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PROFESSOR: So last time we were
starting more complicated

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00:00:24,480 --> 00:00:37,760
mechanisms and we looked
at series reactions.

10
00:00:37,760 --> 00:00:40,220
A goes B, which is
an intermediate.

11
00:00:40,220 --> 00:00:43,830
Goes to C with rates
k1 and k2.

12
00:00:43,830 --> 00:00:49,380
And we solved the
problem exactly.

13
00:00:49,380 --> 00:00:52,470
Which turned out to be a
little bit complicated.

14
00:00:52,470 --> 00:00:58,030
And then we were looking
at special cases.

15
00:00:58,030 --> 00:01:05,350
And I left the special case
of k1 equals k2 to do as a

16
00:01:05,350 --> 00:01:08,510
homework problem.

17
00:01:08,510 --> 00:01:14,390
And then on the board we did
k1 much greater than k2.

18
00:01:14,390 --> 00:01:22,680
Which looks like, if k1 is much
greater than k2, the rate

19
00:01:22,680 --> 00:01:27,660
into k1 is much greater than
k2, so using the analogy of

20
00:01:27,660 --> 00:01:35,860
buckets and pipes, it's a big
pipe between k1 and between A

21
00:01:35,860 --> 00:01:39,090
and B. And a little pipe
between B and C and you

22
00:01:39,090 --> 00:01:41,250
started with a bunch
of liquid on top.

23
00:01:41,250 --> 00:01:45,270
And you basically dump
it into here.

24
00:01:45,270 --> 00:01:49,380
And then you slowly
drip into C here.

25
00:01:49,380 --> 00:01:54,160
So the first thing that happens
is that A gets dumped,

26
00:01:54,160 --> 00:01:56,840
it becomes B very quickly.

27
00:01:56,840 --> 00:01:58,840
With a rate constant k1.

28
00:01:58,840 --> 00:02:02,260
In a first order fashion.

29
00:02:02,260 --> 00:02:07,460
So A is A0 e to the minus
k1 times time.

30
00:02:07,460 --> 00:02:10,710
Then B comes up sharply.

31
00:02:10,710 --> 00:02:18,610
And then drops down with rate
constant k2 slowly.

32
00:02:18,610 --> 00:02:24,750
So B is approximately A0 e to
the minus k2 times time.

33
00:02:24,750 --> 00:02:30,240
And then C has a very small
quadratic rise, followed by

34
00:02:30,240 --> 00:02:33,790
the exponential rise to
saturation that you'd expect.

35
00:02:33,790 --> 00:02:39,300
So C is approximately a0 times
one minus e to the minus k2

36
00:02:39,300 --> 00:02:40,080
times time.

37
00:02:40,080 --> 00:02:43,300
The rate constant k2 dominates
the long times.

38
00:02:43,300 --> 00:02:45,890
And everything it's looking like
it's pseudo first order

39
00:02:45,890 --> 00:02:47,310
at long times.

40
00:02:47,310 --> 00:02:51,680
So you can do this either by
thinking about it, solve the

41
00:02:51,680 --> 00:02:55,650
problem by thinking about it,
which is perfectly fine.

42
00:02:55,650 --> 00:02:58,250
Or you can do it by solving
it exactly, as we

43
00:02:58,250 --> 00:03:00,060
have done last time.

44
00:03:00,060 --> 00:03:02,500
And then putting in the
right approximations.

45
00:03:02,500 --> 00:03:06,320
Which in this case was Taylor's
approximation, if I

46
00:03:06,320 --> 00:03:08,900
remember right.

47
00:03:08,900 --> 00:03:12,110
And know just by looking at the
equation and putting the

48
00:03:12,110 --> 00:03:14,100
right approximation,
you get it out.

49
00:03:14,100 --> 00:03:22,030
And a third approximation, which
is the opposite one, k2

50
00:03:22,030 --> 00:03:28,260
much less than k1, so k1 much
less than k2, the way that

51
00:03:28,260 --> 00:03:39,720
that's going to look, we can
do it by inspection.

52
00:03:39,720 --> 00:03:43,970
So basically if we think of
buckets and pipes, that's a

53
00:03:43,970 --> 00:03:48,440
very thin pipe connecting the
bucket for A and B, so we drip

54
00:03:48,440 --> 00:03:52,560
from A to B, then we have this
very fat pipe connecting B and

55
00:03:52,560 --> 00:03:56,710
C. So as soon as a little B gets
in there, it immediately

56
00:03:56,710 --> 00:04:01,740
becomes C. So, the rate
determining step is the first

57
00:04:01,740 --> 00:04:06,620
one, A goes to B. And so k1, we
expect that to be the rate

58
00:04:06,620 --> 00:04:09,770
that dominates the long time.

59
00:04:09,770 --> 00:04:17,230
And we never expect to get any
significant amount of B to

60
00:04:17,230 --> 00:04:19,620
pile up in the middle bucket,
because this pipe is much

61
00:04:19,620 --> 00:04:22,180
bigger than that one here.

62
00:04:22,180 --> 00:04:28,860
Therefore, what we expect to see
is a slow decay in A, with

63
00:04:28,860 --> 00:04:32,730
rate constant k1 dominating
the long times, A is

64
00:04:32,730 --> 00:04:40,070
approximately A0 e to the
minus k1 times the time.

65
00:04:40,070 --> 00:04:46,910
We don't expect B to amount
to very much at all.

66
00:04:46,910 --> 00:04:49,220
It'll rise up a little bit at
the very early times, and

67
00:04:49,220 --> 00:04:52,210
it'll pretty much stay
very, very small.

68
00:04:52,210 --> 00:04:53,700
Not quite constant.

69
00:04:53,700 --> 00:05:01,450
And in fact if you look at the
full result, and you put in

70
00:05:01,450 --> 00:05:05,680
your approximations, and I'll
let you do that, you would

71
00:05:05,680 --> 00:05:10,675
find that B is essentially
related to A through a

72
00:05:10,675 --> 00:05:15,500
constant term where k1 is very
small compared to k2.

73
00:05:15,500 --> 00:05:17,530
So this is a very small
number here.

74
00:05:17,530 --> 00:05:20,120
B is equal to a very small
number times A. So it

75
00:05:20,120 --> 00:05:24,670
basically follows A, but at
a very very, low level.

76
00:05:24,670 --> 00:05:33,850
And then you would find that C
rises up in approximately e to

77
00:05:33,850 --> 00:05:37,990
the minus k1 times t, so
with rate constant k1.

78
00:05:37,990 --> 00:05:45,700
One minus e to the minus
k1 times the time.

79
00:05:45,700 --> 00:05:49,370
So the dominant rate is k1, and
in this case here this is

80
00:05:49,370 --> 00:05:52,250
an approximation that we're
going to see again.

81
00:05:52,250 --> 00:05:56,120
We're going to revisit this
approximation here when we get

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00:05:56,120 --> 00:05:57,870
to more complicated mechanisms,
it's going to

83
00:05:57,870 --> 00:06:01,340
become this steady state
approximation.

84
00:06:01,340 --> 00:06:03,710
Where the amount of the
intermediate, or the

85
00:06:03,710 --> 00:06:07,250
concentration of intermediate,
is low at all times.

86
00:06:07,250 --> 00:06:09,150
And doesn't change very fast.

87
00:06:09,150 --> 00:06:11,220
The slope here is very slow.

88
00:06:11,220 --> 00:06:14,980
Because k1 is so slow coming
down, and B is related to A,

89
00:06:14,980 --> 00:06:19,850
which means it's related to this
slow decrease in A, the

90
00:06:19,850 --> 00:06:22,350
amount of B is small and the
rate of change in B is also

91
00:06:22,350 --> 00:06:24,070
very small.

92
00:06:24,070 --> 00:06:25,610
And hopefully we'll
get to that very

93
00:06:25,610 --> 00:06:28,530
soon in this lecture.

94
00:06:28,530 --> 00:06:31,900
OK, so that's a summary of
what we did last time.

95
00:06:31,900 --> 00:06:36,540
Any questions?

96
00:06:36,540 --> 00:06:50,050
So the next thing now is to
go to the next mechanism.

97
00:06:50,050 --> 00:06:51,960
Which is, so far we've looked
at reactions that

98
00:06:51,960 --> 00:06:53,910
proceed all the way.

99
00:06:53,910 --> 00:06:55,200
We haven't looked at
equilibrium yet.

100
00:06:55,200 --> 00:06:58,120
But we know thermodynamics
is all about equilibrium.

101
00:06:58,120 --> 00:07:01,970
So now we're going to connect
kinetics and thermodynamics.

102
00:07:01,970 --> 00:07:05,620
We're going to look at a
mechanism which is an

103
00:07:05,620 --> 00:07:09,400
equilibrium mechanism.

104
00:07:09,400 --> 00:07:16,920
Equilibrium or reversible
reactions.

105
00:07:16,920 --> 00:07:21,380
Where you have A goes to B, with
summary constant k1, but

106
00:07:21,380 --> 00:07:24,600
you can have the reverse process
B goes back to A with

107
00:07:24,600 --> 00:07:27,870
the rate constant k minus one.

