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PROFESSOR: So, any
questions from last time?
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00:00:21 --> 00:00:27
We started doing
reaction mechanisms.
11
00:00:27 --> 00:00:33
We did first order
parallel reactions.
12
00:00:33 --> 00:00:38
So one of the things that we'll
be stressing is that the
13
00:00:38 --> 00:00:43
mechanisms that we write down
are ways to try to understand a
14
00:00:43 --> 00:00:46
complicated chemical process.
15
00:00:46 --> 00:00:53
And you take data, you can
find intermediates sometimes.
16
00:00:53 --> 00:00:56
And you get rate
laws from the data.
17
00:00:56 --> 00:00:58
You infer rate laws.
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00:00:58 --> 00:01:02
And from these rate laws,
you think up of a reaction
19
00:01:02 --> 00:01:09
mechanism and you make sure
that what you measure is
20
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consistent with your mechanism
that you make up to explain the
21
00:01:13 --> 00:01:15
complicated chemical process.
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00:01:15 --> 00:01:18
And one really important thing
to remember is that when you
23
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come up with a mechanism and
you've got a bunch of data
24
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that supports it, it's great.
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00:01:25 --> 00:01:29
But it doesn't mean that
you've solve the problem.
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00:01:29 --> 00:01:33
It doesn't mean that you've
proven that the way that the
27
00:01:33 --> 00:01:36
chemistry proceeds is really
the way it proceeds.
28
00:01:36 --> 00:01:41
Because there's always the
possibility that there's an
29
00:01:41 --> 00:01:43
intermediate somewhere in
there that you haven't
30
00:01:43 --> 00:01:45
been able to measure.
31
00:01:45 --> 00:01:49
And a lot of people got tripped
up over the course of the last
32
00:01:49 --> 00:01:55
century, of making up
mechanisms, getting a lot of
33
00:01:55 --> 00:01:56
data that supports the
mechanism and missing
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00:01:56 --> 00:02:01
something really important.
35
00:02:01 --> 00:02:05
So it's a common mistake, that
people think they've proven a
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00:02:05 --> 00:02:09
particular chemical mechanism
just because they have
37
00:02:09 --> 00:02:12
data that supports it.
38
00:02:12 --> 00:02:15
It's up to a certain point
that you know what you have.
39
00:02:15 --> 00:02:20
And we'll see examples of
that today, probably.
40
00:02:20 --> 00:02:23
And then certainly next
time when we start making
41
00:02:23 --> 00:02:25
approximations in our kinetics.
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00:02:25 --> 00:02:28
And see where things can
really go wrong if you
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don't have enough data.
44
00:02:30 --> 00:02:33
If you don't try to fish
out really, really, tiny
45
00:02:33 --> 00:02:37
intermediates somewhere
along the way.
46
00:02:37 --> 00:02:40
So last time we did then
parallel first order reactions.
47
00:02:40 --> 00:02:46
And today we're going to march
through and do parallel
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00:02:46 --> 00:02:47
first and second order.
49
00:02:47 --> 00:02:50
So one is first order, the
other one is second order.
50
00:02:50 --> 00:02:52
Things are going to
get more complicated.
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00:02:52 --> 00:02:54
And what you're going to
get is a flavor of how
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00:02:54 --> 00:02:55
to solve the problems.
53
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Setting up the problems.
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We're not going to do a lot of
the algebra here on the board,
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because otherwise we would
spend the next three
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weeks doing algebra.
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00:03:04 --> 00:03:06
And that would be
no fun at all.
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00:03:06 --> 00:03:08
That doesn't mean that you're
not going to be doing any
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algebra on the homework.
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00:03:10 --> 00:03:15
Because part of it is
figuring how to do it.
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00:03:15 --> 00:03:22
So, we have a reaction
where A goes to B plus C.
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And the first thing we did was,
the first mechanism we wrote
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was a parallel mechanism
where we had A goes
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B and A goes to C.
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00:03:32 --> 00:03:34
That's the mechanism.
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And one other way to write it
is A goes to B or C, in a way
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that is more, sort of describes
the branching process, that
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when we talk about a branching
ratio, the ratio of amount of B
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to C, this looks like the
branching out of two
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different paths.
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And this time we're going to
do where this is first order
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and this is second order.
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With the rate k1 and a rate k2.
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And the way that you do all
these problems, you have to
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be very systematic about it.
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The first thing you do is
you have to write down
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all your rate laws.
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And make sure that you
include everything.
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So you start by writing
the rate law for the
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destruction of A.
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00:04:28 --> 00:04:32
And then the rate law for the
appearance of B and for C.
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00:04:32 --> 00:04:33
So for a, we have dA/dt.
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00:04:33 --> 00:04:36
And again, I'm dropping the
brackets around the A, because
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00:04:36 --> 00:04:41
I don't want to carry around
all these extra symbols.
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So we destroy A by making B.
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So that's going to
be first order.
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00:04:47 --> 00:04:50
And then we also destroy
A by making C, and
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that's second order.
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00:04:50 --> 00:04:55
So we have two paths out of A.
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00:04:55 --> 00:04:57
To get rid of A.
91
00:04:57 --> 00:05:02
Then we write down the
appearance of B, dB/dt.
92
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It's only through one path,
first order in A, and then the
93
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appearance of C, only one
path which is second order.
94
00:05:16 --> 00:05:18
And these are all our
differential equations.
95
00:05:18 --> 00:05:24
And now the trick is, how do
you solve these three
96
00:05:24 --> 00:05:27
differential equations in a
way that you get meaningful
97
00:05:27 --> 00:05:32
information out.
98
00:05:32 --> 00:05:36
Well, the first thing
to do is to, so these
99
00:05:36 --> 00:05:40
two here depend on A.
100
00:05:40 --> 00:05:42
So the appearance of B
depends on A, the appearance
101
00:05:42 --> 00:05:43
of C depends on A.
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00:05:43 --> 00:05:45
But this one here only
depends on A by itself.
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00:05:45 --> 00:05:48
So this is the first one you're
going to start out with.
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00:05:48 --> 00:05:50
Because you can put all of
the A's on one side and
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00:05:50 --> 00:05:53
all the time on the other
side and integrate.
106
00:05:53 --> 00:06:02
So then you solve, and you can
rewrite this as minus dA/dt is
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00:06:02 --> 00:06:14
equal to k1 times A times one
plus k2 over k1 times A.
108
00:06:14 --> 00:06:16
And then you put all the
A's on one side, all the
109
00:06:16 --> 00:06:18
t's on the other side.
110
00:06:18 --> 00:06:28
And integrate from A0 to A
minus dA over A times one
111
00:06:28 --> 00:06:33
plus k2 over k1 times A.
112
00:06:33 --> 00:06:39
That is equal to k1
from zero to t dt.
113
00:06:39 --> 00:06:41
And then you have to solve
this integral here.
114
00:06:41 --> 00:06:44
And there's a reason why I
factored out the A here.
115
00:06:44 --> 00:06:47
It's because this is of the
form you can use to use the
116
00:06:47 --> 00:06:51
trick of partial fractions,
which we mentioned this before.
117
00:06:51 --> 00:07:01
So you use partial
fractions to solve this.
118
00:07:01 --> 00:07:04
And I'm not going to
do it on the board.
119
00:07:04 --> 00:07:08
You turn the crank, you
basically write one over A
120
00:07:08 --> 00:07:12
times one plus k2 over k1 A is
equal to n1 of A plus n2 over
121
00:07:12 --> 00:07:14
this part here, and you solve
for n1 and n2 and you plug it
122
00:07:14 --> 00:07:18
back in, and you redo your
integral, et cetera.
123
00:07:18 --> 00:07:20
And you get your answer.
124
00:07:20 --> 00:07:23
Which I'm going to write down
because I'm going to use it
125
00:07:23 --> 00:07:36
later. k1 A0 over e to the
k1 times t, k1 plus k2 A0.
