1
00:00:01 --> 00:00:04
The following content is
provided by MIT OpenCourseWare

2
00:00:04 --> 00:00:06
under a Creative Commons
license.

3
00:00:06 --> 00:00:10
Additional information about
our license and MIT

4
00:00:10 --> 00:00:15
OpenCourseWare in general is
available at ocw.mit.edu.

5
00:00:15 --> 00:00:21
Last time, we went through the
same arguments that Maxwell and

6
00:00:21 --> 00:00:27
Boltzmann did to understand the
microscopic origin of the ideal

7
00:00:27 --> 00:00:32
gas law, PV equal nRT.

8
00:00:32 --> 00:00:38
And we saw how Maxwell had
hypothesized that the pressure

9
00:00:38 --> 00:00:45
of a gas on the walls of some
container, if the molecules were

10
00:00:45 --> 00:00:51
moving and they collided onto
the walls, that pressure must be

11
00:00:51 --> 00:00:59
due to the individual impacts of
the molecules on the wall.

12
00:00:59 --> 00:01:04
That must be due to the change
in the momentum of the wall when

13
00:01:04 --> 00:01:07
the molecules slammed right into
it.

14
00:01:07 --> 00:01:12
And that change in momentum
over some change in time is this

15
00:01:12 --> 00:01:16
force.
That force divided by the total

16
00:01:16 --> 00:01:20
area is the pressure.
And it was using that argument

17
00:01:20 --> 00:01:25
that we came up with an
expression, just like Maxwell

18
00:01:25 --> 00:01:30
did for the pressure times the
volume.

19
00:01:30 --> 00:01:34
And we see that we were able to
write it in terms of the

20
00:01:34 --> 00:01:39
velocity of the molecules that
were moving in this gas.

21
00:01:39 --> 00:01:44
Well, that is very nice because
if we have this theoretical

22
00:01:44 --> 00:01:48
expression P times V,
and we know experimentally that

23
00:01:48 --> 00:01:52
P times V equal nRT,
if this theory is

24
00:01:52 --> 00:01:57
correct, then this quantity n M,
average of the velocity

25
00:01:57 --> 00:02:01
squared, over three,
that better be equal to nRT.

26
00:02:01 --> 00:02:07
Or, solving for what we call

27
00:02:07 --> 00:02:11
the root mean square velocity,
that better be equal to the

28
00:02:11 --> 00:02:15
square root of 3RT over M.

29
00:02:15 --> 00:02:19
That kinetic theory made a
prediction for what the velocity

30
00:02:19 --> 00:02:22
ought to be.
It took, of course,

31
00:02:22 --> 00:02:26
another hundred years before
somebody could measure that.

32
00:02:26 --> 00:02:28
And, of course,
it is correct.

33
00:02:28 --> 00:02:32
But what is interesting here is
that this model gave us an

34
00:02:32 --> 00:02:37
understanding for what
temperature is.

35
00:02:37 --> 00:02:41
Temperature is a measure of the
speed of the molecules in the

36
00:02:41 --> 00:02:44
gas.
It is also, as we saw last

37
00:02:44 --> 00:02:47
time, a measure of their kinetic
energy.

38
00:02:47 --> 00:02:52
We saw that the kinetic energy
is one-half M average of the

39
00:02:52 --> 00:02:56
velocity squared. We put

40
00:02:56 --> 00:03:00
in our result from kinetic
theory.

41
00:03:00 --> 00:03:05
That shows us that the average
energy is three-halves RT.

42
00:03:05 --> 00:03:08
Temperature is the measure of

43
00:03:08 --> 00:03:12
the kinetic energy of these
molecules in the gas.

44
00:03:12 --> 00:03:15
For the first time,
there was a microscopic

45
00:03:15 --> 00:03:18
understanding of what
temperature was.

46
00:03:18 --> 00:03:22
Now, what we looked at also,
last time, was the

47
00:03:22 --> 00:03:27
Maxwell-Boltzmann distribution
of velocities or speeds and

48
00:03:27 --> 00:03:32
talked about this particular
functional form.

49
00:03:32 --> 00:03:35
How there was this quadratic
dependence right here,

50
00:03:35 --> 00:03:39
a low v because of the v
squared here and then an

51
00:03:39 --> 00:03:44
exponentially decaying tail.
There were two parameters,

52
00:03:44 --> 00:03:48
the mass and the temperature.
We talked about that last time.

53
00:03:48 --> 00:03:52
But, of course,
if there is a distribution here

54
00:03:52 --> 00:03:56
of speeds where this
distribution is a probability of

55
00:03:56 --> 00:04:01
finding a molecule with a
particular speed between v and v

56
00:04:01 --> 00:04:05
plus dv, --
-- then there also has to be a

57
00:04:05 --> 00:04:09
distribution of energies because
the velocity is related to the

58
00:04:09 --> 00:04:11
energy.
What do we have to do?

59
00:04:11 --> 00:04:14
We have to take that
Maxwell-Boltzmann speed

60
00:04:14 --> 00:04:18
distribution and change the
variable to energy.

61
00:04:18 --> 00:04:22
We know how to do that.
We know how to equate those two

62
00:04:22 --> 00:04:26
distribution functions.
And, using the Jacobean of the

63
00:04:26 --> 00:04:29
transformation that we talked
about last time,

64
00:04:29 --> 00:04:34
we can calculate what the
energy distribution is.

65
00:04:34 --> 00:04:37
f of E is our
Maxwell-Boltzmann energy

66
00:04:37 --> 00:04:40
distribution.
It is the probability of

67
00:04:40 --> 00:04:45
finding a molecule with a
kinetic energy E to E plus dE.

68
00:04:45 --> 00:04:48
It also has a decaying
exponential term,

69
00:04:48 --> 00:04:53
here, with the energy in the
argument, and it is multiplied

70
00:04:53 --> 00:04:57
by E of the one-half.
Let's take a look at what that

71
00:04:57 --> 00:05:01
distribution function looks
like.

72
00:05:01 --> 00:05:04
Here it is.
Notice, as we learned last

73
00:05:04 --> 00:05:09
time, that the energy has
nothing to do with the mass of

74
00:05:09 --> 00:05:12
the particle.
The only parameter that is

75
00:05:12 --> 00:05:17
important is the temperature.
At a given temperature,

76
00:05:17 --> 00:05:21
all particles,
it doesn't matter what their

77
00:05:21 --> 00:05:26
mass is, have the same energy.
So, all particles at

78
00:05:26 --> 00:05:30
degrees Kelvin have a
Maxwell-Boltzmann energy

79
00:05:30 --> 00:05:35
distribution that looks like
this.

80
00:05:35 --> 00:05:39
There is a very rapid rise in
that distribution function,

81
00:05:39 --> 00:05:42
unlike the velocity
distribution,

82
00:05:42 --> 00:05:45
which increased as v squared.

83
00:05:45 --> 00:05:48
It peaks rapidly,
and then there is an

84
00:05:48 --> 00:05:52
exponentially decaying term in
the energy.

