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PROFESSOR: Please note that this
is a clicker competition,

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00:00:28,020 --> 00:00:32,660
perhaps the last of the year,
before you put in your vote

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00:00:32,660 --> 00:00:37,070
for this particular
clicker question.

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00:00:37,070 --> 00:00:41,840
And the defending champs are
recitation three, and I

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00:00:41,840 --> 00:00:45,420
believe that no recitation
has won more than once.

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00:00:45,420 --> 00:00:47,080
Is that true?

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00:00:47,080 --> 00:00:52,570
Is there a recitation that
has two victories?

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00:00:52,570 --> 00:00:56,180
So, this might be your last
chance if your recitation has

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00:00:56,180 --> 00:01:01,040
not won, or if you want to be
decisive victors in the

17
00:01:01,040 --> 00:01:04,426
clicker competition to have a
repeat victory, this is your

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00:01:04,426 --> 00:01:04,770
chance -- no pressure.

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00:01:04,770 --> 00:01:24,210
But go ahead and put in
your clicker response.

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00:01:24,210 --> 00:01:44,550
OK, let's just take 10 more
seconds to click in.

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00:01:44,550 --> 00:01:53,410
OK, the answer here is b, and
the trick was to recognize

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00:01:53,410 --> 00:01:57,730
which is the equation for first
order half life, and

23
00:01:57,730 --> 00:02:01,210
which is the equation for
second order half life.

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00:02:01,210 --> 00:02:03,620
Most people got this right.

25
00:02:03,620 --> 00:02:07,420
And we picked this clicker
question to demonstrate a

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00:02:07,420 --> 00:02:11,620
point that will, for the final
exam, which is that you will

27
00:02:11,620 --> 00:02:14,620
have an equation sheet
of all the equations.

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00:02:14,620 --> 00:02:16,520
There'll be a few you have to
memorize, we'll let you know

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00:02:16,520 --> 00:02:20,230
what they are, but that's
a lot of equations.

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00:02:20,230 --> 00:02:23,510
So, so far on each hour exam,
you've had an equation sheet

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00:02:23,510 --> 00:02:26,420
that's been related to that
particular material.

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00:02:26,420 --> 00:02:29,370
On the final, you of an equation
sheet that has the

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00:02:29,370 --> 00:02:34,620
equations from exam 1, exam 2,
exam 3, and also the fourth

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00:02:34,620 --> 00:02:36,650
exam material.

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00:02:36,650 --> 00:02:40,280
And so, it doesn't say on them,
oh, first order half

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00:02:40,280 --> 00:02:44,510
life equation, colon, and then
the equation, it just has all

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00:02:44,510 --> 00:02:45,780
of the equations.

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00:02:45,780 --> 00:02:49,320
So one thing that you have to do
is recognize which equation

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00:02:49,320 --> 00:02:51,120
can be applied where.

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00:02:51,120 --> 00:02:54,060
And this is something that
people, occasionally that's

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00:02:54,060 --> 00:02:56,840
part of knowing the
material is to

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00:02:56,840 --> 00:02:58,510
know/recognize the equations.

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00:02:58,510 --> 00:03:00,610
You don't have to memorize them,
but you do have to be

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00:03:00,610 --> 00:03:03,950
able to recognize them and
know when to apply them.

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00:03:03,950 --> 00:03:06,550
So that's something you should
be thinking about as you're

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00:03:06,550 --> 00:03:09,070
reviewing the material
for the exam.

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00:03:09,070 --> 00:03:11,870
And we're just going to at jump
right in and talk about

48
00:03:11,870 --> 00:03:18,070
reaction mechanisms. So this is
the last type of question

49
00:03:18,070 --> 00:03:21,520
that's on problem-set 10.

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00:03:21,520 --> 00:03:26,010
So, when you're investigating
reaction mechanisms, you often

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00:03:26,010 --> 00:03:30,940
will have an experimentally
determined rate law, and then

52
00:03:30,940 --> 00:03:33,240
you want to try to come up
with a mechanism that's

53
00:03:33,240 --> 00:03:35,430
consistent with that rate law.

54
00:03:35,430 --> 00:03:48,670
So, for this particular
reaction, we have 2 n o plus o

55
00:03:48,670 --> 00:03:52,260
2 going to 2 n o 2 gas.

56
00:03:52,260 --> 00:03:55,920
And experimental rate laws
determined there was a rate

57
00:03:55,920 --> 00:04:00,310
constant called k obs, for k
observed, the observed rate

58
00:04:00,310 --> 00:04:02,250
constant that was calculated.

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00:04:02,250 --> 00:04:08,000
And we know that it's the order
of the reactions here.

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00:04:08,000 --> 00:04:11,880
Overall, what is the overall
order at this

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00:04:11,880 --> 00:04:15,360
particular rate law?

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00:04:15,360 --> 00:04:17,160
Three.

63
00:04:17,160 --> 00:04:23,020
So we have second order in n
o and first order in o 2.

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00:04:23,020 --> 00:04:29,110
So if it's overall order of
three, is it likely to occur

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00:04:29,110 --> 00:04:33,400
in one step?

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00:04:33,400 --> 00:04:34,220
No.

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00:04:34,220 --> 00:04:37,080
What would it be called if it
did occur in one step -- if

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00:04:37,080 --> 00:04:39,530
three things came together
at the same

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00:04:39,530 --> 00:04:42,670
time to form a product?

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00:04:42,670 --> 00:04:44,970
Yeah, thermonuclear.

71
00:04:44,970 --> 00:04:48,730
And we learned last time
that thermonuclear

72
00:04:48,730 --> 00:04:50,930
reactions are rare.

73
00:04:50,930 --> 00:04:55,020
So it's unlikely that those
three things, the two n o's

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00:04:55,020 --> 00:04:58,180
and the one o 2 are going to
come together at the same time

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00:04:58,180 --> 00:05:02,560
to form product, so
it probably has

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00:05:02,560 --> 00:05:04,750
more than one step.

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00:05:04,750 --> 00:05:09,250
So, on the board here and on
your notes, we have a two-step

78
00:05:09,250 --> 00:05:15,700
proposed mechanism, and let's
talk about these two steps and

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00:05:15,700 --> 00:05:20,060
come up with a rate law that's
consistent with this

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00:05:20,060 --> 00:05:23,740
mechanism, and then see if
that agrees with the

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00:05:23,740 --> 00:05:27,970
experimental rate law
that was determined.

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00:05:27,970 --> 00:05:31,160
So, in the first step of this
reaction, we have 2 n o's

83
00:05:31,160 --> 00:05:32,770
coming together.

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00:05:32,770 --> 00:05:35,910
There's a little rate constant,
k 1, above the first

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00:05:35,910 --> 00:05:40,130
arrow, and for the reverse
direction, we have k minus 1,

86
00:05:40,130 --> 00:05:43,060
so this is written as a
reversible reaction, and it's

87
00:05:43,060 --> 00:05:46,830
forming an intermediate
n 2 o 2.

88
00:05:46,830 --> 00:05:51,110
In the second step, we have the
o 2 coming in, reacting

89
00:05:51,110 --> 00:05:55,530
with that intermediate with a
rate constant, k 2, and it's

90
00:05:55,530 --> 00:05:59,700
forming two molecules
of n o 2.

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00:05:59,700 --> 00:06:02,690
So now we can write the
rates for each of

92
00:06:02,690 --> 00:06:04,800
these individual steps.

93
00:06:04,800 --> 00:06:10,400
We can write the rate for the
foward reaction here, and so

94
00:06:10,400 --> 00:06:15,200
that's going to be
equal to k 1.