126
00:07:36 --> 00:07:38
So one of the things we
immediately see is even though
127
00:07:38 --> 00:07:42
this mechanism isn't very
complicated, just two paths,
128
00:07:42 --> 00:07:45
one first order, one second
order, the solutions start to
129
00:07:45 --> 00:07:53
get not so simple
pretty quickly.
130
00:07:53 --> 00:07:55
So it depends, there's an
exponential on the bottom here.
131
00:07:55 --> 00:07:58
Depends on the initial
concentrations and both rate
132
00:07:58 --> 00:08:02
constants are in there.
133
00:08:02 --> 00:08:03
The first thing you want
to do is you want to
134
00:08:03 --> 00:08:09
look at limiting cases.
135
00:08:09 --> 00:08:12
Just like before.
136
00:08:12 --> 00:08:17
We did, and the first limiting
case we're going to look at, so
137
00:08:17 --> 00:08:20
interesting to look at is this
guy right here. k1
138
00:08:20 --> 00:08:22
and k2 A here.
139
00:08:22 --> 00:08:24
If one is bigger than
the other, then things
140
00:08:24 --> 00:08:27
will cancel out.
141
00:08:27 --> 00:08:30
And we can get some intuition.
142
00:08:30 --> 00:08:34
So the first limiting case
we're going to look at is where
143
00:08:34 --> 00:08:44
k2 A0, this term right here,
is much smaller than k1.
144
00:08:44 --> 00:08:47
Now, we want to make sure that
in these limiting cases that
145
00:08:47 --> 00:08:48
when I write something like
this, that the units
146
00:08:48 --> 00:08:50
actually make sense.
147
00:08:50 --> 00:08:52
That I'm not saying,
three apples are less
148
00:08:52 --> 00:08:55
than four oranges.
149
00:08:55 --> 00:09:00
So k1, it's first order, so the
units are one over second. k2,
150
00:09:00 --> 00:09:02
it's a second order rate
constant, so the units are
151
00:09:02 --> 00:09:07
one over second molar,
times moles per liter.
152
00:09:07 --> 00:09:09
So the moles per liters cancel
out and I have one over our
153
00:09:09 --> 00:09:12
second is less than one over
second, so things are great.
154
00:09:12 --> 00:09:15
It I'm making the right kind
of approximation here.
155
00:09:15 --> 00:09:17
What else is this
approximation saying?
156
00:09:17 --> 00:09:23
This is saying that the rate
into B, so this one is the
157
00:09:23 --> 00:09:28
faster one. k1, the rate into
B, is the rate into, is k1, is
158
00:09:28 --> 00:09:35
faster than k2 times A, which
is basically the rate into C.
159
00:09:35 --> 00:09:41
So this is saying that the
rate into B, to form B,
160
00:09:41 --> 00:09:49
is faster than into C.
161
00:09:49 --> 00:09:53
So we can write down a sketch.
162
00:09:53 --> 00:09:57
We can sketch what we expect
this approximation to be, then.
163
00:09:57 --> 00:10:04
And we know it's going to
look something like this.
164
00:10:04 --> 00:10:10
We know that A is going to
come down, exponentially.
165
00:10:10 --> 00:10:12
Without any structure to it.
166
00:10:12 --> 00:10:14
It's going to go either
into B or into C, in some
167
00:10:14 --> 00:10:15
branching fractions.
168
00:10:15 --> 00:10:19
And we know that the rate into
B is faster than into C, so B
169
00:10:19 --> 00:10:22
is going to come up like this.
170
00:10:22 --> 00:10:27
And C is going to
come up, but slower.
171
00:10:27 --> 00:10:34
And there's going to be a
ratio between these two guys.
172
00:10:34 --> 00:10:37
So we know the slope here
is going to be slower
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00:10:37 --> 00:10:38
then the slope for B.
174
00:10:38 --> 00:10:41
And we can check that out,
within our approximation.
175
00:10:41 --> 00:10:47
We can write, at t equals
zero, find out what these
176
00:10:47 --> 00:10:48
initial slopes are.
177
00:10:48 --> 00:10:50
For the creation of B and C.
178
00:10:50 --> 00:10:58
So dB/dt is the slope of the
creation of B at the beginning.
179
00:10:58 --> 00:11:02
At t equals zero.
180
00:11:02 --> 00:11:05
And that's just the
rate law. dB/dt at t
181
00:11:05 --> 00:11:11
equals zero is k1 A0.
182
00:11:11 --> 00:11:21
And dC/dt, the initial slope, t
equals zero, is k1 A0 squared.
183
00:11:21 --> 00:11:28
Or k2 A0 squared, rather.
184
00:11:28 --> 00:11:36
So I'm going to write it as
k2 A0 squared, like this.
185
00:11:36 --> 00:11:40
We said k1 is much bigger
than k2 times A0.
186
00:11:40 --> 00:11:41
There's the same A0 here.
187
00:11:41 --> 00:11:48
So we see, just by writing this
down, that our intuition that
188
00:11:48 --> 00:11:50
because the rate into B is
much faster than into A, that
189
00:11:50 --> 00:11:51
it should look like this.
190
00:11:51 --> 00:11:54
Well it's borne out just like
this very simple sort of
191
00:11:54 --> 00:11:59
looking at the initial rate
where the slope here is much
192
00:11:59 --> 00:12:04
smaller than the slope here.
193
00:12:04 --> 00:12:10
And then we can take the
approximation in here.
194
00:12:10 --> 00:12:13
And see what it implies.
195
00:12:13 --> 00:12:18
So let me do that here, let me
just use my green chalk here.
196
00:12:18 --> 00:12:21
So k2 A0 is much less than k1.
197
00:12:21 --> 00:12:27
So this is going to go away.
k2 A0 is much less than k1.
198
00:12:27 --> 00:12:29
Well, this is building
up from zero.
199
00:12:29 --> 00:12:31
So this is always
bigger than one here.
200
00:12:31 --> 00:12:33
So I can get rid
of this as well.
201
00:12:33 --> 00:12:34
Because there's the
k1 sitting here.
202
00:12:34 --> 00:12:39
And k1 is much
bigger than k2 A0.
203
00:12:39 --> 00:12:44
Once I got rid of all these two
guys, then the k1's cancel out.
204
00:12:44 --> 00:12:45
And this is looking very nice.
205
00:12:45 --> 00:12:48
Because now I can take
that e to the k1 t
206
00:12:48 --> 00:12:50
and put it upstairs.
207
00:12:50 --> 00:12:56
A is equal to approximately
A0 e to the minus k1 times
208
00:12:56 --> 00:13:00
time in this approximation.
209
00:13:00 --> 00:13:03
It looks like it's first
order coming down, with
210
00:13:03 --> 00:13:05
a rate constant k1.
211
00:13:05 --> 00:13:11
It looks like basically the
branching into C is nonexistent
212
00:13:11 --> 00:13:14
as far as, at least in
this approximation, as
213
00:13:14 --> 00:13:14
far A is concerned.
214
00:13:14 --> 00:13:19
It's going to look
like this thing here.
215
00:13:19 --> 00:13:25
It's going to look like first
order A goes to B with
216
00:13:25 --> 00:13:26
rate constant k1.
217
00:13:26 --> 00:13:28
So if you didn't know, if you
didn't know that there was
218
00:13:28 --> 00:13:32
another substance, C, that was
being formed along the way, if
219
00:13:32 --> 00:13:34
you didn't measure it, if you
didn't do the analytical
220
00:13:34 --> 00:13:39
chemistry and sort of fish it
out, and you just looked at the
221
00:13:39 --> 00:13:43
two major components, A and B,
in this case here, you'd
222
00:13:43 --> 00:13:45
measure a rate out of A,
it would look like
223
00:13:45 --> 00:13:48
it's first order.
224
00:13:48 --> 00:13:50
It's great.
225
00:13:50 --> 00:13:51
Got my mechanism.
226
00:13:51 --> 00:13:54
A goes to B, period.