85
00:05:52 --> 00:05:56
At 1500 degrees Kelvin,
the energy distribution looks

86
00:05:56 --> 00:05:59
like this.
The energy of the molecules has

87
00:05:59 --> 00:06:04
increased.
Since this is a probability,

88
00:06:04 --> 00:06:07
and the area under these curves
has to equal to one,

89
00:06:07 --> 00:06:11
if the energy goes up,
that is we have more molecules

90
00:06:11 --> 00:06:16
out here with higher energies,
well, then this maximum value

91
00:06:16 --> 00:06:21
for the probabilities has got to
go down because we have to keep

92
00:06:21 --> 00:06:25
the area under that curve equal
to one, since this is a

93
00:06:25 --> 00:06:28
probability.
What we see here,

94
00:06:28 --> 00:06:33
as we raise the energy,
is that we have more and more

95
00:06:33 --> 00:06:38
molecules with high energies.
There are not a lot of them,

96
00:06:38 --> 00:06:42
but there are some.
And they are going to be really

97
00:06:42 --> 00:06:47
important in the example I am
going to show you in just a

98
00:06:47 --> 00:06:51
moment.
Average energy at 600 Kelvin,

99
00:06:51 --> 00:06:55
just three-halves RT,
is 7.5 kilojoules per mole.

100
00:06:55 --> 00:07:02
Average energy at 1500 Kelvin
is 18.7 kilojoules per mole.

101
00:07:02 --> 00:07:07
But now, I want to show you an
example of the importance of

102
00:07:07 --> 00:07:12
this Maxwell-Boltzmann
distribution function for

103
00:07:12 --> 00:07:18
energies to chemical reactions,
the importance of it in making

104
00:07:18 --> 00:07:23
some kinds of chemical reactions
actually work.

105
00:07:23 --> 00:07:29
And so the example is going to
be this reaction.

106
00:07:29 --> 00:07:32
This reaction is called steam
reforming of natural gas.

107
00:07:32 --> 00:07:35
It is the reaction of methane
plus water.

108
00:07:35 --> 00:07:39
Natural gas is mostly methane.
But this reaction of methane

109
00:07:39 --> 00:07:43
and water makes CO and hydrogen.
It

110
00:07:43 --> 00:07:47
turns out that this reaction
does not work in the gas phase.

111
00:07:47 --> 00:07:51
That is, if you have a methane
molecule and water molecule

112
00:07:51 --> 00:07:56
collide, they are just going to
collide, bounce apart and go in

113
00:07:56 --> 00:07:59
their opposite directions.
They are not going to make CO

114
00:07:59 --> 00:08:04
and hydrogen.
And so what you have to have in

115
00:08:04 --> 00:08:08
this reaction to make it go is a
catalyst.

116
00:08:08 --> 00:08:11
That catalyst is going to be a
nickel surface.

117
00:08:11 --> 00:08:17
A catalyst is something that
will lower the activation energy

118
00:08:17 --> 00:08:21
barrier by changing the
mechanism of a reaction so that

119
00:08:21 --> 00:08:24
the reaction can proceed.
In this case,

120
00:08:24 --> 00:08:29
what happens is that the
methane and the water impinge on

121
00:08:29 --> 00:08:35
this nickel metal catalyst.
And the methane and the water

122
00:08:35 --> 00:08:38
decompose.
They fall apart to their

123
00:08:38 --> 00:08:43
elements on that nickel surface.
And then, once you have carbon,

124
00:08:43 --> 00:08:46
hydrogen and oxygen on that
nickel surface,

125
00:08:46 --> 00:08:50
the atoms rearrange and come
off as CO and molecular

126
00:08:50 --> 00:08:53
hydrogen.
So, that nickel surface is a

127
00:08:53 --> 00:08:56
catalyst.
We call this the catalytic

128
00:08:56 --> 00:08:59
reaction.
In particular,

129
00:08:59 --> 00:09:02
we call it a heterogeneous
catalytic reaction.

130
00:09:02 --> 00:09:08
It is heterogeneous because the
catalyst is in a different phase

131
00:09:08 --> 00:09:12
than the reactants.
The catalyst here is a solid.

132
00:09:12 --> 00:09:16
The reactants are gases.
A homogeneous catalyst is one

133
00:09:16 --> 00:09:21
in which the phase of the
catalyst and the reactants is

134
00:09:21 --> 00:09:24
the same.
This is a heterogeneous

135
00:09:24 --> 00:09:28
catalytic reaction.
It turns out that this reaction

136
00:09:28 --> 00:09:33
here is really an important
reaction from the standpoint of

137
00:09:33 --> 00:09:39
the production of hydrogen.
All of the hydrogen that you

138
00:09:39 --> 00:09:42
use, you know,
if you are doing a laboratory

139
00:09:42 --> 00:09:47
experiment and you go get a tank
of hydrogen, that hydrogen is

140
00:09:47 --> 00:09:51
made by this reaction,
by reacting methane and water

141
00:09:51 --> 00:09:55
to form that hydrogen.
All of our commercial sources

142
00:09:55 --> 00:10:00
of hydrogen come from carrying
out this reaction.

143
00:10:00 --> 00:10:03
For example,
if you are in the business of

144
00:10:03 --> 00:10:08
ammonia synthesis,
which is taking hydrogen and

145
00:10:08 --> 00:10:13
nitrogen to make ammonia,
which is also carried out on an

146
00:10:13 --> 00:10:17
iron surface,
another heterogeneous catalytic

147
00:10:17 --> 00:10:19
reaction.
And ammonia,

148
00:10:19 --> 00:10:24
of course, is the starting
material for lots of chemicals,

149
00:10:24 --> 00:10:30
in particular fertilizers.
If you have an ammonia plant,

150
00:10:30 --> 00:10:35
right next to the ammonia plant
you have a steamer-forming plant

151
00:10:35 --> 00:10:39
to make the hydrogen to feed
into this reaction.

152
00:10:39 --> 00:10:44
And another place where this is
useful is in methanol synthesis.

153
00:10:44 --> 00:10:49
That is, you can take hydrogen
and CO on a copper zinc oxide

154
00:10:49 --> 00:10:54
heterogeneous catalyst and make
methanol, the starting point for

155
00:10:54 --> 00:11:00
gasoline, which now is certainly
economically feasible.

156
00:11:00 --> 00:11:05
But it turns out that this
reaction here is really a hard

157
00:11:05 --> 00:11:09
one to carry out.
Despite the fact that nickel is

158
00:11:09 --> 00:11:15
a catalyst for the reaction and
makes it go, the reaction still

159
00:11:15 --> 00:11:20
has to be carried out at very
high temperatures.

160
00:11:20 --> 00:11:24
1500 degrees kelvin is a very
high temperature.

161
00:11:24 --> 00:11:29
It also needs very high
pressure.

162
00:11:29 --> 00:11:33
And one of the reasons why we
don't have a hydrogen economy is

163
00:11:33 --> 00:11:36
because it is so difficult to
make.