95
00:06:15,200 --> 00:06:21,540
And we have n o, two n
o's, so n o squared.

96
00:06:21,540 --> 00:06:25,380
And again, this is a step or an
elementary reaction so we

97
00:06:25,380 --> 00:06:28,130
can write the reaction
exactly as it occurs.

98
00:06:28,130 --> 00:06:32,000
We can write a rate law for
that step using the

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00:06:32,000 --> 00:06:34,010
stiochiometry of the reaction.

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00:06:34,010 --> 00:06:36,550
For an overall reaction, you
can't do that, it has to be

101
00:06:36,550 --> 00:06:38,120
experimentally determined.

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00:06:38,120 --> 00:06:42,000
But for an elementary reaction
or a step in a mechanism, you

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00:06:42,000 --> 00:06:43,080
can do that.

104
00:06:43,080 --> 00:06:45,960
So we write it exactly
as it occurs.

105
00:06:45,960 --> 00:06:52,290
So, what would be the order of
this particular reaction as

106
00:06:52,290 --> 00:06:54,450
written here?

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00:06:54,450 --> 00:06:56,450
Two.

108
00:06:56,450 --> 00:06:58,870
And what about molecularity?

109
00:06:58,870 --> 00:07:06,250
It would be bimolecular.

110
00:07:06,250 --> 00:07:13,990
So we're just getting you used
to some of these terms.

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00:07:13,990 --> 00:07:17,840
All right, so then for the
reverse direction, the rate of

112
00:07:17,840 --> 00:07:23,250
the reverse direction would be
equal to what rate constant?

113
00:07:23,250 --> 00:07:25,500
k minus 1.

114
00:07:25,500 --> 00:07:28,450
Times what?

115
00:07:28,450 --> 00:07:34,370
Yeah, n 2 o 2.

116
00:07:34,370 --> 00:07:39,180
So what's the order here?

117
00:07:39,180 --> 00:07:40,300
One.

118
00:07:40,300 --> 00:07:43,470
And what's this called?

119
00:07:43,470 --> 00:07:53,800
Yeah, so that's called
unimolecular.

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00:07:53,800 --> 00:07:59,550
All right, what about step 2?

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00:07:59,550 --> 00:08:02,930
What would be the rate
here, and that

122
00:08:02,930 --> 00:08:47,340
is a clicker question.

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00:08:47,340 --> 00:09:01,250
OK, just 10 more seconds.

124
00:09:01,250 --> 00:09:09,450
Yup, so we write it exactly as
written, so we have a k 2

125
00:09:09,450 --> 00:09:13,550
times the concentration
of o 2 times the

126
00:09:13,550 --> 00:09:17,490
concentration of n 2 o 2.

127
00:09:17,490 --> 00:09:24,150
All right, so what is the
overall order here?

128
00:09:24,150 --> 00:09:25,220
Two.

129
00:09:25,220 --> 00:09:38,480
And again, that would
be the bimolecular.

130
00:09:38,480 --> 00:09:42,270
All right, so we have our two
steps, and we've written rates

131
00:09:42,270 --> 00:09:45,310
for all of those particular
steps.

132
00:09:45,310 --> 00:09:48,360
Now we're interested in writing
the rate of the

133
00:09:48,360 --> 00:09:53,500
overall reaction that
forms n o 2.

134
00:09:53,500 --> 00:10:09,440
So, let's think about what the
rate of n o 2 formation is.

135
00:10:09,440 --> 00:10:13,920
So, in the last step, we're
forming n o 2, and we can just

136
00:10:13,920 --> 00:10:17,000
use that last step to
write this out.

137
00:10:17,000 --> 00:10:21,320
We're forming two molecules
of n o 2.

138
00:10:21,320 --> 00:10:25,440
So we're going to put a 2 in
there, because the rate at

139
00:10:25,440 --> 00:10:30,780
which the concentration of this
will increase twice as

140
00:10:30,780 --> 00:10:34,180
much as will decrease
this amount.

141
00:10:34,180 --> 00:10:37,820
And the book is somewhat
inconsistent in its use of 2's

142
00:10:37,820 --> 00:10:38,910
in these equations.

143
00:10:38,910 --> 00:10:41,890
If you're forming two molecules
in your last step,

144
00:10:41,890 --> 00:10:45,560
there should be a 2, but 2 is
not always in the answer key.

145
00:10:45,560 --> 00:10:48,690
So what I've done in the past
is if you put a 2 and there

146
00:10:48,690 --> 00:10:51,660
should be a 2, great, but if you
don't put it, that's OK,

147
00:10:51,660 --> 00:10:55,010
too, so I have not taken off for
this because the book is

148
00:10:55,010 --> 00:10:58,800
not been completely consistent
in its use of that.

149
00:10:58,800 --> 00:11:04,300
But, if you see a two, and 2
things are being formed, you

150
00:11:04,300 --> 00:11:05,960
know where that comes from.

151
00:11:05,960 --> 00:11:10,340
So, 2 times k 2, that's the
rate constant of that last

152
00:11:10,340 --> 00:11:16,930
step, and that we're talking
about the concentration of o 2

153
00:11:16,930 --> 00:11:20,450
and the concentration
of n 2 o 2.

154
00:11:20,450 --> 00:11:25,150
So we're writing out just that
last out there with the 2,

155
00:11:25,150 --> 00:11:28,430
because two molecules of
n o 2 are being formed.

156
00:11:28,430 --> 00:11:32,050
But we're not done with this, we
can't use this, because it

157
00:11:32,050 --> 00:11:36,460
has an intermediate
term in there.

158
00:11:36,460 --> 00:11:39,610
And if we're going to write the
rate of a reaction, the

159
00:11:39,610 --> 00:11:43,500
rate of product formation, we
can't have any intermediates

160
00:11:43,500 --> 00:11:44,910
in our equation.

161
00:11:44,910 --> 00:11:48,790
We need to solve it in terms
of products and reactants.

162
00:11:48,790 --> 00:11:55,850
So, we need to solve
for n 2 o 2.

163
00:11:55,850 --> 00:12:03,110
So we need to solve for the
concentration of n 2 o 2, and

164
00:12:03,110 --> 00:12:06,040
substitute that into
our equation.

165
00:12:06,040 --> 00:12:07,460
So we need to solve
for it in terms of

166
00:12:07,460 --> 00:12:10,690
products and reactants.

167
00:12:10,690 --> 00:12:15,410
So, we need to think about how
this intermediate is formed,

168
00:12:15,410 --> 00:12:18,470
and we need to think about
how the intermediate is

169
00:12:18,470 --> 00:12:20,810
decomposed, and we need to
think about how this

170
00:12:20,810 --> 00:12:22,990
intermediate is consumed.

171
00:12:22,990 --> 00:12:38,150
So, the net formation of n 2 o
2 is going to be equal to the

172
00:12:38,150 --> 00:12:43,280
rate at which it's formed.

173
00:12:43,280 --> 00:12:46,830
So the rate at which
it's formed.

174
00:12:46,830 --> 00:12:51,640
What step is it formed in, the
intermediate formed in?

175
00:12:51,640 --> 00:12:55,230
It's in step one in the forward
direction, so the rate

176
00:12:55,230 --> 00:13:04,020
at which its formed is k
1 times n o squared.

177
00:13:04,020 --> 00:13:09,000
So that's the forward rate
of the first step.