227
00:13:54 --> 00:13:57
And you would know there was a
minor component that was being
228
00:13:57 --> 00:14:02
formed at the same time.
229
00:14:02 --> 00:14:15
OK, let's do another
approximation.
230
00:14:15 --> 00:14:25
Let's do the other case, where
k2 A0 is much bigger than k1.
231
00:14:25 --> 00:14:26
And, in this case here,
you have to be a little
232
00:14:26 --> 00:14:27
bit more careful.
233
00:14:27 --> 00:14:31
You have to add that
you're going to look at
234
00:14:31 --> 00:14:32
this at early times.
235
00:14:32 --> 00:14:36
And we're going to see why
early times is important here.
236
00:14:36 --> 00:14:38
In just five minutes.
237
00:14:38 --> 00:14:41
And early times, and how
you define early times.
238
00:14:41 --> 00:14:44
What does it mean for things to
be close to time equals zero?
239
00:14:44 --> 00:14:48
Well, we have to have a
reference time scale.
240
00:14:48 --> 00:14:51
The reactions, I'm going
at a certain rate.
241
00:14:51 --> 00:14:54
And one second may
be very slow.
242
00:14:54 --> 00:14:55
Or it may be very fast.
243
00:14:55 --> 00:14:57
Depending on the rate
of the reaction.
244
00:14:57 --> 00:15:03
So k1 here defines the time
scale of the problem.
245
00:15:03 --> 00:15:06
It's one over second,
unit of one over second.
246
00:15:06 --> 00:15:10
And early times means that the
times that I'm looking at
247
00:15:10 --> 00:15:15
compare to this rate,
it's very small.
248
00:15:15 --> 00:15:19
So k1 t is less than one.
249
00:15:19 --> 00:15:23
That means that the time,
during the time period that I'm
250
00:15:23 --> 00:15:26
looking at, hardly anything
has happened to the
251
00:15:26 --> 00:15:30
branching of A to B.
252
00:15:30 --> 00:15:32
That that's what I
mean by early time.
253
00:15:32 --> 00:15:38
So B is hardly being built up
yet. k1 is our reference time.
254
00:15:38 --> 00:15:40
k1 is units of one over second.
255
00:15:40 --> 00:15:42
This has units of seconds.
256
00:15:42 --> 00:15:45
So it's all working out fine.
257
00:15:45 --> 00:15:47
In other words, I would have no
idea how to define early times.
258
00:15:47 --> 00:15:50
I could say, it's in a year, it
could be, if you're looking at
259
00:15:50 --> 00:15:53
plutonium decay, it's a
100,000 year half-life.
260
00:15:53 --> 00:15:55
A year would be an early time.
261
00:15:55 --> 00:15:58
So it would be really fast,
compared to the process.
262
00:15:58 --> 00:16:02
But if you're looking at a
reaction that takes a few
263
00:16:02 --> 00:16:04
nanoseconds than a year
would be very, very long.
264
00:16:04 --> 00:16:06
So you really need
your reference time
265
00:16:06 --> 00:16:11
somewhere in there.
266
00:16:11 --> 00:16:13
So this approximation means
that essentially, no
267
00:16:13 --> 00:16:16
B is being created.
268
00:16:16 --> 00:16:19
While we're looking
at the process.
269
00:16:19 --> 00:16:24
And so we might then very well
expect that if we look at A and
270
00:16:24 --> 00:16:27
C, that the answer is going to
be that we're going to see
271
00:16:27 --> 00:16:29
something that's second order.
272
00:16:29 --> 00:16:35
So we expect that the form of
A, the way to write it is
273
00:16:35 --> 00:16:39
going to be one over A
equals k1 t plus A0.
274
00:16:39 --> 00:16:44
So let's go ahead and take our
complete solution and write it
275
00:16:44 --> 00:16:46
in the way that we might
expect the answer to be.
276
00:16:46 --> 00:16:49
In terms of one over A
rather than A here.
277
00:16:49 --> 00:16:56
So let's just invert it. e to
the k1 times time, k1 plus k2
278
00:16:56 --> 00:17:07
A0, minus k2 A0 over k1 A0.
279
00:17:07 --> 00:17:11
This should be a
minus sign here.
280
00:17:11 --> 00:17:14
There should be a
minus sign here.
281
00:17:14 --> 00:17:16
It didn't matter for our
approximation because we got
282
00:17:16 --> 00:17:24
rid of it anyway, but it would
be a proper sign here is minus.
283
00:17:24 --> 00:17:27
And so now we need to make
our approximation and
284
00:17:27 --> 00:17:29
figure out what this here.
285
00:17:29 --> 00:17:33
So if we take the approximation
that k1 t is small, less than
286
00:17:33 --> 00:17:37
one, then what that should
trigger in your mind if that if
287
00:17:37 --> 00:17:39
you have e to the something
that small you should
288
00:17:39 --> 00:17:39
use a Taylor series.
289
00:17:39 --> 00:17:42
And expand this into
a Taylor series.
290
00:17:42 --> 00:17:45
So we're going to see
this multiple times.
291
00:17:45 --> 00:17:51
So we take e to the k1 t and
expand it out as one plus
292
00:17:51 --> 00:17:55
k1 t plus et cetera.
293
00:17:55 --> 00:17:58
We're going to keep the
first order terms in there.
294
00:17:58 --> 00:18:05
So we're going to plug this
approximation into here.
295
00:18:05 --> 00:18:07
That's this
approximation there.
296
00:18:07 --> 00:18:11
And then we're going
to expand it out.
297
00:18:11 --> 00:18:20
So plug it in here, and,
keeping it to first order in
298
00:18:20 --> 00:18:24
time, and multiplying it out,
and I'm not going to do the
299
00:18:24 --> 00:18:27
intermediate steps, I'm just
going to give you the
300
00:18:27 --> 00:18:30
stuff after doing the
multiplications, we
301
00:18:30 --> 00:18:32
get k1 t squared.
302
00:18:32 --> 00:18:37
So that's basically k1 times
one plus k1 t, and then you
303
00:18:37 --> 00:18:45
have plus k1 k2 A0 times t.
304
00:18:45 --> 00:18:50
That's this term here. k1
k2 A0 times t times that.
305
00:18:50 --> 00:18:54
And the k2 A0 times one
gets subtracted out from
306
00:18:54 --> 00:18:56
this minus k2 A0 here.
307
00:18:56 --> 00:18:57
So that's all we have on top.
308
00:18:57 --> 00:19:00
And then there are terms
of higher order in time.
309
00:19:00 --> 00:19:01
But we're not going
to worry about this.
310
00:19:01 --> 00:19:05
We're going to stick to
first order in time.
311
00:19:05 --> 00:19:08
Divided by k1 A0.
312
00:19:08 --> 00:19:11
And now you do your
cancellations.
313
00:19:11 --> 00:19:15
We have our approximation
at k1 times time is
314
00:19:15 --> 00:19:16
much less than one.
315
00:19:16 --> 00:19:20
So here we have k1
times k1 times time.
316
00:19:20 --> 00:19:23
So this term here means it's
much less than this term here,
317
00:19:23 --> 00:19:24
because of our approximations.
318
00:19:24 --> 00:19:28
So we can get rid
of this term here.
319
00:19:28 --> 00:19:29
Get rid of that.
320
00:19:29 --> 00:19:35
There's no reason for us to
get rid of this one here.
321
00:19:35 --> 00:19:40
Because we've got k2 A0 there,
which is much larger than k1.
322
00:19:40 --> 00:19:44
And so when you expand
this out, all the
323
00:19:44 --> 00:19:45
k1's cancel out here.
324
00:19:45 --> 00:19:49
So so k1 cancels out that
k1, we have one plus one.
325
00:19:49 --> 00:19:51
So this is basically
one over A0.
326
00:19:51 --> 00:19:56
And then we have plus k2 A0
time over A0, so this ends
327
00:19:56 --> 00:20:07
up being equal to one over
A0 plus k2 times time.