164
00:11:36 --> 00:11:39
Hydrogen can be done,
and it is done in all of the

165
00:11:39 --> 00:11:43
commercial processes that I
mentioned to you,

166
00:11:43 --> 00:11:46
but it is still hard.
1500 degrees kelvin,

167
00:11:46 --> 00:11:48
high pressures.
The question is,

168
00:11:48 --> 00:11:52
why is it so hard?
Why is there this barrier here?

169
00:11:52 --> 00:11:56
Why do we need to raise the
temperature of the gas in order

170
00:11:56 --> 00:12:02
to get this reaction to go?
Well, let's take a look at

171
00:12:02 --> 00:12:05
that.
What there exists here in this

172
00:12:05 --> 00:12:09
problem is what is called an
activation energy barrier to

173
00:12:09 --> 00:12:13
making the reaction go.
It turns out that this

174
00:12:13 --> 00:12:18
activation energy barrier is
really in the very first step of

175
00:12:18 --> 00:12:22
the reaction,
which is pulling the methane

176
00:12:22 --> 00:12:27
apart, pulling that first
hydrogen atom off of the methane

177
00:12:27 --> 00:12:31
molecule.
What I represent right here is

178
00:12:31 --> 00:12:36
the energy of the reaction as a
function of the reaction,

179
00:12:36 --> 00:12:40
and the reaction here is just
the first step.

180
00:12:40 --> 00:12:45
It is taking methane gas,
pulling the hydrogen off so

181
00:12:45 --> 00:12:50
that you have a methyl radical
stuck to the surface and you

182
00:12:50 --> 00:12:54
have a hydrogen atom stuck to
the surface.

183
00:12:54 --> 00:12:59
It is just breaking that first
C-H bond.

184
00:12:59 --> 00:13:04
What you can see here is that
you have to put 50 kilojoules of

185
00:13:04 --> 00:13:08
energy into that reaction in
order to make it go.

186
00:13:08 --> 00:13:12
You have to put that 50
kilocalories of energy in first,

187
00:13:12 --> 00:13:17
before you get any energy back.
You can see that this is

188
00:13:17 --> 00:13:20
exothermic.
But you have to put this energy

189
00:13:20 --> 00:13:23
in, in order to get that
reaction to go.

190
00:13:23 --> 00:13:29
Well, how do we know that?
Let me back up a minute.

191
00:13:29 --> 00:13:31
Here is where the
Maxwell-Boltzmann energy

192
00:13:31 --> 00:13:35
distribution is important.
What I did was I took those

193
00:13:35 --> 00:13:39
Boltzmann energy distributions
that I plotted for you at

194
00:13:39 --> 00:13:43
Kelvin and 1500 Kelvin and
turned them on their side.

195
00:13:43 --> 00:13:47
This is essentially equal to
the probability of finding a

196
00:13:47 --> 00:13:51
molecule at a particular energy
versus the energy.

197
00:13:51 --> 00:13:55
Here is the Maxwell-Boltzmann
distribution at 600 degrees

198
00:13:55 --> 00:13:57
Kelvin.
It is peaked to very low

199
00:13:57 --> 00:14:01
energies.
And, if you look here in this

200
00:14:01 --> 00:14:05
Maxwell-Boltzmann tail,
there are not very many

201
00:14:05 --> 00:14:10
molecules that have enough
energy to get over that barrier.

202
00:14:10 --> 00:14:14
However, if we raise the
temperature of the gas to 

203
00:14:14 --> 00:14:18
degrees Kelvin,
then now you can see that there

204
00:14:18 --> 00:14:23
are more molecules here in the
high energy part of this tail.

205
00:14:23 --> 00:14:27
And it is these molecules at
these high energies that

206
00:14:27 --> 00:14:32
actually can get over this
barrier to dissociation of the

207
00:14:32 --> 00:14:36
methane molecule.
They are not many of those

208
00:14:36 --> 00:14:39
molecules with high energy,
but there are some.

209
00:14:39 --> 00:14:42
And that is important,
because what happens is when

210
00:14:42 --> 00:14:45
they have enough energy,
they react and they leave.

211
00:14:45 --> 00:14:49
And then this Boltzmann
distribution re-equilibrates.

212
00:14:49 --> 00:14:52
If the high energy molecules
leave, then there are lots of

213
00:14:52 --> 00:14:56
collisions here that kick some
more molecules up to the high

214
00:14:56 --> 00:15:00
energy part of the tail,
and they react.

215
00:15:00 --> 00:15:04
And so it works.
That is why it is important.

216
00:15:04 --> 00:15:09
How do we really know that
there is this barrier here to

217
00:15:09 --> 00:15:15
the dissociation of methane,
and why is there this barrier?

218
00:15:15 --> 00:15:21
What is the physical origin of
this barrier to pulling this C-H

219
00:15:21 --> 00:15:25
bond apart?
Well, one way we could know

220
00:15:25 --> 00:15:30
that there was this barrier
there for sure is to do the

221
00:15:30 --> 00:15:35
following.
If we had a way to take methane

222
00:15:35 --> 00:15:38
gas and make it have just a
single energy,

223
00:15:38 --> 00:15:41
not the Maxwell-Boltzmann
distribution of energies that

224
00:15:41 --> 00:15:46
have lots of different energies,
which have lots of different

225
00:15:46 --> 00:15:49
energies in them,
and you saw how broad those

226
00:15:49 --> 00:15:51
curves were.
But if we had a way to make

227
00:15:51 --> 00:15:56
methane gas with just say energy
E sub 1, whatever that is,

228
00:15:56 --> 00:15:59
we could then take those
molecules with just those

229
00:15:59 --> 00:16:02
energies, aim them at this
nickel surface,

230
00:16:02 --> 00:16:06
and see if the methane fell
apart.

231
00:16:06 --> 00:16:09
And, if it didn't,
then we would prepare molecules

232
00:16:09 --> 00:16:13
at some higher energy and see if
they fell apart.

233
00:16:13 --> 00:16:16
If they didn't then we would go
to higher energy.

234
00:16:16 --> 00:16:20
We would keep going until we
got to the top of the barrier.

235
00:16:20 --> 00:16:24
Well, how do you do that?
How do you make molecules with

236
00:16:24 --> 00:16:28
a particular energy?
Because I just showed you a

237
00:16:28 --> 00:16:32
Maxwell-Boltzmann energy
distribution at 300 Kelvin,

238
00:16:32 --> 00:16:37
600 Kelvin.
There are a lot of energies

239
00:16:37 --> 00:16:41
present.
How do we make molecules with

240
00:16:41 --> 00:16:46
just one energy?
We have a way to do that by

241
00:16:46 --> 00:16:52
using some beam techniques and
high pressure adiabatic

242
00:16:52 --> 00:16:57
expansions.
Basically, the way it works is

243
00:16:57 --> 00:17:01
this.
What we are going to do is take

244
00:17:01 --> 00:17:07
a tube here, which is going to
have a high pressure of methane

245
00:17:07 --> 00:17:10
in it.
Then we are going to punch a

246
00:17:10 --> 00:17:14
little hole in that tube.
This tube is actually sitting

247
00:17:14 --> 00:17:19
in a vacuum, so we are going to
expand the methane from that

248
00:17:19 --> 00:17:24
tube into the vacuum.
We are going to squirt it out.

