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PROFESSOR: Please note that
this is a clicker competition,
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00:00:28 --> 00:00:32
perhaps the last of the year,
before you put in your vote
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for this particular
clicker question.
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And the defending champs are
recitation three, and I believe
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that no recitation has
won more than once.
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Is that true?
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Is there a recitation
that has two victories?
16
00:00:52 --> 00:00:56
So, this might be your last
chance if your recitation has
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00:00:56 --> 00:01:01
not won, or if you want to be
decisive victors in the clicker
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00:01:01 --> 00:01:04
competition to have a repeat
victory, this is your
19
00:01:04 --> 00:01:04
chance -- no pressure.
20
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But go ahead and put in
your clicker response.
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OK, let's just take 10
more seconds to click in.
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OK, the answer here is b, and
the trick was to recognize
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which is the equation for first
order half life, and which is
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the equation for second
order half life.
25
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Most people got this right.
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And we picked this clicker
question to demonstrate a point
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that will, for the final exam,
which is that you will have
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an equation sheet of
all the equations.
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There'll be a few you have to
memorize, we'll let you know
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00:02:16 --> 00:02:20
what they are, but that's
a lot of equations.
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00:02:20 --> 00:02:23
So, so far on each hour exam,
you've had an equation sheet
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that's been related to
that particular material.
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00:02:26 --> 00:02:29
On the final, you of an
equation sheet that has the
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equations from exam 1, exam
2, exam 3, and also the
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00:02:34 --> 00:02:36
fourth exam material.
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00:02:36 --> 00:02:40
And so, it doesn't say on them,
oh, first order half life
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equation, colon, and then the
equation, it just has
38
00:02:44 --> 00:02:45
all of the equations.
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00:02:45 --> 00:02:49
So one thing that you have to
do is recognize which equation
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can be applied where.
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And this is something that
people, occasionally that's
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part of knowing the material
is to know/recognize
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the equations.
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You don't have to memorize
them, but you do have to be
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able to recognize them and
know when to apply them.
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00:03:03 --> 00:03:06
So that's something you should
be thinking about as you're
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reviewing the material
for the exam.
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And we're just going to at
jump right in and talk
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about reaction mechanisms.
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00:03:13 --> 00:03:18
So this is the last type
of question that's
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00:03:18 --> 00:03:21
on problem-set 10.
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00:03:21 --> 00:03:26
So, when you're investigating
reaction mechanisms, you often
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00:03:26 --> 00:03:30
will have an experimentally
determined rate law, and then
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you want to try to come up with
a mechanism that's consistent
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00:03:33 --> 00:03:35
with that rate law.
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00:03:35 --> 00:03:48
So, for this particular
reaction, we have 2 n o plus
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o 2 going to 2 n o 2 gas.
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And experimental rate laws
determined there was a rate
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constant called k obs, for k
observed, the observed rate
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constant that was calculated.
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And we know that it's the
order of the reactions here.
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00:04:08 --> 00:04:11
Overall, what is the
overall order at this
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particular rate law?
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Three.
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00:04:17 --> 00:04:23
So we have second order in n
o and first order in o 2.
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So if it's overall order
of three, is it likely
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to occur in one step?
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No.
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What would it be called if it
did occur in one step -- if
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three things came together at
the same time to
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form a product?
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Yeah, thermonuclear.
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And we learned last time
that thermonuclear
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reactions are rare.
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So it's unlikely that those
three things, the two n o's and
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the one o 2 are going to come
together at the same time to
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form product, so it probably
has more than one step.
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So, on the board here and on
your notes, we have a two-step
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proposed mechanism, and let's
talk about these two steps and
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00:05:15 --> 00:05:21
come up with a rate law that's
consistent with this mechanism,
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00:05:21 --> 00:05:25
and then see if that agrees
with the experimental rate
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00:05:25 --> 00:05:27
law that was determined.
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00:05:27 --> 00:05:30
So, in the first step of
this reaction, we have 2
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n o's coming together.
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00:05:32 --> 00:05:37
There's a little rate constant,
k 1, above the first arrow, and
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00:05:37 --> 00:05:40
for the reverse direction, we
have k minus 1, so this is
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written as a reversible
reaction, and it's forming
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an intermediate n 2 o 2.
89
00:05:46 --> 00:05:51
In the second step, we have the
o 2 coming in, reacting with
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00:05:51 --> 00:05:55
that intermediate with a rate
constant, k 2, and it's forming
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00:05:55 --> 00:05:59
two molecules of n o 2.
92
00:05:59 --> 00:06:02
So now we can write the
rates for each of these
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00:06:02 --> 00:06:04
individual steps.
94
00:06:04 --> 00:06:10
We can write the rate for the
foward reaction here, and
95
00:06:10 --> 00:06:15
so that's going to
be equal to k 1.
96
00:06:15 --> 00:06:21
And we have n o, two n
o's, so n o squared.
97
00:06:21 --> 00:06:25
And again, this is a step or
an elementary reaction so
98
00:06:25 --> 00:06:28
we can write the reaction
exactly as it occurs.
99
00:06:28 --> 00:06:32
We can write a rate law
for that step using the
100
00:06:32 --> 00:06:34
stiochiometry of the reaction.
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00:06:34 --> 00:06:36
For an overall reaction, you
can't do that, it has to be
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00:06:36 --> 00:06:38
experimentally determined.
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00:06:38 --> 00:06:41
But for an elementary reaction
or a step in a mechanism,
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00:06:41 --> 00:06:43
you can do that.
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00:06:43 --> 00:06:45
So we write it exactly
as it occurs.
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00:06:45 --> 00:06:51
So, what would be the
order of this particular
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00:06:51 --> 00:06:54
reaction as written here?
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Two.
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00:06:56 --> 00:06:58
And what about molecularity?
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00:06:58 --> 00:07:06
It would be bimolecular.
111
00:07:06 --> 00:07:13
So we're just getting you
used to some of these terms.
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00:07:13 --> 00:07:17
All right, so then for the
reverse direction, the rate of
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00:07:17 --> 00:07:20
the reverse direction would
be equal to what rate
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00:07:20 --> 00:07:25
constant? k minus 1.
115
00:07:25 --> 00:07:28
Times what?
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00:07:28 --> 00:07:34
Yeah, n 2 o 2.
117
00:07:34 --> 00:07:39
So what's the order here?
