:33:47 --> 00:33:48
pseudo first order problem.
535
00:33:48 --> 00:33:53
With a funny rate that contains
all the elementary rate
536
00:33:53 --> 00:33:55
constants as part of it.
537
00:33:55 --> 00:33:59
And when you do the same
thing for C, dC/dt, you
538
00:33:59 --> 00:34:01
find that depends on A.
539
00:34:01 --> 00:34:07
It's k1, you plug these steady
states in here. k1 k2 divided
540
00:34:07 --> 00:34:11
by k1 plus k2 times A.
541
00:34:11 --> 00:34:13
This is k prime again.
542
00:34:13 --> 00:34:16
So the problem, and
it's first order.
543
00:34:16 --> 00:34:18
So the problem looks
like effectively you're
544
00:34:18 --> 00:34:20
going from A to C.
545
00:34:20 --> 00:34:21
Forget about the intermediate.
546
00:34:21 --> 00:34:24
With an effective rate k prime.
547
00:34:24 --> 00:34:30
Where k prime contains
these rates.
548
00:34:30 --> 00:34:32
So this is steady
state approximation.
549
00:34:32 --> 00:34:36
It's the prototypical problem.
550
00:34:36 --> 00:34:38
Questions on steady state,
before we go to the
551
00:34:38 --> 00:34:44
next approximation.
552
00:34:44 --> 00:34:44
Alright.
553
00:34:44 --> 00:34:53
So now, as promised, the next
approximation is that where are
554
00:34:53 --> 00:34:54
set up the fast equilibrium.
555
00:34:54 --> 00:34:59
And you can use thermo
to help you out.
556
00:34:59 --> 00:35:11
Equilibrium approximation
A goes to B.
557
00:35:11 --> 00:35:13
This is a fast process.
558
00:35:13 --> 00:35:21
And then k2 out of B
is a slow process.
559
00:35:21 --> 00:35:24
Little arrow here, and
two big arrows here.
560
00:35:24 --> 00:35:27
You put your A in your flask.
561
00:35:27 --> 00:35:30
Immediately you set
up the equilibrium.
562
00:35:30 --> 00:35:32
And you slowly
dribble out of that.
563
00:35:32 --> 00:35:40
If you want to do it as a
function of buckets and, so you
564
00:35:40 --> 00:35:42
have a big pipe connecting
two buckets that are just
565
00:35:42 --> 00:35:43
offset from each other.
566
00:35:43 --> 00:35:48
So there's, in this case
here B is favored over A.
567
00:35:48 --> 00:35:51
Because it's a little bit
lower than this guy here.
568
00:35:51 --> 00:35:53
And then there's a little
bucket, there's a little
569
00:35:53 --> 00:35:58
tube, comes out of here.
570
00:35:58 --> 00:36:02
So there's a little
dribble into C over time.
571
00:36:02 --> 00:36:05
First thing that happens is
you set up your equilibrium.
572
00:36:05 --> 00:36:08
And then you slowly
extract stuff through
573
00:36:08 --> 00:36:10
a little tube into C.
574
00:36:10 --> 00:36:14
So what do you expect
this to look like?
575
00:36:14 --> 00:36:22
Well, you expect A to
really slowly come out.
576
00:36:22 --> 00:36:26
Because the rate limiting step
is the rate from B to C.
577
00:36:26 --> 00:36:30
So k2 is going to be the way
that A is going to come out.
578
00:36:30 --> 00:36:32
You expect it to
slowly go away.
579
00:36:32 --> 00:36:35
You expect B to get created
very fast, to some
580
00:36:35 --> 00:36:37
equilibrium amount.
581
00:36:37 --> 00:36:42
And then also to follow
the same rate k2, slow
582
00:36:42 --> 00:36:44
k2, to disappear.
583
00:36:44 --> 00:36:48
And you expect C to come up.
584
00:36:48 --> 00:36:50
You're basically in a
first order process
585
00:36:50 --> 00:36:53
to saturation of A0.
586
00:36:53 --> 00:36:56
So when you expect the dominant
rate to be k2, and the fast
587
00:36:56 --> 00:37:00
dynamics to happen at very,
very early times, and then
588
00:37:00 --> 00:37:05
everything to follow
first order kinetics.
589
00:37:05 --> 00:37:07
So now let's do the math and
make sure that it agrees with
590
00:37:07 --> 00:37:11
what we've just assumed that
it's going to look like.
591
00:37:11 --> 00:37:14
So equilibrium is happening.
592
00:37:14 --> 00:37:20
You can assume that at any time
after the initial very fast
593
00:37:20 --> 00:37:21
process of getting equilibrium.