249
00:17:24 --> 00:17:29
It is going to become a beam of
molecules.

250
00:17:29 --> 00:17:34
But when you expand a gas from
high pressure to low pressure,

251
00:17:34 --> 00:17:38
this expansion is called an
adiabatic expansion,

252
00:17:38 --> 00:17:42
which means the gas cools.
I will explain that in a

253
00:17:42 --> 00:17:47
moment, but the adiabatic
expansion you are going to talk

254
00:17:47 --> 00:17:50
a lot about in 5.60.
And, if you are a chemical

255
00:17:50 --> 00:17:55
engineer, you will talk even
more about it past 5.60.

256
00:17:55 --> 00:17:59
Here is how it works.
You take these molecules and

257
00:17:59 --> 00:18:04
squirt them out.
Because the pressure right here

258
00:18:04 --> 00:18:08
is so high, what happens is
there are lots and lots and lots

259
00:18:08 --> 00:18:11
of collisions.
And so, you can imagine,

260
00:18:11 --> 00:18:15
if you have some slow molecule
just kind of lumbering along and

261
00:18:15 --> 00:18:20
a fast one comes and hits you,
what is going to happen is you

262
00:18:20 --> 00:18:23
are going to speed up,
and the fast one is going to

263
00:18:23 --> 00:18:26
slow down.
And, if you keep doing that

264
00:18:26 --> 00:18:30
again and again and again,
all the molecules are going to

265
00:18:30 --> 00:18:35
end up with the same energy or
the same velocity.

266
00:18:35 --> 00:18:39
If you have so many collisions,
after a while they are going to

267
00:18:39 --> 00:18:44
all have the same velocity or
the same energy because one gets

268
00:18:44 --> 00:18:48
sped up, one gets slowed down.
They are all going to end up

269
00:18:48 --> 00:18:51
with the same energy.
That is what happens.

270
00:18:51 --> 00:18:56
That is how we make a beam of
molecules, a source of molecules

271
00:18:56 --> 00:19:00
with the same kinetic energy.
And basically,

272
00:19:00 --> 00:19:03
what is happening then is that
before the expansion,

273
00:19:03 --> 00:19:07
just look at this curve,
here is an actual Boltzmann

274
00:19:07 --> 00:19:10
distribution of velocities.
It is broad.

275
00:19:10 --> 00:19:14
But after this expansion,
here is the distribution.

276
00:19:14 --> 00:19:18
It is really pretty narrow.
If you put a temperature to

277
00:19:18 --> 00:19:21
that distribution,
you might find that at one

278
00:19:21 --> 00:19:24
degree Kelvin,
you can really make molecules

279
00:19:24 --> 00:19:30
with a single or very narrow
distribution in energies.

280
00:19:30 --> 00:19:33
That is what we do.
We make those molecules with

281
00:19:33 --> 00:19:38
this energy E sub 1.
Then we have some more tricks

282
00:19:38 --> 00:19:41
for changing those energies in a
controlled way.

283
00:19:41 --> 00:19:46
And we just keep cranking those
energies up and watch to see

284
00:19:46 --> 00:19:51
when the methane falls apart.
And so, you get data that kind

285
00:19:51 --> 00:19:54
of looks like this.
This is a dissociation

286
00:19:54 --> 00:19:58
probability of a methane as a
function of its energy.

287
00:19:58 --> 00:20:02
This is in kilocalories per
mole.

288
00:20:02 --> 00:20:05
At some point,
say right about here,

289
00:20:05 --> 00:20:10
15 kilocalories per mole,
the dissociation probability

290
00:20:10 --> 00:20:13
skyrockets.
All of a sudden,

291
00:20:13 --> 00:20:18
you are at a high enough energy
to get that methane to fall

292
00:20:18 --> 00:20:22
apart.
This tells you exactly where

293
00:20:22 --> 00:20:26
that barrier is.
But now, the question is,

294
00:20:26 --> 00:20:32
why is there this barrier?
Physically, why is there a

295
00:20:32 --> 00:20:38
barrier to pulling the hydrogen
off of the carbon when that

296
00:20:38 --> 00:20:43
methane molecule comes close to
the surface?

297
00:20:43 --> 00:20:48
Well, the reason is this.
In order to break the

298
00:20:48 --> 00:20:54
carbon-hydrogen bond in methane,
in order to have enough energy

299
00:20:54 --> 00:21:00
to break that bond,
what you need to do is you need

300
00:21:00 --> 00:21:05
to simultaneously form a
nickel-carbon bond and a

301
00:21:05 --> 00:21:10
nickel-hydrogen bond.
In other words,

302
00:21:10 --> 00:21:17
when this methane molecule here
comes into some nickel surface

303
00:21:17 --> 00:21:23
very slowly, because that
methane is tetrahedral and the

304
00:21:23 --> 00:21:30
carbon is kind of hidden in the
center, the hydrogens interact

305
00:21:30 --> 00:21:35
with the nickel.
But the carbon does not get in

306
00:21:35 --> 00:21:39
close enough to the nickel to
start to form a nickel-carbon

307
00:21:39 --> 00:21:42
bond.
Now, if you speed that methane

308
00:21:42 --> 00:21:46
molecule up and you really ram
it into the surface,

309
00:21:46 --> 00:21:49
upon collision of the molecule
with the surface,

310
00:21:49 --> 00:21:54
the hydrogens are pushed back,
the carbon gets in close enough

311
00:21:54 --> 00:22:00
to the nickel to start to form,
here, this nickel-carbon bond.

312
00:22:00 --> 00:22:05
You form a nickel-hydrogen
bond, and now you can break that

313
00:22:05 --> 00:22:09
C-H bond.
We call this black chemistry.

314
00:22:09 --> 00:22:12
The barrier,
there, to that reaction

315
00:22:12 --> 00:22:18
physically is the amount of
energy that you have to put into

316
00:22:18 --> 00:22:22
the molecule to push those
hydrogens back,

317
00:22:22 --> 00:22:30
to bend those hydrogens back to
distort this methane molecule.

318
00:22:30 --> 00:22:33
Once you do that,
the reaction goes.

319
00:22:33 --> 00:22:35
You are on a roll there.
It goes.

320
00:22:35 --> 00:22:39
So, that is the physical origin
of this barrier,

321
00:22:39 --> 00:22:43
here, to the dissociation of
methane.

322
00:22:43 --> 00:22:47
You need to put enough kinetic
energy into those methane

323
00:22:47 --> 00:22:51
molecules in order to get over
that barrier.