178
00:13:09,000 --> 00:13:12,410
Then we need to think about
the rate at which its

179
00:13:12,410 --> 00:13:16,660
decomposed.

180
00:13:16,660 --> 00:13:21,830
What step does the intermediate
decompose in?

181
00:13:21,830 --> 00:13:23,780
Right, the reverse of
the first step.

182
00:13:23,780 --> 00:13:32,140
So we're going to put a minus
sign here, and k minus 1 times

183
00:13:32,140 --> 00:13:35,860
the concentration of
the intermediate.

184
00:13:35,860 --> 00:13:39,490
And then we also want to
think about the rate

185
00:13:39,490 --> 00:13:45,200
at which it's consumed.

186
00:13:45,200 --> 00:13:51,310
And it's consumed in the second
step, so that's k 2

187
00:13:51,310 --> 00:13:56,130
times the concentration of the
intermediate times the

188
00:13:56,130 --> 00:14:01,010
concentration of o 2.

189
00:14:01,010 --> 00:14:03,890
So that's the net formation.

190
00:14:03,890 --> 00:14:07,000
And now we can use something
called the steady state

191
00:14:07,000 --> 00:14:11,370
approximation, and the steady
state approximation is used in

192
00:14:11,370 --> 00:14:15,350
all sorts of kinetics, including
enzyme kinetics, and

193
00:14:15,350 --> 00:14:18,140
it can be phrased two different
ways, which are

194
00:14:18,140 --> 00:14:19,130
equivalent.

195
00:14:19,130 --> 00:14:22,840
You can say that the steady
state approximation is that

196
00:14:22,840 --> 00:14:27,030
the net formation of the
intermediate equals zero --

197
00:14:27,030 --> 00:14:31,190
i.e. this is the net formation
of the intermediate, so this

198
00:14:31,190 --> 00:14:34,850
entire equation equals zero.

199
00:14:34,850 --> 00:14:38,770
Another way to say it is that
the rate of formation of the

200
00:14:38,770 --> 00:14:42,590
intermediate equals
the rate of decay.

201
00:14:42,590 --> 00:14:46,600
So that's saying the same thing,
that this term is going

202
00:14:46,600 --> 00:14:49,930
to be equal to those
two terms. That's

203
00:14:49,930 --> 00:14:51,480
saying the same thing.

204
00:14:51,480 --> 00:14:55,550
So, if we can set this whole
equation equal to zero, then

205
00:14:55,550 --> 00:14:58,610
we can solve for the
concentrations of the

206
00:14:58,610 --> 00:15:03,240
intermediates in terms of the
rate constants and the

207
00:15:03,240 --> 00:15:06,740
concentrations of the reactants
and the products,

208
00:15:06,740 --> 00:15:09,520
which is what we need to do.

209
00:15:09,520 --> 00:15:12,280
So, let's do that.

210
00:15:12,280 --> 00:15:16,200
So we can rearrange this
expression now that we have

211
00:15:16,200 --> 00:15:26,800
set equal to zero, and so we can
bring n 2 o 2 over to one

212
00:15:26,800 --> 00:15:33,010
side, so we have our
intermediate on one side.

213
00:15:33,010 --> 00:15:40,620
And then we are going to have k
minus 1, and our little k 2

214
00:15:40,620 --> 00:15:44,080
times the concentration
of o 2.

215
00:15:44,080 --> 00:15:49,730
And so, that is the rate in
which the intermediate is

216
00:15:49,730 --> 00:15:54,770
decomposed and consumed, and
that's going to be equal to

217
00:15:54,770 --> 00:15:58,470
the rate at which it's formed,
which is another way of

218
00:15:58,470 --> 00:16:01,310
expressing the steady
state approximation.

219
00:16:01,310 --> 00:16:05,070
So we've just moved the
decomposed and consumed to one

220
00:16:05,070 --> 00:16:08,280
side, and we have the rate at
which it's formed on the other

221
00:16:08,280 --> 00:16:13,970
side, and now we can easily
solve for the concentration of

222
00:16:13,970 --> 00:16:19,920
the n 2 o 2, so we're going
to do that over here.

223
00:16:19,920 --> 00:16:26,360
So that's now going to be equal
to our k 1 times n o

224
00:16:26,360 --> 00:16:42,790
squared over k minus 1
plus k 2 times the

225
00:16:42,790 --> 00:16:45,020
concentration of o 2.

226
00:16:45,020 --> 00:16:49,760
So now we've just solved for n
2 o 2, our intermediate, in

227
00:16:49,760 --> 00:16:55,250
terms of our rate constants and
in terms of our reactants.

228
00:16:55,250 --> 00:16:56,560
So that's good.

229
00:16:56,560 --> 00:17:02,890
Now we need to take this and
substitute it back into here.

230
00:17:02,890 --> 00:17:08,780
And if we substitute it back
into here, let's do that over

231
00:17:08,780 --> 00:17:14,900
here, so now we'll have
k, 2 times k 2, we'll

232
00:17:14,900 --> 00:17:17,760
also have a k 1 term.

233
00:17:17,760 --> 00:17:22,830
We're going to have, we
have n o squared.

234
00:17:22,830 --> 00:17:29,190
We also have an o 2 term from
over here, and this is all

235
00:17:29,190 --> 00:17:43,200
going to be over k minus
1 plus k 2 times our

236
00:17:43,200 --> 00:17:45,970
concentration of o 2.

237
00:17:45,970 --> 00:17:49,500
So we've just substituted that
term in here, and now we have

238
00:17:49,500 --> 00:17:56,490
a new expression for the rate of
our product formation, and

239
00:17:56,490 --> 00:17:57,830
that's lovely.

240
00:17:57,830 --> 00:18:01,860
Except that it doesn't agree
with the experiment.

241
00:18:01,860 --> 00:18:03,240
So we have a problem.

242
00:18:03,240 --> 00:18:08,150
In the experiment, there's n o
squared and o 2, but there is

243
00:18:08,150 --> 00:18:12,170
no o 2 in the bottom, so
something is going on.

244
00:18:12,170 --> 00:18:15,640
So that means that our reaction
as written, which

245
00:18:15,640 --> 00:18:19,980
didn't have any fast steps or
any slow steps is inconsistent

246
00:18:19,980 --> 00:18:22,700
if you write a mechanism for
this, if you write a rate law

247
00:18:22,700 --> 00:18:25,560
for this, it's inconsistent
with the experiment.

248
00:18:25,560 --> 00:18:28,460
So we need to do more to
our proposed mechanism.

249
00:18:28,460 --> 00:18:32,370
We need to add fast steps and
slow steps until we can write

250
00:18:32,370 --> 00:18:36,280
a mechanism that agrees
with the experiment.

251
00:18:36,280 --> 00:18:42,200
And so, if we want to do this,
we're going to assume, now we

252
00:18:42,200 --> 00:18:44,980
can try that the first
step is fast and the

253
00:18:44,980 --> 00:18:47,270
second step is slow.

254
00:18:47,270 --> 00:18:51,890
So let's talk about this slow
step and what happens if you

255
00:18:51,890 --> 00:18:55,780
have a slow step.

256
00:18:55,780 --> 00:19:00,670
So I'm going to introduce this
term of rate determining step.

257
00:19:00,670 --> 00:19:04,720
So the slowest step in an
elementary reaction is going

258
00:19:04,720 --> 00:19:06,500
to determine the overall rate.

259
00:19:06,500 --> 00:19:09,420
If it's slow compared
to the other steps.