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00:07:39 --> 00:07:40
One.
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00:07:40 --> 00:07:43
And what's this called?
120
00:07:43 --> 00:07:53
Yeah, so that's
called unimolecular.
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00:07:53 --> 00:07:59
All right, what about step 2?
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00:07:59 --> 00:08:47
What would be the rate here,
and that is a clicker question.
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00:08:47 --> 00:09:01
OK, just 10 more seconds.
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00:09:01 --> 00:09:09
Yup, so we write it exactly as
written, so we have a k 2 times
125
00:09:09 --> 00:09:17
the concentration of o 2 times
the concentration of n 2 o 2.
126
00:09:17 --> 00:09:24
All right, so what is
the overall order here?
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00:09:24 --> 00:09:25
Two.
128
00:09:25 --> 00:09:38
And again, that would
be the bimolecular.
129
00:09:38 --> 00:09:41
All right, so we have our two
steps, and we've written
130
00:09:41 --> 00:09:45
rates for all of those
particular steps.
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00:09:45 --> 00:09:49
Now we're interested in writing
the rate of the overall
132
00:09:49 --> 00:09:53
reaction that forms n o 2.
133
00:09:53 --> 00:10:09
So, let's think about what the
rate of n o 2 formation is.
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00:10:09 --> 00:10:13
So, in the last step, we're
forming n o 2, and we can
135
00:10:13 --> 00:10:17
just use that last step
to write this out.
136
00:10:17 --> 00:10:21
We're forming two
molecules of n o 2.
137
00:10:21 --> 00:10:25
So we're going to put a 2 in
there, because the rate at
138
00:10:25 --> 00:10:32
which the concentration of this
will increase twice as much as
139
00:10:32 --> 00:10:34
will decrease this amount.
140
00:10:34 --> 00:10:37
And the book is somewhat
inconsistent in its use of
141
00:10:37 --> 00:10:38
2's in these equations.
142
00:10:38 --> 00:10:42
If you're forming two molecules
in your last step, there
143
00:10:42 --> 00:10:45
should be a 2, but 2 is not
always in the answer key.
144
00:10:45 --> 00:10:48
So what I've done in the past
is if you put a 2 and there
145
00:10:48 --> 00:10:51
should be a 2, great, but if
you don't put it, that's OK,
146
00:10:51 --> 00:10:55
too, so I have not taken off
for this because the book is
147
00:10:55 --> 00:10:58
not been completely consistent
in its use of that.
148
00:10:58 --> 00:11:04
But, if you see a two, and 2
things are being formed, you
149
00:11:04 --> 00:11:05
know where that comes from.
150
00:11:05 --> 00:11:10
So, 2 times k 2, that's the
rate constant of that last
151
00:11:10 --> 00:11:16
step, and that we're talking
about the concentration of o
152
00:11:16 --> 00:11:20
2 and the concentration
of n 2 o 2.
153
00:11:20 --> 00:11:25
So we're writing out just that
last out there with the 2,
154
00:11:25 --> 00:11:28
because two molecules of
n o 2 are being formed.
155
00:11:28 --> 00:11:32
But we're not done with this,
we can't use this, because it
156
00:11:32 --> 00:11:36
has an intermediate
term in there.
157
00:11:36 --> 00:11:39
And if we're going to write the
rate of a reaction, the rate of
158
00:11:39 --> 00:11:43
product formation, we can't
have any intermediates
159
00:11:43 --> 00:11:44
in our equation.
160
00:11:44 --> 00:11:48
We need to solve it in terms
of products and reactants.
161
00:11:48 --> 00:11:55
So, we need to
solve for n 2 o 2.
162
00:11:55 --> 00:12:02
So we need to solve for the
concentration of n 2 o 2,
163
00:12:02 --> 00:12:06
and substitute that
into our equation.
164
00:12:06 --> 00:12:08
So we need to solve for
it in terms of products
165
00:12:08 --> 00:12:10
and reactants.
166
00:12:10 --> 00:12:15
So, we need to think about how
this intermediate is formed,
167
00:12:15 --> 00:12:19
and we need to think about how
the intermediate is decomposed,
168
00:12:19 --> 00:12:22
and we need to think about how
this intermediate is consumed.
169
00:12:22 --> 00:12:38
So, the net formation of n 2 o
2 is going to be equal to the
170
00:12:38 --> 00:12:43
rate at which it's formed.
171
00:12:43 --> 00:12:46
So the rate at
which it's formed.
172
00:12:46 --> 00:12:51
What step is it formed in,
the intermediate formed in?
173
00:12:51 --> 00:12:55
It's in step one in the forward
direction, so the rate at
174
00:12:55 --> 00:13:04
which its formed is k
1 times n o squared.
175
00:13:04 --> 00:13:09
So that's the forward
rate of the first step.
176
00:13:09 --> 00:13:16
Then we need to think about the
rate at which its decomposed.
177
00:13:16 --> 00:13:21
What step does the
intermediate decompose in?
178
00:13:21 --> 00:13:23
Right, the reverse
of the first step.
179
00:13:23 --> 00:13:32
So we're going to put a minus
sign here, and k minus 1 times
180
00:13:32 --> 00:13:35
the concentration of
the intermediate.
181
00:13:35 --> 00:13:39
And then we also want to
think about the rate at
182
00:13:39 --> 00:13:45
which it's consumed.
183
00:13:45 --> 00:13:52
And it's consumed in the second
step, so that's k 2 times the
184
00:13:52 --> 00:13:56
concentration of the
intermediate times the
185
00:13:56 --> 00:14:01
concentration of o 2.
186
00:14:01 --> 00:14:03
So that's the net formation.
187
00:14:03 --> 00:14:07
And now we can use something
called the steady state
188
00:14:07 --> 00:14:11
approximation, and the steady
state approximation is used in
189
00:14:11 --> 00:14:15
all sorts of kinetics,
including enzyme kinetics, and
190
00:14:15 --> 00:14:19
it can be phrased two different
ways, which are equivalent.
191
00:14:19 --> 00:14:22
You can say that the steady
state approximation is that
192
00:14:22 --> 00:14:27
the net formation of the
intermediate equals zero --
193
00:14:27 --> 00:14:31
i.e. this is the net formation
of the intermediate, so this
194
00:14:31 --> 00:14:34
entire equation equals zero.