324
00:22:51 --> 00:22:56
And we do that on a practical
scale by raising the temperature

325
00:22:56 --> 00:23:01
of the gas.
And on a microscopic scale,

326
00:23:01 --> 00:23:06
this is what is happening.
You can imagine that my

327
00:23:06 --> 00:23:11
students and I did this
experiment 15 years ago.

328
00:23:11 --> 00:23:16
But then, we said if this
barrier here is the energy

329
00:23:16 --> 00:23:22
required to deform that methane
molecule, then we should be able

330
00:23:22 --> 00:23:26
to do this experiment.
This experiment is the

331
00:23:26 --> 00:23:31
following.
We are going to take our nickel

332
00:23:31 --> 00:23:34
surface, here,
and lower the temperature of

333
00:23:34 --> 00:23:38
the surface to 47 kelvin.
At that temperature,

334
00:23:38 --> 00:23:42
what we can do is we can freeze
a layer of methane on the

335
00:23:42 --> 00:23:44
surface.
We call it fizzy-sorb,

336
00:23:44 --> 00:23:47
a layer of methane on the
surface.

337
00:23:47 --> 00:23:51
It will just stick there.
And then, if this barrier is

338
00:23:51 --> 00:23:55
the energy required to deform or
to start the molecule,

339
00:23:55 --> 00:23:59
then in principle I could take
a hammer and pound that molecule

340
00:23:59 --> 00:24:04
into the correct shape for the
transition state that leads to

341
00:24:04 --> 00:24:09
dissociation.
It sounds like a simple idea,

342
00:24:09 --> 00:24:14
and it is, except that we
cannot really take that hammer.

343
00:24:14 --> 00:24:19
But we can take an argon atom.
We can freeze the methane onto

344
00:24:19 --> 00:24:23
the surface, and now we come in
with an argon atom.

345
00:24:23 --> 00:24:27
That is just a big ball.
Or, a xenon atom or a krypton

346
00:24:27 --> 00:24:30
atom.
And we know how to accelerate

347
00:24:30 --> 00:24:35
xenon or krypton.
We don't accelerate it too

348
00:24:35 --> 00:24:38
much, 50 kilocalories,
70 kilocalories per mole,

349
00:24:38 --> 00:24:42
something like that.
And then we can bring it in.

350
00:24:42 --> 00:24:47
What happens is that the impact
of the collision on that methane

351
00:24:47 --> 00:24:51
causes that methane molecule to
compress, distort,

352
00:24:51 --> 00:24:56
gets it into the configuration
of the transition state that

353
00:24:56 --> 00:25:00
leads to the methane falling
apart.

354
00:25:00 --> 00:25:02
And so, you get the same
result.

355
00:25:02 --> 00:25:05
The question,
there, is just getting the

356
00:25:05 --> 00:25:09
energy into the molecule to
actually distort it,

357
00:25:09 --> 00:25:14
to deform it so that you can
make the nickel-carbon bond and

358
00:25:14 --> 00:25:18
the nickel-hydrogen bond.
That is the key.

359
00:25:18 --> 00:25:22
My students and I,
after we had spent many years

360
00:25:22 --> 00:25:26
doing this experiment and spent
a lot of money on this

361
00:25:26 --> 00:25:31
experiment, said we have proven
for sure that when a bug flies

362
00:25:31 --> 00:25:36
into the windshield of your car,
you do get the same result as

363
00:25:36 --> 00:25:41
if you hit the bug on the
windshield of your car with a

364
00:25:41 --> 00:25:46
bug swatter.
That is what we get.

365
00:25:46 --> 00:25:51
This is an example of one
chemical reaction for which we

366
00:25:51 --> 00:25:55
know the physical origin of the
barrier.

367
00:25:55 --> 00:26:01
There are very few chemical
reactions for which we know

368
00:26:01 --> 00:26:06
anything about the physical
origin of a barrier to a

369
00:26:06 --> 00:26:11
reaction.
Now, what I want to talk about

370
00:26:11 --> 00:26:18
is go back to kinetic theory,
because the kinetic theory is

371
00:26:18 --> 00:26:25
going to allow us to calculate a
couple of more quantities that

372
00:26:25 --> 00:26:31
are of interest to us.
And one of those quantities is

373
00:26:31 --> 00:26:37
the collision frequency.
And let me just write that on

374
00:26:37 --> 00:26:41
the board here.
What I am going to want to

375
00:26:41 --> 00:26:47
calculate is something called
Z1, or I am going to call it Z1.

376
00:26:47 --> 00:26:53
It is the number of collisions
that a molecule makes in a gas

377
00:26:53 --> 00:26:58
per unit time.
And our unit time is going to

378
00:26:58 --> 00:27:06
be seconds.
This is the collision frequency

379
00:27:06 --> 00:27:13
of a single molecule in a gas.

380
00:27:13 --> 00:27:28


381
00:27:28 --> 00:27:29
How am I going to calculate the
quantity using the kinetic

382
00:27:29 --> 00:27:30
theory approach that we have
been talking about?

383
00:27:30 --> 00:27:33
What I am going to do is this.
I am going to take some gas,

384
00:27:33 --> 00:27:38
and the molecules in that gas
are represented by these blue

385
00:27:38 --> 00:27:42
circles right here.
They have a diameter D.

386
00:27:42 --> 00:27:47
And then, within that gas,
I am going to imagine this

387
00:27:47 --> 00:27:52
lighter blue cylinder.
That cylinder is an imaginary

388
00:27:52 --> 00:27:55
construct.
We are going to use it to

389
00:27:55 --> 00:28:00
calculate this collision
frequency.

390
00:28:00 --> 00:28:04
What it is going to be is the
collision volume.

391
00:28:04 --> 00:28:09
I have set the diameter of that
cylinder to be equal to two

392
00:28:09 --> 00:28:12
times the diameter of the
molecule.

393
00:28:12 --> 00:28:17
I have set this upright.
And the bottom line is that all

394
00:28:17 --> 00:28:22
of the molecules that
instantaneously happen to be in

395
00:28:22 --> 00:28:26
this cylinder when a molecule
comes through,

396
00:28:26 --> 00:28:32
all of those molecules are
actually going to be hit.

397
00:28:32 --> 00:28:37
They are going to suffer a
collision with another molecule.

398
00:28:37 --> 00:28:42
And I am going to use that,
the number of molecules in this

399
00:28:42 --> 00:28:46
volume, to calculate this
collision frequency.

400
00:28:46 --> 00:28:51
That is what I am going to do.
Snapshot in time,

401
00:28:51 --> 00:28:56
those molecules are frozen
there, except there is one

402
00:28:56 --> 00:29:02
smart-alecky molecule that comes
cruising on through.

403
00:29:02 --> 00:29:04
Here he comes.
Bam, bam, bam.

404
00:29:04 --> 00:29:10
Hits these three molecules,
here, because they are more

405
00:29:10 --> 00:29:14
than halfway into that collision
cylinder.

406
00:29:14 --> 00:29:20
What I know is this smart-aleck
was moving at some average

407
00:29:20 --> 00:29:24
velocity, which I am going to
call vbar.

