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Last time, we went through the same arguments that Maxwell and
Boltzmann did to understand the microscopic origin of the ideal
gas law, PV equal nRT.
And we saw how Maxwell had hypothesized that the pressure
of a gas on the walls of some container, if the molecules were
moving and they collided onto the walls, that pressure must be
due to the individual impacts of the molecules on the wall.
That must be due to the change in the momentum of the wall when
the molecules slammed right into it.
And that change in momentum over some change in time is this
force. That force divided by the total
area is the pressure. And it was using that argument
that we came up with an expression, just like Maxwell
did for the pressure times the volume.
And we see that we were able to write it in terms of the
velocity of the molecules that were moving in this gas.
Well, that is very nice because if we have this theoretical
expression P times V, and we know experimentally that
P times V equal nRT, if this theory is
correct, then this quantity n M, average of the velocity
squared, over three, that better be equal to nRT.
Or, solving for what we call
the root mean square velocity, that better be equal to the
square root of 3RT over M.
That kinetic theory made a prediction for what the velocity
ought to be. It took, of course,
another hundred years before somebody could measure that.
And, of course, it is correct.
But what is interesting here is that this model gave us an
understanding for what temperature is.
Temperature is a measure of the speed of the molecules in the
gas. It is also, as we saw last
time, a measure of their kinetic energy.
We saw that the kinetic energy is one-half M average of the
velocity squared. We put
in our result from kinetic theory.
That shows us that the average energy is three-halves RT.
Temperature is the measure of
the kinetic energy of these molecules in the gas.
For the first time, there was a microscopic
understanding of what temperature was.
Now, what we looked at also, last time, was the
Maxwell-Boltzmann distribution of velocities or speeds and
talked about this particular functional form.
How there was this quadratic dependence right here,
a low v because of the v squared here and then an
exponentially decaying tail. There were two parameters,
the mass and the temperature. We talked about that last time.
But, of course, if there is a distribution here
of speeds where this distribution is a probability of
finding a molecule with a particular speed between v and v
plus dv, -- -- then there also has to be a
distribution of energies because the velocity is related to the
energy. What do we have to do?
We have to take that Maxwell-Boltzmann speed
distribution and change the variable to energy.
We know how to do that. We know how to equate those two
distribution functions. And, using the Jacobean of the
transformation that we talked about last time,
we can calculate what the energy distribution is.
f of E is our Maxwell-Boltzmann energy
distribution. It is the probability of
finding a molecule with a kinetic energy E to E plus dE.
It also has a decaying exponential term,
here, with the energy in the argument, and it is multiplied
by E of the one-half. Let's take a look at what that
distribution function looks like.
Here it is. Notice, as we learned last
time, that the energy has nothing to do with the mass of
the particle. The only parameter that is
important is the temperature. At a given temperature,
all particles, it doesn't matter what their
mass is, have the same energy. So, all particles at
degrees Kelvin have a Maxwell-Boltzmann energy
distribution that looks like this.
There is a very rapid rise in that distribution function,
unlike the velocity distribution,
which increased as v squared.
It peaks rapidly, and then there is an
exponentially decaying term in the energy.
At 1500 degrees Kelvin, the energy distribution looks
like this. The energy of the molecules has
increased. Since this is a probability,
and the area under these curves has to equal to one,
if the energy goes up, that is we have more molecules
out here with higher energies, well, then this maximum value
for the probabilities has got to go down because we have to keep
the area under that curve equal to one, since this is a
probability. What we see here,
as we raise the energy, is that we have more and more
molecules with high energies. There are not a lot of them,
but there are some. And they are going to be really
important in the example I am going to show you in just a
moment. Average energy at 600 Kelvin,
just three-halves RT, is 7.5 kilojoules per mole.
Average energy at 1500 Kelvin is 18.7 kilojoules per mole.
But now, I want to show you an example of the importance of
this Maxwell-Boltzmann distribution function for
energies to chemical reactions, the importance of it in making
some kinds of chemical reactions actually work.
And so the example is going to be this reaction.
This reaction is called steam reforming of natural gas.
It is the reaction of methane plus water.
Natural gas is mostly methane. But this reaction of methane
and water makes CO and hydrogen. It
turns out that this reaction does not work in the gas phase.
