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Last time, we went through the
same arguments that Maxwell and

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Boltzmann did to understand the
microscopic origin of the ideal

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gas law, PV equal nRT.

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And we saw how Maxwell had
hypothesized that the pressure

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of a gas on the walls of some
container, if the molecules were

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moving and they collided onto
the walls, that pressure must be

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due to the individual impacts of
the molecules on the wall.

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That must be due to the change
in the momentum of the wall when

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the molecules slammed right into
it.

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And that change in momentum
over some change in time is this

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force.
That force divided by the total

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area is the pressure.
And it was using that argument

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that we came up with an
expression, just like Maxwell

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did for the pressure times the
volume.

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And we see that we were able to
write it in terms of the

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velocity of the molecules that
were moving in this gas.

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Well, that is very nice because
if we have this theoretical

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expression P times V,
and we know experimentally that

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P times V equal nRT,
if this theory is

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correct, then this quantity n M,
average of the velocity

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squared, over three,
that better be equal to nRT.

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Or, solving for what we call

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the root mean square velocity,
that better be equal to the

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square root of 3RT over M.

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That kinetic theory made a
prediction for what the velocity

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ought to be.
It took, of course,

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another hundred years before
somebody could measure that.

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And, of course,
it is correct.

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But what is interesting here is
that this model gave us an

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understanding for what
temperature is.

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Temperature is a measure of the
speed of the molecules in the

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gas.
It is also, as we saw last

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time, a measure of their kinetic
energy.

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We saw that the kinetic energy
is one-half M average of the

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velocity squared. We put

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in our result from kinetic
theory.

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That shows us that the average
energy is three-halves RT.

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Temperature is the measure of

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the kinetic energy of these
molecules in the gas.

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For the first time,
there was a microscopic

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understanding of what
temperature was.

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Now, what we looked at also,
last time, was the

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Maxwell-Boltzmann distribution
of velocities or speeds and

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talked about this particular
functional form.

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How there was this quadratic
dependence right here,

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a low v because of the v
squared here and then an

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exponentially decaying tail.
There were two parameters,

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the mass and the temperature.
We talked about that last time.

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But, of course,
if there is a distribution here

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of speeds where this
distribution is a probability of

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finding a molecule with a
particular speed between v and v

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plus dv, --
-- then there also has to be a

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distribution of energies because
the velocity is related to the

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energy.
What do we have to do?

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We have to take that
Maxwell-Boltzmann speed

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distribution and change the
variable to energy.

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We know how to do that.
We know how to equate those two

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distribution functions.
And, using the Jacobean of the

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transformation that we talked
about last time,

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we can calculate what the
energy distribution is.

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f of E is our
Maxwell-Boltzmann energy

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distribution.
It is the probability of

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finding a molecule with a
kinetic energy E to E plus dE.

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It also has a decaying
exponential term,

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here, with the energy in the
argument, and it is multiplied

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by E of the one-half.
Let's take a look at what that

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distribution function looks
like.

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Here it is.
Notice, as we learned last

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time, that the energy has
nothing to do with the mass of

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the particle.
The only parameter that is

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important is the temperature.
At a given temperature,

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all particles,
it doesn't matter what their

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mass is, have the same energy.
So, all particles at

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degrees Kelvin have a
Maxwell-Boltzmann energy

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distribution that looks like
this.

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There is a very rapid rise in
that distribution function,

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unlike the velocity
distribution,

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which increased as v squared.

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It peaks rapidly,
and then there is an

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exponentially decaying term in
the energy.

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At 1500 degrees Kelvin,
the energy distribution looks

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like this.
The energy of the molecules has

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increased.
Since this is a probability,

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and the area under these curves
has to equal to one,

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if the energy goes up,
that is we have more molecules

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out here with higher energies,
well, then this maximum value

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for the probabilities has got to
go down because we have to keep

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the area under that curve equal
to one, since this is a

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probability.
What we see here,

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as we raise the energy,
is that we have more and more

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molecules with high energies.
There are not a lot of them,

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but there are some.
And they are going to be really

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important in the example I am
going to show you in just a

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moment.
Average energy at 600 Kelvin,

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just three-halves RT,
is 7.5 kilojoules per mole.

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Average energy at 1500 Kelvin
is 18.7 kilojoules per mole.

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But now, I want to show you an
example of the importance of

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this Maxwell-Boltzmann
distribution function for

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energies to chemical reactions,
the importance of it in making

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some kinds of chemical reactions
actually work.

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And so the example is going to
be this reaction.

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This reaction is called steam
reforming of natural gas.

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It is the reaction of methane
plus water.

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Natural gas is mostly methane.
But this reaction of methane

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and water makes CO and hydrogen.
It

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turns out that this reaction
does not work in the gas phase.

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That is, if you have a methane
molecule and water molecule

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collide, they are just going to
collide, bounce apart and go in

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their opposite directions.
They are not going to make CO

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and hydrogen.
And so what you have to have in

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this reaction to make it go is a
catalyst.

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That catalyst is going to be a
nickel surface.

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A catalyst is something that
will lower the activation energy

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barrier by changing the
mechanism of a reaction so that

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the reaction can proceed.
In this case,

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what happens is that the
methane and the water impinge on

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this nickel metal catalyst.
And the methane and the water

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decompose.
They fall apart to their

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elements on that nickel surface.
And then, once you have carbon,

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hydrogen and oxygen on that
nickel surface,

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the atoms rearrange and come
off as CO and molecular

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hydrogen.
So, that nickel surface is a

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catalyst.
We call this the catalytic

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reaction.
In particular,

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we call it a heterogeneous
catalytic reaction.

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It is heterogeneous because the
catalyst is in a different phase

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than the reactants.
The catalyst here is a solid.

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The reactants are gases.
A homogeneous catalyst is one

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in which the phase of the
catalyst and the reactants is

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the same.
This is a heterogeneous

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catalytic reaction.
It turns out that this reaction

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here is really an important
reaction from the standpoint of

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the production of hydrogen.
All of the hydrogen that you

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use, you know,
if you are doing a laboratory

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experiment and you go get a tank
of hydrogen, that hydrogen is

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made by this reaction,
by reacting methane and water

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to form that hydrogen.
All of our commercial sources

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of hydrogen come from carrying
out this reaction.

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For example,
if you are in the business of

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ammonia synthesis,
which is taking hydrogen and

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nitrogen to make ammonia,
which is also carried out on an

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iron surface,
another heterogeneous catalytic

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reaction.
And ammonia,

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of course, is the starting
material for lots of chemicals,

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in particular fertilizers.
If you have an ammonia plant,

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right next to the ammonia plant
you have a steamer-forming plant

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to make the hydrogen to feed
into this reaction.