328
00:20:07 --> 00:20:09
Exactly what we expected.
329
00:20:09 --> 00:20:13
That if you put in the right
approximation in the math, it
330
00:20:13 --> 00:20:16
behaves as your intuition
would tell you.
331
00:20:16 --> 00:20:20
Which is that the branching
into B nonexistent.
332
00:20:20 --> 00:20:23
And that you're looking at
everything going to C.
333
00:20:23 --> 00:20:27
It looks second order in C.
334
00:20:27 --> 00:20:30
But in this case here,
it's only early times.
335
00:20:30 --> 00:20:31
So if you keep going,
what happens?
336
00:20:31 --> 00:20:35
If you keep going in time,
beyond the time scale of this
337
00:20:35 --> 00:20:44
first order rate constant, and
you wait long enough, then the
338
00:20:44 --> 00:20:46
amount of A keeps decreasing.
339
00:20:46 --> 00:20:50
Keeps getting smaller
and smaller.
340
00:20:50 --> 00:20:53
So if you start the clock
again, a little bit later,
341
00:20:53 --> 00:20:58
where A is much smaller, then
k2 times the amount of A you
342
00:20:58 --> 00:21:04
have there, is no longer going
to be much greater than k1.
343
00:21:04 --> 00:21:10
At some point along the way,
k2 times A is going to
344
00:21:10 --> 00:21:13
be much less than k1.
345
00:21:13 --> 00:21:16
So at some point along the way,
it's not going to be this
346
00:21:16 --> 00:21:17
approximation any more.
347
00:21:17 --> 00:21:19
It's going to be the
previous one that we did.
348
00:21:19 --> 00:21:22
The one up there.
349
00:21:22 --> 00:21:24
So at early times we start
with something that may
350
00:21:24 --> 00:21:27
look second order for A.
351
00:21:27 --> 00:21:31
Time goes on, time goes on,
time goes on, A gets depleted.
352
00:21:31 --> 00:21:35
A gets depleted, and the rate
k2 times A, which is sitting
353
00:21:35 --> 00:21:40
right here, this part here
becomes smaller and smaller.
354
00:21:40 --> 00:21:43
And pretty soon this part wins.
355
00:21:43 --> 00:21:46
And it becomes first order.
356
00:21:46 --> 00:21:52
So what you'd expect to see,
then, is if you were to plot as
357
00:21:52 --> 00:22:01
a function of A, you expect to
see something that starts out
358
00:22:01 --> 00:22:04
as first order and then as A
gets depleted, switches
359
00:22:04 --> 00:22:07
over to second order.
360
00:22:07 --> 00:22:09
Something that's
second order here.
361
00:22:09 --> 00:22:11
And first order.
362
00:22:11 --> 00:22:12
This is the concentration of A.
363
00:22:12 --> 00:22:18
So if you were to plot
it, here you're plotting
364
00:22:18 --> 00:22:20
the concentration of A.
365
00:22:20 --> 00:22:26
If you were to plot the log, of
the concentration of A, at long
366
00:22:26 --> 00:22:29
times you'd expect it
to be first order.
367
00:22:29 --> 00:22:31
So something linear.
368
00:22:31 --> 00:22:35
But at early times you'd
expect it to be second order.
369
00:22:35 --> 00:22:37
So you'd expect to see
something that's nonlinear,
370
00:22:37 --> 00:22:40
then switching to
something linear.
371
00:22:40 --> 00:22:48
And if you were to plot it as
one over A, then you'd expect
372
00:22:48 --> 00:22:50
to start off with something
that's linear, at second
373
00:22:50 --> 00:22:53
order, early times.
374
00:22:53 --> 00:22:58
But instead of keeping, being
second order, as you deplete
375
00:22:58 --> 00:23:05
the amount of A, the branching
into B becomes important.
376
00:23:05 --> 00:23:09
And you become second order.
377
00:23:09 --> 00:23:14
So this is second order
at the beginning.
378
00:23:14 --> 00:23:23
And this becomes first order.
379
00:23:23 --> 00:23:26
So, any questions here?
380
00:23:26 --> 00:23:29
The importance of doing this
problem here was to lay out
381
00:23:29 --> 00:23:32
basically a systematic way
of looking at the problem.
382
00:23:32 --> 00:23:39
You lay out your
rate equations.
383
00:23:39 --> 00:23:40
Like this.
384
00:23:40 --> 00:23:45
You solve what you can, in
this case you solve for A.
385
00:23:45 --> 00:23:46
And it's complicated.
386
00:23:46 --> 00:23:49
So you want to learn something
about what the data
387
00:23:49 --> 00:23:50
might look like.
388
00:23:50 --> 00:23:51
And you look at limiting cases.
389
00:23:51 --> 00:23:53
Two obvious limiting cases.
390
00:23:53 --> 00:23:56
One rate is faster
than the other.
391
00:23:56 --> 00:23:59
You pick the first one
first, then go through it.
392
00:23:59 --> 00:24:00
And then use your intuition.
393
00:24:00 --> 00:24:04
Every point along the way, your
intuition can tell you what
394
00:24:04 --> 00:24:06
you expect the result to be.
395
00:24:06 --> 00:24:12
So in this case here, our
intuition told us that if we
396
00:24:12 --> 00:24:15
took the rate into B to be much
faster than that into C, than
397
00:24:15 --> 00:24:19
we expected to see something
that was largely first order.
398
00:24:19 --> 00:24:22
And when you we the
approximation in our exact
399
00:24:22 --> 00:24:25
solution, in fact it does
look like it's first order.
400
00:24:25 --> 00:24:26
So always use your intuition.
401
00:24:26 --> 00:24:29
Because the math is going
to get pretty hairy.
402
00:24:29 --> 00:24:32
The algebra might
make it complicated.
403
00:24:32 --> 00:24:34
And it's really easy to make a
plus sign into a minus sign
404
00:24:34 --> 00:24:35
somewhere along the way.
405
00:24:35 --> 00:24:37
Just like I did here.
406
00:24:37 --> 00:24:43
So if I hadn't fixed my mistake
here, and I had gone and put in
407
00:24:43 --> 00:24:46
my Taylor series here, I would
have gotten in trouble.
408
00:24:46 --> 00:24:48
Because this minus sign
would have been a plus.
409
00:24:48 --> 00:24:50
And these k2 A0's would
not have canceled out.
410
00:24:50 --> 00:24:53
And I wouldn't have gotten
the right result, which my
411
00:24:53 --> 00:24:59
intuition had told me should
be a second order process.
412
00:24:59 --> 00:25:03
And so if you have a problem,
let's say you're on an exam and
413
00:25:03 --> 00:25:05
you're doing algebra, you're
trying to, and you end up with
414
00:25:05 --> 00:25:08
a result that doesn't match
what your intuition tells you,
415
00:25:08 --> 00:25:10
and you don't have
time to fix it.
416
00:25:10 --> 00:25:10
Tell us.
417
00:25:10 --> 00:25:13
You know, I know this is wrong
because I know it's supposed
418
00:25:13 --> 00:25:14
to be second order.
419
00:25:14 --> 00:25:15
But I don't know
where I went wrong.
420
00:25:15 --> 00:25:18
And that's really important.
421
00:25:18 --> 00:25:21
Just tell us.
422
00:25:21 --> 00:25:27
That means that
you're thinking.
423
00:25:27 --> 00:25:28
Let's get a little bit
more complicated.
424
00:25:28 --> 00:25:35
Any questions?
425
00:25:35 --> 00:25:38
Consecutive series reactions.
426
00:25:38 --> 00:25:41
These were parallel
reactions and now.
427
00:25:41 --> 00:25:43
Basically what we're doing
here is we're building up a
428
00:25:43 --> 00:25:47
toolkit of simple mechanisms.
429
00:25:47 --> 00:25:51
And then we'll be able to put
these mechanisms together
430
00:25:51 --> 00:25:57
to make something
more complicated.