195
00:14:34 --> 00:14:38
Another way to say it is that
the rate of formation of the
196
00:14:38 --> 00:14:42
intermediate equals
the rate of decay.
197
00:14:42 --> 00:14:46
So that's saying the same
thing, that this term is going
198
00:14:46 --> 00:14:49
to be equal to those two terms.
199
00:14:49 --> 00:14:51
That's saying the same thing.
200
00:14:51 --> 00:14:55
So, if we can set this whole
equation equal to zero, then
201
00:14:55 --> 00:14:58
we can solve for the
concentrations of the
202
00:14:58 --> 00:15:03
intermediates in terms of the
rate constants and the
203
00:15:03 --> 00:15:06
concentrations of the reactants
and the products, which
204
00:15:06 --> 00:15:09
is what we need to do.
205
00:15:09 --> 00:15:12
So, let's do that.
206
00:15:12 --> 00:15:16
So we can rearrange this
expression now that we have set
207
00:15:16 --> 00:15:26
equal to zero, and so we can
bring n 2 o 2 over to one
208
00:15:26 --> 00:15:33
side, so we have our
intermediate on one side.
209
00:15:33 --> 00:15:40
And then we are going to have k
minus 1, and our little k 2
210
00:15:40 --> 00:15:44
times the concentration of o 2.
211
00:15:44 --> 00:15:49
And so, that is the rate in
which the intermediate is
212
00:15:49 --> 00:15:54
decomposed and consumed, and
that's going to be equal to the
213
00:15:54 --> 00:15:58
rate at which it's formed,
which is another way of
214
00:15:58 --> 00:16:01
expressing the steady
state approximation.
215
00:16:01 --> 00:16:05
So we've just moved the
decomposed and consumed to one
216
00:16:05 --> 00:16:08
side, and we have the rate at
which it's formed on the other
217
00:16:08 --> 00:16:13
side, and now we can easily
solve for the concentration of
218
00:16:13 --> 00:16:19
the n 2 o 2, so we're going
to do that over here.
219
00:16:19 --> 00:16:28
So that's now going to be equal
to our k 1 times n o squared
220
00:16:28 --> 00:16:45
over k minus 1 plus k 2 times
the concentration of o 2.
221
00:16:45 --> 00:16:49
So now we've just solved for n
2 o 2, our intermediate, in
222
00:16:49 --> 00:16:55
terms of our rate constants and
in terms of our reactants.
223
00:16:55 --> 00:16:56
So that's good.
224
00:16:56 --> 00:17:02
Now we need to take this and
substitute it back into here.
225
00:17:02 --> 00:17:08
And if we substitute it back
into here, let's do that over
226
00:17:08 --> 00:17:15
here, so now we'll have k, 2
times k 2, we'll also
227
00:17:15 --> 00:17:17
have a k 1 term.
228
00:17:17 --> 00:17:22
We're going to have,
we have n o squared.
229
00:17:22 --> 00:17:29
We also have an o 2 term from
over here, and this is all
230
00:17:29 --> 00:17:43
going to be over k minus
1 plus k 2 times our
231
00:17:43 --> 00:17:45
concentration of o 2.
232
00:17:45 --> 00:17:49
So we've just substituted that
term in here, and now we have a
233
00:17:49 --> 00:17:54
new expression for the rate of
our product formation,
234
00:17:54 --> 00:17:57
and that's lovely.
235
00:17:57 --> 00:18:01
Except that it doesn't
agree with the experiment.
236
00:18:01 --> 00:18:03
So we have a problem.
237
00:18:03 --> 00:18:08
In the experiment, there's n o
squared and o 2, but there is
238
00:18:08 --> 00:18:12
no o 2 in the bottom, so
something is going on.
239
00:18:12 --> 00:18:16
So that means that our reaction
as written, which didn't have
240
00:18:16 --> 00:18:20
any fast steps or any slow
steps is inconsistent if you
241
00:18:20 --> 00:18:22
write a mechanism for this, if
you write a rate law for
242
00:18:22 --> 00:18:25
this, it's inconsistent
with the experiment.
243
00:18:25 --> 00:18:28
So we need to do more to
our proposed mechanism.
244
00:18:28 --> 00:18:32
We need to add fast steps and
slow steps until we can write
245
00:18:32 --> 00:18:36
a mechanism that agrees
with the experiment.
246
00:18:36 --> 00:18:42
And so, if we want to do this,
we're going to assume, now we
247
00:18:42 --> 00:18:45
can try that the first step is
fast and the second
248
00:18:45 --> 00:18:47
step is slow.
249
00:18:47 --> 00:18:51
So let's talk about this slow
step and what happens if
250
00:18:51 --> 00:18:55
you have a slow step.
251
00:18:55 --> 00:19:00
So I'm going to introduce this
term of rate determining step.
252
00:19:00 --> 00:19:04
So the slowest step in an
elementary reaction is going to
253
00:19:04 --> 00:19:06
determine the overall rate.
254
00:19:06 --> 00:19:09
If it's slow compared
to the other steps.
255
00:19:09 --> 00:19:13
And so, the overall rate, it
can't be any faster than that
256
00:19:13 --> 00:19:17
slow step, and that slow step
is going to govern how fast
257
00:19:17 --> 00:19:18
the overall reaction is.
258
00:19:18 --> 00:19:21
All right, so let me give
you an example of this.
259
00:19:21 --> 00:19:25
I know they're you're all very
anxious to complete problem-set
260
00:19:25 --> 00:19:28
10, the last problem-set that
will be graded in this course.
261
00:19:28 --> 00:19:31
And you know, because I told
you at the beginning of class,
262
00:19:31 --> 00:19:36
that after today's lecture, you
will know all the material that
263
00:19:36 --> 00:19:39
will allow you to complete
that problem-set.
264
00:19:39 --> 00:19:41
So some of you are going to get
really antsy at the end of
265
00:19:41 --> 00:19:45
class thinking OK, I can go
and do that problem-set.
266
00:19:45 --> 00:19:49
So, you're going to be ready,
you may have already started
267
00:19:49 --> 00:19:52
packing up your backpack as the
class is winding down, you see
268
00:19:52 --> 00:19:54
we're getting close to the end
of the handouts saying, all
269
00:19:54 --> 00:19:56
right, I've got to get out of
here, I've got to get to
270
00:19:56 --> 00:20:00
the library and finish the
rest of problem-set 10.