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And another place where this is
useful is in methanol synthesis.

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That is, you can take hydrogen
and CO on a copper zinc oxide

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heterogeneous catalyst and make
methanol, the starting point for

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gasoline, which now is certainly
economically feasible.

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But it turns out that this
reaction here is really a hard

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one to carry out.
Despite the fact that nickel is

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a catalyst for the reaction and
makes it go, the reaction still

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has to be carried out at very
high temperatures.

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1500 degrees kelvin is a very
high temperature.

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It also needs very high
pressure.

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And one of the reasons why we
don't have a hydrogen economy is

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because it is so difficult to
make.

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Hydrogen can be done,
and it is done in all of the

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commercial processes that I
mentioned to you,

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but it is still hard.
1500 degrees kelvin,

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high pressures.
The question is,

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why is it so hard?
Why is there this barrier here?

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Why do we need to raise the
temperature of the gas in order

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to get this reaction to go?
Well, let's take a look at

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that.
What there exists here in this

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problem is what is called an
activation energy barrier to

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making the reaction go.
It turns out that this

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activation energy barrier is
really in the very first step of

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the reaction,
which is pulling the methane

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apart, pulling that first
hydrogen atom off of the methane

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molecule.
What I represent right here is

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the energy of the reaction as a
function of the reaction,

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and the reaction here is just
the first step.

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It is taking methane gas,
pulling the hydrogen off so

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that you have a methyl radical
stuck to the surface and you

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have a hydrogen atom stuck to
the surface.

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It is just breaking that first
C-H bond.

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What you can see here is that
you have to put 50 kilojoules of

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energy into that reaction in
order to make it go.

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You have to put that 50
kilocalories of energy in first,

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before you get any energy back.
You can see that this is

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exothermic.
But you have to put this energy

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in, in order to get that
reaction to go.

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Well, how do we know that?
Let me back up a minute.

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Here is where the
Maxwell-Boltzmann energy

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distribution is important.
What I did was I took those

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Boltzmann energy distributions
that I plotted for you at

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Kelvin and 1500 Kelvin and
turned them on their side.

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This is essentially equal to
the probability of finding a

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molecule at a particular energy
versus the energy.

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Here is the Maxwell-Boltzmann
distribution at 600 degrees

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00:13:55,000 --> 00:13:57,000
Kelvin.
It is peaked to very low

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00:13:57,000 --> 00:14:01,000
energies.
And, if you look here in this

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00:14:01,000 --> 00:14:05,000
Maxwell-Boltzmann tail,
there are not very many

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00:14:05,000 --> 00:14:10,000
molecules that have enough
energy to get over that barrier.

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00:14:10,000 --> 00:14:14,000
However, if we raise the
temperature of the gas to

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degrees Kelvin,
then now you can see that there

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are more molecules here in the
high energy part of this tail.

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00:14:23,000 --> 00:14:27,000
And it is these molecules at
these high energies that

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00:14:27,000 --> 00:14:32,000
actually can get over this
barrier to dissociation of the

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00:14:32,000 --> 00:14:36,000
methane molecule.
They are not many of those

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00:14:36,000 --> 00:14:39,000
molecules with high energy,
but there are some.

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00:14:39,000 --> 00:14:42,000
And that is important,
because what happens is when

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they have enough energy,
they react and they leave.

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00:14:45,000 --> 00:14:49,000
And then this Boltzmann
distribution re-equilibrates.

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00:14:49,000 --> 00:14:52,000
If the high energy molecules
leave, then there are lots of

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00:14:52,000 --> 00:14:56,000
collisions here that kick some
more molecules up to the high

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00:14:56,000 --> 00:15:00,000
energy part of the tail,
and they react.

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00:15:00,000 --> 00:15:04,000
And so it works.
That is why it is important.

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00:15:04,000 --> 00:15:09,000
How do we really know that
there is this barrier here to

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00:15:09,000 --> 00:15:15,000
the dissociation of methane,
and why is there this barrier?

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00:15:15,000 --> 00:15:21,000
What is the physical origin of
this barrier to pulling this C-H

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00:15:21,000 --> 00:15:25,000
bond apart?
Well, one way we could know

220
00:15:25,000 --> 00:15:30,000
that there was this barrier
there for sure is to do the

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00:15:30,000 --> 00:15:35,000
following.
If we had a way to take methane

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00:15:35,000 --> 00:15:38,000
gas and make it have just a
single energy,

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not the Maxwell-Boltzmann
distribution of energies that

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00:15:41,000 --> 00:15:46,000
have lots of different energies,
which have lots of different

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00:15:46,000 --> 00:15:49,000
energies in them,
and you saw how broad those

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00:15:49,000 --> 00:15:51,000
curves were.
But if we had a way to make

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00:15:51,000 --> 00:15:56,000
methane gas with just say energy
E sub 1, whatever that is,

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00:15:56,000 --> 00:15:59,000
we could then take those
molecules with just those

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00:15:59,000 --> 00:16:02,000
energies, aim them at this
nickel surface,

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00:16:02,000 --> 00:16:06,000
and see if the methane fell
apart.

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00:16:06,000 --> 00:16:09,000
And, if it didn't,
then we would prepare molecules

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00:16:09,000 --> 00:16:13,000
at some higher energy and see if
they fell apart.

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00:16:13,000 --> 00:16:16,000
If they didn't then we would go
to higher energy.

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00:16:16,000 --> 00:16:20,000
We would keep going until we
got to the top of the barrier.

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00:16:20,000 --> 00:16:24,000
Well, how do you do that?
How do you make molecules with

236
00:16:24,000 --> 00:16:28,000
a particular energy?
Because I just showed you a

237
00:16:28,000 --> 00:16:32,000
Maxwell-Boltzmann energy
distribution at 300 Kelvin,

238
00:16:32,000 --> 00:16:37,000
600 Kelvin.
There are a lot of energies

239
00:16:37,000 --> 00:16:41,000
present.
How do we make molecules with

240
00:16:41,000 --> 00:16:46,000
just one energy?
We have a way to do that by

241
00:16:46,000 --> 00:16:52,000
using some beam techniques and
high pressure adiabatic

242
00:16:52,000 --> 00:16:57,000
expansions.
Basically, the way it works is

243
00:16:57,000 --> 00:17:01,000
this.
What we are going to do is take

244
00:17:01,000 --> 00:17:07,000
a tube here, which is going to
have a high pressure of methane

245
00:17:07,000 --> 00:17:10,000
in it.
Then we are going to punch a

246
00:17:10,000 --> 00:17:14,000
little hole in that tube.
This tube is actually sitting

247
00:17:14,000 --> 00:17:19,000
in a vacuum, so we are going to
expand the methane from that

248
00:17:19,000 --> 00:17:24,000
tube into the vacuum.
We are going to squirt it out.