108
00:07:27,870 --> 00:07:35,240
Or you can write is as A k1, k
minus one to B, where k1 could

109
00:07:35,240 --> 00:07:39,600
be k forward and k minus one
could be k backwards.

110
00:07:39,600 --> 00:07:42,330
Also written as k sub
f, and this is also

111
00:07:42,330 --> 00:07:46,110
written as k sub b.

112
00:07:46,110 --> 00:07:47,290
So now we have more

113
00:07:47,290 --> 00:07:49,360
information than we had before.

114
00:07:49,360 --> 00:07:51,930
Because now we know this is
an equilibrium problem.

115
00:07:51,930 --> 00:07:54,830
And we already know about
equilibrium from

116
00:07:54,830 --> 00:07:55,020
thermodynamics.

117
00:07:55,020 --> 00:08:00,840
And we know that at equilibrium,
in addition to

118
00:08:00,840 --> 00:08:03,580
all the rate laws that we're
going to write down to solve

119
00:08:03,580 --> 00:08:06,940
the problem, we also know
that at equilibrium, the

120
00:08:06,940 --> 00:08:09,900
equilibrium constant is related
to the equilibrium

121
00:08:09,900 --> 00:08:11,980
concentrations.

122
00:08:11,980 --> 00:08:14,130
To the ratio of the equilibrium
concentrations.

123
00:08:14,130 --> 00:08:19,040
This is something additional
that we have.

124
00:08:19,040 --> 00:08:20,900
So let's write down everything
we know.

125
00:08:20,900 --> 00:08:24,290
And our goal is to find
out the time, the

126
00:08:24,290 --> 00:08:25,420
dynamics of this problem.

127
00:08:25,420 --> 00:08:28,400
Suppose you start at some time
t equals zero, at a certain

128
00:08:28,400 --> 00:08:32,030
amount of A0 in your pot, a
certain amount of B0, how do

129
00:08:32,030 --> 00:08:33,270
you get to equilibrium.

130
00:08:33,270 --> 00:08:37,240
What's the rate, what
does it look like?

131
00:08:37,240 --> 00:08:41,790
So the first thing is to
write down the rates.

132
00:08:41,790 --> 00:08:47,300
So the forward rate minus dA/dt,
which is the forward

133
00:08:47,300 --> 00:08:53,000
rate, R sub forward is equal
to k1 times A. And the

134
00:08:53,000 --> 00:08:58,200
backwards rate, minus dB/dt,
which we can write as R sub

135
00:08:58,200 --> 00:09:15,750
backwards, that's k minus one
times B. At equilibrium, we

136
00:09:15,750 --> 00:09:16,790
have no dynamics.

137
00:09:16,790 --> 00:09:19,450
Or no ensemble dynamics.

138
00:09:19,450 --> 00:09:21,610
The pot looks constant.

139
00:09:21,610 --> 00:09:24,330
It doesn't look like anything's
going on in there.

140
00:09:24,330 --> 00:09:26,300
The concentration of
A doesn't change.

141
00:09:26,300 --> 00:09:27,930
The concentration of
B doesn't change.

142
00:09:27,930 --> 00:09:32,510
They're constant at the
equilibrium concentration.

143
00:09:32,510 --> 00:09:36,600
So that implies that the rate
of formation of B must be

144
00:09:36,600 --> 00:09:39,620
equal to the rate of destruction
of B. Or, in other

145
00:09:39,620 --> 00:09:44,890
words, that the rate forward at
equilibrium has to be equal

146
00:09:44,890 --> 00:09:46,280
the rate backwards.

147
00:09:46,280 --> 00:09:50,020
Otherwise, there'd be changes
in concentrations.

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00:09:50,020 --> 00:09:54,930
So if that's true, then that
means that at equilibrium, k1

149
00:09:54,930 --> 00:10:02,050
A equilibrium has to be equal to
k minus one B equilibrium.

150
00:10:02,050 --> 00:10:10,990
Which means that we can get this
ratio of B to A. And see

151
00:10:10,990 --> 00:10:16,830
that this is equal to
k1 over k minus one.

152
00:10:16,830 --> 00:10:19,970
So we've just, this is
pretty major here.

153
00:10:19,970 --> 00:10:22,170
Looks pretty simple,
but we've taken an

154
00:10:22,170 --> 00:10:24,350
equilibrium property here.

155
00:10:24,350 --> 00:10:27,900
And related to kinetics.

156
00:10:27,900 --> 00:10:34,440
To a rate property.

157
00:10:34,440 --> 00:10:37,070
So here we have kinetics.

158
00:10:37,070 --> 00:10:46,440
Here we have thermo.

159
00:10:46,440 --> 00:10:49,960
And this is going to turn out
to be, this kind of ratio

160
00:10:49,960 --> 00:10:52,530
between forward rates and
backwards rates, related to

161
00:10:52,530 --> 00:10:55,180
equilibrium constants is going
to be valid, not just for this

162
00:10:55,180 --> 00:10:58,730
very simple mechanism we have
here with just two species.

163
00:10:58,730 --> 00:11:01,270
But it's going to be valid
for more complicated.

164
00:11:01,270 --> 00:11:03,510
And we have two species reacting
to form another

165
00:11:03,510 --> 00:11:09,050
species, or bimolecular
stuff, et cetera.

166
00:11:09,050 --> 00:11:11,460
OK, so now we've laid
the groundwork,

167
00:11:11,460 --> 00:11:12,990
let's look at the dynamics.

168
00:11:12,990 --> 00:11:14,820
Let's look at the
time evolution.

169
00:11:14,820 --> 00:11:19,210
So that means that we need
to take, well this

170
00:11:19,210 --> 00:11:19,980
is not quite right.

171
00:11:19,980 --> 00:11:22,230
I shouldn't write minus dA/dt.

172
00:11:22,230 --> 00:11:32,380
This is minus dA/dt rate forward
in the absence of

173
00:11:32,380 --> 00:11:38,290
reverse process.

174
00:11:38,290 --> 00:11:41,550
It says just the destruction of
A and not the creation of

175
00:11:41,550 --> 00:11:48,550
A. So if you look at the full
kinetics, the full mechanism,

176
00:11:48,550 --> 00:11:51,560
this also, the rate backwards is
minus dB/dt, in the absence

177
00:11:51,560 --> 00:11:55,360
of forming B back through
the reverse process.

178
00:11:55,360 --> 00:11:58,590
So both of these are
in the absence of

179
00:11:58,590 --> 00:11:59,700
their reverse process.

180
00:11:59,700 --> 00:12:07,150
The full kinetics, for the
destruction of A. A is the

181
00:12:07,150 --> 00:12:11,470
rate backwards, or the rate
forwards, which is k1 times A,

182
00:12:11,470 --> 00:12:14,030
and then the rate backwards,
which is the creation of A.

183
00:12:14,030 --> 00:12:15,840
That's the two rates together.

184
00:12:15,840 --> 00:12:19,860
Minus k minus one times B.
There's the formation of A

185
00:12:19,860 --> 00:12:23,050
here, this is the destruction
of A. This is the full

186
00:12:23,050 --> 00:12:24,960
differential equation for
the full mechanism.

187
00:12:24,960 --> 00:12:26,760
Not just one part of it.

188
00:12:26,760 --> 00:12:28,990
This one here is just for the
first step, and that one here

189
00:12:28,990 --> 00:12:31,090
is for the second step.

190
00:12:31,090 --> 00:12:34,920
This is what we need to solve.

191
00:12:34,920 --> 00:12:40,690
But now we have a lot of help
by what we've already done.

192
00:12:40,690 --> 00:12:44,320
So, what we can do is well, the
problem here is that's it

193
00:12:44,320 --> 00:12:47,080
looks hard because there's
B included in here.

194
00:12:47,080 --> 00:12:48,970
This is not just A,
there's B in here.

195
00:12:48,970 --> 00:12:50,380
But there's only two species.

196
00:12:50,380 --> 00:12:52,580
And stoichiometry is going
to help us out.

197
00:12:52,580 --> 00:12:55,500
Remember last time, or a few
times ago, we said the last

198
00:12:55,500 --> 00:12:57,990
species is the easy one.

199
00:12:57,990 --> 00:13:00,340
Because it's always related to
the concentration of all the

200
00:13:00,340 --> 00:13:02,790
other species by
stoichiometry.

201
00:13:02,790 --> 00:13:06,400
And the last one here is B. So
we know that B is related to

202
00:13:06,400 --> 00:13:11,130
A. B is whatever you
started out with.

203
00:13:11,130 --> 00:13:15,450
Plus whatever you used
up for A. That's one

204
00:13:15,450 --> 00:13:17,040
way of writing it.

205
00:13:17,040 --> 00:13:23,310
So you start with this amount,
and you create this amount.

206
00:13:23,310 --> 00:13:24,950
By having A react.

207
00:13:24,950 --> 00:13:27,480
This is what you started with,
A, and this is what's left.