271
00:20:00 --> 00:20:04
All right, so let's say it
takes you five seconds to
272
00:20:04 --> 00:20:06
pack up and be out the door.
273
00:20:06 --> 00:20:09
Or maybe there's some people
coming in, OK, maybe ten
274
00:20:09 --> 00:20:10
seconds to get out the door.
275
00:20:10 --> 00:20:12
Then you're running down toward
the library and you're moving
276
00:20:12 --> 00:20:15
pretty fast, you're jumping
over people that are stopping
277
00:20:15 --> 00:20:18
and talking to their friends,
you're moving very quickly, you
278
00:20:18 --> 00:20:21
might slide down the banisters
to get to the first floor, and
279
00:20:21 --> 00:20:25
you get to the library,
but you were too slow.
280
00:20:25 --> 00:20:26
Even though you did it really
fast, all the tables are
281
00:20:26 --> 00:20:29
busy, everyone else is
completing problem-set 10.
282
00:20:29 --> 00:20:33
So you go up and down, you're
looking for a table, there's
283
00:20:33 --> 00:20:36
no tables, they're all busy,
everyone has their chemistry
284
00:20:36 --> 00:20:37
textbooks out and
they're going.
285
00:20:37 --> 00:20:38
All right.
286
00:20:38 --> 00:20:41
So, then you have to leave the
library, you go to building
287
00:20:41 --> 00:20:44
2, that's close by, they're
classrooms, they're not in use,
288
00:20:44 --> 00:20:47
the first couple you check
there's recitations going on.
289
00:20:47 --> 00:20:48
Finally you find the
one that's empty.
290
00:20:48 --> 00:20:51
When you find it that's empty,
your backpack is off, your
291
00:20:51 --> 00:20:53
books are out, your
calculator's out
292
00:20:53 --> 00:20:54
and you're going.
293
00:20:54 --> 00:20:56
Right, maybe another
ten seconds.
294
00:20:56 --> 00:20:59
But it took you about
20 minutes to find
295
00:20:59 --> 00:21:01
that free table.
296
00:21:01 --> 00:21:05
So, maybe ten seconds to get
out of this classroom, ten
297
00:21:05 --> 00:21:08
seconds to get your books out
of your backpack, but then
298
00:21:08 --> 00:21:11
20 minutes overall to
find that free table.
299
00:21:11 --> 00:21:15
So that's the rate determining
step, that 20 minutes to find
300
00:21:15 --> 00:21:17
that table is going to control
the overall rate
301
00:21:17 --> 00:21:20
of the reaction.
302
00:21:20 --> 00:21:23
And so, many of you have
probably experienced rate
303
00:21:23 --> 00:21:26
determining steps in your life,
some of you may have friends
304
00:21:26 --> 00:21:31
that are always your rate
determining step, and so
305
00:21:31 --> 00:21:33
you know about this.
306
00:21:33 --> 00:21:34
You know about this.
307
00:21:34 --> 00:21:39
And so rate determining steps
allow us to come up with
308
00:21:39 --> 00:21:44
different kinds of mechanisms
and simplify different
309
00:21:44 --> 00:21:46
expressions.
310
00:21:46 --> 00:21:48
All right, so let's look at
what this does for us here.
311
00:21:48 --> 00:21:52
If we make assumptions now
about a step being fast
312
00:21:52 --> 00:21:55
and a step being slow.
313
00:21:55 --> 00:22:00
All right, so we're going to
be able to -- this expression
314
00:22:00 --> 00:22:04
here, we're going to
be able to simplify.
315
00:22:04 --> 00:22:08
All right, so we're going to
ask the question then, here,
316
00:22:08 --> 00:22:11
we're assuming that we're going
to say, all right, let's just
317
00:22:11 --> 00:22:14
say the first step is fast
and the second step is slow.
318
00:22:14 --> 00:22:17
We're going to see whether that
gives us an answer that's
319
00:22:17 --> 00:22:20
consistent with the
experimentally
320
00:22:20 --> 00:22:22
determined rate law.
321
00:22:22 --> 00:22:28
So here, if we say this is fast
and this is slow, basically
322
00:22:28 --> 00:22:32
we're asking the question, is
the decomposition of the
323
00:22:32 --> 00:22:35
intermediate or it's
consumption faster?
324
00:22:35 --> 00:22:39
So as we wrote this,
then, we're saying
325
00:22:39 --> 00:22:41
decomposition is fast.
326
00:22:41 --> 00:22:43
The first step is fast
and reversible, so the
327
00:22:43 --> 00:22:45
decomposition is fast.
328
00:22:45 --> 00:22:48
The second step is slow,
the consumption of that
329
00:22:48 --> 00:22:50
intermediate is slow.
330
00:22:50 --> 00:22:56
So that allows us to
change our rate.
331
00:22:56 --> 00:23:00
So, what we're then saying in
terms of the here's the words,
332
00:23:00 --> 00:23:04
here's sort of the equations,
we're saying that this term
333
00:23:04 --> 00:23:09
here, this k minus 1 times the
intermediate term, is a lot
334
00:23:09 --> 00:23:11
bigger, because this
is faster than this.
335
00:23:11 --> 00:23:13
This is slow, the rate
of consumption is slow.
336
00:23:13 --> 00:23:17
So k minus 1 is going
to be a bigger number
337
00:23:17 --> 00:23:21
than k 2 times o 2.
338
00:23:21 --> 00:23:24
So, if we say that, if we say
this is fast, that's a big
339
00:23:24 --> 00:23:28
number, this is a fast step, k
2 is a small number, it's slow,
340
00:23:28 --> 00:23:33
if this is a lot smaller than k
minus 1, and these are both
341
00:23:33 --> 00:23:38
appearing in the bottom of the
equation here, that allows
342
00:23:38 --> 00:23:40
you to get rid of this term.
343
00:23:40 --> 00:23:43
So if you have something really
big and something pretty small,
344
00:23:43 --> 00:23:46
on the bottom of this equation,
the really small thing is
345
00:23:46 --> 00:23:48
kind of insignificant.
346
00:23:48 --> 00:23:50
And so we can get
rid of that here.
347
00:23:50 --> 00:23:55
So we can get rid of this term.
348
00:23:55 --> 00:23:59
So that allows us to
simplify this expression.
349
00:23:59 --> 00:24:06
Now we have k 1 times n o
squared over k minus 1.