249
00:17:24,000 --> 00:17:29,000
It is going to become a beam of
molecules.

250
00:17:29,000 --> 00:17:34,000
But when you expand a gas from
high pressure to low pressure,

251
00:17:34,000 --> 00:17:38,000
this expansion is called an
adiabatic expansion,

252
00:17:38,000 --> 00:17:42,000
which means the gas cools.
I will explain that in a

253
00:17:42,000 --> 00:17:47,000
moment, but the adiabatic
expansion you are going to talk

254
00:17:47,000 --> 00:17:50,000
a lot about in 5.60.
And, if you are a chemical

255
00:17:50,000 --> 00:17:55,000
engineer, you will talk even
more about it past 5.60.

256
00:17:55,000 --> 00:17:59,000
Here is how it works.
You take these molecules and

257
00:17:59,000 --> 00:18:04,000
squirt them out.
Because the pressure right here

258
00:18:04,000 --> 00:18:08,000
is so high, what happens is
there are lots and lots and lots

259
00:18:08,000 --> 00:18:11,000
of collisions.
And so, you can imagine,

260
00:18:11,000 --> 00:18:15,000
if you have some slow molecule
just kind of lumbering along and

261
00:18:15,000 --> 00:18:20,000
a fast one comes and hits you,
what is going to happen is you

262
00:18:20,000 --> 00:18:23,000
are going to speed up,
and the fast one is going to

263
00:18:23,000 --> 00:18:26,000
slow down.
And, if you keep doing that

264
00:18:26,000 --> 00:18:30,000
again and again and again,
all the molecules are going to

265
00:18:30,000 --> 00:18:35,000
end up with the same energy or
the same velocity.

266
00:18:35,000 --> 00:18:39,000
If you have so many collisions,
after a while they are going to

267
00:18:39,000 --> 00:18:44,000
all have the same velocity or
the same energy because one gets

268
00:18:44,000 --> 00:18:48,000
sped up, one gets slowed down.
They are all going to end up

269
00:18:48,000 --> 00:18:51,000
with the same energy.
That is what happens.

270
00:18:51,000 --> 00:18:56,000
That is how we make a beam of
molecules, a source of molecules

271
00:18:56,000 --> 00:19:00,000
with the same kinetic energy.
And basically,

272
00:19:00,000 --> 00:19:03,000
what is happening then is that
before the expansion,

273
00:19:03,000 --> 00:19:07,000
just look at this curve,
here is an actual Boltzmann

274
00:19:07,000 --> 00:19:10,000
distribution of velocities.
It is broad.

275
00:19:10,000 --> 00:19:14,000
But after this expansion,
here is the distribution.

276
00:19:14,000 --> 00:19:18,000
It is really pretty narrow.
If you put a temperature to

277
00:19:18,000 --> 00:19:21,000
that distribution,
you might find that at one

278
00:19:21,000 --> 00:19:24,000
degree Kelvin,
you can really make molecules

279
00:19:24,000 --> 00:19:30,000
with a single or very narrow
distribution in energies.

280
00:19:30,000 --> 00:19:33,000
That is what we do.
We make those molecules with

281
00:19:33,000 --> 00:19:38,000
this energy E sub 1.
Then we have some more tricks

282
00:19:38,000 --> 00:19:41,000
for changing those energies in a
controlled way.

283
00:19:41,000 --> 00:19:46,000
And we just keep cranking those
energies up and watch to see

284
00:19:46,000 --> 00:19:51,000
when the methane falls apart.
And so, you get data that kind

285
00:19:51,000 --> 00:19:54,000
of looks like this.
This is a dissociation

286
00:19:54,000 --> 00:19:58,000
probability of a methane as a
function of its energy.

287
00:19:58,000 --> 00:20:02,000
This is in kilocalories per
mole.

288
00:20:02,000 --> 00:20:05,000
At some point,
say right about here,

289
00:20:05,000 --> 00:20:10,000
15 kilocalories per mole,
the dissociation probability

290
00:20:10,000 --> 00:20:13,000
skyrockets.
All of a sudden,

291
00:20:13,000 --> 00:20:18,000
you are at a high enough energy
to get that methane to fall

292
00:20:18,000 --> 00:20:22,000
apart.
This tells you exactly where

293
00:20:22,000 --> 00:20:26,000
that barrier is.
But now, the question is,

294
00:20:26,000 --> 00:20:32,000
why is there this barrier?
Physically, why is there a

295
00:20:32,000 --> 00:20:38,000
barrier to pulling the hydrogen
off of the carbon when that

296
00:20:38,000 --> 00:20:43,000
methane molecule comes close to
the surface?

297
00:20:43,000 --> 00:20:48,000
Well, the reason is this.
In order to break the

298
00:20:48,000 --> 00:20:54,000
carbon-hydrogen bond in methane,
in order to have enough energy

299
00:20:54,000 --> 00:21:00,000
to break that bond,
what you need to do is you need

300
00:21:00,000 --> 00:21:05,000
to simultaneously form a
nickel-carbon bond and a

301
00:21:05,000 --> 00:21:10,000
nickel-hydrogen bond.
In other words,

302
00:21:10,000 --> 00:21:17,000
when this methane molecule here
comes into some nickel surface

303
00:21:17,000 --> 00:21:23,000
very slowly, because that
methane is tetrahedral and the

304
00:21:23,000 --> 00:21:30,000
carbon is kind of hidden in the
center, the hydrogens interact

305
00:21:30,000 --> 00:21:35,000
with the nickel.
But the carbon does not get in

306
00:21:35,000 --> 00:21:39,000
close enough to the nickel to
start to form a nickel-carbon

307
00:21:39,000 --> 00:21:42,000
bond.
Now, if you speed that methane

308
00:21:42,000 --> 00:21:46,000
molecule up and you really ram
it into the surface,

309
00:21:46,000 --> 00:21:49,000
upon collision of the molecule
with the surface,

310
00:21:49,000 --> 00:21:54,000
the hydrogens are pushed back,
the carbon gets in close enough

311
00:21:54,000 --> 00:22:00,000
to the nickel to start to form,
here, this nickel-carbon bond.