208
00:13:27,480 --> 00:13:31,560
The difference has to go into
B. So stroke stoichiometry

209
00:13:31,560 --> 00:13:38,480
give you a relationship, which
you can use to plug into your

210
00:13:38,480 --> 00:13:41,230
differential equation here.

211
00:13:41,230 --> 00:13:43,740
And now, after rearranging your
differential equation,

212
00:13:43,740 --> 00:13:49,150
putting B in there, you can
rewrite this as k1 plus k

213
00:13:49,150 --> 00:14:01,310
minus one times A minus k minus
one times B0 plus A0.

214
00:14:01,310 --> 00:14:03,380
Something which depends
on A, something

215
00:14:03,380 --> 00:14:07,900
which is constant here.

216
00:14:07,900 --> 00:14:08,770
OK, and what else do we know?

217
00:14:08,770 --> 00:14:10,180
We know something about
equilibrium.

218
00:14:10,180 --> 00:14:14,000
We know something about when
we reach a state where the

219
00:14:14,000 --> 00:14:16,830
rate forward is equal to
the rate backward.

220
00:14:16,830 --> 00:14:21,510
When the rate forward is equal
to the rate backwards, the

221
00:14:21,510 --> 00:14:24,760
rate of change of A is zero.

222
00:14:24,760 --> 00:14:25,590
A is not changing.

223
00:14:25,590 --> 00:14:27,070
The concentration of
A is not changing.

224
00:14:27,070 --> 00:14:38,170
So at equilibrium, dA/dt
is equal to zero.

225
00:14:38,170 --> 00:14:42,850
And just like in this example
here, this is going to turn

226
00:14:42,850 --> 00:14:44,980
out to be related to an
approximation called the

227
00:14:44,980 --> 00:14:48,170
steady state approximation,
which we're going to see later

228
00:14:48,170 --> 00:14:51,560
this morning.

229
00:14:51,560 --> 00:14:55,070
Writing at equilibrium that
the change in the

230
00:14:55,070 --> 00:14:57,770
concentration of one of the
species is equal to zero,

231
00:14:57,770 --> 00:15:00,320
we're going to use that as an
approximation also later, when

232
00:15:00,320 --> 00:15:02,570
we get to more complicated
mechanisms.

233
00:15:02,570 --> 00:15:04,830
This is going to be the
equilibrium approximation.

234
00:15:04,830 --> 00:15:08,740
And this is going to be the
steady state approximation.

235
00:15:08,740 --> 00:15:12,500
Let's keep going with our, we're
solving this guy here.

236
00:15:12,500 --> 00:15:14,660
So at equilibrium, dA/dt
is equal to zero.

237
00:15:14,660 --> 00:15:27,280
And then we can replace our
concentrations here with, and

238
00:15:27,280 --> 00:15:30,890
so if you write equal to zero
here, then we can write it as

239
00:15:30,890 --> 00:15:32,040
equilibrium right here.

240
00:15:32,040 --> 00:15:35,250
And we'll be able to solve for
A equilibrium in terms of k1,

241
00:15:35,250 --> 00:15:40,580
k minus one, B0, and A0.

242
00:15:40,580 --> 00:15:44,250
And then if you do that, you get
A equilibrium is equal to

243
00:15:44,250 --> 00:15:53,000
k minus one over k1 plus k minus
one times B0 plus A0.

244
00:15:53,000 --> 00:16:00,870
And we're going to need that.

245
00:16:00,870 --> 00:16:05,700
Because now we're going to be
able to plug this, so there's

246
00:16:05,700 --> 00:16:07,540
A0 plus B0 sitting here.

247
00:16:07,540 --> 00:16:10,880
And we're going to replace
A0 plus B0 in terms of A

248
00:16:10,880 --> 00:16:12,030
equilibrium.

249
00:16:12,030 --> 00:16:15,320
There's a minus sign here.

250
00:16:15,320 --> 00:16:21,730
And between this A and this
A equilibrium here, so by

251
00:16:21,730 --> 00:16:27,060
rewriting, by plugging in A B0
A equilibrium instead of B0

252
00:16:27,060 --> 00:16:36,830
plus A0, we can rewrite that
equation as dA/dt is equal to

253
00:16:36,830 --> 00:16:45,310
k1 plus k minus one times
A minus A equilibrium.

254
00:16:45,310 --> 00:16:48,020
This is nice because, this
quantity, A minus A

255
00:16:48,020 --> 00:16:52,890
equilibrium, describes the
difference between where are

256
00:16:52,890 --> 00:16:55,070
you are and where
you want to be.

257
00:16:55,070 --> 00:16:56,820
It's a nice variable to have.

258
00:16:56,820 --> 00:16:59,960
It's going to change in time and
at infinite time, this is

259
00:16:59,960 --> 00:17:01,560
going to go to zero.

260
00:17:01,560 --> 00:17:03,510
A is going to go
to equilibrium.

261
00:17:03,510 --> 00:17:05,500
A equilibrium is a constant.

262
00:17:05,500 --> 00:17:08,110
Now we have a differential
equation that relates A to

263
00:17:08,110 --> 00:17:10,440
this difference.

264
00:17:10,440 --> 00:17:12,500
But A equilibrium
is a constant.

265
00:17:12,500 --> 00:17:14,960
So there's nothing that forbids
us from just writing

266
00:17:14,960 --> 00:17:20,210
minus A equilibrium here. d
of a constant dt is zero.

267
00:17:20,210 --> 00:17:24,300
So I'm just basically
subtracting zero here.

268
00:17:24,300 --> 00:17:26,590
So and I have something of the
form dx/dt is equal to

269
00:17:26,590 --> 00:17:28,370
constant times x.

270
00:17:28,370 --> 00:17:30,430
And I know how to write that.

271
00:17:30,430 --> 00:17:32,790
That looks just like a
first order process.

272
00:17:32,790 --> 00:17:39,630
I know the solution is A minus
A equilibrium is equal to my

273
00:17:39,630 --> 00:17:42,540
initial A, my initial state.

274
00:17:42,540 --> 00:17:52,510
A0 times e to the minus k1 plus
k minus one times time.

275
00:17:52,510 --> 00:17:55,270
And I've solved the problem.

276
00:17:55,270 --> 00:18:00,990
I've described how, as a
function of time, how A, the

277
00:18:00,990 --> 00:18:04,100
concentration of A,
gets equilibrium.

278
00:18:04,100 --> 00:18:06,680
And if you want to know B,
it's just the same thing.

279
00:18:06,680 --> 00:18:09,360
You replace B here
and B0 here.

280
00:18:09,360 --> 00:18:11,940
You put k minus one here, k1
here, but it's the same thing.

281
00:18:17,010 --> 00:18:32,560
So if you were to sketch it
out as a function of time,

282
00:18:32,560 --> 00:18:36,500
there's, eventually you'll get
to the concentration A

283
00:18:36,500 --> 00:18:37,120
equilibrium.

284
00:18:37,120 --> 00:18:40,120
If you're above it, you're
going to decay in a first

285
00:18:40,120 --> 00:18:44,840
order fashion, with a
rate constant k1.

286
00:18:44,840 --> 00:18:51,810
A minus A equilibrium is A0
minus A equilibrium e to the

287
00:18:51,810 --> 00:18:58,220
minus k prime times t, where k
prime is k1 plus k minus one.

288
00:18:58,220 --> 00:19:02,340
And this is when A is greater
than A0, A equilibrium.

289
00:19:02,340 --> 00:19:03,940
A0 is greater than
equilibrium.

290
00:19:03,940 --> 00:19:08,290
And if you start below, you're
going to come up like this.

291
00:19:08,290 --> 00:19:11,800
A0 is less than A equilibrium.

292
00:19:11,800 --> 00:19:14,210
And both rates come into here,
in a very simple way.

293
00:19:14,210 --> 00:19:17,300
Just the sum of the two.

294
00:19:17,300 --> 00:19:22,090
So experimental, if you want to
find out these rates, you

295
00:19:22,090 --> 00:19:27,750
can measure their equilibrium
concentrations.

296
00:19:27,750 --> 00:19:34,370
Measure k equilibrium by
obtaining the equilibrium

297
00:19:34,370 --> 00:19:39,350
concentrations of B and A. And
that gives you the ratio of k1

298
00:19:39,350 --> 00:19:40,850
plus k minus one.

299
00:19:40,850 --> 00:19:43,930
And then you can just start
the process with some

300
00:19:43,930 --> 00:19:46,380
concentration of A. Which is
different in equilibrium.

301
00:19:46,380 --> 00:19:55,350
Then watching it in time,
extract out, observe and

302
00:19:55,350 --> 00:19:58,600
measure k1 plus k minus one.

303
00:19:58,600 --> 00:19:59,960
In this kinetic equation.

304
00:19:59,960 --> 00:20:01,570
At kinetic relation.

305
00:20:01,570 --> 00:20:06,330
And then you have two results,
two pieces of data.