350
00:24:06 --> 00:24:10
We can also rearrange that and
bring our concentration terms
351
00:24:10 --> 00:24:14
on one side, so we have our
intermediate over our n o
352
00:24:14 --> 00:24:17
squared and have that now is
going to be equal to
353
00:24:17 --> 00:24:20
k 1 over k minus 1.
354
00:24:20 --> 00:24:30
What is little rate constant k
1 over rate constant k minus 1?
355
00:24:30 --> 00:24:32
Big k, right.
356
00:24:32 --> 00:24:36
So that's another way of saying
the equilibrium expression.
357
00:24:36 --> 00:24:40
So it's the equilibrium
constant for the first step is
358
00:24:40 --> 00:24:43
equal to k 1 over k minus 1.
359
00:24:43 --> 00:24:48
So basically what we're saying
then, if we have the first step
360
00:24:48 --> 00:24:51
being fast and reversible, and
the second step being slow,
361
00:24:51 --> 00:24:55
we're saying that the first
step is really an equilibrium.
362
00:24:55 --> 00:24:59
So what allows us to say that?
363
00:24:59 --> 00:25:03
Well, if you have a fast
reversible step followed by a
364
00:25:03 --> 00:25:07
slow step, that means that not
much of your intermediate is
365
00:25:07 --> 00:25:10
being siphoned off by the
second step, allowing you
366
00:25:10 --> 00:25:12
to reach equilibrium.
367
00:25:12 --> 00:25:15
So you can think about that
in terms of this plot here.
368
00:25:15 --> 00:25:18
If you have reactants going to
intermediate, and this is fast
369
00:25:18 --> 00:25:22
and reversible, so they're
going back and forth, fast,
370
00:25:22 --> 00:25:27
reversible, and very little of
that intermediate is getting
371
00:25:27 --> 00:25:29
siphoned off, this
is very slow.
372
00:25:29 --> 00:25:33
So very, very little of this is
being siphoned off to products.
373
00:25:33 --> 00:25:37
So that allows that first step
to really reach equilibrium
374
00:25:37 --> 00:25:40
conditions, because this
doesn't really factor in,
375
00:25:40 --> 00:25:43
and so you reach an
equilibrium here.
376
00:25:43 --> 00:25:50
And that allows you to
simplify your expressions.
377
00:25:50 --> 00:25:54
So now, we can go back with our
simplified expression, and we
378
00:25:54 --> 00:25:58
can express it as rate constant
k 1 over rate constant k minus
379
00:25:58 --> 00:26:02
1 times n o squared, or we can
write it as the equilibrium
380
00:26:02 --> 00:26:06
constant 1 times n o squared,
and we can plug that in
381
00:26:06 --> 00:26:10
solving for our intermediate.
382
00:26:10 --> 00:26:16
And so, if we put that now back
into this expression, we have
383
00:26:16 --> 00:26:23
the 2 k 1, k 2, we have our
oxygen concentration and our n
384
00:26:23 --> 00:26:28
o squared over k minus 1, or we
can have equilibrium constant
385
00:26:28 --> 00:26:31
1 with our k 2 in our 2.
386
00:26:31 --> 00:26:35
And now, we're consistent.
387
00:26:35 --> 00:26:39
If we realize that the k
observed that was given in the
388
00:26:39 --> 00:26:43
experimental rate law is a
combination of all
389
00:26:43 --> 00:26:44
of our k terms.
390
00:26:44 --> 00:26:47
So it's a combination, you
can't, at least in this
391
00:26:47 --> 00:26:50
experiment, you weren't able to
come up with the individual
392
00:26:50 --> 00:26:53
rate constants, they just came
up with overall observed rate,
393
00:26:53 --> 00:26:56
and so that's all of
our terms in there.
394
00:26:56 --> 00:27:00
And so if we put that in here,
we realize that this is
395
00:27:00 --> 00:27:02
consistent with the experiment.
396
00:27:02 --> 00:27:06
And in doing these problems,
it's OK to leave your little
397
00:27:06 --> 00:27:10
rate constants, it's OK to use
equilibrium constants in there,
398
00:27:10 --> 00:27:15
you should show all of your
work, so that if you have a k
399
00:27:15 --> 00:27:18
obs in your final answer,
we can look back and see
400
00:27:18 --> 00:27:20
what that was equal to.
401
00:27:20 --> 00:27:23
So it's hard not to show all
your work as you're writing
402
00:27:23 --> 00:27:27
through all of these on a test,
but you can end your answer,
403
00:27:27 --> 00:27:32
it would be correct end your
answer here or here or here.
404
00:27:32 --> 00:27:36
So I don't care which k is
in your final answer, but
405
00:27:36 --> 00:27:38
definitely show all your work.
406
00:27:38 --> 00:27:42
So that agrees with
the experiment.
407
00:27:42 --> 00:27:45
All right, so let's look
at another example now.
408
00:27:45 --> 00:27:49
And now we have, again, a
2-step reaction, and the first
409
00:27:49 --> 00:27:53
step is fast and reversible,
and the second step is slow.
410
00:27:53 --> 00:27:58
So, with those predictions of
the mechanism, let's just go
411
00:27:58 --> 00:28:03
right through and try to solve
for the rate here without going
412
00:28:03 --> 00:28:06
through all of the steps we did
the first time and see if we
413
00:28:06 --> 00:28:09
can do this more efficiently.
414
00:28:09 --> 00:28:14
And again, this is a reaction
with ozone, and depletion of
415
00:28:14 --> 00:28:18
ozone is a major problem that
will be facing us, that has
416
00:28:18 --> 00:28:23
faced us, and will continue
to face us in the future.
417
00:28:23 --> 00:28:24
All right.
418
00:28:24 --> 00:28:29
So lets first think about the
rates of the individual steps.
419
00:28:29 --> 00:28:32
So the rate of the
forward reaction here
420
00:28:32 --> 00:28:36
is going to be what?
421
00:28:36 --> 00:28:38
Just yell it out.
422
00:28:38 --> 00:28:43
Yup, k 1 times the
concentration of o 3.
423
00:28:43 --> 00:28:49
What about for the
reverse step?
424
00:28:49 --> 00:28:53
Yup, it's like clear and then
it gets mumbly. k minus 1 times
425
00:28:53 --> 00:28:56
the concentration of o 2 times
this concentration of o,
426
00:28:56 --> 00:28:58
which is an intermediate.