260
00:19:09,420 --> 00:19:13,490
And so, the overall rate, it
can't be any faster than that

261
00:19:13,490 --> 00:19:17,100
slow step, and that slow step
is going to govern how fast

262
00:19:17,100 --> 00:19:18,520
the overall reaction is.

263
00:19:18,520 --> 00:19:21,550
All right, so let me give
you an example of this.

264
00:19:21,550 --> 00:19:24,570
I know they're you're all
very anxious to complete

265
00:19:24,570 --> 00:19:27,330
problem-set 10, the last
problem-set that will be

266
00:19:27,330 --> 00:19:28,740
graded in this course.

267
00:19:28,740 --> 00:19:31,960
And you know, because I told you
at the beginning of class,

268
00:19:31,960 --> 00:19:36,100
that after today's lecture, you
will know all the material

269
00:19:36,100 --> 00:19:39,200
that will allow you to complete
that problem-set.

270
00:19:39,200 --> 00:19:41,900
So some of you are going to get
really antsy at the end of

271
00:19:41,900 --> 00:19:45,830
class thinking OK, I can go
and do that problem-set.

272
00:19:45,830 --> 00:19:49,080
So, you're going to be ready,
you may have already started

273
00:19:49,080 --> 00:19:52,490
packing up your backpack as the
class is winding down, you

274
00:19:52,490 --> 00:19:54,580
see we're getting close to the
end of the handouts saying,

275
00:19:54,580 --> 00:19:56,530
all right, I've got to get out
of here, I've got to get to

276
00:19:56,530 --> 00:20:00,750
the library and finish the
rest of problem-set 10.

277
00:20:00,750 --> 00:20:05,220
All right, so let's say it takes
you five seconds to pack

278
00:20:05,220 --> 00:20:06,680
up and be out the door.

279
00:20:06,680 --> 00:20:09,220
Or maybe there's some people
coming in, OK, maybe ten

280
00:20:09,220 --> 00:20:10,550
seconds to get out the door.

281
00:20:10,550 --> 00:20:12,610
Then you're running down toward
the library and you're

282
00:20:12,610 --> 00:20:14,850
moving pretty fast, you're
jumping over people that are

283
00:20:14,850 --> 00:20:17,510
stopping and talking to their
friends, you're moving very

284
00:20:17,510 --> 00:20:19,850
quickly, you might slide down
the banisters to get to the

285
00:20:19,850 --> 00:20:23,400
first floor, and you get
to the library, but

286
00:20:23,400 --> 00:20:25,420
you were too slow.

287
00:20:25,420 --> 00:20:27,280
Even though you did it really
fast, all the tables are busy,

288
00:20:27,280 --> 00:20:29,620
everyone else is completing
problem-set 10.

289
00:20:29,620 --> 00:20:33,040
So you go up and down, you're
looking for a table, there's

290
00:20:33,040 --> 00:20:36,090
no tables, they're all busy,
everyone has their chemistry

291
00:20:36,090 --> 00:20:37,850
textbooks out and
they're going.

292
00:20:37,850 --> 00:20:38,960
All right.

293
00:20:38,960 --> 00:20:42,130
So, then you have to leave the
library, you go to building 2,

294
00:20:42,130 --> 00:20:44,340
that's close by, they're
classrooms, they're not in

295
00:20:44,340 --> 00:20:46,040
use, the first couple
you check there's

296
00:20:46,040 --> 00:20:47,210
recitations going on.

297
00:20:47,210 --> 00:20:48,910
Finally you find the
one that's empty.

298
00:20:48,910 --> 00:20:51,870
When you find it that's empty,
your backpack is off, your

299
00:20:51,870 --> 00:20:53,280
books are out, your
calculator's

300
00:20:53,280 --> 00:20:54,600
out and you're going.

301
00:20:54,600 --> 00:20:56,700
Right, maybe another
ten seconds.

302
00:20:56,700 --> 00:20:59,360
But it took you about
20 minutes to

303
00:20:59,360 --> 00:21:01,370
find that free table.

304
00:21:01,370 --> 00:21:05,380
So, maybe ten seconds to get
out of this classroom, ten

305
00:21:05,380 --> 00:21:08,590
seconds to get your books out of
your backpack, but then 20

306
00:21:08,590 --> 00:21:11,150
minutes overall to find
that free table.

307
00:21:11,150 --> 00:21:15,030
So that's the rate determining
step, that 20 minutes to find

308
00:21:15,030 --> 00:21:17,020
that table is going
to control the

309
00:21:17,020 --> 00:21:20,080
overall rate of the reaction.

310
00:21:20,080 --> 00:21:23,100
And so, many of you have
probably experienced rate

311
00:21:23,100 --> 00:21:26,020
determining steps in your life,
some of you may have

312
00:21:26,020 --> 00:21:31,130
friends that are always your
rate determining step, and so

313
00:21:31,130 --> 00:21:33,270
you know about this.

314
00:21:33,270 --> 00:21:34,850
You know about this.

315
00:21:34,850 --> 00:21:39,740
And so rate determining steps
allow us to come up with

316
00:21:39,740 --> 00:21:44,040
different kinds of mechanisms
and simplify different

317
00:21:44,040 --> 00:21:46,010
expressions.

318
00:21:46,010 --> 00:21:48,840
All right, so let's look at what
this does for us here.

319
00:21:48,840 --> 00:21:52,970
If we make assumptions now about
a step being fast and a

320
00:21:52,970 --> 00:21:55,900
step being slow.

321
00:21:55,900 --> 00:22:00,130
All right, so we're going to be
able to -- this expression

322
00:22:00,130 --> 00:22:04,210
here, we're going to be
able to simplify.

323
00:22:04,210 --> 00:22:08,550
All right, so we're going to ask
the question then, here,

324
00:22:08,550 --> 00:22:11,330
we're assuming that we're going
to say, all right, let's

325
00:22:11,330 --> 00:22:13,410
just say the first step
is fast and the

326
00:22:13,410 --> 00:22:14,610
second step is slow.

327
00:22:14,610 --> 00:22:17,730
We're going to see whether that
gives us an answer that's

328
00:22:17,730 --> 00:22:20,270
consistent with the
experimentally

329
00:22:20,270 --> 00:22:22,760
determined rate law.

330
00:22:22,760 --> 00:22:26,860
So here, if we say this is
fast and this is slow,

331
00:22:26,860 --> 00:22:31,850
basically we're asking the
question, is the decomposition

332
00:22:31,850 --> 00:22:35,370
of the intermediate or it's
consumption faster?

333
00:22:35,370 --> 00:22:40,560
So as we wrote this, then, we're
saying decomposition is

334
00:22:40,560 --> 00:22:43,820
fast. The first step is fast
and reversible, so the

335
00:22:43,820 --> 00:22:47,820
decomposition is fast. The
second step is slow, the

336
00:22:47,820 --> 00:22:50,810
consumption of that intermediate
is slow.

337
00:22:50,810 --> 00:22:56,470
So that allows us to
change our rate.

338
00:22:56,470 --> 00:23:00,580
So, what we're then saying in
terms of the here's the words,

339
00:23:00,580 --> 00:23:04,860
here's sort of the equations,
we're saying that this term

340
00:23:04,860 --> 00:23:09,100
here, this k minus 1 times the
intermediate term, is a lot

341
00:23:09,100 --> 00:23:11,460
bigger, because this is
faster than this.

342
00:23:11,460 --> 00:23:13,990
This is slow, the rate of
consumption is slow.

343
00:23:13,990 --> 00:23:17,320
So k minus 1 is going to
be a bigger number

344
00:23:17,320 --> 00:23:21,600
than k 2 times o 2.