306
00:20:06,330 --> 00:20:09,610
You've measured k prime, which
gives you the sum of the two.

307
00:20:09,610 --> 00:20:11,490
And you've measured K
equilibrium, which gives you

308
00:20:11,490 --> 00:20:12,660
the ratio of the two.

309
00:20:12,660 --> 00:20:16,250
And you've got your two
rate constants out.

310
00:20:16,250 --> 00:20:21,020
OK, any questions for the
equilibrium problem?

311
00:20:21,020 --> 00:20:22,550
We're slowly -- yes.

312
00:20:22,550 --> 00:20:27,870
STUDENT: [INAUDIBLE]

313
00:20:27,870 --> 00:20:29,990
PROFESSOR: There is not yet
a relationship between

314
00:20:29,990 --> 00:20:31,110
k1 and k minus one.

315
00:20:31,110 --> 00:20:38,170
It depends on the problem.

316
00:20:38,170 --> 00:20:45,470
And we're going to study the,
when we get to potential

317
00:20:45,470 --> 00:20:52,570
barriers, we look at, we'll
look at this issue.

318
00:20:52,570 --> 00:20:54,800
OK, it's a good question.

319
00:20:54,800 --> 00:20:59,110
We're going to do that
probably next time.

320
00:20:59,110 --> 00:21:01,190
So let's get going, then, with
reversible reactions.

321
00:21:01,190 --> 00:21:03,910
So now we have, let's make it a
little bit more complicated.

322
00:21:03,910 --> 00:21:07,760
Let's, instead of two species,
let's make it three species.

323
00:21:07,760 --> 00:21:14,010
So we have A plus B goes to C,
where the rate constant k1,

324
00:21:14,010 --> 00:21:22,200
and then C goes to A plus to B
with a rate constant k minus

325
00:21:22,200 --> 00:21:31,720
one that you can rewrite as A
plus B goes to C, k1, k minus

326
00:21:31,720 --> 00:21:36,850
one, and then you write down
your, you want to solve this.

327
00:21:36,850 --> 00:21:38,760
So the first thing you do
is your write down your

328
00:21:38,760 --> 00:21:43,680
differential equations. dA/dt is
equal to, put all the rates

329
00:21:43,680 --> 00:21:50,040
in there. k1 A B minus k minus
one times C. And at

330
00:21:50,040 --> 00:21:57,940
equilibrium, we set that
equal to zero.

331
00:21:57,940 --> 00:22:01,670
So you can write equilibrium
here, equilibrium here,

332
00:22:01,670 --> 00:22:04,630
equilibrium here.

333
00:22:04,630 --> 00:22:11,980
And when you bring this term
over to the other side, you

334
00:22:11,980 --> 00:22:17,380
find, then that at equilibrium
the ratio of the forward rate

335
00:22:17,380 --> 00:22:22,320
to the backward rates is equal
to C equilibrium divided by A

336
00:22:22,320 --> 00:22:23,750
equilibrium.

337
00:22:23,750 --> 00:22:24,840
The equilibrium, which,
you know is

338
00:22:24,840 --> 00:22:26,590
the equilibrium constant.

339
00:22:26,590 --> 00:22:30,800
So again, for this slightly more
complicated problem, the

340
00:22:30,800 --> 00:22:32,920
ratio of the forward and
backward rates are related to

341
00:22:32,920 --> 00:22:39,220
the equilibrium constant.

342
00:22:39,220 --> 00:22:41,150
So now you want to solve this.

343
00:22:41,150 --> 00:22:42,790
Well, we're not even
going to try.

344
00:22:42,790 --> 00:22:44,810
Because it's going to
be too complicated.

345
00:22:44,810 --> 00:22:48,540
So as soon as you get away from
first order kinetics and

346
00:22:48,540 --> 00:22:50,200
go to second order kinetics
with multiple

347
00:22:50,200 --> 00:22:55,860
steps, it's a mess.

348
00:22:55,860 --> 00:22:59,090
So instead you try to
find approximations.

349
00:22:59,090 --> 00:23:00,880
In this case here there's
one we can use.

350
00:23:00,880 --> 00:23:02,910
Or a limiting case, at least.

351
00:23:02,910 --> 00:23:04,830
One we can use, which
is an obvious

352
00:23:04,830 --> 00:23:09,110
one, which is flooding.

353
00:23:09,110 --> 00:23:11,000
As a limiting case.

354
00:23:11,000 --> 00:23:16,640
If we want to isolate a small
part of the problem, we can

355
00:23:16,640 --> 00:23:17,580
use flooding.

356
00:23:17,580 --> 00:23:22,490
Because we can overwhelm the
system with either A or B.

357
00:23:22,490 --> 00:23:27,280
Take B0 much greater
than A0 and C0.

358
00:23:27,280 --> 00:23:30,070
So over the course of the
reaction of the process, the

359
00:23:30,070 --> 00:23:33,170
concentration of B
hardly changes.

360
00:23:33,170 --> 00:23:37,430
And so when we write our kinetic
equation, our rate

361
00:23:37,430 --> 00:23:45,730
law, k1 A B minus k minus one
C, we could put a little

362
00:23:45,730 --> 00:23:46,890
naught here, because
we know the

363
00:23:46,890 --> 00:23:49,270
concentration of B is not changing.

364
00:23:49,270 --> 00:23:53,260
And so now we're left with the
problem we had before.

365
00:23:53,260 --> 00:23:57,220
Which was a reversible first
order process which we have

366
00:23:57,220 --> 00:24:00,210
just solved.

367
00:24:00,210 --> 00:24:04,340
And you go through the
experiment and you extract out

368
00:24:04,340 --> 00:24:08,680
k1 times B0 and k minus one.

369
00:24:08,680 --> 00:24:10,990
You do the experiment again with
a different concentration

370
00:24:10,990 --> 00:24:12,430
of B0 to start out with.

371
00:24:12,430 --> 00:24:14,650
And you can extract out k1.

372
00:24:14,650 --> 00:24:18,940
So this is a process of getting
the rates by using a

373
00:24:18,940 --> 00:24:21,530
simple approximation here.

374
00:24:21,530 --> 00:24:26,950
Simple limiting case.

375
00:24:26,950 --> 00:24:30,030
Any questions?

376
00:24:30,030 --> 00:24:32,790
So we've just finished putting
all the building blocks

377
00:24:32,790 --> 00:24:35,630
together now.

378
00:24:35,630 --> 00:24:43,630
Let me remind you what these
building blocks are.

379
00:24:43,630 --> 00:24:45,870
We have a bunch of
approximations under our belt.

380
00:24:45,870 --> 00:24:52,960
Flooding, we've looked
at rates being

381
00:24:52,960 --> 00:24:55,310
much faster than others.

382
00:24:55,310 --> 00:25:01,260
And we have three simple
mechanisms, we've looked at

383
00:25:01,260 --> 00:25:03,810
parallel reactions.

384
00:25:03,810 --> 00:25:06,660
We looked at series reactions.

385
00:25:06,660 --> 00:25:13,220
And we've looked at reversible
reactions.

386
00:25:13,220 --> 00:25:16,340
So a complicated mechanism will
basically put these three

387
00:25:16,340 --> 00:25:19,210
building blocks together
in a series.

388
00:25:19,210 --> 00:25:22,960
And they'll all be happening
at the same time somehow.

389
00:25:22,960 --> 00:25:27,890
And obviously, since it was,
since we threw up our hands,

390
00:25:27,890 --> 00:25:31,540
just with a simple reversible
process like here, it's clear

391
00:25:31,540 --> 00:25:35,670
that we're going to throw
up our hands a lot.

392
00:25:35,670 --> 00:25:37,270
When we write down
these mechanisms.

393
00:25:37,270 --> 00:25:38,510
So we're going to need
approximations.

394
00:25:38,510 --> 00:25:41,090
We're going to need something
that we'll automatically go

395
00:25:41,090 --> 00:25:43,530
to, and say this is going
to be too hard.

396
00:25:43,530 --> 00:25:45,880
I'm not even going to try.

397
00:25:45,880 --> 00:25:50,380
Let's do an approximation.

398
00:25:50,380 --> 00:25:52,300
And two approximations that
we're going to talk about are

399
00:25:52,300 --> 00:25:55,570
ones I already mentioned
at the beginning.

400
00:25:55,570 --> 00:26:01,810
Which is the steady state
approximation, where one rate,

401
00:26:01,810 --> 00:26:07,055
the rate to go into the
intermediate is very slow, or

402
00:26:07,055 --> 00:26:10,680
the rate to get out of the
intermediate is very fast.

403
00:26:10,680 --> 00:26:14,040
And the equilibrium
approximation where the system

404
00:26:14,040 --> 00:26:17,920
sets up a very fast equilibrium
and you're allowed

405
00:26:17,920 --> 00:26:20,650
to use thermodynamics
to help you out

406
00:26:20,650 --> 00:26:22,550
in solving the problem.

407
00:26:22,550 --> 00:26:24,700
These are the two approximations
that we'll use

408
00:26:24,700 --> 00:26:27,920
when we put these
things together.