408
00:29:24 --> 00:29:27
In a time T,
this molecule,

409
00:29:27 --> 00:29:32
since it is moving with an
average velocity vbar,

410
00:29:32 --> 00:29:36
in a time t,
it is traveling a length vbar

411
00:29:36 --> 00:29:41
times t.

412
00:29:41 --> 00:29:44
Velocity times time,
that is going to give you a

413
00:29:44 --> 00:29:47
length.
And so, I am going to set,

414
00:29:47 --> 00:29:50
for convenience,
this time equal to one second

415
00:29:50 --> 00:29:53
because that is going to be my
unit of time.

416
00:29:53 --> 00:29:58
I want to do this per second.
I am going to set this equal to

417
00:29:58 --> 00:30:02
one second.
Therefore, this molecule is

418
00:30:02 --> 00:30:07
going to travel a length vbar in
one second.

419
00:30:07 --> 00:30:13
And that is the length that I
am going to make the collision

420
00:30:13 --> 00:30:16
cylinder.
I am going to make that

421
00:30:16 --> 00:30:22
collision cylinder be vbar long.
That is the distance traveled

422
00:30:22 --> 00:30:28
in one second by this
smart-alecky molecule.

423
00:30:28 --> 00:30:34
Now, all I have to do is I have
got to take the volume of the

424
00:30:34 --> 00:30:40
collision cylinder and multiply
it by all the molecules that

425
00:30:40 --> 00:30:45
happen to be at the volume at
that particular time.

426
00:30:45 --> 00:30:51
Because that volume represents
the volume swept out in one

427
00:30:51 --> 00:30:55
second by this smart-alecky
molecule.

428
00:30:55 --> 00:31:00
What is the volume of the
cylinder?

429
00:31:00 --> 00:31:03
I set the diameter to be equal
to 2d.

430
00:31:03 --> 00:31:08
The volume of the cylinder is
the cross-sectional area times

431
00:31:08 --> 00:31:11
the length.
That cross-sectional area,

432
00:31:11 --> 00:31:16
then, is pi r squared,
but r here is equal to d,

433
00:31:16 --> 00:31:21
just to make it confusing.
The area is pi d squared.

434
00:31:21 --> 00:31:24
The length is vbar.

435
00:31:24 --> 00:31:28
That is the volume of the
cylinder, pi d squared vbar.

436
00:31:28 --> 00:31:34
So, we have the volume of the

437
00:31:34 --> 00:31:37
cylinder.
Now what we are going to need

438
00:31:37 --> 00:31:42
to know is the density of the
molecules in this cylinder,

439
00:31:42 --> 00:31:46
which is the density of the
molecules in the gas.

440
00:31:46 --> 00:31:50
The cylinder is not special.
The cylinder is an imaginary

441
00:31:50 --> 00:31:55
construct that I put in there to
help me calculate the collision

442
00:31:55 --> 00:31:58
frequency.
I need the density of the

443
00:31:58 --> 00:32:03
molecules in the gas or in the
cylinder.

444
00:32:03 --> 00:32:07
The density of the molecules I
am going to set as equal to N,

445
00:32:07 --> 00:32:10
the total number of molecules
in this volume,

446
00:32:10 --> 00:32:13
divided by the volume of the
gas.

447
00:32:13 --> 00:32:17
I am going to call that rho.
N over V is going to be equal

448
00:32:17 --> 00:32:21
to rho.
That is the number of molecules

449
00:32:21 --> 00:32:25
per cubic meter.
That is the density of the gas.

450
00:32:25 --> 00:32:29
This is the symbol that I am
going to use from now on for

451
00:32:29 --> 00:32:34
density defined as molecules per
cubic meter.

452
00:32:34 --> 00:32:39
Then the collision frequency is
simply going to be the density

453
00:32:39 --> 00:32:43
of the molecules times the
volume of the cylinder.

454
00:32:43 --> 00:32:47
The collision frequency Z1,
here, is this volume of the

455
00:32:47 --> 00:32:52
cylinder, pi d squared vbar
times the density.

456
00:32:52 --> 00:32:53
Why?

457
00:32:53 --> 00:32:58
Because I set it up so that
every molecule in that cylinder

458
00:32:58 --> 00:33:04
would suffer a collision.
And that cylinder is the length

459
00:33:04 --> 00:33:08
that a molecule travels in one
second.

460
00:33:08 --> 00:33:11
There is the density times the
volume.

461
00:33:11 --> 00:33:16
The meters cubed disappears.
What I have left is rho,

462
00:33:16 --> 00:33:20
the density,
pi d squared vbar.

463
00:33:20 --> 00:33:24
This is collisions per second.

464
00:33:24 --> 00:33:27
So, I have my collision
frequency.

465
00:33:27 --> 00:33:33
This is the single-molecule
collision frequency.

466
00:33:33 --> 00:33:37
This is the frequency of
collisions that one molecule

467
00:33:37 --> 00:33:40
makes with the other gas
molecules.

468
00:33:40 --> 00:33:44
That is important.
We are going to do a different

469
00:33:44 --> 00:33:47
kind of collision frequency in
just a moment.

470
00:33:47 --> 00:33:52
But now, it turns out that
there is another factor here

471
00:33:52 --> 00:33:57
that I had left out because I
haven't done as sophisticated of

472
00:33:57 --> 00:34:02
an analysis as I could.
And that is that I made the

473
00:34:02 --> 00:34:07
assumption in my picture before
that all the other molecules

474
00:34:07 --> 00:34:11
were frozen in time and that
there was only one smart-aleck

475
00:34:11 --> 00:34:15
that was moving around,
cruising through.

476
00:34:15 --> 00:34:17
The reality is they are all
moving.

477
00:34:17 --> 00:34:23
And what I really have to do is
I have to take into account the

478
00:34:23 --> 00:34:27
relative velocities of the
molecules, the relative speeds.

479
00:34:27 --> 00:34:31
I can do that.
There is a little more

480
00:34:31 --> 00:34:36
sophisticated analysis in doing
that, but I could do that.

481
00:34:36 --> 00:34:40
And, if we do that,
this makes a difference of the

482
00:34:40 --> 00:34:44
square root of two.
I am just going to put that

483
00:34:44 --> 00:34:49
square root of two in there
right now, and later on you will

484
00:34:49 --> 00:34:54
be able to see where that comes
from, in a later course.

485
00:34:54 --> 00:34:59
That actually is the collision
frequency of a single molecule

486
00:34:59 --> 00:35:01
in the gas.