345
00:23:21,600 --> 00:23:24,920
So, if we say that, if we say
this is fast, that's a big

346
00:23:24,920 --> 00:23:28,070
number, this is a fast step,
k 2 is a small number, it's

347
00:23:28,070 --> 00:23:33,370
slow, if this is a lot smaller
than k minus 1, and these are

348
00:23:33,370 --> 00:23:37,870
both appearing in the bottom
of the equation here, that

349
00:23:37,870 --> 00:23:40,360
allows you to get rid
of this term.

350
00:23:40,360 --> 00:23:42,460
So if you have something really
big and something

351
00:23:42,460 --> 00:23:45,930
pretty small, on the bottom of
this equation, the really

352
00:23:45,930 --> 00:23:48,250
small thing is kind
of insignificant.

353
00:23:48,250 --> 00:23:50,510
And so we can get rid
of that here.

354
00:23:50,510 --> 00:23:55,980
So we can get rid
of this term.

355
00:23:55,980 --> 00:23:59,240
So that allows us to simplify
this expression.

356
00:23:59,240 --> 00:24:06,650
Now we have k 1 times n o
squared over k minus 1.

357
00:24:06,650 --> 00:24:10,570
We can also rearrange that and
bring our concentration terms

358
00:24:10,570 --> 00:24:14,930
on one side, so we have our
intermediate over our n o

359
00:24:14,930 --> 00:24:18,100
squared and have that now is
going to be equal to k 1

360
00:24:18,100 --> 00:24:20,660
over k minus 1.

361
00:24:20,660 --> 00:24:24,670
What is little rate constant
k 1 over rate

362
00:24:24,670 --> 00:24:30,430
constant k minus 1?

363
00:24:30,430 --> 00:24:32,710
Big k, right.

364
00:24:32,710 --> 00:24:35,260
So that's another way of
saying the equilibrium

365
00:24:35,260 --> 00:24:36,800
expression.

366
00:24:36,800 --> 00:24:40,600
So it's the equilibrium constant
for the first step is

367
00:24:40,600 --> 00:24:43,890
equal to k 1 over k minus 1.

368
00:24:43,890 --> 00:24:47,780
So basically what we're saying
then, if we have the first

369
00:24:47,780 --> 00:24:49,870
step being fast and reversible,
and the second

370
00:24:49,870 --> 00:24:53,400
step being slow, we're saying
that the first step is really

371
00:24:53,400 --> 00:24:55,330
an equilibrium.

372
00:24:55,330 --> 00:24:59,430
So what allows us to say that?

373
00:24:59,430 --> 00:25:03,100
Well, if you have a fast
reversible step followed by a

374
00:25:03,100 --> 00:25:07,220
slow step, that means that not
much of your intermediate is

375
00:25:07,220 --> 00:25:10,990
being siphoned off by the second
step, allowing you to

376
00:25:10,990 --> 00:25:12,630
reach equilibrium.

377
00:25:12,630 --> 00:25:15,920
So you can think about that in
terms of this plot here.

378
00:25:15,920 --> 00:25:18,970
If you have reactants going to
intermediate, and this is fast

379
00:25:18,970 --> 00:25:22,650
and reversible, so they're going
back and forth, fast,

380
00:25:22,650 --> 00:25:27,020
reversible, and very little of
that intermediate is getting

381
00:25:27,020 --> 00:25:29,410
siphoned off, this
is very slow.

382
00:25:29,410 --> 00:25:31,390
So very, very little
of this is being

383
00:25:31,390 --> 00:25:33,260
siphoned off to products.

384
00:25:33,260 --> 00:25:37,250
So that allows that first step
to really reach equilibrium

385
00:25:37,250 --> 00:25:41,070
conditions, because this doesn't
really factor in, and

386
00:25:41,070 --> 00:25:43,300
so you reach an equilibrium
here.

387
00:25:43,300 --> 00:25:50,020
And that allows you to simplify
your expressions.

388
00:25:50,020 --> 00:25:54,060
So now, we can go back with our
simplified expression, and

389
00:25:54,060 --> 00:25:57,360
we can express it as rate
constant k 1 over rate

390
00:25:57,360 --> 00:26:01,710
constant k minus 1 times n o
squared, or we can write it as

391
00:26:01,710 --> 00:26:05,890
the equilibrium constant 1 times
n o squared, and we can

392
00:26:05,890 --> 00:26:10,210
plug that in solving for
our intermediate.

393
00:26:10,210 --> 00:26:16,090
And so, if we put that now back
into this expression, we

394
00:26:16,090 --> 00:26:22,890
have the 2 k 1, k 2, we have our
oxygen concentration and

395
00:26:22,890 --> 00:26:28,140
our n o squared over k minus 1,
or we can have equilibrium

396
00:26:28,140 --> 00:26:31,620
constant 1 with our
k 2 in our 2.

397
00:26:31,620 --> 00:26:35,810
And now, we're consistent.

398
00:26:35,810 --> 00:26:39,330
If we realize that the k
observed that was given in the

399
00:26:39,330 --> 00:26:43,890
experimental rate law is a
combination of all of our k

400
00:26:43,890 --> 00:26:47,420
terms. So it's a combination,
you can't, at least in this

401
00:26:47,420 --> 00:26:50,450
experiment, you weren't able to
come up with the individual

402
00:26:50,450 --> 00:26:53,930
rate constants, they just came
up with overall observed rate,

403
00:26:53,930 --> 00:26:56,650
and so that's all of
our terms in there.

404
00:26:56,650 --> 00:27:00,260
And so if we put that in here,
we realize that this is

405
00:27:00,260 --> 00:27:02,880
consistent with the
experiment.

406
00:27:02,880 --> 00:27:06,280
And in doing these problems,
it's OK to leave your little

407
00:27:06,280 --> 00:27:10,340
rate constants, it's OK to use
equilibrium constants in

408
00:27:10,340 --> 00:27:15,190
there, you should show all of
your work, so that if you have

409
00:27:15,190 --> 00:27:18,960
a k obs in your final answer, we
can look back and see what

410
00:27:18,960 --> 00:27:20,410
that was equal to.

411
00:27:20,410 --> 00:27:23,630
So it's hard not to show all
your work as you're writing

412
00:27:23,630 --> 00:27:26,690
through all of these on a test,
but you can end your

413
00:27:26,690 --> 00:27:29,680
answer, it would be correct
end your answer

414
00:27:29,680 --> 00:27:32,200
here or here or here.

415
00:27:32,200 --> 00:27:36,150
So I don't care which k is
in your final answer, but

416
00:27:36,150 --> 00:27:38,380
definitely show all your work.

417
00:27:38,380 --> 00:27:42,350
So that agrees with
the experiment.

418
00:27:42,350 --> 00:27:45,340
All right, so let's look
at another example now.

419
00:27:45,340 --> 00:27:49,780
And now we have, again, a 2-step
reaction, and the first

420
00:27:49,780 --> 00:27:53,540
step is fast and reversible, and
the second step is slow.

421
00:27:53,540 --> 00:27:58,890
So, with those predictions of
the mechanism, let's just go

422
00:27:58,890 --> 00:28:02,920
right through and try to solve
for the rate here without

423
00:28:02,920 --> 00:28:05,820
going through all of the steps
we did the first time and see

424
00:28:05,820 --> 00:28:09,360
if we can do this more
efficiently.