431
00:25:57 --> 00:26:04
So the next kind of
mechanism, series mechanism,
432
00:26:04 --> 00:26:09
series reactions, so we
have our reaction.
433
00:26:09 --> 00:26:12
Which is A goes to C.
434
00:26:12 --> 00:26:15
And the mechanism that's been
thought for this reaction is
435
00:26:15 --> 00:26:18
that there's an intermediate.
436
00:26:18 --> 00:26:21
That first you have A goes to
B, some intermediate, which
437
00:26:21 --> 00:26:26
gets used up to form the
final product, C, with
438
00:26:26 --> 00:26:28
a rate, k2, here.
439
00:26:28 --> 00:26:33
And we can either write the
mechanism in two steps.
440
00:26:33 --> 00:26:41
Or we can write it as A goes
to B goes to C, like this.
441
00:26:41 --> 00:26:45
And see both ways
of writing it.
442
00:26:45 --> 00:26:46
And we want to solve this.
443
00:26:46 --> 00:26:48
And we want to look at special
cases, and we just want to
444
00:26:48 --> 00:26:51
understand the implications
of this mechanism.
445
00:26:51 --> 00:26:53
So the first thing to do, just
like we did before, is to
446
00:26:53 --> 00:26:57
write all the rate laws.
447
00:26:57 --> 00:27:01
Before we got going
solving anything.
448
00:27:01 --> 00:27:05
So we start with
A, minus dA/dt.
449
00:27:05 --> 00:27:07
There's only one way
that gets used up.
450
00:27:07 --> 00:27:11
Through the formation of B
in a first order process.
451
00:27:11 --> 00:27:12
Easy.
452
00:27:12 --> 00:27:13
This is easy to solve.
453
00:27:13 --> 00:27:17
We know the answer is going
to be exponential in it.
454
00:27:17 --> 00:27:24
For B, dB/dt is equal to, well
it can be formed through my
455
00:27:24 --> 00:27:27
first order in A, so
that's k1 times A.
456
00:27:27 --> 00:27:31
But it also could get
destroyed by forming C.
457
00:27:31 --> 00:27:35
So we've got to get all the
paths out of A, here on the
458
00:27:35 --> 00:27:37
right side of this equation.
459
00:27:37 --> 00:27:41
So it can be destroyed
in a process, which
460
00:27:41 --> 00:27:42
is first order in B.
461
00:27:42 --> 00:27:46
So we're going to take both
k1 and k2 to be first
462
00:27:46 --> 00:27:50
order initially.
463
00:27:50 --> 00:27:51
Then we'll make one of them
second order and things will
464
00:27:51 --> 00:27:54
become very complicated and
we'll throw up our hands.
465
00:27:54 --> 00:27:57
But for now, let's not throw
up our hands quite yet.
466
00:27:57 --> 00:28:02
OK and for dC/dt, there's only
one way it can be formed is
467
00:28:02 --> 00:28:05
first order by destroying B.
468
00:28:05 --> 00:28:08
And we have a couple of
differential equations here.
469
00:28:08 --> 00:28:16
This differential equation
for C, depends on B here.
470
00:28:16 --> 00:28:21
The differential for
B depends on A here.
471
00:28:21 --> 00:28:22
So that means that things
are going to get a little
472
00:28:22 --> 00:28:28
bit more complicated.
473
00:28:28 --> 00:28:30
The easy one to solve is
always the first one here.
474
00:28:30 --> 00:28:31
That's first order.
475
00:28:31 --> 00:28:33
So we just write
down the answer.
476
00:28:33 --> 00:28:38
A is equal to A0, e
to the minus k1 t.
477
00:28:38 --> 00:28:42
We know the answer to that one.
478
00:28:42 --> 00:28:46
Then we have to work on B.
479
00:28:46 --> 00:28:48
So we want to find out
integrated rate laws for every
480
00:28:48 --> 00:28:59
one of these chemical species.
481
00:28:59 --> 00:29:01
And always there are
tricks involved.
482
00:29:01 --> 00:29:03
Partial fractions, whatever.
483
00:29:03 --> 00:29:09
So we write down B now. dB/dt
plus, let me rearrange it a
484
00:29:09 --> 00:29:12
little bit, plus k2 times
B, putting the minus k2 B
485
00:29:12 --> 00:29:14
on the other side here.
486
00:29:14 --> 00:29:16
Is equal to k1 times A.
487
00:29:16 --> 00:29:18
But I'm not going to keep this
A there, because now I know
488
00:29:18 --> 00:29:20
what A is in terms of time.
489
00:29:20 --> 00:29:28
And a constant. k1 times A0 e
to the minus k1 times time.
490
00:29:28 --> 00:29:31
OK, that's my differential
equation for B.
491
00:29:31 --> 00:29:35
Got to solve this thing.
492
00:29:35 --> 00:29:39
Well, the last time I did
differential equations in
493
00:29:39 --> 00:29:40
college was a century ago.
494
00:29:40 --> 00:29:42
And I wasn't kidding.
495
00:29:42 --> 00:29:43
It really was last century.
496
00:29:43 --> 00:29:46
In fact, it was
last millennium.
497
00:29:46 --> 00:29:49
That was a long time ago.
498
00:29:49 --> 00:29:50
So how do you do this?
499
00:29:50 --> 00:29:52
Well, the trick here,
there's a trick.
500
00:29:52 --> 00:29:56
And the trick is to multiply
the whole thing, both
501
00:29:56 --> 00:30:01
sides, by e to the k2 t.
502
00:30:01 --> 00:30:04
And you multiply this side by e
to the k2 t and you multiply
503
00:30:04 --> 00:30:09
this side by e to the k2 t,
to solve this equation.
504
00:30:09 --> 00:30:16
So once you do that, then you
have e to the k2 t times
505
00:30:16 --> 00:30:21
dB/dt plus k2 times B.
506
00:30:21 --> 00:30:22
This side here.
507
00:30:22 --> 00:30:30
And then you have is equal to
k1 times A0 e to the minus, e
508
00:30:30 --> 00:30:39
to the k2 minus k1 times time.
509
00:30:39 --> 00:30:43
OK, so the reason why this
trick works is because this
510
00:30:43 --> 00:30:51
guy right here can be
nicely written as d/dt of
511
00:30:51 --> 00:30:56
B times e to the k2 t.
512
00:30:56 --> 00:30:58
It's in nice simple form, so
when you integrate, it'll
513
00:30:58 --> 00:31:01
be very easy to integrate.
514
00:31:01 --> 00:31:09
And then the other side, you
still have k1 A0 e to the k2
515
00:31:09 --> 00:31:12
minus k1 times the time.
516
00:31:12 --> 00:31:15
So now you integrate both sides
with respect to time, The
517
00:31:15 --> 00:31:19
integral of something d/dt with
respect to time, it's just
518
00:31:19 --> 00:31:20
going to be itself right here.
519
00:31:20 --> 00:31:22
So it's very simple.
520
00:31:22 --> 00:31:24
So you integrate both sides
from zero to time, dt.
521
00:31:24 --> 00:31:27
522
00:31:27 --> 00:31:31
Zero to t, dt on both sides.
523
00:31:31 --> 00:31:37
And on this side here, you end
up with B, e to the k2 times
524
00:31:37 --> 00:31:42
time minus B, e to
the k2 times zero.
525
00:31:42 --> 00:31:47
Just taking the upper limit
minus the lower limit.
526
00:31:47 --> 00:31:53
And that's just equal to
B0, at time to equal zero.
527
00:31:53 --> 00:31:58
And so, and at time to equal
zero, B0 is equal to zero
528
00:31:58 --> 00:32:01
because we haven't made
any intermediates.
529
00:32:01 --> 00:32:06
So this goes away.
530
00:32:06 --> 00:32:08
And then we have
an equals sign.
531
00:32:08 --> 00:32:15
And on the other side, we have
k1 A0 over k2 minus k1, it's
532
00:32:15 --> 00:32:18
just the integral of this
thing, here with a k2 minus k1
533
00:32:18 --> 00:32:24
goes in the denominator. e to
the k2 minus k1 times
534
00:32:24 --> 00:32:28
time minus one.