594
00:37:21 --> 00:37:25
So after some initial time, the
ratio of B over A is always
595
00:37:25 --> 00:37:27
going to be a constant.
596
00:37:27 --> 00:37:30
The dribble out of B here is
not going to be fast enough
597
00:37:30 --> 00:37:33
to change that ratio.
598
00:37:33 --> 00:37:39
As soon as you get a little bit
of B out here, immediately the
599
00:37:39 --> 00:37:44
ratios, the amounts rearrange
to keep the ratio constant.
600
00:37:44 --> 00:37:48
So now when I look at the rate
of the reaction, the rate of
601
00:37:48 --> 00:37:53
formation of C, k2 times B,
well, in terms of A, it's
602
00:37:53 --> 00:37:57
k2 times K equilibrium.
603
00:37:57 --> 00:38:00
Times A.
604
00:38:00 --> 00:38:06
Which is k2 times k1 over
k minus one times A.
605
00:38:06 --> 00:38:11
And you can immediately see
that C is behaving, or the
606
00:38:11 --> 00:38:15
reaction is behaving, like
a first order process.
607
00:38:15 --> 00:38:18
With an effective rate which
contains all three rates
608
00:38:18 --> 00:38:23
in this ratio here.
609
00:38:23 --> 00:38:25
So it looks like A goes
to C, with some k prime
610
00:38:25 --> 00:38:28
where this is k prime.
611
00:38:28 --> 00:38:33
It looks like A goes to C
with an effective rate
612
00:38:33 --> 00:38:36
constant k prime.
613
00:38:36 --> 00:38:36
OK.
614
00:38:36 --> 00:38:39
Questions?
615
00:38:39 --> 00:38:40
We're going to do
some examples.
616
00:38:40 --> 00:38:43
And then we're going to do
chain reactions next time.
617
00:38:43 --> 00:38:50
We're one lecture behind.
618
00:38:50 --> 00:38:50
Alright.
619
00:38:50 --> 00:38:53
Let's do some reactions.
620
00:38:53 --> 00:39:01
Some examples.
621
00:39:01 --> 00:39:04
I'm going to skip the
first example, which is
622
00:39:04 --> 00:39:08
the example with the
chaperone, in the notes.
623
00:39:08 --> 00:39:12
And I'm going to go directly
to the gas decomposition.
624
00:39:12 --> 00:39:15
The two examples are basically
very similar in their use of
625
00:39:15 --> 00:39:18
the steady state approximation.
626
00:39:18 --> 00:39:23
So I'll let you read
over the first example.
627
00:39:23 --> 00:39:26
So this example here is going
to give us the Lindemann
628
00:39:26 --> 00:39:29
mechanism, for which Mr.
Lindemann got a Nobel Prize
629
00:39:29 --> 00:39:33
many many, years ago.
630
00:39:33 --> 00:39:35
Basically, it's looking
at decomposition of a
631
00:39:35 --> 00:39:44
gas phase molecule.
632
00:39:44 --> 00:39:52
Where you have a molecule, A,
that breaks down into products.
633
00:39:52 --> 00:39:58
And the observation before Mr.
Lindemann got around, was
634
00:39:58 --> 00:40:06
that it looked like a
first order process.
635
00:40:06 --> 00:40:10
It looked like A is just
falling apart on its own
636
00:40:10 --> 00:40:13
into these products.
637
00:40:13 --> 00:40:16
Mr. Lindemann got around and
said, well, why would A
638
00:40:16 --> 00:40:18
just want to fall apart.
639
00:40:18 --> 00:40:21
There's something a little
bit odd about that.
640
00:40:21 --> 00:40:24
Stable molecule, why would
it want to fall apart.
641
00:40:24 --> 00:40:29
And so he hypothesized
a mechanism.
642
00:40:29 --> 00:40:33
And it went like this.
643
00:40:33 --> 00:40:36
That A actually collides with
another molecule, which
644
00:40:36 --> 00:40:39
could be a bystander.
645
00:40:39 --> 00:40:42
Could be a chaperone molecule
that just happens to be there.
646
00:40:42 --> 00:40:43
Or it could be another
molecule of A.
647
00:40:43 --> 00:40:46
There's a collision.
648
00:40:46 --> 00:40:50
There's a collision that
creates a vibrationally
649
00:40:50 --> 00:40:52
excited version of A.
650
00:40:52 --> 00:40:53
Of A sitting there.
651
00:40:53 --> 00:40:53
It collides.
652
00:40:53 --> 00:40:56
It suddenly starts to be
really vibrationally excited.