That is, if you have a methane molecule and water molecule
collide, they are just going to collide, bounce apart and go in
their opposite directions. They are not going to make CO
and hydrogen. And so what you have to have in
this reaction to make it go is a catalyst.
That catalyst is going to be a nickel surface.
A catalyst is something that will lower the activation energy
barrier by changing the mechanism of a reaction so that
the reaction can proceed. In this case,
what happens is that the methane and the water impinge on
this nickel metal catalyst. And the methane and the water
decompose. They fall apart to their
elements on that nickel surface. And then, once you have carbon,
hydrogen and oxygen on that nickel surface,
the atoms rearrange and come off as CO and molecular
hydrogen. So, that nickel surface is a
catalyst. We call this the catalytic
reaction. In particular,
we call it a heterogeneous catalytic reaction.
It is heterogeneous because the catalyst is in a different phase
than the reactants. The catalyst here is a solid.
The reactants are gases. A homogeneous catalyst is one
in which the phase of the catalyst and the reactants is
the same. This is a heterogeneous
catalytic reaction. It turns out that this reaction
here is really an important reaction from the standpoint of
the production of hydrogen. All of the hydrogen that you
use, you know, if you are doing a laboratory
experiment and you go get a tank of hydrogen, that hydrogen is
made by this reaction, by reacting methane and water
to form that hydrogen. All of our commercial sources
of hydrogen come from carrying out this reaction.
For example, if you are in the business of
ammonia synthesis, which is taking hydrogen and
nitrogen to make ammonia, which is also carried out on an
iron surface, another heterogeneous catalytic
reaction. And ammonia,
of course, is the starting material for lots of chemicals,
in particular fertilizers. If you have an ammonia plant,
right next to the ammonia plant you have a steamer-forming plant
to make the hydrogen to feed into this reaction.
And another place where this is useful is in methanol synthesis.
That is, you can take hydrogen and CO on a copper zinc oxide
heterogeneous catalyst and make methanol, the starting point for
gasoline, which now is certainly economically feasible.
But it turns out that this reaction here is really a hard
one to carry out. Despite the fact that nickel is
a catalyst for the reaction and makes it go, the reaction still
has to be carried out at very high temperatures.
1500 degrees kelvin is a very high temperature.
It also needs very high pressure.
And one of the reasons why we don't have a hydrogen economy is
because it is so difficult to make.
Hydrogen can be done, and it is done in all of the
commercial processes that I mentioned to you,
but it is still hard. 1500 degrees kelvin,
high pressures. The question is,
why is it so hard? Why is there this barrier here?
Why do we need to raise the temperature of the gas in order
to get this reaction to go? Well, let's take a look at
that. What there exists here in this
problem is what is called an activation energy barrier to
making the reaction go. It turns out that this
activation energy barrier is really in the very first step of
the reaction, which is pulling the methane
apart, pulling that first hydrogen atom off of the methane
molecule. What I represent right here is
the energy of the reaction as a function of the reaction,
and the reaction here is just the first step.
It is taking methane gas, pulling the hydrogen off so
that you have a methyl radical stuck to the surface and you
have a hydrogen atom stuck to the surface.
It is just breaking that first C-H bond.
What you can see here is that you have to put 50 kilojoules of
energy into that reaction in order to make it go.
You have to put that 50 kilocalories of energy in first,
before you get any energy back. You can see that this is
exothermic. But you have to put this energy
in, in order to get that reaction to go.
Well, how do we know that? Let me back up a minute.
Here is where the Maxwell-Boltzmann energy
distribution is important. What I did was I took those
Boltzmann energy distributions that I plotted for you at
Kelvin and 1500 Kelvin and turned them on their side.
This is essentially equal to the probability of finding a
molecule at a particular energy versus the energy.
Here is the Maxwell-Boltzmann distribution at 600 degrees
Kelvin. It is peaked to very low
energies. And, if you look here in this
Maxwell-Boltzmann tail, there are not very many
molecules that have enough energy to get over that barrier.
However, if we raise the temperature of the gas to
degrees Kelvin, then now you can see that there
are more molecules here in the high energy part of this tail.
And it is these molecules at these high energies that
actually can get over this barrier to dissociation of the
methane molecule. They are not many of those
molecules with high energy, but there are some.