312
00:22:00,000 --> 00:22:05,000
You form a nickel-hydrogen
bond, and now you can break that

313
00:22:05,000 --> 00:22:09,000
C-H bond.
We call this black chemistry.

314
00:22:09,000 --> 00:22:12,000
The barrier,
there, to that reaction

315
00:22:12,000 --> 00:22:18,000
physically is the amount of
energy that you have to put into

316
00:22:18,000 --> 00:22:22,000
the molecule to push those
hydrogens back,

317
00:22:22,000 --> 00:22:30,000
to bend those hydrogens back to
distort this methane molecule.

318
00:22:30,000 --> 00:22:33,000
Once you do that,
the reaction goes.

319
00:22:33,000 --> 00:22:35,000
You are on a roll there.
It goes.

320
00:22:35,000 --> 00:22:39,000
So, that is the physical origin
of this barrier,

321
00:22:39,000 --> 00:22:43,000
here, to the dissociation of
methane.

322
00:22:43,000 --> 00:22:47,000
You need to put enough kinetic
energy into those methane

323
00:22:47,000 --> 00:22:51,000
molecules in order to get over
that barrier.

324
00:22:51,000 --> 00:22:56,000
And we do that on a practical
scale by raising the temperature

325
00:22:56,000 --> 00:23:01,000
of the gas.
And on a microscopic scale,

326
00:23:01,000 --> 00:23:06,000
this is what is happening.
You can imagine that my

327
00:23:06,000 --> 00:23:11,000
students and I did this
experiment 15 years ago.

328
00:23:11,000 --> 00:23:16,000
But then, we said if this
barrier here is the energy

329
00:23:16,000 --> 00:23:22,000
required to deform that methane
molecule, then we should be able

330
00:23:22,000 --> 00:23:26,000
to do this experiment.
This experiment is the

331
00:23:26,000 --> 00:23:31,000
following.
We are going to take our nickel

332
00:23:31,000 --> 00:23:34,000
surface, here,
and lower the temperature of

333
00:23:34,000 --> 00:23:38,000
the surface to 47 kelvin.
At that temperature,

334
00:23:38,000 --> 00:23:42,000
what we can do is we can freeze
a layer of methane on the

335
00:23:42,000 --> 00:23:44,000
surface.
We call it fizzy-sorb,

336
00:23:44,000 --> 00:23:47,000
a layer of methane on the
surface.

337
00:23:47,000 --> 00:23:51,000
It will just stick there.
And then, if this barrier is

338
00:23:51,000 --> 00:23:55,000
the energy required to deform or
to start the molecule,

339
00:23:55,000 --> 00:23:59,000
then in principle I could take
a hammer and pound that molecule

340
00:23:59,000 --> 00:24:04,000
into the correct shape for the
transition state that leads to

341
00:24:04,000 --> 00:24:09,000
dissociation.
It sounds like a simple idea,

342
00:24:09,000 --> 00:24:14,000
and it is, except that we
cannot really take that hammer.

343
00:24:14,000 --> 00:24:19,000
But we can take an argon atom.
We can freeze the methane onto

344
00:24:19,000 --> 00:24:23,000
the surface, and now we come in
with an argon atom.

345
00:24:23,000 --> 00:24:27,000
That is just a big ball.
Or, a xenon atom or a krypton

346
00:24:27,000 --> 00:24:30,000
atom.
And we know how to accelerate

347
00:24:30,000 --> 00:24:35,000
xenon or krypton.
We don't accelerate it too

348
00:24:35,000 --> 00:24:38,000
much, 50 kilocalories,
70 kilocalories per mole,

349
00:24:38,000 --> 00:24:42,000
something like that.
And then we can bring it in.

350
00:24:42,000 --> 00:24:47,000
What happens is that the impact
of the collision on that methane

351
00:24:47,000 --> 00:24:51,000
causes that methane molecule to
compress, distort,

352
00:24:51,000 --> 00:24:56,000
gets it into the configuration
of the transition state that

353
00:24:56,000 --> 00:25:00,000
leads to the methane falling
apart.

354
00:25:00,000 --> 00:25:02,000
And so, you get the same
result.

355
00:25:02,000 --> 00:25:05,000
The question,
there, is just getting the

356
00:25:05,000 --> 00:25:09,000
energy into the molecule to
actually distort it,

357
00:25:09,000 --> 00:25:14,000
to deform it so that you can
make the nickel-carbon bond and

358
00:25:14,000 --> 00:25:18,000
the nickel-hydrogen bond.
That is the key.

359
00:25:18,000 --> 00:25:22,000
My students and I,
after we had spent many years

360
00:25:22,000 --> 00:25:26,000
doing this experiment and spent
a lot of money on this

361
00:25:26,000 --> 00:25:31,000
experiment, said we have proven
for sure that when a bug flies

362
00:25:31,000 --> 00:25:36,000
into the windshield of your car,
you do get the same result as

363
00:25:36,000 --> 00:25:41,000
if you hit the bug on the
windshield of your car with a

364
00:25:41,000 --> 00:25:46,000
bug swatter.
That is what we get.

365
00:25:46,000 --> 00:25:51,000
This is an example of one
chemical reaction for which we

366
00:25:51,000 --> 00:25:55,000
know the physical origin of the
barrier.

367
00:25:55,000 --> 00:26:01,000
There are very few chemical
reactions for which we know

368
00:26:01,000 --> 00:26:06,000
anything about the physical
origin of a barrier to a

369
00:26:06,000 --> 00:26:11,000
reaction.
Now, what I want to talk about

370
00:26:11,000 --> 00:26:18,000
is go back to kinetic theory,
because the kinetic theory is

371
00:26:18,000 --> 00:26:25,000
going to allow us to calculate a
couple of more quantities that

372
00:26:25,000 --> 00:26:31,000
are of interest to us.
And one of those quantities is

373
00:26:31,000 --> 00:26:37,000
the collision frequency.
And let me just write that on

374
00:26:37,000 --> 00:26:41,000
the board here.
What I am going to want to

375
00:26:41,000 --> 00:26:47,000
calculate is something called
Z1, or I am going to call it Z1.

376
00:26:47,000 --> 00:26:53,000
It is the number of collisions
that a molecule makes in a gas

377
00:26:53,000 --> 00:26:58,000
per unit time.
And our unit time is going to

378
00:26:58,000 --> 00:27:06,000
be seconds.
This is the collision frequency

379
00:27:06,000 --> 00:27:13,000
of a single molecule in a gas.