535
00:32:28 --> 00:32:33
So now you have your integral
form for the amount of B that
536
00:32:33 --> 00:32:36
gets created at any time.
537
00:32:36 --> 00:32:38
And I want to keep
that on the board.
538
00:32:38 --> 00:32:45
Because I'm going to use it
later. k1 A0 over k1 minus
539
00:32:45 --> 00:32:54
k2 times e to the minus
k1 times time minus e to
540
00:32:54 --> 00:33:01
the minus k2 times time.
541
00:33:01 --> 00:33:03
OK, we're two thirds
of the way through.
542
00:33:03 --> 00:33:07
Now we've got to find C.
543
00:33:07 --> 00:33:10
And for the last component,
you don't want to be
544
00:33:10 --> 00:33:12
doing this kind of stuff.
545
00:33:12 --> 00:33:14
You don't want to waste your
time solving differential
546
00:33:14 --> 00:33:20
equations, because
stoichiometry can help you out.
547
00:33:20 --> 00:33:22
At the end of the day, you know
that everything that started
548
00:33:22 --> 00:33:25
out as A, in terms of
quantities, is going
549
00:33:25 --> 00:33:27
to end up at C.
550
00:33:27 --> 00:33:29
You know that if you end up
with a concentration A0 at the
551
00:33:29 --> 00:33:33
beginning, at the end of the
process the concentration of C,
552
00:33:33 --> 00:33:37
at infinite time, is
also going to be A0.
553
00:33:37 --> 00:33:41
If you know that there's a
correlation, there's a
554
00:33:41 --> 00:33:46
stoichiometry that relates
A and C together.
555
00:33:46 --> 00:33:47
And B.
556
00:33:47 --> 00:33:53
You have conservation of
mass here, basically.
557
00:33:53 --> 00:33:55
Conservation of atoms.
558
00:33:55 --> 00:33:58
That has to come into play
559
00:33:58 --> 00:34:07
So, for C, we're just going to
do algebra instead of calculus.
560
00:34:07 --> 00:34:10
So let's write down the
stoichiometry here.
561
00:34:10 --> 00:34:15
The amount of C at any
time is what it's going
562
00:34:15 --> 00:34:17
to be at the end.
563
00:34:17 --> 00:34:19
Which is A0, very
end of the process.
564
00:34:19 --> 00:34:21
It's going to be A0.
565
00:34:21 --> 00:34:25
But before we get to the
end, there's still stuff
566
00:34:25 --> 00:34:28
that's left in A and B.
567
00:34:28 --> 00:34:30
And that's going to
decrease the amount of C.
568
00:34:30 --> 00:34:35
So minus whatever's
still in A and B.
569
00:34:35 --> 00:34:37
That's the stoichiometry.
570
00:34:37 --> 00:34:38
That's one way of writing it.
571
00:34:38 --> 00:34:40
You could also write it
in a different way.
572
00:34:40 --> 00:34:47
Which is that the amount of
C is equal to the amount of
573
00:34:47 --> 00:34:50
A that you have used up.
574
00:34:50 --> 00:34:53
A0 is what you
started out with.
575
00:34:53 --> 00:34:54
A is what's left over.
576
00:34:54 --> 00:34:57
So this is the amount
of A used up.
577
00:34:57 --> 00:34:59
It's not all quite in C yet.
578
00:34:59 --> 00:35:02
Some of it is stuck in B.
579
00:35:02 --> 00:35:05
So it's the amount of A
used up, minus the amount
580
00:35:05 --> 00:35:08
that's stuck in B.
581
00:35:08 --> 00:35:09
Those two are the same.
582
00:35:09 --> 00:35:10
I hope.
583
00:35:10 --> 00:35:12
A0 minus A, yeah, that's right.
584
00:35:12 --> 00:35:13
Minus B, yeah, that's right.
585
00:35:13 --> 00:35:18
So this is used up.
586
00:35:18 --> 00:35:24
And then stuck in B.
587
00:35:24 --> 00:35:26
That's the hard part.
588
00:35:26 --> 00:35:29
Putting down the stoichiometry.
589
00:35:29 --> 00:35:31
Once you've done the hard
part, then if you solve the
590
00:35:31 --> 00:35:32
other two then it's easy.
591
00:35:32 --> 00:35:32
You just plug in.
592
00:35:32 --> 00:35:35
It's just algebra.
593
00:35:35 --> 00:35:37
And you get something
complicated.
594
00:35:37 --> 00:35:43
C is equal to complicated.
595
00:35:43 --> 00:35:49
It's in the notes, I'm not
going to write it down.
596
00:35:49 --> 00:35:51
So we've solved the
problem exactly.
597
00:35:51 --> 00:35:53
And now you're going
to do an experiment.
598
00:35:53 --> 00:35:55
And the experiment you're going
to do is not going to follow
599
00:35:55 --> 00:35:57
this whole thing, exactly.
600
00:35:57 --> 00:35:59
It's going to look
at limited cases.
601
00:35:59 --> 00:36:02
That's what you can do.
602
00:36:02 --> 00:36:04
And so the first thing to
look at is to look at
603
00:36:04 --> 00:36:06
the initial times.
604
00:36:06 --> 00:36:08
The very beginning of
the process, how are
605
00:36:08 --> 00:36:11
things appearing.
606
00:36:11 --> 00:36:33
And then we'll look
at the late times.
607
00:36:33 --> 00:36:35
So initial times.
608
00:36:35 --> 00:36:46
Near t equals zero.
609
00:36:46 --> 00:36:50
Well, every one of these things
has an exponential in it.
610
00:36:50 --> 00:36:52
And whenever we see
exponentials and we look at
611
00:36:52 --> 00:36:56
something at the early times,
times that are faster than one
612
00:36:56 --> 00:36:59
over k1, or faster than one
over k2, it tells us
613
00:36:59 --> 00:37:02
use a Taylor series.
614
00:37:02 --> 00:37:07
So use the approximation e to
the minus kt is approximately
615
00:37:07 --> 00:37:15
equal to one minus kt plus kt
squared over two
616
00:37:15 --> 00:37:16
plus et cetera.
617
00:37:16 --> 00:37:20
And keep as many terms as you
need in the Taylor series
618
00:37:20 --> 00:37:21
to get something sensible.
619
00:37:21 --> 00:37:23
So if everything comes out of
zero at the end, you know you
620
00:37:23 --> 00:37:26
haven't kept enough terms in t.
621
00:37:26 --> 00:37:28
Things have canceled
out too much.
622
00:37:28 --> 00:37:31
And you want something
that is time dependent.
623
00:37:31 --> 00:37:33
Because you've got a
change in time, right?
624
00:37:33 --> 00:37:36
It's a very common mistake to
say only keep one here and then
625
00:37:36 --> 00:37:39
everything cancels out you get
zero, and you say, well, B is
626
00:37:39 --> 00:37:41
equal to zero at all times.
627
00:37:41 --> 00:37:46
That's nonsensical.
628
00:37:46 --> 00:37:49
And I don't know a priori
how much time to keep.
629
00:37:49 --> 00:37:53
So let's go to second order.
630
00:37:53 --> 00:38:00
So if you plug this Taylor
series into A, and get
631
00:38:00 --> 00:38:02
something that's time
dependent, you only need
632
00:38:02 --> 00:38:04
to go to first order.
633
00:38:04 --> 00:38:08
One minus k1 times
time plus et cetera.
634
00:38:08 --> 00:38:14
Plug it into b, plug your
Taylor series into these two
635
00:38:14 --> 00:38:22
guys here, get k1 A0 over
k2 minus k1, you only need
636
00:38:22 --> 00:38:24
it to go to first order.
637
00:38:24 --> 00:38:27
The one's cancel out but the
times don't cancel out.