And that is important, because what happens is when
they have enough energy, they react and they leave.
And then this Boltzmann distribution re-equilibrates.
If the high energy molecules leave, then there are lots of
collisions here that kick some more molecules up to the high
energy part of the tail, and they react.
And so it works. That is why it is important.
How do we really know that there is this barrier here to
the dissociation of methane, and why is there this barrier?
What is the physical origin of this barrier to pulling this C-H
bond apart? Well, one way we could know
that there was this barrier there for sure is to do the
following. If we had a way to take methane
gas and make it have just a single energy,
not the Maxwell-Boltzmann distribution of energies that
have lots of different energies, which have lots of different
energies in them, and you saw how broad those
curves were. But if we had a way to make
methane gas with just say energy E sub 1, whatever that is,
we could then take those molecules with just those
energies, aim them at this nickel surface,
and see if the methane fell apart.
And, if it didn't, then we would prepare molecules
at some higher energy and see if they fell apart.
If they didn't then we would go to higher energy.
We would keep going until we got to the top of the barrier.
Well, how do you do that? How do you make molecules with
a particular energy? Because I just showed you a
Maxwell-Boltzmann energy distribution at 300 Kelvin,
600 Kelvin. There are a lot of energies
present. How do we make molecules with
just one energy? We have a way to do that by
using some beam techniques and high pressure adiabatic
expansions. Basically, the way it works is
this. What we are going to do is take
a tube here, which is going to have a high pressure of methane
in it. Then we are going to punch a
little hole in that tube. This tube is actually sitting
in a vacuum, so we are going to expand the methane from that
tube into the vacuum. We are going to squirt it out.
It is going to become a beam of molecules.
But when you expand a gas from high pressure to low pressure,
this expansion is called an adiabatic expansion,
which means the gas cools. I will explain that in a
moment, but the adiabatic expansion you are going to talk
a lot about in 5.60. And, if you are a chemical
engineer, you will talk even more about it past 5.60.
Here is how it works. You take these molecules and
squirt them out. Because the pressure right here
is so high, what happens is there are lots and lots and lots
of collisions. And so, you can imagine,
if you have some slow molecule just kind of lumbering along and
a fast one comes and hits you, what is going to happen is you
are going to speed up, and the fast one is going to
slow down. And, if you keep doing that
again and again and again, all the molecules are going to
end up with the same energy or the same velocity.
If you have so many collisions, after a while they are going to
all have the same velocity or the same energy because one gets
sped up, one gets slowed down. They are all going to end up
with the same energy. That is what happens.
That is how we make a beam of molecules, a source of molecules
with the same kinetic energy. And basically,
what is happening then is that before the expansion,
just look at this curve, here is an actual Boltzmann
distribution of velocities. It is broad.
But after this expansion, here is the distribution.
It is really pretty narrow. If you put a temperature to
that distribution, you might find that at one
degree Kelvin, you can really make molecules
with a single or very narrow distribution in energies.
That is what we do. We make those molecules with
this energy E sub 1. Then we have some more tricks
for changing those energies in a controlled way.
And we just keep cranking those energies up and watch to see
when the methane falls apart. And so, you get data that kind
of looks like this. This is a dissociation
probability of a methane as a function of its energy.
This is in kilocalories per mole.
At some point, say right about here,
15 kilocalories per mole, the dissociation probability
skyrockets. All of a sudden,
you are at a high enough energy to get that methane to fall
apart. This tells you exactly where
that barrier is. But now, the question is,
why is there this barrier? Physically, why is there a
barrier to pulling the hydrogen off of the carbon when that
methane molecule comes close to the surface?
Well, the reason is this. In order to break the
carbon-hydrogen bond in methane, in order to have enough energy
to break that bond, what you need to do is you need
to simultaneously form a nickel-carbon bond and a
nickel-hydrogen bond. In other words,
when this methane molecule here comes into some nickel surface
very slowly, because that methane is tetrahedral and the
carbon is kind of hidden in the center, the hydrogens interact
with the nickel. But the carbon does not get in
close enough to the nickel to start to form a nickel-carbon
bond. Now, if you speed that methane
molecule up and you really ram it into the surface,
upon collision of the molecule with the surface,
the hydrogens are pushed back, the carbon gets in close enough
to the nickel to start to form, here, this nickel-carbon bond.