380
00:27:28,000 --> 00:27:29,000
How am I going to calculate the
quantity using the kinetic

381
00:27:29,000 --> 00:27:30,000
theory approach that we have
been talking about?

382
00:27:30,000 --> 00:27:33,000
What I am going to do is this.
I am going to take some gas,

383
00:27:33,000 --> 00:27:38,000
and the molecules in that gas
are represented by these blue

384
00:27:38,000 --> 00:27:42,000
circles right here.
They have a diameter D.

385
00:27:42,000 --> 00:27:47,000
And then, within that gas,
I am going to imagine this

386
00:27:47,000 --> 00:27:52,000
lighter blue cylinder.
That cylinder is an imaginary

387
00:27:52,000 --> 00:27:55,000
construct.
We are going to use it to

388
00:27:55,000 --> 00:28:00,000
calculate this collision
frequency.

389
00:28:00,000 --> 00:28:04,000
What it is going to be is the
collision volume.

390
00:28:04,000 --> 00:28:09,000
I have set the diameter of that
cylinder to be equal to two

391
00:28:09,000 --> 00:28:12,000
times the diameter of the
molecule.

392
00:28:12,000 --> 00:28:17,000
I have set this upright.
And the bottom line is that all

393
00:28:17,000 --> 00:28:22,000
of the molecules that
instantaneously happen to be in

394
00:28:22,000 --> 00:28:26,000
this cylinder when a molecule
comes through,

395
00:28:26,000 --> 00:28:32,000
all of those molecules are
actually going to be hit.

396
00:28:32,000 --> 00:28:37,000
They are going to suffer a
collision with another molecule.

397
00:28:37,000 --> 00:28:42,000
And I am going to use that,
the number of molecules in this

398
00:28:42,000 --> 00:28:46,000
volume, to calculate this
collision frequency.

399
00:28:46,000 --> 00:28:51,000
That is what I am going to do.
Snapshot in time,

400
00:28:51,000 --> 00:28:56,000
those molecules are frozen
there, except there is one

401
00:28:56,000 --> 00:29:02,000
smart-alecky molecule that comes
cruising on through.

402
00:29:02,000 --> 00:29:04,000
Here he comes.
Bam, bam, bam.

403
00:29:04,000 --> 00:29:10,000
Hits these three molecules,
here, because they are more

404
00:29:10,000 --> 00:29:14,000
than halfway into that collision
cylinder.

405
00:29:14,000 --> 00:29:20,000
What I know is this smart-aleck
was moving at some average

406
00:29:20,000 --> 00:29:24,000
velocity, which I am going to
call vbar.

407
00:29:24,000 --> 00:29:27,000
In a time T,
this molecule,

408
00:29:27,000 --> 00:29:32,000
since it is moving with an
average velocity vbar,

409
00:29:32,000 --> 00:29:36,000
in a time t,
it is traveling a length vbar

410
00:29:36,000 --> 00:29:41,000
times t.

411
00:29:41,000 --> 00:29:44,000
Velocity times time,
that is going to give you a

412
00:29:44,000 --> 00:29:47,000
length.
And so, I am going to set,

413
00:29:47,000 --> 00:29:50,000
for convenience,
this time equal to one second

414
00:29:50,000 --> 00:29:53,000
because that is going to be my
unit of time.

415
00:29:53,000 --> 00:29:58,000
I want to do this per second.
I am going to set this equal to

416
00:29:58,000 --> 00:30:02,000
one second.
Therefore, this molecule is

417
00:30:02,000 --> 00:30:07,000
going to travel a length vbar in
one second.

418
00:30:07,000 --> 00:30:13,000
And that is the length that I
am going to make the collision

419
00:30:13,000 --> 00:30:16,000
cylinder.
I am going to make that

420
00:30:16,000 --> 00:30:22,000
collision cylinder be vbar long.
That is the distance traveled

421
00:30:22,000 --> 00:30:28,000
in one second by this
smart-alecky molecule.

422
00:30:28,000 --> 00:30:34,000
Now, all I have to do is I have
got to take the volume of the

423
00:30:34,000 --> 00:30:40,000
collision cylinder and multiply
it by all the molecules that

424
00:30:40,000 --> 00:30:45,000
happen to be at the volume at
that particular time.

425
00:30:45,000 --> 00:30:51,000
Because that volume represents
the volume swept out in one

426
00:30:51,000 --> 00:30:55,000
second by this smart-alecky
molecule.

427
00:30:55,000 --> 00:31:00,000
What is the volume of the
cylinder?

428
00:31:00,000 --> 00:31:03,000
I set the diameter to be equal
to 2d.

429
00:31:03,000 --> 00:31:08,000
The volume of the cylinder is
the cross-sectional area times

430
00:31:08,000 --> 00:31:11,000
the length.
That cross-sectional area,

431
00:31:11,000 --> 00:31:16,000
then, is pi r squared,
but r here is equal to d,

432
00:31:16,000 --> 00:31:21,000
just to make it confusing.
The area is pi d squared.

433
00:31:21,000 --> 00:31:24,000
The length is vbar.

434
00:31:24,000 --> 00:31:28,000
That is the volume of the
cylinder, pi d squared vbar.

435
00:31:28,000 --> 00:31:34,000
So, we have the volume of the

436
00:31:34,000 --> 00:31:37,000
cylinder.
Now what we are going to need

437
00:31:37,000 --> 00:31:42,000
to know is the density of the
molecules in this cylinder,

438
00:31:42,000 --> 00:31:46,000
which is the density of the
molecules in the gas.

439
00:31:46,000 --> 00:31:50,000
The cylinder is not special.
The cylinder is an imaginary

440
00:31:50,000 --> 00:31:55,000
construct that I put in there to
help me calculate the collision

441
00:31:55,000 --> 00:31:58,000
frequency.
I need the density of the

442
00:31:58,000 --> 00:32:03,000
molecules in the gas or in the
cylinder.

443
00:32:03,000 --> 00:32:07,000
The density of the molecules I
am going to set as equal to N,

444
00:32:07,000 --> 00:32:10,000
the total number of molecules
in this volume,

445
00:32:10,000 --> 00:32:13,000
divided by the volume of the
gas.

446
00:32:13,000 --> 00:32:17,000
I am going to call that rho.
N over V is going to be equal

447
00:32:17,000 --> 00:32:21,000
to rho.
That is the number of molecules

448
00:32:21,000 --> 00:32:25,000
per cubic meter.
That is the density of the gas.