487
00:35:01 --> 00:35:07


488
00:35:07 --> 00:35:12
I am also interested in another
quantity, which is the total

489
00:35:12 --> 00:35:16
collision frequency.
That was the single collision

490
00:35:16 --> 00:35:21
frequency, but I am also
interested in knowing how many

491
00:35:21 --> 00:35:26
collisions are occurring in the
entire gas per unit time,

492
00:35:26 --> 00:35:32
the total collision frequency.
You know why I am interested in

493
00:35:32 --> 00:35:36
that number?
I am interested in that number

494
00:35:36 --> 00:35:41
because that is going to be the
upper limit to any reaction

495
00:35:41 --> 00:35:44
rate.
A reaction in the gas phase or

496
00:35:44 --> 00:35:49
in solution cannot happen any
faster than the molecules

497
00:35:49 --> 00:35:52
collide.
They have to collide before a

498
00:35:52 --> 00:35:57
reaction is going to occur.
The total collision frequency,

499
00:35:57 --> 00:36:03
the importance of that number
is that it is the upper limit to

500
00:36:03 --> 00:36:08
a reaction rate.
Let's calculate that.

501
00:36:08 --> 00:36:13
Z is going to be equal to the
collision frequency of one

502
00:36:13 --> 00:36:17
molecule, Z1,
times the total number of

503
00:36:17 --> 00:36:21
molecules in the gas, N.

504
00:36:21 --> 00:36:25
That is what N stands for.
But since each collision

505
00:36:25 --> 00:36:31
involves two molecules,
I am going to have to multiply

506
00:36:31 --> 00:36:36
this by one-half.
Otherwise, I am going to over

507
00:36:36 --> 00:36:41
count the number of collisions
because each collision involves

508
00:36:41 --> 00:36:44
two molecules.
The total collision frequency

509
00:36:44 --> 00:36:48
here is one-half times N times
Z1.

510
00:36:48 --> 00:36:52
Therefore, if I go and plug in
my expression for Z1 and N and

511
00:36:52 --> 00:36:56
simplify things,
my total collision frequency

512
00:36:56 --> 00:37:01
here is one over the square root
of two times N times rho pi d

513
00:37:01 --> 00:37:06
squared times the average speed.

514
00:37:06 --> 00:37:11
That is my total collisions per

515
00:37:11 --> 00:37:15
second.
This is not rocket science.

516
00:37:15 --> 00:37:20
This is easy.
What I want you to realize is

517
00:37:20 --> 00:37:26
that you do have to understand
what N is, what rho is,

518
00:37:26 --> 00:37:32
what d is, what vbar is.
And I am always surprised on an

519
00:37:32 --> 00:37:38
exam, when we are going to give
you this equation,

520
00:37:38 --> 00:37:43
that students don't actually
know what rho is,

521
00:37:43 --> 00:37:48
or N is, or v is,
or d is.

522
00:37:48 --> 00:37:51
This is easy.
You do just have to understand,

523
00:37:51 --> 00:37:54
this is the total number of
molecules in the gas,

524
00:37:54 --> 00:37:58
this is the density of the gas
in molecules per cubic meter,

525
00:37:58 --> 00:38:02
the diameter of the molecule,
the average speed.

526
00:38:02 --> 00:38:07
That was just a helpful hint.
This is the upper limit to the

527
00:38:07 --> 00:38:10
reaction rate.
When somebody tells you,

528
00:38:10 --> 00:38:15
the rate of this reaction is so
and so many molecules per cubic

529
00:38:15 --> 00:38:18
meter per second,
what you can do is a

530
00:38:18 --> 00:38:24
back-of-the-envelope calculation
to see whether or not they are

531
00:38:24 --> 00:38:30
telling you the reaction rate is
greater than this number.

532
00:38:30 --> 00:38:35
If it is, you can say ah-ha,
got you, it cannot possibly be.

533
00:38:35 --> 00:38:40
Really important.
If a reaction has this rate,

534
00:38:40 --> 00:38:44
it will mean that the
probability of the reaction

535
00:38:44 --> 00:38:49
occurring is one.
If the reaction has that rate,

536
00:38:49 --> 00:38:52
we call that the gas kinetic
rate.

537
00:38:52 --> 00:38:55
That is a term that we also
use.

538
00:38:55 --> 00:38:59
Then, finally,
one other quantity from the gas

539
00:38:59 --> 00:39:03
kinetic theory.
Yes?

540
00:39:03 --> 00:39:10


541
00:39:10 --> 00:39:13
She asked, what velocity,
here, would you use?

542
00:39:13 --> 00:39:17
You actually are going to,
in that particular case,

543
00:39:17 --> 00:39:21
have to take the average of the
average velocities.

544
00:39:21 --> 00:39:24
In other words,
if you had two different

545
00:39:24 --> 00:39:29
molecules that were reacting for
this average velocity here,

546
00:39:29 --> 00:39:33
you are going to have to use
the average of the average

547
00:39:33 --> 00:39:38
velocity.
And then your error bars on the

548
00:39:38 --> 00:39:44
experiment will always be large
enough to take that into

549
00:39:44 --> 00:39:47
consideration.
Good question.

550
00:39:47 --> 00:39:52
One other quantity,
something called the mean free

551
00:39:52 --> 00:39:55
path.
What I want to know,

552
00:39:55 --> 00:40:00
here, is on the average,
how far does the molecule

553
00:40:00 --> 00:40:07
travel in the gas before it
suffers a collision?

554
00:40:07 --> 00:40:09
That is what a mean free path
is.

555
00:40:09 --> 00:40:13
A mean free path in solution.
Sometimes you will do a solid

556
00:40:13 --> 00:40:17
state physics course and will
hear about the electron mean

557
00:40:17 --> 00:40:20
free path.
A mean free path is always the

558
00:40:20 --> 00:40:22
distance traveled between
collisions.

559
00:40:22 --> 00:40:26
That is what that is.
And we are going to call it

560
00:40:26 --> 00:40:30
lambda.
This is not wavelength anymore.

561
00:40:30 --> 00:40:33
We just changed our definition
of lambda.

562
00:40:33 --> 00:40:36
Lambda, here,
is the average distance between

563
00:40:36 --> 00:40:40
collisions.
How are we going to calculate

564
00:40:40 --> 00:40:43
that?
What we are going to do is take

565
00:40:43 --> 00:40:47
the average distance that a
molecule travels per unit time,

566
00:40:47 --> 00:40:52
per second, and are going to
divide it by the number of

567
00:40:52 --> 00:40:55
collisions that occur per
second.

568
00:40:55 --> 00:40:59
And you are going to see that
the per seconds are going to

569
00:40:59 --> 00:41:03
cancel here.
We are going to have the

570
00:41:03 --> 00:41:08
average distance per collision.
That is what we are after.

571
00:41:08 --> 00:41:13
This is easy to do because the
average distance traveled per

572
00:41:13 --> 00:41:16
second is simply the average
velocity.

573
00:41:16 --> 00:41:20
It is meters per second.
The average distance traveled

574
00:41:20 --> 00:41:24
per second.
The number of collisions per

575
00:41:24 --> 00:41:27
second that happened,
we just calculated that.

576
00:41:27 --> 00:41:33
That was Z1.
We can substitute in Z1 vbar,

577
00:41:33 --> 00:41:38
vbar cancels,
and what we are left with is

578
00:41:38 --> 00:41:46
one over the square root of 2
rho pi d squared.

579
00:41:46 --> 00:41:49
That is meters.