425
00:28:09,360 --> 00:28:14,380
And again, this is a reaction
with ozone, and depletion of

426
00:28:14,380 --> 00:28:18,520
ozone is a major problem that
will be facing us, that has

427
00:28:18,520 --> 00:28:23,190
faced us, and will continue
to face us in the future.

428
00:28:23,190 --> 00:28:24,180
All right.

429
00:28:24,180 --> 00:28:29,030
So lets first think about the
rates of the individual steps.

430
00:28:29,030 --> 00:28:32,700
So the rate of the forward
reaction here

431
00:28:32,700 --> 00:28:36,190
is going to be what?

432
00:28:36,190 --> 00:28:38,570
Just yell it out.

433
00:28:38,570 --> 00:28:43,610
Yup, k 1 times the concentration
of o 3.

434
00:28:43,610 --> 00:28:49,690
What about for the
reverse step?

435
00:28:49,690 --> 00:28:53,040
Yup, it's like clear and then
it gets mumbly. k minus 1

436
00:28:53,040 --> 00:28:56,800
times the concentration of o 2
times this concentration of o,

437
00:28:56,800 --> 00:28:58,810
which is an intermediate.

438
00:28:58,810 --> 00:29:01,610
All right, so let's look at
the second step here.

439
00:29:01,610 --> 00:29:06,770
So the rate here is going to be
k 2 times this intermediate

440
00:29:06,770 --> 00:29:09,620
-- concentration of this
intermediate, o times the

441
00:29:09,620 --> 00:29:13,620
concentration of o 3.

442
00:29:13,620 --> 00:29:17,430
All right, so now we know that
this second step is going to

443
00:29:17,430 --> 00:29:20,390
be controlling the rate, it's
a slow step, at least that's

444
00:29:20,390 --> 00:29:25,070
our prediction that we're
going to write a rate

445
00:29:25,070 --> 00:29:26,870
expression based on.

446
00:29:26,870 --> 00:29:29,950
And so, that's going to be
our slow step, our rate

447
00:29:29,950 --> 00:29:32,010
determining step.

448
00:29:32,010 --> 00:29:36,990
And again, we're forming two
molecules of o 2 in this step.

449
00:29:36,990 --> 00:29:39,810
So we can have a two in there.

450
00:29:39,810 --> 00:29:44,350
And that the rate of formation
of o 2 in this last step, 2

451
00:29:44,350 --> 00:29:49,450
times k 2 times our intermediate
o times o 3.

452
00:29:49,450 --> 00:29:50,000
Are we done?

453
00:29:50,000 --> 00:29:54,080
Was it that easy?

454
00:29:54,080 --> 00:29:56,260
Are we done?

455
00:29:56,260 --> 00:29:57,390
No.

456
00:29:57,390 --> 00:30:01,630
Why isn't this finished?

457
00:30:01,630 --> 00:30:03,440
There's an intermediate.

458
00:30:03,440 --> 00:30:05,690
Intermediates will not be your
friend in doing these

459
00:30:05,690 --> 00:30:06,280
problems.

460
00:30:06,280 --> 00:30:10,060
Yes, we have an intermediate
so we're not complete.

461
00:30:10,060 --> 00:30:13,640
We need to solve for o in terms
of products, reactants,

462
00:30:13,640 --> 00:30:15,040
and rate constants.

463
00:30:15,040 --> 00:30:17,550
So we're not done.

464
00:30:17,550 --> 00:30:20,210
So we need to get rid
of the intermediate.

465
00:30:20,210 --> 00:30:25,240
But now, we can do that more
simply because we've set this

466
00:30:25,240 --> 00:30:27,660
up that we have a fast
reversible step

467
00:30:27,660 --> 00:30:29,430
followed by a slow step.

468
00:30:29,430 --> 00:30:32,590
So we can solve for the
concentration of that

469
00:30:32,590 --> 00:30:36,780
intermediate in terms of
equilibrium expressions.

470
00:30:36,780 --> 00:30:41,000
So, if you haven't learned how
to write an equilibrium

471
00:30:41,000 --> 00:30:44,830
expression yet, remember you
need it for the unit on

472
00:30:44,830 --> 00:30:48,550
chemical equilibrium, the unit
of acid base, the unit of

473
00:30:48,550 --> 00:30:52,890
oxidation reduction, and here
you need it again in kinetics,

474
00:30:52,890 --> 00:30:57,980
so definitely want to be able to
write those for the final.

475
00:30:57,980 --> 00:31:00,540
So we can go ahead
and write those.

476
00:31:00,540 --> 00:31:04,930
So, for the first step, again,
it's products over reactants.

477
00:31:04,930 --> 00:31:09,030
You can also express that in
terms of rate constant k 1

478
00:31:09,030 --> 00:31:13,300
over rate constant k minus 1,
and that's all equal to our

479
00:31:13,300 --> 00:31:17,000
big k, our equilibrium
constant, k 1.

480
00:31:17,000 --> 00:31:22,210
And so now we can solve in terms
of o, our intermediate,

481
00:31:22,210 --> 00:31:26,400
and so here, we pull that out,
here we've solved for it in

482
00:31:26,400 --> 00:31:32,930
terms of k 1 times o 3 over k
minus 1 times o 2, so we just

483
00:31:32,930 --> 00:31:34,570
brought those to
the other side.

484
00:31:34,570 --> 00:31:37,960
You could have also used the
equilibrium constant here

485
00:31:37,960 --> 00:31:42,360
instead of the rate constants,
either one is OK.

486
00:31:42,360 --> 00:31:45,970
So, now we've solved for it
here, and we can put it down

487
00:31:45,970 --> 00:31:47,630
into this expression here.

488
00:31:47,630 --> 00:31:49,860
So we are going back
to our overall

489
00:31:49,860 --> 00:31:51,860
expression that we wrote.

490
00:31:51,860 --> 00:31:56,150
So the expression that we wrote
based on this was 2, k 2

491
00:31:56,150 --> 00:31:58,790
times the concentration of our
intermediate, times the

492
00:31:58,790 --> 00:32:00,730
concentration of o 3.

493
00:32:00,730 --> 00:32:04,800
Now we substitute in for the
intermediate, so we have 2

494
00:32:04,800 --> 00:32:11,280
times k 2, k 1, we have o 3 and
an o 3, so that squared,

495
00:32:11,280 --> 00:32:17,590
over k minus 1 times o 2.

496
00:32:17,590 --> 00:32:21,690
And this can also be expressed
in terms of k observed,

497
00:32:21,690 --> 00:32:27,800
putting those terms together,
o 3 squared over o 2.

498
00:32:27,800 --> 00:32:31,850
And that would be the answer to
what the rate is for that

499
00:32:31,850 --> 00:32:35,090
particular proposed mechanism.

500
00:32:35,090 --> 00:32:39,320
So, let's think about if that's
true, what we should

501
00:32:39,320 --> 00:32:42,740
expect if we do some
experiments, what we should

502
00:32:42,740 --> 00:32:45,950
expect about the order of the
reactions, and what would

503
00:32:45,950 --> 00:32:48,960
happen if you double
concentrations of things and

504
00:32:48,960 --> 00:32:50,560
look for the effect.

505
00:32:50,560 --> 00:32:53,170
So if you're going to
test this proposal,

506
00:32:53,170 --> 00:32:54,860
what would we get?

507
00:32:54,860 --> 00:32:59,960
So, what is the overall order
of this rate law here.

508
00:32:59,960 --> 00:33:04,720
Oh, sorry, the order first
let's do of o 3.

509
00:33:04,720 --> 00:33:06,370
2.