653
00:40:56 --> 00:40:58
Bonds vibrate all
over the place.
654
00:40:58 --> 00:41:01
Atoms wiggle.
655
00:41:01 --> 00:41:04
There's the collision
partner that goes away.
656
00:41:04 --> 00:41:08
So kinetic energy is
transferred from M and A to
657
00:41:08 --> 00:41:11
the vibrational loads of A.
658
00:41:11 --> 00:41:15
In a process which is
reversible. k1, k minus one,
659
00:41:15 --> 00:41:18
because this excited A, this
vibrationally excited A,
660
00:41:18 --> 00:41:21
could also collide with a
molecule and cool down.
661
00:41:21 --> 00:41:25
The energy in the vibrations
could be transferred back to
662
00:41:25 --> 00:41:29
kinetic energy for
a reverse process.
663
00:41:29 --> 00:41:34
But if you wait long enough, or
if this vibrationally excited
664
00:41:34 --> 00:41:39
A, with these atoms widely
moving around, that has more of
665
00:41:39 --> 00:41:42
a chance of falling apart
than this guy here.
666
00:41:42 --> 00:41:49
So A star could fall
apart into products.
667
00:41:49 --> 00:41:51
With a rate k2.
668
00:41:51 --> 00:41:52
So he said, this is actually
an important step.
669
00:41:52 --> 00:41:54
It's not that A just
suddenly falls apart.
670
00:41:54 --> 00:41:58
But this has to happen.
671
00:41:58 --> 00:42:01
So all the observations that
supported a first order
672
00:42:01 --> 00:42:05
process, which basically only
supported one elementary
673
00:42:05 --> 00:42:13
reactions, might not be right.
674
00:42:13 --> 00:42:16
So let's assume his mechanism,
and let's look at what we need
675
00:42:16 --> 00:42:18
to do as an approximation.
676
00:42:18 --> 00:42:21
And where that leads us
in terms of trying to
677
00:42:21 --> 00:42:28
experimentally confirm that
this mechanism is plausible.
678
00:42:28 --> 00:42:31
Not prove it, but just to make
sure that it's consistent with,
679
00:42:31 --> 00:42:43
the data's consistent
with the hypothesis.
680
00:42:43 --> 00:42:50
So let's compare
our rates here.
681
00:42:50 --> 00:42:51
So we have a collision.
682
00:42:51 --> 00:42:55
A and M collide, and there's a
certain probability on the
683
00:42:55 --> 00:42:59
collision that kinetic energy
will be transferred to
684
00:42:59 --> 00:43:00
vibrational energy.
685
00:43:00 --> 00:43:02
That turns out to be a
moderate probability.
686
00:43:02 --> 00:43:09
So k1 is reasonably fast.
687
00:43:09 --> 00:43:11
The reverse process, you have
something that's vibrationally
688
00:43:11 --> 00:43:14
excited colliding
with a molecule.
689
00:43:14 --> 00:43:18
There's a probability that that
excitation is going to get
690
00:43:18 --> 00:43:20
turned back into
kinetic energy.
691
00:43:20 --> 00:43:22
And then these two molecules
will fly apart with
692
00:43:22 --> 00:43:23
high velocity.
693
00:43:23 --> 00:43:25
And that tends to be much
more highly probable
694
00:43:25 --> 00:43:28
then the first process.
695
00:43:28 --> 00:43:30
So this is faster.
696
00:43:30 --> 00:43:32
Then the excited molecule's
waiting around.
697
00:43:32 --> 00:43:34
And there's a probability
that there's it's going
698
00:43:34 --> 00:43:35
to just fall apart.
699
00:43:35 --> 00:43:38
And that turns out
to be really slow.
700
00:43:38 --> 00:43:49
So we're going to assume that
the last step is a slow step.
701
00:43:49 --> 00:43:50
So what does it look like?
702
00:43:50 --> 00:43:55
Well, we have the
creation is fast.
703
00:43:55 --> 00:43:57
So the intermediate is fast.
704
00:43:57 --> 00:44:02
But the destruction through the
reverse process is much faster.
705
00:44:02 --> 00:44:05
We have at least one step, out
of the intermediate, which
706
00:44:05 --> 00:44:07
is faster than the step
into the intermediate.
707
00:44:07 --> 00:44:08
That's all we care about.
708
00:44:08 --> 00:44:11
It doesn't really matter that
this is slow in terms of using
709
00:44:11 --> 00:44:12
the steady state approximation.