580
00:41:49 --> 00:41:56
That is the average distance
the molecule traveled per

581
00:41:56 --> 00:42:01
collision.
This is another general generic

582
00:42:01 --> 00:42:06
quantity useful in many cases,
also in other kinds of cases

583
00:42:06 --> 00:42:10
that I just mentioned in solid
state physics.

584
00:42:10 --> 00:42:16
What I want to do now is that
we said we were going to start

585
00:42:16 --> 00:42:19
talking about the motions of
molecules.

586
00:42:19 --> 00:42:24
We have taken care of,
now, the translational motion.

587
00:42:24 --> 00:42:29
Now I want to start talking
about the internal degrees of

588
00:42:29 --> 00:42:33
freedom.
And let's do that.

589
00:42:33 --> 00:42:38
Let me just see if I have
something, here.

590
00:42:38 --> 00:42:43
Let me raise the board,
here.

591
00:42:43 --> 00:42:58


592
00:42:58 --> 00:43:02
We have talked about this
molecule, which is a set of

593
00:43:02 --> 00:43:07
atoms that are bonded together.
And, as we always say,

594
00:43:07 --> 00:43:12
they are bonded together for
the mutual comfort of their

595
00:43:12 --> 00:43:15
electrons.
And we saw that it is not

596
00:43:15 --> 00:43:17
frozen.
It moves.

597
00:43:17 --> 00:43:20
It has kinetic energy.
It has velocity.

598
00:43:20 --> 00:43:25
The velocity generates this
macroscopic property of

599
00:43:25 --> 00:43:30
pressure.
But molecules also jiggle.

600
00:43:30 --> 00:43:33
For example,
we are going to take here

601
00:43:33 --> 00:43:39
nitrogen, triple bond.
That triple bond actually

602
00:43:39 --> 00:43:44
functions like a spring.
We can think of that bond as a

603
00:43:44 --> 00:43:49
spring between the two atoms.
That spring can stretch,

604
00:43:49 --> 00:43:54
and that spring can compress.
When we tell you the

605
00:43:54 --> 00:44:00
equilibrium bond length of some
molecule is some number,

606
00:44:00 --> 00:44:05
it is the equilibrium bond
length.

607
00:44:05 --> 00:44:09
It is not the bond length when
the molecule is stretched.

608
00:44:09 --> 00:44:14
It is not the bond length when
the molecule is compressed.

609
00:44:14 --> 00:44:19
This motion is the vibrational
motion of the molecule.

610
00:44:19 --> 00:44:24


611
00:44:24 --> 00:44:30
So, we have that kind of
internal motion in a molecule.

612
00:44:30 --> 00:44:33
Molecules also tumble.
For example,

613
00:44:33 --> 00:44:38
let's take our nitrogen
molecule, again.

614
00:44:38 --> 00:44:45
It rotates around an axis.
Going through the center of the

615
00:44:45 --> 00:44:49
mass, it rotates around this
axis.

616
00:44:49 --> 00:44:54
This nitrogen molecule also
rotates around an axis,

617
00:44:54 --> 00:45:00
here, perpendicular to the
board.

618
00:45:00 --> 00:45:06
It rotates in that direction.
This is the rotational motion

619
00:45:06 --> 00:45:11
of the molecule.
The different ways in which we

620
00:45:11 --> 00:45:15
can have this motion are called
modes.

621
00:45:15 --> 00:45:21
This is a rotational mode.
This is vibrational mode.

622
00:45:21 --> 00:45:28
We did not use the word before
in talking about translation,

623
00:45:28 --> 00:45:33
but those are translational
modes.

624
00:45:33 --> 00:45:40
Sometimes, we also call these
modes degrees of freedom.

625
00:45:40 --> 00:45:49
A molecule has translational
modes or degrees of freedom,

626
00:45:49 --> 00:45:57
vibrational degrees of freedom,
and rotational degrees of

627
00:45:57 --> 00:46:00
freedom.
Now, in general,

628
00:46:00 --> 00:46:10
if you have an N-atom molecule,
you have 3N total modes.

629
00:46:10 --> 00:46:17
When I say total modes that
means the sum of translation,

630
00:46:17 --> 00:46:24
rotation and vibration.
Of those 3N total modes,

631
00:46:24 --> 00:46:30
three of them are translational
modes.

632
00:46:30 --> 00:46:35


633
00:46:35 --> 00:46:38
Why do we have three
translational modes?

634
00:46:38 --> 00:46:43
We have three translational
modes because we are operating

635
00:46:43 --> 00:46:48
in a three-dimensional space.
If we have a molecule here,

636
00:46:48 --> 00:46:53
that molecule has motion in the
x direction, in the y direction,

637
00:46:53 --> 00:46:57
and also in the z direction.
Those are the three

638
00:46:57 --> 00:47:02
translational modes of the
molecule.

639
00:47:02 --> 00:47:08
What that means then,
if three of the modes are used

640
00:47:08 --> 00:47:13
up for translation,
then we must have 3N minus 3

641
00:47:13 --> 00:47:20
internal modes.
3N minus 3 must be the number

642
00:47:20 --> 00:47:27
of modes that is leftover for
the internal degrees of freedom

643
00:47:27 --> 00:47:33
for rotation and also for
vibration.

644
00:47:33 --> 00:47:39
And what we are going to see,
next time, is that these

645
00:47:39 --> 00:47:46
internal degrees of freedom,
these internal modes are going

646
00:47:46 --> 00:47:51
to be quantized.
That is, this molecule is going

647
00:47:51 --> 00:47:59
to be able to vibrate with only
certain amounts of energy.

648
00:47:59 --> 00:48:02
There is going to be a ground
vibrational state.

649
00:48:02 --> 00:48:07
There is going to be a first
excited vibrational state.

650
00:48:07 --> 00:48:12
A second excited vibrational
state, just like we talked about

651
00:48:12 --> 00:48:17
the energy levels of a hydrogen
atom, where the electron was

652
00:48:17 --> 00:48:22
bound with different amounts of
energy dictated by a principle

653
00:48:22 --> 00:48:26
quantum number.
The vibrational degree of

654
00:48:26 --> 00:48:30
freedom is also going to be
dictated by a vibrational

655
00:48:30 --> 00:48:33
quantum number.
In addition,

656
00:48:33 --> 00:48:38
the rotations of the molecules,
they are also quantized.

657
00:48:38 --> 00:48:43
A molecule is going to be able
to rotate with only one energy,

658
00:48:43 --> 00:48:46
or another energy,
or another energy,

659
00:48:46 --> 00:48:49
but not any energies in
between.

660
00:48:49 --> 00:48:52
There are discrete rotational
states.

661
00:48:52 --> 00:48:56
We are going to be talking
about another kind of quantum

662
00:48:56 --> 00:48:58
number.
In that case,

663
00:48:58 --> 00:49:04
a rotational quantum number.
That is where we are going to

664
00:49:04.451 --> 49:07
pick up on Monday.
See you then.