427
00:28:58 --> 00:29:01
All right, so let's look
at the second step here.
428
00:29:01 --> 00:29:06
So the rate here is going to be
k 2 times this intermediate --
429
00:29:06 --> 00:29:09
concentration of this
intermediate, o times the
430
00:29:09 --> 00:29:13
concentration of o 3.
431
00:29:13 --> 00:29:17
All right, so now we know that
this second step is going to be
432
00:29:17 --> 00:29:21
controlling the rate, it's a
slow step, at least that's our
433
00:29:21 --> 00:29:25
prediction that we're
going to write a rate
434
00:29:25 --> 00:29:26
expression based on.
435
00:29:26 --> 00:29:29
And so, that's going to
be our slow step, our
436
00:29:29 --> 00:29:32
rate determining step.
437
00:29:32 --> 00:29:36
And again, we're forming two
molecules of o 2 in this step.
438
00:29:36 --> 00:29:39
So we can have a two in there.
439
00:29:39 --> 00:29:44
And that the rate of formation
of o 2 in this last step,
440
00:29:44 --> 00:29:49
2 times k 2 times our
intermediate o times o 3.
441
00:29:49 --> 00:29:50
Are we done?
442
00:29:50 --> 00:29:54
Was it that easy?
443
00:29:54 --> 00:29:56
Are we done?
444
00:29:56 --> 00:29:57
No.
445
00:29:57 --> 00:30:01
Why isn't this finished?
446
00:30:01 --> 00:30:03
There's an intermediate.
447
00:30:03 --> 00:30:06
Intermediates will not be your
friend in doing these problems.
448
00:30:06 --> 00:30:10
Yes, we have an intermediate
so we're not complete.
449
00:30:10 --> 00:30:13
We need to solve for o in
terms of products, reactants,
450
00:30:13 --> 00:30:15
and rate constants.
451
00:30:15 --> 00:30:17
So we're not done.
452
00:30:17 --> 00:30:20
So we need to get rid
of the intermediate.
453
00:30:20 --> 00:30:25
But now, we can do that more
simply because we've set this
454
00:30:25 --> 00:30:27
up that we have a fast
reversible step followed
455
00:30:27 --> 00:30:29
by a slow step.
456
00:30:29 --> 00:30:32
So we can solve for the
concentration of that
457
00:30:32 --> 00:30:36
intermediate in terms of
equilibrium expressions.
458
00:30:36 --> 00:30:41
So, if you haven't learned how
to write an equilibrium
459
00:30:41 --> 00:30:44
expression yet, remember you
need it for the unit on
460
00:30:44 --> 00:30:48
chemical equilibrium, the unit
of acid base, the unit of
461
00:30:48 --> 00:30:52
oxidation reduction, and here
you need it again in kinetics,
462
00:30:52 --> 00:30:57
so definitely want to be able
to write those for the final.
463
00:30:57 --> 00:31:00
So we can go ahead
and write those.
464
00:31:00 --> 00:31:04
So, for the first step, again,
it's products over reactants.
465
00:31:04 --> 00:31:09
You can also express that in
terms of rate constant k 1 over
466
00:31:09 --> 00:31:14
rate constant k minus 1, and
that's all equal to our big k,
467
00:31:14 --> 00:31:17
our equilibrium constant, k 1.
468
00:31:17 --> 00:31:22
And so now we can solve in
terms of o, our intermediate,
469
00:31:22 --> 00:31:26
and so here, we pull that out,
here we've solved for it in
470
00:31:26 --> 00:31:32
terms of k 1 times o 3 over k
minus 1 times o 2, so we just
471
00:31:32 --> 00:31:34
brought those to
the other side.
472
00:31:34 --> 00:31:37
You could have also used the
equilibrium constant here
473
00:31:37 --> 00:31:42
instead of the rate
constants, either one is OK.
474
00:31:42 --> 00:31:45
So, now we've solved for it
here, and we can put it down
475
00:31:45 --> 00:31:47
into this expression here.
476
00:31:47 --> 00:31:51
So we are going back to
our overall expression
477
00:31:51 --> 00:31:51
that we wrote.
478
00:31:51 --> 00:31:56
So the expression that we wrote
based on this was 2, k 2 times
479
00:31:56 --> 00:31:58
the concentration of our
intermediate, times the
480
00:31:58 --> 00:32:00
concentration of o 3.
481
00:32:00 --> 00:32:04
Now we substitute in for the
intermediate, so we have 2
482
00:32:04 --> 00:32:11
times k 2, k 1, we have o 3
and an o 3, so that squared,
483
00:32:11 --> 00:32:17
over k minus 1 times o 2.
484
00:32:17 --> 00:32:21
And this can also be expressed
in terms of k observed, putting
485
00:32:21 --> 00:32:27
those terms together,
o 3 squared over o 2.
486
00:32:27 --> 00:32:31
And that would be the answer
to what the rate is for that
487
00:32:31 --> 00:32:35
particular proposed mechanism.
488
00:32:35 --> 00:32:40
So, let's think about if that's
true, what we should expect if
489
00:32:40 --> 00:32:44
we do some experiments, what we
should expect about the order
490
00:32:44 --> 00:32:46
of the reactions, and what
would happen if you double
491
00:32:46 --> 00:32:50
concentrations of things
and look for the effect.
492
00:32:50 --> 00:32:54
So if you're going to test this
proposal, what would we get?
493
00:32:54 --> 00:32:59
So, what is the overall order
of this rate law here.
494
00:32:59 --> 00:33:04
Oh, sorry, the order
first let's do of o 3.
495
00:33:04 --> 00:33:06
2.
496
00:33:06 --> 00:33:11
So if we double the
concentration of o 3, what
497
00:33:11 --> 00:33:16
would you expect to see
in terms of the rate?
498
00:33:16 --> 00:33:18
It will what?
499
00:33:18 --> 00:33:21
Yup, quadruple.
500
00:33:21 --> 00:33:30
What is the overall order of o
2, or order of o 2, and that is
501
00:33:30 --> 00:33:33
a clicker question, actually.
502
00:33:33 --> 00:33:39
So you know what
some people think.
503
00:33:39 --> 00:33:42
And not only tell me the
order, but tell me the
504
00:33:42 --> 00:34:29
effect of doubling.
505
00:34:29 --> 00:34:45
OK, let's just take
10 more seconds.