638
00:38:27 --> 00:38:33
Minus k1 plus k2, that is,
unless k1 and k2 are the same.
639
00:38:33 --> 00:38:36
So we're making the
approximation that k1
640
00:38:36 --> 00:38:38
is different than k2.
641
00:38:38 --> 00:38:41
Times the time.
642
00:38:41 --> 00:38:44
And when we look at C, this
complicated thing that I didn't
643
00:38:44 --> 00:38:48
write down, if you were just to
stop at first order in C,
644
00:38:48 --> 00:38:50
you get C equals zero.
645
00:38:50 --> 00:38:52
You know that can't be right.
646
00:38:52 --> 00:38:55
You know that C is going
to increase in time.
647
00:38:55 --> 00:38:56
It can't be equal to
zero at all times.
648
00:38:56 --> 00:39:00
So we have to go to
second order for C0.
649
00:39:00 --> 00:39:08
And it's approximately equal
to A0 times k1 k2 over two
650
00:39:08 --> 00:39:11
times the time squared.
651
00:39:11 --> 00:39:13
Let me rewrite this a little
bit closer in, because
652
00:39:13 --> 00:39:15
I need some room here.
653
00:39:15 --> 00:39:23
So c is approximately equal
to A0 k1 is k2 over two
654
00:39:23 --> 00:39:24
times time squared.
655
00:39:24 --> 00:39:26
OK, and then we can start
plotting out what this is
656
00:39:26 --> 00:39:27
going to look like then.
657
00:39:27 --> 00:39:30
So we have the concentrations
time on the axis here.
658
00:39:30 --> 00:39:34
Let's start with the
concentrating of A.
659
00:39:34 --> 00:39:38
A is dropping down
linearly in time.
660
00:39:38 --> 00:39:41
It's dropping down
linearly in time here.
661
00:39:41 --> 00:39:46
B is increasing linearly in
time, at the beginning.
662
00:39:46 --> 00:39:51
So B is going up linearly
in time, like this.
663
00:39:51 --> 00:39:55
And C is increasing
quadratically in time.
664
00:39:55 --> 00:39:58
So it's pretty flat,
coming up first.
665
00:39:58 --> 00:40:03
And then it starts to curve up.
666
00:40:03 --> 00:40:05
It's our first approximation.
667
00:40:05 --> 00:40:10
Then, we go to late times.
t equals infinity.
668
00:40:10 --> 00:40:11
So t equals infinity where
am I going to write this?
669
00:40:11 --> 00:40:33
Have to move this up.
670
00:40:33 --> 00:40:35
Let's go, t goes to infinity.
671
00:40:35 --> 00:40:38
Well, here I know that I'm
allowed to go to zero.
672
00:40:38 --> 00:40:42
Or allowed to go to constant,
because infinity is forever.
673
00:40:42 --> 00:40:46
So at t infinity, I know that
I'm going to use up all of A.
674
00:40:46 --> 00:40:48
So I know that A is
going to be equal to 0.
675
00:40:48 --> 00:40:50
I know I'm going to
use up all of B.
676
00:40:50 --> 00:40:53
So I know B is going
to go to zero.
677
00:40:53 --> 00:40:54
Somehow.
678
00:40:54 --> 00:40:58
And I know that C is
going to go to A0.
679
00:40:58 --> 00:41:07
So C is going to go
to A0 at infinity.
680
00:41:07 --> 00:41:17
So now on my graph here, I know
that at infinity, A is going to
681
00:41:17 --> 00:41:20
go down to zero like
this, somehow.
682
00:41:20 --> 00:41:23
Probably exponentially, because
it is exponential in time.
683
00:41:23 --> 00:41:25
So it's going to go to
zero exponentially.
684
00:41:25 --> 00:41:27
B is going to go down
to zero exponentially.
685
00:41:27 --> 00:41:30
There are the
exponentials right here.
686
00:41:30 --> 00:41:33
So B is going to go down
to zero exponentially.
687
00:41:33 --> 00:41:35
Something like this.
688
00:41:35 --> 00:41:39
And C is going to
saturate to A0.
689
00:41:39 --> 00:41:40
Exponentially rise up to A0.
690
00:41:40 --> 00:41:42
So this was A0 right here.
691
00:41:42 --> 00:41:45
It's going to go like this.
692
00:41:45 --> 00:41:47
So then I connect the dots.
693
00:41:47 --> 00:41:50
And I know that C is going
to start quadratically.
694
00:41:50 --> 00:41:53
And then it's going to rise
up an exponent like this.
695
00:41:53 --> 00:41:57
I know that B is going to
start off linearly, and
696
00:41:57 --> 00:41:59
then it has to go to zero.
697
00:41:59 --> 00:42:03
So it has to go through
some sort of maximum.
698
00:42:03 --> 00:42:08
And A is going to go
down exponentially,
699
00:42:08 --> 00:42:10
exponentially like this.
700
00:42:10 --> 00:42:12
So the interesting part
of this diagram is that
701
00:42:12 --> 00:42:13
B goes to a maximum.
702
00:42:13 --> 00:42:21
There is some point, there's
some time, that we can call t
703
00:42:21 --> 00:42:24
sub B max, where B is a maxima.
704
00:42:24 --> 00:42:27
And that's an
interesting point.
705
00:42:27 --> 00:42:29
That's something that we
can try to figure out
706
00:42:29 --> 00:42:31
experimentally what it is.
707
00:42:31 --> 00:42:34
And we could get some
parameters, we could maybe
708
00:42:34 --> 00:42:36
extract some information.
709
00:42:36 --> 00:42:42
There's a maximum B,
concentration of B,
710
00:42:42 --> 00:42:45
that gets formed.
711
00:42:45 --> 00:42:47
And how do you solve that?
712
00:42:47 --> 00:42:52
You set dB/dt equal to zero.
713
00:42:52 --> 00:42:55
So you have your equation
for B right here.
714
00:42:55 --> 00:42:58
You set it equal to
zero. and you solve.
715
00:42:58 --> 00:43:00
You get your maximum time.
716
00:43:00 --> 00:43:02
It's made up of rate constant.
717
00:43:02 --> 00:43:04
You get your maximum
concentration, which is also
718
00:43:04 --> 00:43:07
made up of rate constants.
719
00:43:07 --> 00:43:09
And you can get rate
constants out of this.
720
00:43:09 --> 00:43:10
Question.
721
00:43:10 --> 00:43:16
STUDENT: [INAUDIBLE]
722
00:43:16 --> 00:43:17
PROFESSOR: Yes.
723
00:43:17 --> 00:43:23
STUDENT: [INAUDIBLE]
724
00:43:23 --> 00:43:24
PROFESSOR: That is
a coincidence.
725
00:43:24 --> 00:43:31
That is my poor sketching.
726
00:43:31 --> 00:43:37
It should probably be somewhere
like, in the middle here.
727
00:43:37 --> 00:43:41
Actually, it completely
depends on the rates.
728
00:43:41 --> 00:43:44
And we're going to see limiting
cases where this maximum can
729
00:43:44 --> 00:43:51
occur here, or it can occur
somewhere down here.
730
00:43:51 --> 00:43:57
So this is the complete case.
731
00:43:57 --> 00:44:04
So, limiting cases. the
important limiting cases.
732
00:44:04 --> 00:44:06
We've made an approximation
along the way that k1
733
00:44:06 --> 00:44:07
is different than k2.
734
00:44:07 --> 00:44:10
So that has to be a case that
needs to be worked out.
735
00:44:10 --> 00:44:17
And it's likely to be easier
than the full solution.
736
00:44:17 --> 00:44:22
So one of the limiting cases
is k1 is equal to k2 and I'll
737
00:44:22 --> 00:44:27
leave that for homework.
738
00:44:27 --> 00:44:31
You can do that yourselves.
739
00:44:31 --> 00:44:34
It's good exercise.
740
00:44:34 --> 00:44:35
What about another
limiting case?
741
00:44:35 --> 00:44:37
Well, you have two
other limiting cases.