449
00:32:25,000 --> 00:32:29,000
This is the symbol that I am
going to use from now on for

450
00:32:29,000 --> 00:32:34,000
density defined as molecules per
cubic meter.

451
00:32:34,000 --> 00:32:39,000
Then the collision frequency is
simply going to be the density

452
00:32:39,000 --> 00:32:43,000
of the molecules times the
volume of the cylinder.

453
00:32:43,000 --> 00:32:47,000
The collision frequency Z1,
here, is this volume of the

454
00:32:47,000 --> 00:32:52,000
cylinder, pi d squared vbar
times the density.

455
00:32:52,000 --> 00:32:53,000
Why?

456
00:32:53,000 --> 00:32:58,000
Because I set it up so that
every molecule in that cylinder

457
00:32:58,000 --> 00:33:04,000
would suffer a collision.
And that cylinder is the length

458
00:33:04,000 --> 00:33:08,000
that a molecule travels in one
second.

459
00:33:08,000 --> 00:33:11,000
There is the density times the
volume.

460
00:33:11,000 --> 00:33:16,000
The meters cubed disappears.
What I have left is rho,

461
00:33:16,000 --> 00:33:20,000
the density,
pi d squared vbar.

462
00:33:20,000 --> 00:33:24,000
This is collisions per second.

463
00:33:24,000 --> 00:33:27,000
So, I have my collision
frequency.

464
00:33:27,000 --> 00:33:33,000
This is the single-molecule
collision frequency.

465
00:33:33,000 --> 00:33:37,000
This is the frequency of
collisions that one molecule

466
00:33:37,000 --> 00:33:40,000
makes with the other gas
molecules.

467
00:33:40,000 --> 00:33:44,000
That is important.
We are going to do a different

468
00:33:44,000 --> 00:33:47,000
kind of collision frequency in
just a moment.

469
00:33:47,000 --> 00:33:52,000
But now, it turns out that
there is another factor here

470
00:33:52,000 --> 00:33:57,000
that I had left out because I
haven't done as sophisticated of

471
00:33:57,000 --> 00:34:02,000
an analysis as I could.
And that is that I made the

472
00:34:02,000 --> 00:34:07,000
assumption in my picture before
that all the other molecules

473
00:34:07,000 --> 00:34:11,000
were frozen in time and that
there was only one smart-aleck

474
00:34:11,000 --> 00:34:15,000
that was moving around,
cruising through.

475
00:34:15,000 --> 00:34:17,000
The reality is they are all
moving.

476
00:34:17,000 --> 00:34:23,000
And what I really have to do is
I have to take into account the

477
00:34:23,000 --> 00:34:27,000
relative velocities of the
molecules, the relative speeds.

478
00:34:27,000 --> 00:34:31,000
I can do that.
There is a little more

479
00:34:31,000 --> 00:34:36,000
sophisticated analysis in doing
that, but I could do that.

480
00:34:36,000 --> 00:34:40,000
And, if we do that,
this makes a difference of the

481
00:34:40,000 --> 00:34:44,000
square root of two.
I am just going to put that

482
00:34:44,000 --> 00:34:49,000
square root of two in there
right now, and later on you will

483
00:34:49,000 --> 00:34:54,000
be able to see where that comes
from, in a later course.

484
00:34:54,000 --> 00:34:59,000
That actually is the collision
frequency of a single molecule

485
00:34:59,000 --> 00:35:01,000
in the gas.

486
00:35:07,000 --> 00:35:12,000
I am also interested in another
quantity, which is the total

487
00:35:12,000 --> 00:35:16,000
collision frequency.
That was the single collision

488
00:35:16,000 --> 00:35:21,000
frequency, but I am also
interested in knowing how many

489
00:35:21,000 --> 00:35:26,000
collisions are occurring in the
entire gas per unit time,

490
00:35:26,000 --> 00:35:32,000
the total collision frequency.
You know why I am interested in

491
00:35:32,000 --> 00:35:36,000
that number?
I am interested in that number

492
00:35:36,000 --> 00:35:41,000
because that is going to be the
upper limit to any reaction

493
00:35:41,000 --> 00:35:44,000
rate.
A reaction in the gas phase or

494
00:35:44,000 --> 00:35:49,000
in solution cannot happen any
faster than the molecules

495
00:35:49,000 --> 00:35:52,000
collide.
They have to collide before a

496
00:35:52,000 --> 00:35:57,000
reaction is going to occur.
The total collision frequency,

497
00:35:57,000 --> 00:36:03,000
the importance of that number
is that it is the upper limit to

498
00:36:03,000 --> 00:36:08,000
a reaction rate.
Let's calculate that.

499
00:36:08,000 --> 00:36:13,000
Z is going to be equal to the
collision frequency of one

500
00:36:13,000 --> 00:36:17,000
molecule, Z1,
times the total number of

501
00:36:17,000 --> 00:36:21,000
molecules in the gas, N.

502
00:36:21,000 --> 00:36:25,000
That is what N stands for.
But since each collision

503
00:36:25,000 --> 00:36:31,000
involves two molecules,
I am going to have to multiply

504
00:36:31,000 --> 00:36:36,000
this by one-half.
Otherwise, I am going to over

505
00:36:36,000 --> 00:36:41,000
count the number of collisions
because each collision involves

506
00:36:41,000 --> 00:36:44,000
two molecules.
The total collision frequency

507
00:36:44,000 --> 00:36:48,000
here is one-half times N times
Z1.

508
00:36:48,000 --> 00:36:52,000
Therefore, if I go and plug in
my expression for Z1 and N and

509
00:36:52,000 --> 00:36:56,000
simplify things,
my total collision frequency

510
00:36:56,000 --> 00:37:01,000
here is one over the square root
of two times N times rho pi d

511
00:37:01,000 --> 00:37:06,000
squared times the average speed.

512
00:37:06,000 --> 00:37:11,000
That is my total collisions per

513
00:37:11,000 --> 00:37:15,000
second.
This is not rocket science.

514
00:37:15,000 --> 00:37:20,000
This is easy.
What I want you to realize is

515
00:37:20,000 --> 00:37:26,000
that you do have to understand
what N is, what rho is,

516
00:37:26,000 --> 00:37:32,000
what d is, what vbar is.
And I am always surprised on an

517
00:37:32,000 --> 00:37:38,000
exam, when we are going to give
you this equation,

518
00:37:38,000 --> 00:37:43,000
that students don't actually
know what rho is,

519
00:37:43,000 --> 00:37:48,000
or N is, or v is,
or d is.