You form a nickel-hydrogen bond, and now you can break that
C-H bond. We call this black chemistry.
The barrier, there, to that reaction
physically is the amount of energy that you have to put into
the molecule to push those hydrogens back,
to bend those hydrogens back to distort this methane molecule.
Once you do that, the reaction goes.
You are on a roll there. It goes.
So, that is the physical origin of this barrier,
here, to the dissociation of methane.
You need to put enough kinetic energy into those methane
molecules in order to get over that barrier.
And we do that on a practical scale by raising the temperature
of the gas. And on a microscopic scale,
this is what is happening. You can imagine that my
students and I did this experiment 15 years ago.
But then, we said if this barrier here is the energy
required to deform that methane molecule, then we should be able
to do this experiment. This experiment is the
following. We are going to take our nickel
surface, here, and lower the temperature of
the surface to 47 kelvin. At that temperature,
what we can do is we can freeze a layer of methane on the
surface. We call it fizzy-sorb,
a layer of methane on the surface.
It will just stick there. And then, if this barrier is
the energy required to deform or to start the molecule,
then in principle I could take a hammer and pound that molecule
into the correct shape for the transition state that leads to
dissociation. It sounds like a simple idea,
and it is, except that we cannot really take that hammer.
But we can take an argon atom. We can freeze the methane onto
the surface, and now we come in with an argon atom.
That is just a big ball. Or, a xenon atom or a krypton
atom. And we know how to accelerate
xenon or krypton. We don't accelerate it too
much, 50 kilocalories, 70 kilocalories per mole,
something like that. And then we can bring it in.
What happens is that the impact of the collision on that methane
causes that methane molecule to compress, distort,
gets it into the configuration of the transition state that
leads to the methane falling apart.
And so, you get the same result.
The question, there, is just getting the
energy into the molecule to actually distort it,
to deform it so that you can make the nickel-carbon bond and
the nickel-hydrogen bond. That is the key.
My students and I, after we had spent many years
doing this experiment and spent a lot of money on this
experiment, said we have proven for sure that when a bug flies
into the windshield of your car, you do get the same result as
if you hit the bug on the windshield of your car with a
bug swatter. That is what we get.
This is an example of one chemical reaction for which we
know the physical origin of the barrier.
There are very few chemical reactions for which we know
anything about the physical origin of a barrier to a
reaction. Now, what I want to talk about
is go back to kinetic theory, because the kinetic theory is
going to allow us to calculate a couple of more quantities that
are of interest to us. And one of those quantities is
the collision frequency. And let me just write that on
the board here. What I am going to want to
calculate is something called Z1, or I am going to call it Z1.
It is the number of collisions that a molecule makes in a gas
per unit time. And our unit time is going to
be seconds. This is the collision frequency
of a single molecule in a gas.
How am I going to calculate the quantity using the kinetic
theory approach that we have been talking about?
What I am going to do is this. I am going to take some gas,
and the molecules in that gas are represented by these blue
circles right here. They have a diameter D.
And then, within that gas, I am going to imagine this
lighter blue cylinder. That cylinder is an imaginary
construct. We are going to use it to
calculate this collision frequency.
What it is going to be is the collision volume.
I have set the diameter of that cylinder to be equal to two
times the diameter of the molecule.
I have set this upright. And the bottom line is that all
of the molecules that instantaneously happen to be in
this cylinder when a molecule comes through,
all of those molecules are actually going to be hit.
They are going to suffer a collision with another molecule.
And I am going to use that, the number of molecules in this
volume, to calculate this collision frequency.
That is what I am going to do. Snapshot in time,
those molecules are frozen there, except there is one
smart-alecky molecule that comes cruising on through.
Here he comes. Bam, bam, bam.
Hits these three molecules, here, because they are more
than halfway into that collision cylinder.
What I know is this smart-aleck was moving at some average
velocity, which I am going to call vbar.
In a time T, this molecule,
since it is moving with an average velocity vbar,
in a time t, it is traveling a length vbar
times t.
Velocity times time, that is going to give you a
length. And so, I am going to set,
for convenience, this time equal to one second
because that is going to be my unit of time.