742
00:44:37 --> 00:44:39
One, you've got two rates.
743
00:44:39 --> 00:44:45
So, one is bigger
than the other.
744
00:44:45 --> 00:44:50
No information on
this board here.
745
00:44:50 --> 00:44:54
You start off with one case.
746
00:44:54 --> 00:45:03
So k1 greater than k2.
k1 greater than k2.
747
00:45:03 --> 00:45:05
What does it mean?
k1 greater than k2.
748
00:45:05 --> 00:45:08
It means that the rate
determining step, or the
749
00:45:08 --> 00:45:12
limiting step, in this
reaction, is the second step.
750
00:45:12 --> 00:45:16
k2 is very slow compared to k1.
751
00:45:16 --> 00:45:21
And sometimes I like to think
of these things as pipes.
752
00:45:21 --> 00:45:25
And connecting vessels.
753
00:45:25 --> 00:45:31
So let's say we have the amount
of liquid in the top vessel
754
00:45:31 --> 00:45:32
is the concentration of A.
755
00:45:32 --> 00:45:36
The amount of liquid in the
middle vessel, let me get my
756
00:45:36 --> 00:45:46
color scheme to be
consistent here.
757
00:45:46 --> 00:45:48
Then we have a middle vessel,
which is the concentration of
758
00:45:48 --> 00:45:53
B, and a final vessel, which
is the concentration of C.
759
00:45:53 --> 00:45:59
And we start out with some
sort of amount of A in here.
760
00:45:59 --> 00:46:03
Started with a lot of A,
and then these vessels are
761
00:46:03 --> 00:46:04
connecting with pipes.
762
00:46:04 --> 00:46:10
So there's a very thick,
big pipe that connects
763
00:46:10 --> 00:46:11
A and B together.
764
00:46:11 --> 00:46:16
The rate is fast,
going from A to B.
765
00:46:16 --> 00:46:19
And the rate going from B to C
is slow, so there's a skinny
766
00:46:19 --> 00:46:26
pipe connecting B
and C together.
767
00:46:26 --> 00:46:29
Then I turn on the system,
I set time to equal zero,
768
00:46:29 --> 00:46:31
poof, what happens?
769
00:46:31 --> 00:46:33
There's the big pipe
connecting A and B.
770
00:46:33 --> 00:46:37
The whole amount of a here gets
transferred to B, suddenly.
771
00:46:37 --> 00:46:38
Then it gets stuck here.
772
00:46:38 --> 00:46:41
And then it dribbles
out from B to C.
773
00:46:41 --> 00:46:47
So what do we expect, if we
were to plot our quantities
774
00:46:47 --> 00:46:49
as a function of time.
775
00:46:49 --> 00:46:52
What we'd expect then, then
we're going to make sure that
776
00:46:52 --> 00:46:57
the math works out, is that A
is going to decrease
777
00:46:57 --> 00:46:58
really fast.
778
00:46:58 --> 00:47:02
It's going to go, poof.
779
00:47:02 --> 00:47:08
First order in time
with a very fast rate.
780
00:47:08 --> 00:47:11
And all of this is going
to go straight into B.
781
00:47:11 --> 00:47:13
So we expect B to go linearly.
782
00:47:13 --> 00:47:14
Remember, at the
beginning it's linear.
783
00:47:14 --> 00:47:17
Linear goes up.
784
00:47:17 --> 00:47:20
Almost all the way up to A0.
785
00:47:20 --> 00:47:22
Because it can't
hardly get out.
786
00:47:22 --> 00:47:26
Reaches its maximum, then it
goes from B to C through
787
00:47:26 --> 00:47:28
this skinny, skinny pipe.
788
00:47:28 --> 00:47:32
First order rate, with a rate
k2, so we're exponentially
789
00:47:32 --> 00:47:38
going down, slowly, slowly,
slowly, with a rate k2.
790
00:47:38 --> 00:47:44
And then C starts
quadratically.
791
00:47:44 --> 00:47:46
And then very quickly,
once everything's into
792
00:47:46 --> 00:47:49
B, A is forgotten.
793
00:47:49 --> 00:47:51
A doesn't matter any more.
794
00:47:51 --> 00:47:54
Basically what you're looking
at is a transfer of B into
795
00:47:54 --> 00:47:56
C through this little pipe.
796
00:47:56 --> 00:47:59
And you know that's going to
be a first order of rate.
797
00:47:59 --> 00:48:01
First order of process,
B to C's first order of
798
00:48:01 --> 00:48:02
process with rate k2.
799
00:48:02 --> 00:48:05
So you expect, then, C to
go up in the first order
800
00:48:05 --> 00:48:13
process to A0 with rate k2.
801
00:48:13 --> 00:48:18
There's the rate k2 here.
802
00:48:18 --> 00:48:22
And the rate k1 here.
803
00:48:22 --> 00:48:25
So when you turn the crank on
your approximation in the math,
804
00:48:25 --> 00:48:32
what you'd better see is that
at the end, B as a function of
805
00:48:32 --> 00:48:36
time should look like a first
order process with rate k2.
806
00:48:36 --> 00:48:39
C as a function of time should
look like a first order process
807
00:48:39 --> 00:48:41
with rate k2, going to A0.
808
00:48:41 --> 00:48:45
And you can write the answer,
you should be able to write the
809
00:48:45 --> 00:48:51
answer, for C, it's going to be
C is approximately, it's going
810
00:48:51 --> 00:48:53
to go to A0, there's
no other choice.
811
00:48:53 --> 00:48:56
Everything that was in A gets
transferred to B0, so at time
812
00:48:56 --> 00:48:58
is infinity it's going to be
A0, and it's going to be a
813
00:48:58 --> 00:49:01
first order process
with rate k2.
814
00:49:01 --> 00:49:06
One minus e to the
minus k2 times times.
815
00:49:06 --> 00:49:07
I don't even have
to do the math.
816
00:49:07 --> 00:49:10
I know that's the answer.
817
00:49:10 --> 00:49:14
If you don't believe
me, then do the math.
818
00:49:14 --> 00:49:16
You should do it.
819
00:49:16 --> 00:49:16
But that's the answer.
820
00:49:16 --> 00:49:18
It has to be the answer.
821
00:49:18 --> 00:49:26
And if you look at B, the
answer has to be that it's
822
00:49:26 --> 00:49:32
approximately looking like B
is disappearing with a rate
823
00:49:32 --> 00:49:37
constant that's k2, e to the
minus k2, and the maximum
824
00:49:37 --> 00:49:38
here is very close to A0.
825
00:49:38 --> 00:49:41
So I'm going to be
putting A0 here.
826
00:49:41 --> 00:49:44
Close enough.
827
00:49:44 --> 00:49:48
So as long as I ignore the very
initial time where A suddenly
828
00:49:48 --> 00:49:51
dumps into B, then everything
after that very early time
829
00:49:51 --> 00:49:55
here is just looking
like B goes to C.
830
00:49:55 --> 00:49:58
This is not zero, this is t.
831
00:49:58 --> 00:50:00
And A just disappears quickly.
832
00:50:00 --> 00:50:06
And we can write it as A
is equal to A0 e to the
833
00:50:06 --> 00:50:09
minus k1 times the time.
834
00:50:09 --> 00:50:12
That's an easy approximation.
835
00:50:12 --> 00:50:17
And you've just got to
make sure that this fits.
836
00:50:17 --> 00:50:23
So the other limiting case,
which I'm also going to leave
837
00:50:23 --> 00:50:27
as a homework, because it's so
straightforward, is to have k2
838
00:50:27 --> 00:50:32
greater than k1, and you should
make sure that you can predict
839
00:50:32 --> 00:50:38
it. and that the
math works out.
840
00:50:38 --> 00:50:39
OK, any questions?
841
00:50:39 --> 00:50:42
Next time we'll do
reversible reactions.
842
00:50:42 --> 00:50:46
And some more approximations.
843
00:50:46 --> 00:50:47