506
00:34:45 --> 00:34:57
Very good.
507
00:34:57 --> 00:35:00
All right, so the overall minus
1, that's what it means if it's
508
00:35:00 --> 00:35:03
on the bottom of that
equation there.
509
00:35:03 --> 00:35:08
And so if we double it
then it should half.
510
00:35:08 --> 00:35:13
So overall then, what is
the order of the reaction?
511
00:35:13 --> 00:35:15
It's 1.
512
00:35:15 --> 00:35:20
And so, remember you can sum up
the order of the individual
513
00:35:20 --> 00:35:24
ones to get the overall order,
so the overall order is 1.
514
00:35:24 --> 00:35:29
And what about if you double
both things, what's going
515
00:35:29 --> 00:36:02
to happen to the rate?
516
00:36:02 --> 00:36:19
OK, let's just take
10 more seconds.
517
00:36:19 --> 00:36:24
OK, people are doing
quite well on this.
518
00:36:24 --> 00:36:29
All right, so you double
both and you double.
519
00:36:29 --> 00:36:32
So, everyone's in very good
shape on these types of
520
00:36:32 --> 00:36:34
problems, which is excellent
because this is worth
521
00:36:34 --> 00:36:36
a lot on the final.
522
00:36:36 --> 00:36:42
All right, so let's do a final
example, and in this case we're
523
00:36:42 --> 00:36:47
given an experimental rate law,
which is up here, k observed
524
00:36:47 --> 00:36:52
times the concentration of n o
and the concentration of b r 2,
525
00:36:52 --> 00:36:58
and we want to figure out from
this, we'll look at if we have
526
00:36:58 --> 00:37:02
one step is being fast and the
other step is being fast, and
527
00:37:02 --> 00:37:06
see which would be consistent
with this experimentally
528
00:37:06 --> 00:37:07
determined rate law.
529
00:37:07 --> 00:37:09
So these are some of the types
of problems you have, your
530
00:37:09 --> 00:37:12
given experimental rate law and
say tell me which step has to
531
00:37:12 --> 00:37:17
be fast and which has to be
slow to have a rate that's
532
00:37:17 --> 00:37:20
consistent with the
observed rate.
533
00:37:20 --> 00:37:26
So, in this reaction we have 2
n o plus b r 2 going 2 n o b r.
534
00:37:26 --> 00:37:29
So if we write this in the
first step, we have n o plus b
535
00:37:29 --> 00:37:35
r 2 going to an intermediate, n
o b r 2 with forward rate
536
00:37:35 --> 00:37:39
constant, k 1, and reverse
rate constant, k minus 1.
537
00:37:39 --> 00:37:42
And in the second step we have
the intermediate interacting
538
00:37:42 --> 00:37:45
with another molecule of n o
forming two molecules of
539
00:37:45 --> 00:37:49
n o b r, our product.
540
00:37:49 --> 00:37:52
So let's consider what the rate
of the forward reaction is
541
00:37:52 --> 00:38:00
here, and so that would be --
you can yell it out. k 1
542
00:38:00 --> 00:38:05
times -- yes, excellent.
543
00:38:05 --> 00:38:08
And for the reverse reaction
then we're going to have
544
00:38:08 --> 00:38:14
our k minus 1 times our
intermediate concentration.
545
00:38:14 --> 00:38:19
Now we can look at step 2 where
our intermediate is being
546
00:38:19 --> 00:38:26
consumed, and the rate would
equal then k 2 times the
547
00:38:26 --> 00:38:30
concentration of the
intermediate, n o b r 2 times
548
00:38:30 --> 00:38:33
the concentration of n o.
549
00:38:33 --> 00:38:37
All right, so we know nothing
now about fast or slow steps,
550
00:38:37 --> 00:38:41
so we can write the formation
of our product just in
551
00:38:41 --> 00:38:44
terms of this second step.
552
00:38:44 --> 00:38:47
And again, there are two
molecules being formed, so we
553
00:38:47 --> 00:38:53
have a 2, and 2 times this rate
for this second step, k 2
554
00:38:53 --> 00:38:56
times intermediate times n o.
555
00:38:56 --> 00:38:58
So again, now we have an
intermediate that we
556
00:38:58 --> 00:38:59
have to get rid of.
557
00:38:59 --> 00:39:02
We can't have an intermediate
in our final expression.
558
00:39:02 --> 00:39:05
But we can't use the
equilibrium constant right now
559
00:39:05 --> 00:39:08
because we don't know anything
about what are the fast
560
00:39:08 --> 00:39:09
or the slow steps.
561
00:39:09 --> 00:39:12
So we're going to write it the
long way with no assumption
562
00:39:12 --> 00:39:16
about fast or slow steps.
563
00:39:16 --> 00:39:20
So to do this, so we have our
intermediate, and so we're
564
00:39:20 --> 00:39:24
going to solve for it in terms
of products and reactants.
565
00:39:24 --> 00:39:28
So we're going to write about
the change in the concentration
566
00:39:28 --> 00:39:30
of that intermediate.
567
00:39:30 --> 00:39:34
So, the intermediate's being
formed in the first step.
568
00:39:34 --> 00:39:38
So we first write the rate
at which it's being formed.
569
00:39:38 --> 00:39:43
So it's being formed here, so
that's k 1 times n o times the
570
00:39:43 --> 00:39:49
concentration of b r 2, so this
is how it's being formed.
571
00:39:49 --> 00:39:51
And then we're going to
consider how it's being
572
00:39:51 --> 00:39:55
decomposed, and you told me
before it gets decomposed in
573
00:39:55 --> 00:39:57
the reverse of the first step.
574
00:39:57 --> 00:40:01
So we have a minus sign,
because this is a negative
575
00:40:01 --> 00:40:06
change in concentration, it's
being decomposed, and it's
576
00:40:06 --> 00:40:10
being decomposed with a rate
constant of k minus 1 times
577
00:40:10 --> 00:40:13
the concentration of
that intermediate.
578
00:40:13 --> 00:40:17
And the intermediate's also
being consumed, so that's
579
00:40:17 --> 00:40:20
decreasing the concentration
of the intermediate,
580
00:40:20 --> 00:40:23
and it's being consumed
in this second step.
581
00:40:23 --> 00:40:27
So it's being consumed here
with the rate constant
582
00:40:27 --> 00:40:30
of k 2 times the
intermediate times n o.