I want to do this per second. I am going to set this equal to
one second. Therefore, this molecule is
going to travel a length vbar in one second.
And that is the length that I am going to make the collision
cylinder. I am going to make that
collision cylinder be vbar long. That is the distance traveled
in one second by this smart-alecky molecule.
Now, all I have to do is I have got to take the volume of the
collision cylinder and multiply it by all the molecules that
happen to be at the volume at that particular time.
Because that volume represents the volume swept out in one
second by this smart-alecky molecule.
What is the volume of the cylinder?
I set the diameter to be equal to 2d.
The volume of the cylinder is the cross-sectional area times
the length. That cross-sectional area,
then, is pi r squared, but r here is equal to d,
just to make it confusing. The area is pi d squared.
The length is vbar.
That is the volume of the cylinder, pi d squared vbar.
So, we have the volume of the
cylinder. Now what we are going to need
to know is the density of the molecules in this cylinder,
which is the density of the molecules in the gas.
The cylinder is not special. The cylinder is an imaginary
construct that I put in there to help me calculate the collision
frequency. I need the density of the
molecules in the gas or in the cylinder.
The density of the molecules I am going to set as equal to N,
the total number of molecules in this volume,
divided by the volume of the gas.
I am going to call that rho. N over V is going to be equal
to rho. That is the number of molecules
per cubic meter. That is the density of the gas.
This is the symbol that I am going to use from now on for
density defined as molecules per cubic meter.
Then the collision frequency is simply going to be the density
of the molecules times the volume of the cylinder.
The collision frequency Z1, here, is this volume of the
cylinder, pi d squared vbar times the density.
Why?
Because I set it up so that every molecule in that cylinder
would suffer a collision. And that cylinder is the length
that a molecule travels in one second.
There is the density times the volume.
The meters cubed disappears. What I have left is rho,
the density, pi d squared vbar.
This is collisions per second.
So, I have my collision frequency.
This is the single-molecule collision frequency.
This is the frequency of collisions that one molecule
makes with the other gas molecules.
That is important. We are going to do a different
kind of collision frequency in just a moment.
But now, it turns out that there is another factor here
that I had left out because I haven't done as sophisticated of
an analysis as I could. And that is that I made the
assumption in my picture before that all the other molecules
were frozen in time and that there was only one smart-aleck
that was moving around, cruising through.
The reality is they are all moving.
And what I really have to do is I have to take into account the
relative velocities of the molecules, the relative speeds.
I can do that. There is a little more
sophisticated analysis in doing that, but I could do that.
And, if we do that, this makes a difference of the
square root of two. I am just going to put that
square root of two in there right now, and later on you will
be able to see where that comes from, in a later course.
That actually is the collision frequency of a single molecule
in the gas.
I am also interested in another quantity, which is the total
collision frequency. That was the single collision
frequency, but I am also interested in knowing how many
collisions are occurring in the entire gas per unit time,
the total collision frequency. You know why I am interested in
that number? I am interested in that number
because that is going to be the upper limit to any reaction
rate. A reaction in the gas phase or
in solution cannot happen any faster than the molecules
collide. They have to collide before a
reaction is going to occur. The total collision frequency,
the importance of that number is that it is the upper limit to
a reaction rate. Let's calculate that.
Z is going to be equal to the collision frequency of one
molecule, Z1, times the total number of
molecules in the gas, N.
That is what N stands for. But since each collision
involves two molecules, I am going to have to multiply
this by one-half. Otherwise, I am going to over
count the number of collisions because each collision involves
two molecules. The total collision frequency
here is one-half times N times Z1.
Therefore, if I go and plug in my expression for Z1 and N and
simplify things, my total collision frequency
here is one over the square root of two times N times rho pi d
squared times the average speed.
That is my total collisions per
second. This is not rocket science.
This is easy. What I want you to realize is
that you do have to understand what N is, what rho is,
what d is, what vbar is. And I am always surprised on an
exam, when we are going to give you this equation,
that students don't actually know what rho is,
or N is, or v is, or d is.
This is easy. You do just have to understand,
this is the total number of molecules in the gas,
this is the density of the gas in molecules per cubic meter,
the diameter of the molecule, the average speed.