510
00:33:06,370 --> 00:33:11,030
So if we double the
concentration of o 3, what

511
00:33:11,030 --> 00:33:16,990
would you expect to see
in terms of the rate?

512
00:33:16,990 --> 00:33:18,960
It will what?

513
00:33:18,960 --> 00:33:21,440
Yup, quadruple.

514
00:33:21,440 --> 00:33:29,895
What is the overall order of o
2, or order of o 2, and that

515
00:33:29,895 --> 00:33:33,260
is a clicker question,
actually.

516
00:33:33,260 --> 00:33:39,750
So you know what some
people think.

517
00:33:39,750 --> 00:33:42,495
And not only tell me the order,
but tell me the effect

518
00:33:42,495 --> 00:34:29,610
of doubling.

519
00:34:29,610 --> 00:34:45,280
OK, let's just take
10 more seconds.

520
00:34:45,280 --> 00:34:57,290
Very good.

521
00:34:57,290 --> 00:35:00,540
All right, so the overall minus
1, that's what it means

522
00:35:00,540 --> 00:35:03,100
if it's on the bottom of
that equation there.

523
00:35:03,100 --> 00:35:08,210
And so if we double it
then it should half.

524
00:35:08,210 --> 00:35:13,960
So overall then, what is the
order of the reaction?

525
00:35:13,960 --> 00:35:15,930
It's 1.

526
00:35:15,930 --> 00:35:20,280
And so, remember you can sum up
the order of the individual

527
00:35:20,280 --> 00:35:24,700
ones to get the overall order,
so the overall order is 1.

528
00:35:24,700 --> 00:35:29,850
And what about if you double
both things, what's going to

529
00:35:29,850 --> 00:36:02,320
happen to the rate?

530
00:36:02,320 --> 00:36:19,140
OK, let's just take
10 more seconds.

531
00:36:19,140 --> 00:36:24,130
OK, people are doing
quite well on this.

532
00:36:24,130 --> 00:36:29,600
All right, so you double
both and you double.

533
00:36:29,600 --> 00:36:32,340
So, everyone's in very good
shape on these types of

534
00:36:32,340 --> 00:36:34,350
problems, which is excellent
because this is worth

535
00:36:34,350 --> 00:36:36,340
a lot on the final.

536
00:36:36,340 --> 00:36:42,720
All right, so let's do a final
example, and in this case

537
00:36:42,720 --> 00:36:46,200
we're given an experimental rate
law, which is up here, k

538
00:36:46,200 --> 00:36:50,090
observed times the concentration
of n o and the

539
00:36:50,090 --> 00:36:56,210
concentration of b r 2, and we
want to figure out from this,

540
00:36:56,210 --> 00:37:00,100
we'll look at if we have one
step is being fast and the

541
00:37:00,100 --> 00:37:04,800
other step is being fast, and
see which would be consistent

542
00:37:04,800 --> 00:37:07,860
with this experimentally
determined rate law.

543
00:37:07,860 --> 00:37:09,890
So these are some of the types
of problems you have, your

544
00:37:09,890 --> 00:37:12,790
given experimental rate law and
say tell me which step has

545
00:37:12,790 --> 00:37:17,030
to be fast and which has to be
slow to have a rate that's

546
00:37:17,030 --> 00:37:20,270
consistent with the
observed rate.

547
00:37:20,270 --> 00:37:24,120
So, in this reaction we
have 2 n o plus b r 2

548
00:37:24,120 --> 00:37:26,550
going 2 n o b r.

549
00:37:26,550 --> 00:37:29,550
So if we write this in the first
step, we have n o plus b

550
00:37:29,550 --> 00:37:35,210
r 2 going to an intermediate,
n o b r 2 with forward rate

551
00:37:35,210 --> 00:37:39,020
constant, k 1, and reverse
rate constant, k minus 1.

552
00:37:39,020 --> 00:37:42,180
And in the second step we have
the intermediate interacting

553
00:37:42,180 --> 00:37:46,790
with another molecule of n o
forming two molecules of n o b

554
00:37:46,790 --> 00:37:49,620
r, our product.

555
00:37:49,620 --> 00:37:52,540
So let's consider what the rate
of the forward reaction

556
00:37:52,540 --> 00:38:00,520
is here, and so that would be
-- you can yell it out. k 1

557
00:38:00,520 --> 00:38:05,120
times -- yes, excellent.

558
00:38:05,120 --> 00:38:09,020
And for the reverse reaction
then we're going to have our k

559
00:38:09,020 --> 00:38:14,980
minus 1 times our intermediate
concentration.

560
00:38:14,980 --> 00:38:19,250
Now we can look at step 2 where
our intermediate is

561
00:38:19,250 --> 00:38:26,120
being consumed, and the rate
would equal then k 2 times the

562
00:38:26,120 --> 00:38:30,250
concentration of the
intermediate, n o b r 2 times

563
00:38:30,250 --> 00:38:33,300
the concentration of n o.

564
00:38:33,300 --> 00:38:37,580
All right, so we know nothing
now about fast or slow steps,

565
00:38:37,580 --> 00:38:41,540
so we can write the formation
of our product just in terms

566
00:38:41,540 --> 00:38:44,280
of this second step.

567
00:38:44,280 --> 00:38:47,250
And again, there are two
molecules being formed, so we

568
00:38:47,250 --> 00:38:53,230
have a 2, and 2 times this rate
for this second step, k 2

569
00:38:53,230 --> 00:38:56,020
times intermediate times n o.

570
00:38:56,020 --> 00:38:58,980
So again, now we have an
intermediate that we have to

571
00:38:58,980 --> 00:38:59,640
get rid of.

572
00:38:59,640 --> 00:39:02,610
We can't have an intermediate
in our final expression.

573
00:39:02,610 --> 00:39:05,430
But we can't use the equilibrium
constant right now

574
00:39:05,430 --> 00:39:08,380
because we don't know anything
about what are the fast or the

575
00:39:08,380 --> 00:39:09,420
slow steps.

576
00:39:09,420 --> 00:39:12,840
So we're going to write it the
long way with no assumption

577
00:39:12,840 --> 00:39:16,300
about fast or slow steps.

578
00:39:16,300 --> 00:39:20,660
So to do this, so we have our
intermediate, and so we're

579
00:39:20,660 --> 00:39:24,610
going to solve for it in terms
of products and reactants.

580
00:39:24,610 --> 00:39:27,140
So we're going to write about
the change in the

581
00:39:27,140 --> 00:39:30,740
concentration of that
intermediate.

582
00:39:30,740 --> 00:39:34,610
So, the intermediate's being
formed in the first step.

583
00:39:34,610 --> 00:39:38,810
So we first write the rate at
which it's being formed.

584
00:39:38,810 --> 00:39:43,860
So it's being formed here, so
that's k 1 times n o times the

585
00:39:43,860 --> 00:39:49,150
concentration of b r 2, so this
is how it's being formed.

586
00:39:49,150 --> 00:39:51,690
And then we're going to consider
how it's being

587
00:39:51,690 --> 00:39:55,150
decomposed, and you told me
before it gets decomposed in

588
00:39:55,150 --> 00:39:57,800
the reverse of the first step.

589
00:39:57,800 --> 00:40:01,400
So we have a minus sign, because
this is a negative

590
00:40:01,400 --> 00:40:06,180
change in concentration, it's
being decomposed, and it's

591
00:40:06,180 --> 00:40:10,750
being decomposed with a rate
constant of k minus 1 times

592
00:40:10,750 --> 00:40:13,570
the concentration of
that intermediate.