520
00:37:48,000 --> 00:37:51,000
This is easy.
You do just have to understand,

521
00:37:51,000 --> 00:37:54,000
this is the total number of
molecules in the gas,

522
00:37:54,000 --> 00:37:58,000
this is the density of the gas
in molecules per cubic meter,

523
00:37:58,000 --> 00:38:02,000
the diameter of the molecule,
the average speed.

524
00:38:02,000 --> 00:38:07,000
That was just a helpful hint.
This is the upper limit to the

525
00:38:07,000 --> 00:38:10,000
reaction rate.
When somebody tells you,

526
00:38:10,000 --> 00:38:15,000
the rate of this reaction is so
and so many molecules per cubic

527
00:38:15,000 --> 00:38:18,000
meter per second,
what you can do is a

528
00:38:18,000 --> 00:38:24,000
back-of-the-envelope calculation
to see whether or not they are

529
00:38:24,000 --> 00:38:30,000
telling you the reaction rate is
greater than this number.

530
00:38:30,000 --> 00:38:35,000
If it is, you can say ah-ha,
got you, it cannot possibly be.

531
00:38:35,000 --> 00:38:40,000
Really important.
If a reaction has this rate,

532
00:38:40,000 --> 00:38:44,000
it will mean that the
probability of the reaction

533
00:38:44,000 --> 00:38:49,000
occurring is one.
If the reaction has that rate,

534
00:38:49,000 --> 00:38:52,000
we call that the gas kinetic
rate.

535
00:38:52,000 --> 00:38:55,000
That is a term that we also
use.

536
00:38:55,000 --> 00:38:59,000
Then, finally,
one other quantity from the gas

537
00:38:59,000 --> 00:39:03,000
kinetic theory.
Yes?

538
00:39:10,000 --> 00:39:13,000
She asked, what velocity,
here, would you use?

539
00:39:13,000 --> 00:39:17,000
You actually are going to,
in that particular case,

540
00:39:17,000 --> 00:39:21,000
have to take the average of the
average velocities.

541
00:39:21,000 --> 00:39:24,000
In other words,
if you had two different

542
00:39:24,000 --> 00:39:29,000
molecules that were reacting for
this average velocity here,

543
00:39:29,000 --> 00:39:33,000
you are going to have to use
the average of the average

544
00:39:33,000 --> 00:39:38,000
velocity.
And then your error bars on the

545
00:39:38,000 --> 00:39:44,000
experiment will always be large
enough to take that into

546
00:39:44,000 --> 00:39:47,000
consideration.
Good question.

547
00:39:47,000 --> 00:39:52,000
One other quantity,
something called the mean free

548
00:39:52,000 --> 00:39:55,000
path.
What I want to know,

549
00:39:55,000 --> 00:40:00,000
here, is on the average,
how far does the molecule

550
00:40:00,000 --> 00:40:07,000
travel in the gas before it
suffers a collision?

551
00:40:07,000 --> 00:40:09,000
That is what a mean free path
is.

552
00:40:09,000 --> 00:40:13,000
A mean free path in solution.
Sometimes you will do a solid

553
00:40:13,000 --> 00:40:17,000
state physics course and will
hear about the electron mean

554
00:40:17,000 --> 00:40:20,000
free path.
A mean free path is always the

555
00:40:20,000 --> 00:40:22,000
distance traveled between
collisions.

556
00:40:22,000 --> 00:40:26,000
That is what that is.
And we are going to call it

557
00:40:26,000 --> 00:40:30,000
lambda.
This is not wavelength anymore.

558
00:40:30,000 --> 00:40:33,000
We just changed our definition
of lambda.

559
00:40:33,000 --> 00:40:36,000
Lambda, here,
is the average distance between

560
00:40:36,000 --> 00:40:40,000
collisions.
How are we going to calculate

561
00:40:40,000 --> 00:40:43,000
that?
What we are going to do is take

562
00:40:43,000 --> 00:40:47,000
the average distance that a
molecule travels per unit time,

563
00:40:47,000 --> 00:40:52,000
per second, and are going to
divide it by the number of

564
00:40:52,000 --> 00:40:55,000
collisions that occur per
second.

565
00:40:55,000 --> 00:40:59,000
And you are going to see that
the per seconds are going to

566
00:40:59,000 --> 00:41:03,000
cancel here.
We are going to have the

567
00:41:03,000 --> 00:41:08,000
average distance per collision.
That is what we are after.

568
00:41:08,000 --> 00:41:13,000
This is easy to do because the
average distance traveled per

569
00:41:13,000 --> 00:41:16,000
second is simply the average
velocity.

570
00:41:16,000 --> 00:41:20,000
It is meters per second.
The average distance traveled

571
00:41:20,000 --> 00:41:24,000
per second.
The number of collisions per

572
00:41:24,000 --> 00:41:27,000
second that happened,
we just calculated that.

573
00:41:27,000 --> 00:41:33,000
That was Z1.
We can substitute in Z1 vbar,

574
00:41:33,000 --> 00:41:38,000
vbar cancels,
and what we are left with is

575
00:41:38,000 --> 00:41:46,000
one over the square root of 2
rho pi d squared.

576
00:41:46,000 --> 00:41:49,000
That is meters.

577
00:41:49,000 --> 00:41:56,000
That is the average distance
the molecule traveled per

578
00:41:56,000 --> 00:42:01,000
collision.
This is another general generic

579
00:42:01,000 --> 00:42:06,000
quantity useful in many cases,
also in other kinds of cases

580
00:42:06,000 --> 00:42:10,000
that I just mentioned in solid
state physics.

581
00:42:10,000 --> 00:42:16,000
What I want to do now is that
we said we were going to start

582
00:42:16,000 --> 00:42:19,000
talking about the motions of
molecules.

583
00:42:19,000 --> 00:42:24,000
We have taken care of,
now, the translational motion.

584
00:42:24,000 --> 00:42:29,000
Now I want to start talking
about the internal degrees of

585
00:42:29,000 --> 00:42:33,000
freedom.
And let's do that.

586
00:42:33,000 --> 00:42:38,000
Let me just see if I have
something, here.

587
00:42:38,000 --> 00:42:43,000
Let me raise the board,
here.