710
00:44:12 --> 00:44:17
We just want the step getting
out of the intermediate, a
711
00:44:17 --> 00:44:20
backwards step in this case
here, to be faster than the
712
00:44:20 --> 00:44:22
state into the intermediate.
713
00:44:22 --> 00:44:26
So this allows this, this fact
that this is faster, that
714
00:44:26 --> 00:44:30
allows us to use a steady
state approximation.
715
00:44:30 --> 00:44:33
So now we can go ahead
and solve the problem.
716
00:44:33 --> 00:44:38
So we write down all
the rates minus dA/dt.
717
00:44:38 --> 00:44:44
It's k1 A.
718
00:44:44 --> 00:44:46
Let's start with the
rate of the action.
719
00:44:46 --> 00:44:47
Let's take a product.
720
00:44:47 --> 00:44:48
Let's take C.
721
00:44:48 --> 00:44:50
Let's say this goes
to product C here.
722
00:44:50 --> 00:44:55
So dC/dt is equal to k2
times A, that's the
723
00:44:55 --> 00:45:00
rate of the reaction.
724
00:45:00 --> 00:45:03
And we're going to look at
this rate of reaction.
725
00:45:03 --> 00:45:04
And see how it changes.
726
00:45:04 --> 00:45:08
What it looks like under
limiting conditions.
727
00:45:08 --> 00:45:13
The rate of reaction.
728
00:45:13 --> 00:45:15
Let's look at the intermediate.
729
00:45:15 --> 00:45:16
Because this is what we're
going to solve first, because
730
00:45:16 --> 00:45:19
this is going to turn out to
be an algebraic equation.
731
00:45:19 --> 00:45:20
Because we're going
to apply the steady
732
00:45:20 --> 00:45:23
state approximations.
733
00:45:23 --> 00:45:28
Intermediate d A star / dt
is, we can create it through
734
00:45:28 --> 00:45:30
the forward process.
735
00:45:30 --> 00:45:33
We destroy it through
the backwards process.
736
00:45:33 --> 00:45:35
A star times M.
737
00:45:35 --> 00:45:37
And we destroy it through
the final process.
738
00:45:37 --> 00:45:40
A star.
739
00:45:40 --> 00:45:43
We apply the steady state
approximation, steady
740
00:45:43 --> 00:45:46
state, steady state.
741
00:45:46 --> 00:45:54
We solve for A steady state
algebraically. k1 A M over
742
00:45:54 --> 00:46:00
M times k minus 1 plus k2.
743
00:46:00 --> 00:46:04
And then we plug that back into
the rate of the reaction.
744
00:46:04 --> 00:46:05
The rate of appearance of C.
745
00:46:05 --> 00:46:07
Which is what we're
measuring experimentally.
746
00:46:07 --> 00:46:08
We're looking at the products.
747
00:46:08 --> 00:46:10
Measuring the appearance
of products.
748
00:46:10 --> 00:46:11
So they're destruction.
749
00:46:11 --> 00:46:21
This is also equal to
the destruction of A.
750
00:46:21 --> 00:46:31
And so we plug this into the,
we write our rate of reaction.
751
00:46:31 --> 00:46:34
Which we measure,
experimentally.
752
00:46:34 --> 00:46:37
We find this is k1 times k2.
753
00:46:37 --> 00:46:44
Times, we plug in for, and
here, what we do here, we solve
754
00:46:44 --> 00:46:48
A in terms of A steady state.
755
00:46:48 --> 00:46:53
A is equal to something
A steady state.
756
00:46:53 --> 00:47:04
Well, let's see, A is equal to
M k minus one plus k2 times,
757
00:47:04 --> 00:47:06
well this is actually, I have
this wrong up here,
758
00:47:06 --> 00:47:11
this is d A star.
759
00:47:11 --> 00:47:15
So we don't need
to do this here.
760
00:47:15 --> 00:47:18
The appearance of C
is, A star here.
761
00:47:18 --> 00:47:24
So we plug in A
steady state here.
762
00:47:24 --> 00:47:29
A star steady state, and this
is A star steady state.
763
00:47:29 --> 00:47:36
The intermediate is what we're
solving a steady state for.
764
00:47:36 --> 00:47:43
So this is k1 k2 A M over
M k minus one plus k2.
765
00:47:43 --> 00:47:48
We just have the extra k2 that
appears when we multiply
766
00:47:48 --> 00:47:52
A steady state star
with the k2 there.
767
00:47:52 --> 00:47:55
And then we can look at
the limiting cases.
768
00:47:55 --> 00:47:58
There are two limiting
cases, which we can tell by
769
00:47:58 --> 00:47:59
looking at the denominator.