583
00:40:30 --> 00:40:33
So that second step.
584
00:40:33 --> 00:40:36
So this is how we write it if
we don't know anything about
585
00:40:36 --> 00:40:39
the rate at which -- what's
slow and what's fast.
586
00:40:39 --> 00:40:43
So this will always, always
work to write things this way.
587
00:40:43 --> 00:40:45
So that's the overall change
in the concentration of the
588
00:40:45 --> 00:40:49
intermediate, how it's being
formed, decomposed,
589
00:40:49 --> 00:40:51
and consumed.
590
00:40:51 --> 00:40:54
Now we can use the steady state
approximation, and you can
591
00:40:54 --> 00:40:56
always use the steady state
approximation in these
592
00:40:56 --> 00:41:00
problems, and that says that
this net change is going
593
00:41:00 --> 00:41:02
to be equal to zero.
594
00:41:02 --> 00:41:05
So the steady state
approximation, then, let's
595
00:41:05 --> 00:41:10
you set this entire
term equal to zero.
596
00:41:10 --> 00:41:13
And sometimes, a question you
might get is to say what the
597
00:41:13 --> 00:41:15
steady state approximation is.
598
00:41:15 --> 00:41:18
So you should be able to
articulate that in words as
599
00:41:18 --> 00:41:20
well as use it on a problem.
600
00:41:20 --> 00:41:23
So now we can set this all
equal to zero, and we can
601
00:41:23 --> 00:41:27
solve for our concentration
of our intermediate.
602
00:41:27 --> 00:41:31
So, rearranging this then,
bringing our intermediate terms
603
00:41:31 --> 00:41:34
on one side of the expression,
and again, here's the rate at
604
00:41:34 --> 00:41:37
which it's decomposed and
consumed being equal to the
605
00:41:37 --> 00:41:40
rate at which it's
being formed.
606
00:41:40 --> 00:41:43
And we can pull out our
intermediate, so then that's
607
00:41:43 --> 00:41:48
times k minus 1, k 2 times n o,
and that's equal to the rate at
608
00:41:48 --> 00:41:50
which it's being formed.
609
00:41:50 --> 00:41:54
And so now we can take that,
divide by that term, and we
610
00:41:54 --> 00:41:57
get an expression for the
concentration of the
611
00:41:57 --> 00:42:01
intermediate in terms of our
reactants and our
612
00:42:01 --> 00:42:02
rate constants.
613
00:42:02 --> 00:42:06
So now we need to plug that
back in to the original
614
00:42:06 --> 00:42:09
equation that we had.
615
00:42:09 --> 00:42:13
So this is what we solved for,
and now we can put it into
616
00:42:13 --> 00:42:17
here, and if you rearrange
that, we have the 2, we have
617
00:42:17 --> 00:42:21
the k 1, we have the k 2, we
have an n o here, we have an n
618
00:42:21 --> 00:42:26
o there, so it's squared, b r 2
here, and then on the bottom we
619
00:42:26 --> 00:42:31
have k minus 1 plus
k 2 times n o.
620
00:42:31 --> 00:42:35
OK, so that's great, but that's
not consistent with the
621
00:42:35 --> 00:42:39
experimentally determined rate
law, so something has to be
622
00:42:39 --> 00:42:42
fast and something
has to be slow.
623
00:42:42 --> 00:42:46
So let's first consider if the
first step is slow and the
624
00:42:46 --> 00:42:52
second step is fast, which
would mean that the second step
625
00:42:52 --> 00:42:55
is bigger, it's faster, that's
k 2 times n o would be
626
00:42:55 --> 00:42:57
bigger than k minus 1.
627
00:42:57 --> 00:43:02
So the second step then is
fast, faster than this.
628
00:43:02 --> 00:43:05
So now why don't you go
ahead and tell me how this
629
00:43:05 --> 00:44:01
simplifies if that is true.
630
00:44:01 --> 00:44:17
All right, let's just
take 10 more seconds.
631
00:44:17 --> 00:44:18
Yup.
632
00:44:18 --> 00:44:23
All right, let's look at why.
633
00:44:23 --> 00:44:26
All right, so that's the answer
and let's consider why.
634
00:44:26 --> 00:44:29
So that means if this is
much bigger than this
635
00:44:29 --> 00:44:31
term, that cancels out.
636
00:44:31 --> 00:44:37
If that cancels out, what else
cancels? k 2 also cancels, and
637
00:44:37 --> 00:44:41
what else cancels? one n o
cancels giving you
638
00:44:41 --> 00:44:42
this expression.
639
00:44:42 --> 00:44:45
And that is consistent
with the answer up there.
640
00:44:45 --> 00:44:49
So you could also see
that as the k observed.
641
00:44:49 --> 00:44:50
So that's consistent.
642
00:44:50 --> 00:44:52
And let's just quick
-- oh, first tell me
643
00:44:52 --> 00:44:55
the overall order.
644
00:44:55 --> 00:44:55
2.
645
00:44:55 --> 00:44:59
And now let's just look at what
would have been true if we had
646
00:44:59 --> 00:45:01
done it the other way around.
647
00:45:01 --> 00:45:03
So if we did it the other way
around, we'd say that the
648
00:45:03 --> 00:45:06
first step is fast so that's
a lot bigger than this.
649
00:45:06 --> 00:45:09
If this is a lot bigger
than that, we'd cancel out
650
00:45:09 --> 00:45:11
this second term here.
651
00:45:11 --> 00:45:13
Can anything else cancel?
652
00:45:13 --> 00:45:14
No.
653
00:45:14 --> 00:45:18
So, we're left with this
expression, and that is not
654
00:45:18 --> 00:45:21
consistent with this up
here, because we have
655
00:45:21 --> 00:45:23
this squared term.
656
00:45:23 --> 00:45:25
So even if you write it
as k observed, that's
657
00:45:25 --> 00:45:27
not consistent.
658
00:45:27 --> 00:45:31
And so, the overall order
for this would have been
659
00:45:31 --> 00:45:36
3, which is not consistent
with the experiment.
660
00:45:36 --> 00:45:39
All right, so now you know
everything you need to know
661
00:45:39 --> 00:45:43
to finish problem-set 10.
662
00:45:43 --> 00:45:47
Don't let anything be your
rate determining step.
663
00:45:47 --> 00:45:48