588
00:42:58,000 --> 00:43:02,000
We have talked about this
molecule, which is a set of

589
00:43:02,000 --> 00:43:07,000
atoms that are bonded together.
And, as we always say,

590
00:43:07,000 --> 00:43:12,000
they are bonded together for
the mutual comfort of their

591
00:43:12,000 --> 00:43:15,000
electrons.
And we saw that it is not

592
00:43:15,000 --> 00:43:17,000
frozen.
It moves.

593
00:43:17,000 --> 00:43:20,000
It has kinetic energy.
It has velocity.

594
00:43:20,000 --> 00:43:25,000
The velocity generates this
macroscopic property of

595
00:43:25,000 --> 00:43:30,000
pressure.
But molecules also jiggle.

596
00:43:30,000 --> 00:43:33,000
For example,
we are going to take here

597
00:43:33,000 --> 00:43:39,000
nitrogen, triple bond.
That triple bond actually

598
00:43:39,000 --> 00:43:44,000
functions like a spring.
We can think of that bond as a

599
00:43:44,000 --> 00:43:49,000
spring between the two atoms.
That spring can stretch,

600
00:43:49,000 --> 00:43:54,000
and that spring can compress.
When we tell you the

601
00:43:54,000 --> 00:44:00,000
equilibrium bond length of some
molecule is some number,

602
00:44:00,000 --> 00:44:05,000
it is the equilibrium bond
length.

603
00:44:05,000 --> 00:44:09,000
It is not the bond length when
the molecule is stretched.

604
00:44:09,000 --> 00:44:14,000
It is not the bond length when
the molecule is compressed.

605
00:44:14,000 --> 00:44:19,000
This motion is the vibrational
motion of the molecule.

606
00:44:24,000 --> 00:44:30,000
So, we have that kind of
internal motion in a molecule.

607
00:44:30,000 --> 00:44:33,000
Molecules also tumble.
For example,

608
00:44:33,000 --> 00:44:38,000
let's take our nitrogen
molecule, again.

609
00:44:38,000 --> 00:44:45,000
It rotates around an axis.
Going through the center of the

610
00:44:45,000 --> 00:44:49,000
mass, it rotates around this
axis.

611
00:44:49,000 --> 00:44:54,000
This nitrogen molecule also
rotates around an axis,

612
00:44:54,000 --> 00:45:00,000
here, perpendicular to the
board.

613
00:45:00,000 --> 00:45:06,000
It rotates in that direction.
This is the rotational motion

614
00:45:06,000 --> 00:45:11,000
of the molecule.
The different ways in which we

615
00:45:11,000 --> 00:45:15,000
can have this motion are called
modes.

616
00:45:15,000 --> 00:45:21,000
This is a rotational mode.
This is vibrational mode.

617
00:45:21,000 --> 00:45:28,000
We did not use the word before
in talking about translation,

618
00:45:28,000 --> 00:45:33,000
but those are translational
modes.

619
00:45:33,000 --> 00:45:40,000
Sometimes, we also call these
modes degrees of freedom.

620
00:45:40,000 --> 00:45:49,000
A molecule has translational
modes or degrees of freedom,

621
00:45:49,000 --> 00:45:57,000
vibrational degrees of freedom,
and rotational degrees of

622
00:45:57,000 --> 00:46:00,000
freedom.
Now, in general,

623
00:46:00,000 --> 00:46:10,000
if you have an N-atom molecule,
you have 3N total modes.

624
00:46:10,000 --> 00:46:17,000
When I say total modes that
means the sum of translation,

625
00:46:17,000 --> 00:46:24,000
rotation and vibration.
Of those 3N total modes,

626
00:46:24,000 --> 00:46:30,000
three of them are translational
modes.

627
00:46:35,000 --> 00:46:38,000
Why do we have three
translational modes?

628
00:46:38,000 --> 00:46:43,000
We have three translational
modes because we are operating

629
00:46:43,000 --> 00:46:48,000
in a three-dimensional space.
If we have a molecule here,

630
00:46:48,000 --> 00:46:53,000
that molecule has motion in the
x direction, in the y direction,

631
00:46:53,000 --> 00:46:57,000
and also in the z direction.
Those are the three

632
00:46:57,000 --> 00:47:02,000
translational modes of the
molecule.

633
00:47:02,000 --> 00:47:08,000
What that means then,
if three of the modes are used

634
00:47:08,000 --> 00:47:13,000
up for translation,
then we must have 3N minus 3

635
00:47:13,000 --> 00:47:20,000
internal modes.
3N minus 3 must be the number

636
00:47:20,000 --> 00:47:27,000
of modes that is leftover for
the internal degrees of freedom

637
00:47:27,000 --> 00:47:33,000
for rotation and also for
vibration.

638
00:47:33,000 --> 00:47:39,000
And what we are going to see,
next time, is that these

639
00:47:39,000 --> 00:47:46,000
internal degrees of freedom,
these internal modes are going

640
00:47:46,000 --> 00:47:51,000
to be quantized.
That is, this molecule is going

641
00:47:51,000 --> 00:47:59,000
to be able to vibrate with only
certain amounts of energy.

642
00:47:59,000 --> 00:48:02,000
There is going to be a ground
vibrational state.

643
00:48:02,000 --> 00:48:07,000
There is going to be a first
excited vibrational state.

644
00:48:07,000 --> 00:48:12,000
A second excited vibrational
state, just like we talked about

645
00:48:12,000 --> 00:48:17,000
the energy levels of a hydrogen
atom, where the electron was

646
00:48:17,000 --> 00:48:22,000
bound with different amounts of
energy dictated by a principle

647
00:48:22,000 --> 00:48:26,000
quantum number.
The vibrational degree of

648
00:48:26,000 --> 00:48:30,000
freedom is also going to be
dictated by a vibrational

649
00:48:30,000 --> 00:48:33,000
quantum number.
In addition,

650
00:48:33,000 --> 00:48:38,000
the rotations of the molecules,
they are also quantized.

651
00:48:38,000 --> 00:48:43,000
A molecule is going to be able
to rotate with only one energy,

652
00:48:43,000 --> 00:48:46,000
or another energy,
or another energy,

653
00:48:46,000 --> 00:48:49,000
but not any energies in
between.

654
00:48:49,000 --> 00:48:52,000
There are discrete rotational
states.

655
00:48:52,000 --> 00:48:56,000
We are going to be talking
about another kind of quantum

656
00:48:56,000 --> 00:48:58,000
number.
In that case,

657
00:48:58,000 --> 00:49:04,000
a rotational quantum number.
That is where we are going to

658
00:49:04,451 --> 00:49:07,000
pick up on Monday.
See you then.