770
00:47:59 --> 00:48:01
If we have one term
bigger than the other.
771
00:48:01 --> 00:48:03
Only two choices here.
772
00:48:03 --> 00:48:07
So the first case is
where M k minus one is
773
00:48:07 --> 00:48:10
much bigger than k2.
774
00:48:10 --> 00:48:11
Experimentally, what
does that mean?
775
00:48:11 --> 00:48:14
It means that this in
the gas phase now.
776
00:48:14 --> 00:48:17
It means that regardless of
what these rates are here,
777
00:48:17 --> 00:48:21
we've managed to make the
concentration of M high enough
778
00:48:21 --> 00:48:23
so that this becomes true.
779
00:48:23 --> 00:48:26
There's always an M that's
going to be high enough so that
780
00:48:26 --> 00:48:28
the multiplication of these
two is going to be
781
00:48:28 --> 00:48:29
bigger than that.
782
00:48:29 --> 00:48:29
What does that mean?
783
00:48:29 --> 00:48:33
It means that the
pressure is high.
784
00:48:33 --> 00:48:34
Concentration of M is high.
785
00:48:34 --> 00:48:40
This is a high pressure case.
786
00:48:40 --> 00:48:45
And then the other one is M k
minus one is much less than k2.
787
00:48:45 --> 00:48:47
M is very small.
788
00:48:47 --> 00:48:51
Low concentration,
low pressure.
789
00:48:51 --> 00:48:54
The partial pressure
of M is very low.
790
00:48:54 --> 00:48:55
This could be A.
791
00:48:55 --> 00:48:57
M could be A, or it could
be some other molecule
792
00:48:57 --> 00:49:00
that's in the mix.
793
00:49:00 --> 00:49:02
And by high pressure we mean
something on the order
794
00:49:02 --> 00:49:04
of one bar, let's say.
795
00:49:04 --> 00:49:07
That by low pressure we
mean something around
796
00:49:07 --> 00:49:12
10 to the minus 4 bar.
797
00:49:12 --> 00:49:16
So now we can go and put these
limiting cases in our equation
798
00:49:16 --> 00:49:19
here. k minus one is much
bigger than k2, so we
799
00:49:19 --> 00:49:21
can ignore k2 here.
800
00:49:21 --> 00:49:25
And then we have something that
looks like the M's disappear.
801
00:49:25 --> 00:49:30
We have something that looks
like k1 k2 over k minus
802
00:49:30 --> 00:49:37
one times A is the
rate of reaction.
803
00:49:37 --> 00:49:38
First order.
804
00:49:38 --> 00:49:39
Looks like it's
first order in A.
805
00:49:39 --> 00:49:41
This is what people
had observed.
806
00:49:41 --> 00:49:42
Great.
807
00:49:42 --> 00:49:45
The problem is that they didn't
know to go to low pressure.
808
00:49:45 --> 00:49:48
They were doing all
their experiments at
809
00:49:48 --> 00:49:49
atmospheric pressure.
810
00:49:49 --> 00:49:50
They were getting a
first order rate.
811
00:49:50 --> 00:49:52
They thought they had
solved the problem.
812
00:49:52 --> 00:49:54
It's just A falling
apart by itself.
813
00:49:54 --> 00:49:57
Until you go to low pressure,
where now this term
814
00:49:57 --> 00:49:59
dominates the denominator.
815
00:49:59 --> 00:50:03
Or rather, this one
dominates the denominator.
816
00:50:03 --> 00:50:12
And now you have something
of the form k1 A times M.
817
00:50:12 --> 00:50:14
This dominates, this
term goes away.
818
00:50:14 --> 00:50:19
The k2 goes away and you have
k1 times A times M left over
819
00:50:19 --> 00:50:25
as the rate of the reaction.
820
00:50:25 --> 00:50:25
Second order.
821
00:50:25 --> 00:50:27
There are two species here.
822
00:50:27 --> 00:50:27
A times M.
823
00:50:27 --> 00:50:30
M could be A or it could
be some chaperone.
824
00:50:30 --> 00:50:31
A squared.
825
00:50:31 --> 00:50:33
Second order process.
826
00:50:33 --> 00:50:34
So if you're at low
pressure, you see
827
00:50:34 --> 00:50:35
something at second order.
828
00:50:35 --> 00:50:37
You've discovered something
about the mechanisms
829
00:50:37 --> 00:50:39
you didn't expect.
830
00:50:39 --> 00:50:42
And then you get a Nobel Prize.
831
00:50:42 --> 00:50:44
Alright, next time we'll
do chain reactions.