409
00:26:27,920 --> 00:26:30,480
OK, so let's look at the
first, simple, more

410
00:26:30,480 --> 00:26:32,040
complicated mechanism.

411
00:26:32,040 --> 00:26:34,300
To see where these
approximations come in.

412
00:26:34,300 --> 00:26:36,310
So the first one is going
to be a series.

413
00:26:36,310 --> 00:26:38,130
We're going to start putting
these things together.

414
00:26:38,130 --> 00:26:40,300
First one is a series
reversible.

415
00:26:40,300 --> 00:26:41,740
Those two together.

416
00:26:41,740 --> 00:26:46,620
Series reversible.

417
00:26:46,620 --> 00:26:50,750
So we have first, a reversible
process.

418
00:26:50,750 --> 00:26:54,430
Everything is first order here.
k minus one goes to B.

419
00:26:54,430 --> 00:26:57,670
And then B goes to C with
some rate constant

420
00:26:57,670 --> 00:27:01,130
k2, all first order.

421
00:27:01,130 --> 00:27:03,090
If we were to turn the crank,
we'd say, oh, I've got to

422
00:27:03,090 --> 00:27:04,650
write down all my
rate laws here.

423
00:27:04,650 --> 00:27:09,200
Minus dA/dt is all the ways that
I destroy and create B,

424
00:27:09,200 --> 00:27:17,990
so there's a k1 times A minus
k minus one times B. dB/dt,

425
00:27:17,990 --> 00:27:22,225
all the ways that create and
destroy B is that you create

426
00:27:22,225 --> 00:27:26,730
it through destruction of A.
I destroy it through the

427
00:27:26,730 --> 00:27:30,700
backwards rate to make A. And
I destroy it by making the

428
00:27:30,700 --> 00:27:33,290
product, C. There's
the intermediate.

429
00:27:33,290 --> 00:27:35,130
And then for C, that's
pretty simple.

430
00:27:35,130 --> 00:27:38,660
There's only one channel into C,
and that's the destruction

431
00:27:38,660 --> 00:27:42,110
of B to form C, k2 B.
All the rate laws.

432
00:27:42,110 --> 00:27:45,400
I write everything
I know here.

433
00:27:45,400 --> 00:27:48,290
Couple of differential
equations.

434
00:27:48,290 --> 00:27:52,410
We know it's going to
be hard to solve.

435
00:27:52,410 --> 00:27:58,640
So, let's remind ourselves
of what I just erased.

436
00:27:58,640 --> 00:28:09,700
Which was the case where the
intermediate concentration was

437
00:28:09,700 --> 00:28:14,130
always very small and didn't
change very fast.

438
00:28:14,130 --> 00:28:20,140
And the process was dominated
by k1 here.

439
00:28:20,140 --> 00:28:22,040
That was the rate
limiting step.

440
00:28:22,040 --> 00:28:29,260
Remember that our first
example this morning.

441
00:28:29,260 --> 00:28:34,750
So this is where the
concentration of B is roughly

442
00:28:34,750 --> 00:28:36,800
constant over time.

443
00:28:36,800 --> 00:28:41,190
It's small.

444
00:28:41,190 --> 00:28:45,950
And over any small time period,
it's roughly constant.

445
00:28:45,950 --> 00:28:52,210
Which means that dB/dt is
roughly equal to zero.

446
00:28:52,210 --> 00:28:54,320
If you look at a long term,
obviously it's going to change

447
00:28:54,320 --> 00:28:56,060
as you go from this point
to that point.

448
00:28:56,060 --> 00:28:58,640
In fact, there's a relationship
between A which

449
00:28:58,640 --> 00:29:01,600
is changing in time and B. But
A is also changing in time

450
00:29:01,600 --> 00:29:02,240
very slowly.

451
00:29:02,240 --> 00:29:05,270
Because the rate k1
is very small.

452
00:29:05,270 --> 00:29:08,910
So dB/dt, which is related to
this rate k1, is going to be

453
00:29:08,910 --> 00:29:10,120
very slowly changing in time.

454
00:29:10,120 --> 00:29:16,470
We can approximate it at zero.

455
00:29:16,470 --> 00:29:19,850
What are the ways in terms of
matching these rate constants

456
00:29:19,850 --> 00:29:22,120
that we can get to the
approximation.

457
00:29:22,120 --> 00:29:28,180
To a diagram that
looks like that?

458
00:29:28,180 --> 00:29:34,730
But we don't want
B to pile up.

459
00:29:34,730 --> 00:29:38,670
That means we have to have the
rates out of the intermediate

460
00:29:38,670 --> 00:29:41,810
to be much faster, at least one
of the rates out of the

461
00:29:41,810 --> 00:29:44,070
intermediate to be much faster
than the rate that creates the

462
00:29:44,070 --> 00:29:45,690
intermediate.

463
00:29:45,690 --> 00:29:49,780
So the different ways of
creating that approximation

464
00:29:49,780 --> 00:29:53,350
are if, and the length of the
arrows now is going to be

465
00:29:53,350 --> 00:29:55,360
proportional to the rate.

466
00:29:55,360 --> 00:29:59,900
We want it to be very hard to
make B. And as soon as we make

467
00:29:59,900 --> 00:30:03,830
B, we want it to go away.

468
00:30:03,830 --> 00:30:06,490
So one of the ways to do that
is to have the reverse rate

469
00:30:06,490 --> 00:30:09,450
much faster than the
initial rate.

470
00:30:09,450 --> 00:30:12,660
And we could have a slow rate
into C, that's fine.

471
00:30:12,660 --> 00:30:16,900
We could have, again, a very
slow rate into B. We could

472
00:30:16,900 --> 00:30:19,060
have a slow reverse rate
and a fast rate into

473
00:30:19,060 --> 00:30:21,150
C. Perfectly fine.

474
00:30:21,150 --> 00:30:22,680
We could have both.

475
00:30:22,680 --> 00:30:28,790
Fast rate out of B through A, or
out of B through C. As long

476
00:30:28,790 --> 00:30:31,240
as this first rate into the B
is small compared to one of

477
00:30:31,240 --> 00:30:33,795
those two rates, we're never
going to pile up B. It's going

478
00:30:33,795 --> 00:30:37,140
to go away as soon
as we make it.

479
00:30:37,140 --> 00:30:42,380
So in all these cases, k2
plus k minus one is

480
00:30:42,380 --> 00:30:46,770
much bigger than k1.

481
00:30:46,770 --> 00:30:49,980
When we have that situation,
then we can make the

482
00:30:49,980 --> 00:30:51,800
approximation up here.

483
00:30:51,800 --> 00:30:54,180
That the rate of change
in B is very small.

484
00:30:54,180 --> 00:30:56,410
Almost zero, which means it's
basically a constant.

485
00:30:56,410 --> 00:31:01,130
And instead of writing B in this
case here, we're going to

486
00:31:01,130 --> 00:31:04,670
write it as B steady state.

487
00:31:04,670 --> 00:31:08,270
Which is basically a constant
in terms of solving the

488
00:31:08,270 --> 00:31:10,370
differential equations.

489
00:31:10,370 --> 00:31:14,090
And that's going to make
our life much easier.

490
00:31:14,090 --> 00:31:19,540
Because instead of having
to solve these coupled

491
00:31:19,540 --> 00:31:22,230
differential equations, we're
just going to have to solve

492
00:31:22,230 --> 00:31:24,450
coupled algebraic equations.

493
00:31:24,450 --> 00:31:29,140
Which is really messy,
but less hard.

494
00:31:29,140 --> 00:31:34,460
If you're a bean-counter
than this is heaven.

495
00:31:34,460 --> 00:31:37,240
So now we're going to put steady
state, this is going to

496
00:31:37,240 --> 00:31:38,260
be a constant.

497
00:31:38,260 --> 00:31:40,660
And this is going to
be equal to zero.

498
00:31:40,660 --> 00:31:42,940
And then here this is going
to be steady state here.

499
00:31:42,940 --> 00:31:44,910
This is going to be
steady state here.

500
00:31:44,910 --> 00:31:47,420
And the first thing to do now is
that we've eliminated this

501
00:31:47,420 --> 00:31:48,980
differential equation.

502
00:31:48,980 --> 00:31:57,780
We now have an algebraic
equation where we can solve

503
00:31:57,780 --> 00:32:01,830
for B steady state.

504
00:32:01,830 --> 00:32:07,260
So if you recognize that you're
in this situation here,

505
00:32:07,260 --> 00:32:08,380
forget about trying to solve.

506
00:32:08,380 --> 00:32:12,340
Immediately go to the process
of putting in a constant for

507
00:32:12,340 --> 00:32:16,530
B. Setting dB/dt for the
intermediate equal to zero.

508
00:32:16,530 --> 00:32:22,020
And then starting to turn the
crank on the algebra.

509
00:32:22,020 --> 00:32:26,580
So let me go ahead and
turn the crank here.