593
00:40:13,570 --> 00:40:17,120
And the intermediate's also
being consumed, so that's

594
00:40:17,120 --> 00:40:20,630
decreasing the concentration of
the intermediate, and it's

595
00:40:20,630 --> 00:40:23,480
being consumed in this
second step.

596
00:40:23,480 --> 00:40:28,000
So it's being consumed here with
the rate constant of k 2

597
00:40:28,000 --> 00:40:30,320
times the intermediate
times n o.

598
00:40:30,320 --> 00:40:33,570
So that second step.

599
00:40:33,570 --> 00:40:36,550
So this is how we write it if
we don't know anything about

600
00:40:36,550 --> 00:40:39,940
the rate at which -- what's slow
and what's fast. So this

601
00:40:39,940 --> 00:40:43,550
will always, always work to
write things this way.

602
00:40:43,550 --> 00:40:45,900
So that's the overall change
in the concentration of the

603
00:40:45,900 --> 00:40:48,130
intermediate, how it's
being formed,

604
00:40:48,130 --> 00:40:51,440
decomposed, and consumed.

605
00:40:51,440 --> 00:40:54,100
Now we can use the steady state
approximation, and you

606
00:40:54,100 --> 00:40:56,570
can always use the steady state
approximation in these

607
00:40:56,570 --> 00:41:00,570
problems, and that says that
this net change is going to be

608
00:41:00,570 --> 00:41:02,750
equal to zero.

609
00:41:02,750 --> 00:41:05,560
So the steady state
approximation, then, let's you

610
00:41:05,560 --> 00:41:10,030
set this entire term
equal to zero.

611
00:41:10,030 --> 00:41:13,670
And sometimes, a question you
might get is to say what the

612
00:41:13,670 --> 00:41:15,450
steady state approximation is.

613
00:41:15,450 --> 00:41:18,110
So you should be able to
articulate that in words as

614
00:41:18,110 --> 00:41:20,990
well as use it on a problem.

615
00:41:20,990 --> 00:41:23,650
So now we can set this all
equal to zero, and we can

616
00:41:23,650 --> 00:41:27,260
solve for our concentration
of our intermediate.

617
00:41:27,260 --> 00:41:30,830
So, rearranging this then,
bringing our intermediate

618
00:41:30,830 --> 00:41:34,420
terms on one side of the
expression, and again, here's

619
00:41:34,420 --> 00:41:37,300
the rate at which it's
decomposed and consumed being

620
00:41:37,300 --> 00:41:40,940
equal to the rate at which
it's being formed.

621
00:41:40,940 --> 00:41:43,530
And we can pull out our
intermediate, so then that's

622
00:41:43,530 --> 00:41:48,090
times k minus 1, k 2 times n
o, and that's equal to the

623
00:41:48,090 --> 00:41:50,040
rate at which it's
being formed.

624
00:41:50,040 --> 00:41:54,370
And so now we can take that,
divide by that term, and we

625
00:41:54,370 --> 00:41:57,340
get an expression for the
concentration of the

626
00:41:57,340 --> 00:42:00,720
intermediate in terms
of our reactants

627
00:42:00,720 --> 00:42:02,650
and our rate constants.

628
00:42:02,650 --> 00:42:06,380
So now we need to plug that
back in to the original

629
00:42:06,380 --> 00:42:09,260
equation that we had.

630
00:42:09,260 --> 00:42:13,540
So this is what we solved for,
and now we can put it into

631
00:42:13,540 --> 00:42:17,690
here, and if you rearrange that,
we have the 2, we have

632
00:42:17,690 --> 00:42:21,470
the k 1, we have the k 2, we
have an n o here, we have an n

633
00:42:21,470 --> 00:42:26,130
o there, so it's squared, b r 2
here, and then on the bottom

634
00:42:26,130 --> 00:42:31,700
we have k minus 1 plus
k 2 times n o.

635
00:42:31,700 --> 00:42:35,770
OK, so that's great, but that's
not consistent with the

636
00:42:35,770 --> 00:42:39,410
experimentally determined rate
law, so something has to be

637
00:42:39,410 --> 00:42:42,770
fast and something
has to be slow.

638
00:42:42,770 --> 00:42:46,590
So let's first consider if the
first step is slow and the

639
00:42:46,590 --> 00:42:51,540
second step is fast, which would
mean that the second

640
00:42:51,540 --> 00:42:55,970
step is bigger, it's faster,
that's k 2 times n o would be

641
00:42:55,970 --> 00:42:57,710
bigger than k minus 1.

642
00:42:57,710 --> 00:43:02,430
So the second step then is
fast, faster than this.

643
00:43:02,430 --> 00:43:05,200
So now why don't you go ahead
and tell me how this

644
00:43:05,200 --> 00:44:01,530
simplifies if that is true.

645
00:44:01,530 --> 00:44:17,160
All right, let's just take
10 more seconds.

646
00:44:17,160 --> 00:44:18,060
Yup.

647
00:44:18,060 --> 00:44:23,350
All right, let's look at why.

648
00:44:23,350 --> 00:44:26,790
All right, so that's the answer
and let's consider why.

649
00:44:26,790 --> 00:44:30,310
So that means if this is much
bigger than this term, that

650
00:44:30,310 --> 00:44:31,380
cancels out.

651
00:44:31,380 --> 00:44:37,170
If that cancels out, what else
cancels? k 2 also cancels, and

652
00:44:37,170 --> 00:44:41,380
what else cancels? one n o
cancels giving you this

653
00:44:41,380 --> 00:44:42,390
expression.

654
00:44:42,390 --> 00:44:45,690
And that is consistent with
the answer up there.

655
00:44:45,690 --> 00:44:49,200
So you could also see that
as the k observed.

656
00:44:49,200 --> 00:44:50,570
So that's consistent.

657
00:44:50,570 --> 00:44:52,320
And let's just quick
-- oh, first tell

658
00:44:52,320 --> 00:44:55,130
me the overall order.

659
00:44:55,130 --> 00:44:55,970
2.

660
00:44:55,970 --> 00:44:58,950
And now let's just look at what
would have been true if

661
00:44:58,950 --> 00:45:01,060
we had done it the
other way around.

662
00:45:01,060 --> 00:45:03,560
So if we did it the other way
around, we'd say that the

663
00:45:03,560 --> 00:45:06,630
first step is fast so that's
a lot bigger than this.

664
00:45:06,630 --> 00:45:09,610
If this is a lot bigger than
that, we'd cancel out this

665
00:45:09,610 --> 00:45:11,050
second term here.

666
00:45:11,050 --> 00:45:13,200
Can anything else cancel?

667
00:45:13,200 --> 00:45:14,110
No.

668
00:45:14,110 --> 00:45:18,640
So, we're left with this
expression, and that is not

669
00:45:18,640 --> 00:45:21,810
consistent with this up
here, because we have

670
00:45:21,810 --> 00:45:23,360
this squared term.

671
00:45:23,360 --> 00:45:26,250
So even if you write it as
k observed, that's not

672
00:45:26,250 --> 00:45:27,500
consistent.

673
00:45:27,500 --> 00:45:33,770
And so, the overall order for
this would have been 3, which

674
00:45:33,770 --> 00:45:36,570
is not consistent with
the experiment.

675
00:45:36,570 --> 00:45:39,970
All right, so now you know
everything you need to know to

676
00:45:39,970 --> 00:45:43,970
finish problem-set 10.

677
00:45:43,970 --> 00:45:47,170
Don't let anything be your
rate determining step.