510
00:32:26,580 --> 00:32:29,620
Go through the steps,
which is basically

511
00:32:29,620 --> 00:32:37,010
typical of these problems.

512
00:32:37,010 --> 00:32:38,350
So you turn the crank.

513
00:32:38,350 --> 00:32:42,960
You solve for the steady
state concentration.

514
00:32:42,960 --> 00:32:50,220
You get, in terms of A, over
k minus one, plus k2.

515
00:32:50,220 --> 00:32:55,410
Then you plug this
back in here.

516
00:32:55,410 --> 00:33:01,180
And you get a new differential
equation, minus dA/dt is equal

517
00:33:01,180 --> 00:33:06,636
to k1 times A minus k minus one
times B steady state, so

518
00:33:06,636 --> 00:33:09,340
we plug this in here.

519
00:33:09,340 --> 00:33:12,000
And you get, so there's
A sitting here.

520
00:33:12,000 --> 00:33:13,190
A sitting here.

521
00:33:13,190 --> 00:33:16,660
It's going to be of the form,
effective rate times A. It's

522
00:33:16,660 --> 00:33:18,700
going to be basically
a first order form.

523
00:33:18,700 --> 00:33:21,400
Which is what we'd expect
from sketching the

524
00:33:21,400 --> 00:33:24,050
diagram just like that.

525
00:33:24,050 --> 00:33:31,560
So you get minus dA/dt is an
effective rate. k1 times k2

526
00:33:31,560 --> 00:33:34,330
over k1 plus k2.

527
00:33:34,330 --> 00:33:39,900
Times A. This is k prime.

528
00:33:39,900 --> 00:33:41,810
So if you were to do an
experiment under these

529
00:33:41,810 --> 00:33:47,370
conditions, you'd find that A
behaves, it's basically a

530
00:33:47,370 --> 00:33:48,750
pseudo first order problem.

531
00:33:48,750 --> 00:33:53,250
With a funny rate that contains
all the elementary

532
00:33:53,250 --> 00:33:55,690
rate constants as part of it.

533
00:33:55,690 --> 00:33:59,990
And when you do the same thing
for C, dC/dt, you find that

534
00:33:59,990 --> 00:34:05,410
depends on A. It's k1, you plug
these steady states in

535
00:34:05,410 --> 00:34:11,250
here. k1 k2 divided by
k1 plus k2 times A.

536
00:34:11,250 --> 00:34:13,500
This is k prime again.

537
00:34:13,500 --> 00:34:16,050
So the problem, and
it's first order.

538
00:34:16,050 --> 00:34:18,930
So the problem looks like
effectively you're going from

539
00:34:18,930 --> 00:34:21,710
A to C. Forget about
the intermediate.

540
00:34:21,710 --> 00:34:24,270
With an effective
rate k prime.

541
00:34:24,270 --> 00:34:30,070
Where k prime contains
these rates.

542
00:34:30,070 --> 00:34:32,310
So this is steady state
approximation.

543
00:34:32,310 --> 00:34:36,420
It's the prototypical problem.

544
00:34:36,420 --> 00:34:38,990
Questions on steady state,
before we go to the next

545
00:34:38,990 --> 00:34:44,340
approximation.

546
00:34:44,340 --> 00:34:44,690
Alright.

547
00:34:44,690 --> 00:34:53,240
So now, as promised, the next
approximation is that where

548
00:34:53,240 --> 00:34:54,890
are set up the fast
equilibrium.

549
00:34:54,890 --> 00:34:59,180
And you can use thermo
to help you out.

550
00:34:59,180 --> 00:35:11,250
Equilibrium approximation
A goes to B.

551
00:35:11,250 --> 00:35:13,740
This is a fast process.

552
00:35:13,740 --> 00:35:21,030
And then k2 out of B
is a slow process.

553
00:35:21,030 --> 00:35:24,220
Little arrow here, and
two big arrows here.

554
00:35:24,220 --> 00:35:27,480
You put your A in your flask.

555
00:35:27,480 --> 00:35:30,230
Immediately you set up
the equilibrium.

556
00:35:30,230 --> 00:35:32,540
And you slowly dribble
out of that.

557
00:35:32,540 --> 00:35:40,130
If you want to do it as a
function of buckets and, so

558
00:35:40,130 --> 00:35:42,950
you have a big pipe connecting
two buckets that are just

559
00:35:42,950 --> 00:35:43,820
offset from each other.

560
00:35:43,820 --> 00:35:48,510
So there's, in this case here
B is favored over A. Because

561
00:35:48,510 --> 00:35:51,260
it's a little bit lower
than this guy here.

562
00:35:51,260 --> 00:35:54,880
And then there's a little
bucket, there's a little tube,

563
00:35:54,880 --> 00:35:58,870
comes out of here.

564
00:35:58,870 --> 00:36:02,710
So there's a little dribble
into C over time.

565
00:36:02,710 --> 00:36:05,340
First thing that happens is you
set up your equilibrium.

566
00:36:05,340 --> 00:36:08,690
And then you slowly extract
stuff through a little tube

567
00:36:08,690 --> 00:36:10,280
into C.

568
00:36:10,280 --> 00:36:14,410
So what do you expect
this to look like?

569
00:36:14,410 --> 00:36:22,770
Well, you expect A to really
slowly come out.

570
00:36:22,770 --> 00:36:27,690
Because the rate limiting step
is the rate from B to C. So k2

571
00:36:27,690 --> 00:36:30,540
is going to be the way that
A is going to come out.

572
00:36:30,540 --> 00:36:32,340
You expect it to
slowly go away.

573
00:36:32,340 --> 00:36:35,220
You expect B to get created
very fast, to

574
00:36:35,220 --> 00:36:37,550
some equilibrium amount.

575
00:36:37,550 --> 00:36:40,770
And then also to follow
the same rate

576
00:36:40,770 --> 00:36:44,230
k2, slow k2, to disappear.

577
00:36:44,230 --> 00:36:48,700
And you expect C to come up.

578
00:36:48,700 --> 00:36:51,190
You're basically in a first
order process to

579
00:36:51,190 --> 00:36:53,710
saturation of A0.

580
00:36:53,710 --> 00:36:56,240
So when you expect the dominant
rate to be k2, and

581
00:36:56,240 --> 00:37:00,380
the fast dynamics to happen at
very, very early times, and

582
00:37:00,380 --> 00:37:05,090
then everything to follow
first order kinetics.

583
00:37:05,090 --> 00:37:07,140
So now let's do the math and
make sure that it agrees with

584
00:37:07,140 --> 00:37:11,400
what we've just assumed that
it's going to look like.

585
00:37:11,400 --> 00:37:14,280
So equilibrium is happening.

586
00:37:14,280 --> 00:37:19,000
You can assume that at any time
after the initial very

587
00:37:19,000 --> 00:37:21,610
fast process of getting
equilibrium.

588
00:37:21,610 --> 00:37:25,420
So after some initial time,
the ratio of B over A is

589
00:37:25,420 --> 00:37:27,430
always going to be a constant.

590
00:37:27,430 --> 00:37:31,560
The dribble out of B here is not
going to be fast enough to

591
00:37:31,560 --> 00:37:33,010
change that ratio.

592
00:37:33,010 --> 00:37:39,280
As soon as you get a little bit
of B out here, immediately

593
00:37:39,280 --> 00:37:41,720
the ratios, the amounts
rearrange to

594
00:37:41,720 --> 00:37:44,460
keep the ratio constant.

595
00:37:44,460 --> 00:37:48,120
So now when I look at the rate
of the reaction, the rate of

596
00:37:48,120 --> 00:37:54,780
formation of C, k2 times B,
well, in terms of A, it's k2

597
00:37:54,780 --> 00:37:57,050
times K equilibrium.

598
00:37:57,050 --> 00:38:07,020
Times A. Which is k2 times k1
over k minus one times A. And

599
00:38:07,020 --> 00:38:11,800
you can immediately see that C
is behaving, or the reaction

600
00:38:11,800 --> 00:38:15,650
is behaving, like a first
order process.

601
00:38:15,650 --> 00:38:18,260
With an effective rate which
contains all three rates in

602
00:38:18,260 --> 00:38:23,150
this ratio here.

603
00:38:23,150 --> 00:38:26,140
So it looks like A goes to C,
with some k prime where this

604
00:38:26,140 --> 00:38:28,640
is k prime.

605
00:38:28,640 --> 00:38:32,480
It looks like A goes to
C with an effective

606
00:38:32,480 --> 00:38:36,280
rate constant k prime.

607
00:38:36,280 --> 00:38:36,820
OK.

608
00:38:36,820 --> 00:38:39,100
Questions?

609
00:38:39,100 --> 00:38:40,510
We're going to do
some examples.

610
00:38:40,510 --> 00:38:43,980
And then we're going to do chain
reactions next time.

611
00:38:43,980 --> 00:38:50,000
We're one lecture behind.

612
00:38:50,000 --> 00:38:50,330
Alright.

613
00:38:50,330 --> 00:38:53,020
Let's do some reactions.

614
00:38:53,020 --> 00:39:01,530
Some examples.

615
00:39:01,530 --> 00:39:05,800
I'm going to skip the first
example, which is the example

616
00:39:05,800 --> 00:39:08,260
with the chaperone,
in the notes.

617
00:39:08,260 --> 00:39:12,350
And I'm going to go directly
to the gas decomposition.

618
00:39:12,350 --> 00:39:15,750
The two examples are basically
very similar in their use of

619
00:39:15,750 --> 00:39:18,700
the steady state
approximation.

620
00:39:18,700 --> 00:39:23,820
So I'll let you read over
the first example.

621
00:39:23,820 --> 00:39:26,120
So this example here is going
to give us the Lindemann

622
00:39:26,120 --> 00:39:29,720
mechanism, for which Mr.
Lindemann got a Nobel Prize

623
00:39:29,720 --> 00:39:33,530
many many, years ago.

624
00:39:33,530 --> 00:39:35,820
Basically, it's looking
at decomposition

625
00:39:35,820 --> 00:39:44,910
of a gas phase molecule.

626
00:39:44,910 --> 00:39:47,350
Where you have a molecule,
A, that

627
00:39:47,350 --> 00:39:52,280
breaks down into products.

628
00:39:52,280 --> 00:40:01,460
And the observation before Mr.
Lindemann got around, was that

629
00:40:01,460 --> 00:40:06,410
it looked like a first
order process.

630
00:40:06,410 --> 00:40:10,670
It looked like A is just falling
apart on its own into

631
00:40:10,670 --> 00:40:13,320
these products.

632
00:40:13,320 --> 00:40:17,180
Mr. Lindemann got around and
said, well, why would A just

633
00:40:17,180 --> 00:40:18,420
want to fall apart.

634
00:40:18,420 --> 00:40:21,040
There's something a little
bit odd about that.

635
00:40:21,040 --> 00:40:24,600
Stable molecule, why would
it want to fall apart.

636
00:40:24,600 --> 00:40:29,270
And so he hypothesized
a mechanism.

637
00:40:29,270 --> 00:40:33,040
And it went like this.

638
00:40:33,040 --> 00:40:36,270
That A actually collides with
another molecule, which could

639
00:40:36,270 --> 00:40:39,530
be a bystander.

640
00:40:39,530 --> 00:40:42,080
Could be a chaperone molecule
that just happens to be there.

641
00:40:42,080 --> 00:40:42,930
Or it could be another
molecule of

642
00:40:42,930 --> 00:40:46,110
A. There's a collision.

643
00:40:46,110 --> 00:40:50,210
There's a collision that
creates a vibrationally

644
00:40:50,210 --> 00:40:53,560
excited version of A.
Of A sitting there.

645
00:40:53,560 --> 00:40:53,970
It collides.

646
00:40:53,970 --> 00:40:56,850
It suddenly starts to be really
vibrationally excited.

647
00:40:56,850 --> 00:40:58,840
Bonds vibrate all
over the place.

648
00:40:58,840 --> 00:41:01,670
Atoms wiggle.

649
00:41:01,670 --> 00:41:04,040
There's the collision partner
that goes away.

650
00:41:04,040 --> 00:41:08,870
So kinetic energy is transferred
from M and A to

651
00:41:08,870 --> 00:41:14,240
the vibrational loads of A. In a
process which is reversible.

652
00:41:14,240 --> 00:41:18,020
k1, k minus one, because this
excited A, this vibrationally

653
00:41:18,020 --> 00:41:21,880
excited A, could also collide
with a molecule and cool down.

654
00:41:21,880 --> 00:41:25,850
The energy in the vibrations
could be transferred back to

655
00:41:25,850 --> 00:41:29,120
kinetic energy for a
reverse process.

656
00:41:29,120 --> 00:41:33,980
But if you wait long enough,
or if this vibrationally

657
00:41:33,980 --> 00:41:39,000
excited A, with these atoms
widely moving around, that has

658
00:41:39,000 --> 00:41:42,700
more of a chance of falling
apart than this guy here.

659
00:41:42,700 --> 00:41:49,160
So A star could fall apart
into products.

660
00:41:49,160 --> 00:41:51,010
With a rate k2.

661
00:41:51,010 --> 00:41:52,860
So he said, this is actually
an important step.

662
00:41:52,860 --> 00:41:54,800
It's not that A just suddenly
falls apart.

663
00:41:54,800 --> 00:41:58,520
But this has to happen.

664
00:41:58,520 --> 00:42:01,990
So all the observations that
supported a first order

665
00:42:01,990 --> 00:42:05,960
process, which basically only
supported one elementary

666
00:42:05,960 --> 00:42:13,230
reactions, might not be right.

667
00:42:13,230 --> 00:42:16,870
So let's assume his mechanism,
and let's look at what we need

668
00:42:16,870 --> 00:42:18,800
to do as an approximation.

669
00:42:18,800 --> 00:42:21,930
And where that leads us
in terms of trying to

670
00:42:21,930 --> 00:42:28,320
experimentally confirm that this
mechanism is plausible.

671
00:42:28,320 --> 00:42:30,830
Not prove it, but just to make
sure that it's consistent

672
00:42:30,830 --> 00:42:43,230
with, the data's consistent
with the hypothesis.

673
00:42:43,230 --> 00:42:50,390
So let's compare
our rates here.

674
00:42:50,390 --> 00:42:51,530
So we have a collision.

675
00:42:51,530 --> 00:42:55,230
A and M collide, and there's a
certain probability on the

676
00:42:55,230 --> 00:42:59,280
collision that kinetic energy
will be transferred to

677
00:42:59,280 --> 00:43:00,100
vibrational energy.

678
00:43:00,100 --> 00:43:02,870
That turns out to be a
moderate probability.

679
00:43:02,870 --> 00:43:09,950
So k1 is reasonably fast.

680
00:43:09,950 --> 00:43:11,870
The reverse process, you have
something that's vibrationally

681
00:43:11,870 --> 00:43:14,610
excited colliding
with a molecule.

682
00:43:14,610 --> 00:43:17,700
There's a probability that that
excitation is going to

683
00:43:17,700 --> 00:43:20,320
get turned back into
kinetic energy.

684
00:43:20,320 --> 00:43:21,760
And then these two molecules
will fly

685
00:43:21,760 --> 00:43:23,460
apart with high velocity.

686
00:43:23,460 --> 00:43:25,690
And that tends to be much
more highly probable

687
00:43:25,690 --> 00:43:28,070
then the first process.

688
00:43:28,070 --> 00:43:30,810
So this is faster.

689
00:43:30,810 --> 00:43:32,950
Then the excited molecule's
waiting around.

690
00:43:32,950 --> 00:43:34,380
And there's a probability that
there's it's going to just

691
00:43:34,380 --> 00:43:35,900
fall apart.

692
00:43:35,900 --> 00:43:38,700
And that turns out to
be really slow.

693
00:43:38,700 --> 00:43:49,330
So we're going to assume that
the last step is a slow step.

694
00:43:49,330 --> 00:43:50,680
So what does it look like?

695
00:43:50,680 --> 00:43:55,895
Well, we have the creation
is fast.

696
00:43:55,895 --> 00:43:57,420
So the intermediate is fast.

697
00:43:57,420 --> 00:43:59,160
But the destruction through
the reverse

698
00:43:59,160 --> 00:44:02,340
process is much faster.

699
00:44:02,340 --> 00:44:05,280
We have at least one step, out
of the intermediate, which is

700
00:44:05,280 --> 00:44:07,190
faster than the step into
the intermediate.

701
00:44:07,190 --> 00:44:08,480
That's all we care about.

702
00:44:08,480 --> 00:44:11,050
It doesn't really matter that
this is slow in terms of using

703
00:44:11,050 --> 00:44:12,550
the steady state
approximation.

704
00:44:12,550 --> 00:44:17,075
We just want the step getting
out of the intermediate, a

705
00:44:17,075 --> 00:44:20,470
backwards step in this case
here, to be faster than the

706
00:44:20,470 --> 00:44:22,730
state into the intermediate.

707
00:44:22,730 --> 00:44:26,080
So this allows this, this fact
that this is faster, that

708
00:44:26,080 --> 00:44:30,250
allows us to use a steady
state approximation.

709
00:44:30,250 --> 00:44:33,020
So now we can go ahead and
solve the problem.

710
00:44:33,020 --> 00:44:38,130
So we write down all the
rates minus dA/dt.

711
00:44:38,130 --> 00:44:46,170
It's k1 A. Let's start with
the rate of the action.

712
00:44:46,170 --> 00:44:47,390
Let's take a product.

713
00:44:47,390 --> 00:44:50,050
Let's take C. Let's say this
goes to product C here.

714
00:44:50,050 --> 00:44:55,410
So dC/dt is equal to
k2 times A, that's

715
00:44:55,410 --> 00:45:00,110
the rate of the reaction.

716
00:45:00,110 --> 00:45:03,330
And we're going to look at
this rate of reaction.

717
00:45:03,330 --> 00:45:04,650
And see how it changes.

718
00:45:04,650 --> 00:45:08,830
What it looks like under
limiting conditions.

719
00:45:08,830 --> 00:45:13,990
The rate of reaction.

720
00:45:13,990 --> 00:45:15,220
Let's look at the
intermediate.

721
00:45:15,220 --> 00:45:16,940
Because this is what we're going
to solve first, because

722
00:45:16,940 --> 00:45:19,660
this is going to turn out to
be an algebraic equation.

723
00:45:19,660 --> 00:45:21,550
Because we're going to apply
the steady state

724
00:45:21,550 --> 00:45:23,080
approximations.

725
00:45:23,080 --> 00:45:28,656
Intermediate d A star / dt is,
we can create it through the

726
00:45:28,656 --> 00:45:30,430
forward process.

727
00:45:30,430 --> 00:45:33,400
We destroy it through the
backwards process.

728
00:45:33,400 --> 00:45:37,930
A star times M. And we destroy
it through the final process.

729
00:45:37,930 --> 00:45:40,840
A star.

730
00:45:40,840 --> 00:45:44,390
We apply the steady state
approximation, steady state,

731
00:45:44,390 --> 00:45:46,530
steady state.

732
00:45:46,530 --> 00:45:55,180
We solve for A steady state
algebraically. k1 A M over M

733
00:45:55,180 --> 00:46:00,960
times k minus 1 plus k2.

734
00:46:00,960 --> 00:46:04,330
And then we plug that back into
the rate of the reaction.

735
00:46:04,330 --> 00:46:06,530
The rate of appearance of C.
Which is what we're measuring

736
00:46:06,530 --> 00:46:07,140
experimentally.

737
00:46:07,140 --> 00:46:08,850
We're looking at the products.

738
00:46:08,850 --> 00:46:10,240
Measuring the appearance
of products.

739
00:46:10,240 --> 00:46:11,170
So they're destruction.

740
00:46:11,170 --> 00:46:21,920
This is also equal to the
destruction of A.

741
00:46:21,920 --> 00:46:31,300
And so we plug this into the, we
write our rate of reaction.

742
00:46:31,300 --> 00:46:34,340
Which we measure,
experimentally.

743
00:46:34,340 --> 00:46:37,860
We find this is k1 times k2.

744
00:46:37,860 --> 00:46:42,640
Times, we plug in for, and
here, what we do here, we

745
00:46:42,640 --> 00:46:48,030
solve A in terms of
A steady state.

746
00:46:48,030 --> 00:46:53,330
A is equal to something
A steady state.

747
00:46:53,330 --> 00:47:04,210
Well, let's see, A is equal to
M k minus one plus k2 times,

748
00:47:04,210 --> 00:47:06,545
well this is actually,
I have this wrong up

749
00:47:06,545 --> 00:47:11,670
here, this is d A star.

750
00:47:11,670 --> 00:47:15,600
So we don't need to
do this here.

751
00:47:15,600 --> 00:47:18,590
The appearance of C
is, A star here.

752
00:47:18,590 --> 00:47:24,740
So we plug in A steady
state here.

753
00:47:24,740 --> 00:47:29,325
A star steady state, and this
is A star steady state.

754
00:47:29,325 --> 00:47:36,270
The intermediate is what we're
solving a steady state for.

755
00:47:36,270 --> 00:47:43,900
So this is k1 k2 A M over
M k minus one plus k2.

756
00:47:43,900 --> 00:47:48,130
We just have the extra k2 that
appears when we multiply A

757
00:47:48,130 --> 00:47:52,330
steady state star with
the k2 there.

758
00:47:52,330 --> 00:47:55,590
And then we can look at
the limiting cases.

759
00:47:55,590 --> 00:47:58,920
There are two limiting cases,
which we can tell by looking

760
00:47:58,920 --> 00:47:59,410
at the denominator.

761
00:47:59,410 --> 00:48:01,850
If we have one term bigger
than the other.

762
00:48:01,850 --> 00:48:03,520
Only two choices here.

763
00:48:03,520 --> 00:48:07,270
So the first case is where
M k minus one is

764
00:48:07,270 --> 00:48:10,320
much bigger than k2.

765
00:48:10,320 --> 00:48:11,630
Experimentally, what
does that mean?

766
00:48:11,630 --> 00:48:14,320
It means that this in
the gas phase now.

767
00:48:14,320 --> 00:48:17,560
It means that regardless of
what these rates are here,

768
00:48:17,560 --> 00:48:21,790
we've managed to make the
concentration of M high enough

769
00:48:21,790 --> 00:48:23,980
so that this becomes true.

770
00:48:23,980 --> 00:48:26,310
There's always an M that's going
to be high enough so

771
00:48:26,310 --> 00:48:28,600
that the multiplication of
these two is going to be

772
00:48:28,600 --> 00:48:29,050
bigger than that.

773
00:48:29,050 --> 00:48:29,910
What does that mean?

774
00:48:29,910 --> 00:48:33,180
It means that the pressure
is high.

775
00:48:33,180 --> 00:48:34,680
Concentration of M is high.

776
00:48:34,680 --> 00:48:40,020
This is a high pressure case.

777
00:48:40,020 --> 00:48:43,400
And then the other one
is M k minus one is

778
00:48:43,400 --> 00:48:45,730
much less than k2.

779
00:48:45,730 --> 00:48:47,820
M is very small.

780
00:48:47,820 --> 00:48:51,210
Low concentration,
low pressure.

781
00:48:51,210 --> 00:48:54,910
The partial pressure
of M is very low.

782
00:48:54,910 --> 00:48:57,220
This could be A. M could be A,
or it could be some other

783
00:48:57,220 --> 00:49:00,720
molecule that's in the mix.

784
00:49:00,720 --> 00:49:03,080
And by high pressure we mean
something on the order of one

785
00:49:03,080 --> 00:49:04,990
bar, let's say.

786
00:49:04,990 --> 00:49:07,350
That by low pressure we mean
something around 10 to the

787
00:49:07,350 --> 00:49:12,410
minus 4 bar.

788
00:49:12,410 --> 00:49:16,850
So now we can go and put these
limiting cases in our equation

789
00:49:16,850 --> 00:49:20,120
here. k minus one is much bigger
than k2, so we can

790
00:49:20,120 --> 00:49:21,970
ignore k2 here.

791
00:49:21,970 --> 00:49:23,280
And then we have something
that looks

792
00:49:23,280 --> 00:49:25,380
like the M's disappear.

793
00:49:25,380 --> 00:49:31,880
We have something that looks
like k1 k2 over k minus one

794
00:49:31,880 --> 00:49:37,120
times A is the rate
of reaction.

795
00:49:37,120 --> 00:49:38,030
First order.

796
00:49:38,030 --> 00:49:39,380
Looks like it's first
order in A. This is

797
00:49:39,380 --> 00:49:41,600
what people had observed.

798
00:49:41,600 --> 00:49:42,380
Great.

799
00:49:42,380 --> 00:49:43,810
The problem is that
they didn't know

800
00:49:43,810 --> 00:49:45,860
to go to low pressure.

801
00:49:45,860 --> 00:49:48,290
They were doing all
their experiments

802
00:49:48,290 --> 00:49:49,340
at atmospheric pressure.

803
00:49:49,340 --> 00:49:50,700
They were getting a
first order rate.

804
00:49:50,700 --> 00:49:52,310
They thought they had
solved the problem.

805
00:49:52,310 --> 00:49:54,590
It's just A falling
apart by itself.

806
00:49:54,590 --> 00:49:58,350
Until you go to low pressure,
where now this term dominates

807
00:49:58,350 --> 00:49:59,630
the denominator.

808
00:49:59,630 --> 00:50:03,690
Or rather, this one dominates
the denominator.

809
00:50:03,690 --> 00:50:12,740
And now you have something of
the form k1 A times M. This

810
00:50:12,740 --> 00:50:14,830
dominates, this term
goes away.

811
00:50:14,830 --> 00:50:19,760
The k2 goes away and you have
k1 times A times M left over

812
00:50:19,760 --> 00:50:25,270
as the rate of the reaction.

813
00:50:25,270 --> 00:50:25,970
Second order.

814
00:50:25,970 --> 00:50:27,040
There are two species here.

815
00:50:27,040 --> 00:50:30,290
A times M. M could be A or it
could be some chaperone.

816
00:50:30,290 --> 00:50:31,290
A squared.

817
00:50:31,290 --> 00:50:33,000
Second order process.

818
00:50:33,000 --> 00:50:34,790
So if you're at low
pressure, you see

819
00:50:34,790 --> 00:50:35,740
something at second order.

820
00:50:35,740 --> 00:50:37,630
You've discovered something
about the mechanisms you

821
00:50:37,630 --> 00:50:39,210
didn't expect.

822
00:50:39,210 --> 00:50:42,130
And then you get
a Nobel Prize.

823
00:50:42,130 --> 00:50:44,490
Alright, next time we'll
do chain reactions.