That was just a helpful hint. This is the upper limit to the
reaction rate. When somebody tells you,
the rate of this reaction is so and so many molecules per cubic
meter per second, what you can do is a
back-of-the-envelope calculation to see whether or not they are
telling you the reaction rate is greater than this number.
If it is, you can say ah-ha, got you, it cannot possibly be.
Really important. If a reaction has this rate,
it will mean that the probability of the reaction
occurring is one. If the reaction has that rate,
we call that the gas kinetic rate.
That is a term that we also use.
Then, finally, one other quantity from the gas
kinetic theory. Yes?
She asked, what velocity, here, would you use?
You actually are going to, in that particular case,
have to take the average of the average velocities.
In other words, if you had two different
molecules that were reacting for this average velocity here,
you are going to have to use the average of the average
velocity. And then your error bars on the
experiment will always be large enough to take that into
consideration. Good question.
One other quantity, something called the mean free
path. What I want to know,
here, is on the average, how far does the molecule
travel in the gas before it suffers a collision?
That is what a mean free path is.
A mean free path in solution. Sometimes you will do a solid
state physics course and will hear about the electron mean
free path. A mean free path is always the
distance traveled between collisions.
That is what that is. And we are going to call it
lambda. This is not wavelength anymore.
We just changed our definition of lambda.
Lambda, here, is the average distance between
collisions. How are we going to calculate
that? What we are going to do is take
the average distance that a molecule travels per unit time,
per second, and are going to divide it by the number of
collisions that occur per second.
And you are going to see that the per seconds are going to
cancel here. We are going to have the
average distance per collision. That is what we are after.
This is easy to do because the average distance traveled per
second is simply the average velocity.
It is meters per second. The average distance traveled
per second. The number of collisions per
second that happened, we just calculated that.
That was Z1. We can substitute in Z1 vbar,
vbar cancels, and what we are left with is
one over the square root of 2 rho pi d squared.
That is meters.
That is the average distance the molecule traveled per
collision. This is another general generic
quantity useful in many cases, also in other kinds of cases
that I just mentioned in solid state physics.
What I want to do now is that we said we were going to start
talking about the motions of molecules.
We have taken care of, now, the translational motion.
Now I want to start talking about the internal degrees of
freedom. And let's do that.
Let me just see if I have something, here.
Let me raise the board, here.
We have talked about this molecule, which is a set of
atoms that are bonded together. And, as we always say,
they are bonded together for the mutual comfort of their
electrons. And we saw that it is not
frozen. It moves.
It has kinetic energy. It has velocity.
The velocity generates this macroscopic property of
pressure. But molecules also jiggle.
For example, we are going to take here
nitrogen, triple bond. That triple bond actually
functions like a spring. We can think of that bond as a
spring between the two atoms. That spring can stretch,
and that spring can compress. When we tell you the
equilibrium bond length of some molecule is some number,
it is the equilibrium bond length.
It is not the bond length when the molecule is stretched.
It is not the bond length when the molecule is compressed.
This motion is the vibrational motion of the molecule.
So, we have that kind of internal motion in a molecule.
Molecules also tumble. For example,
let's take our nitrogen molecule, again.
It rotates around an axis. Going through the center of the
mass, it rotates around this axis.
This nitrogen molecule also rotates around an axis,
here, perpendicular to the board.
It rotates in that direction. This is the rotational motion
of the molecule. The different ways in which we
can have this motion are called modes.
This is a rotational mode. This is vibrational mode.
We did not use the word before in talking about translation,
but those are translational modes.
Sometimes, we also call these modes degrees of freedom.
A molecule has translational modes or degrees of freedom,
vibrational degrees of freedom, and rotational degrees of
freedom. Now, in general,
if you have an N-atom molecule, you have 3N total modes.
When I say total modes that means the sum of translation,
rotation and vibration. Of those 3N total modes,
three of them are translational modes.
Why do we have three translational modes?
We have three translational modes because we are operating
in a three-dimensional space. If we have a molecule here,
that molecule has motion in the x direction, in the y direction,
and also in the z direction. Those are the three
translational modes of the molecule.
What that means then, if three of the modes are used
up for translation, then we must have 3N minus 3
internal modes. 3N minus 3 must be the number
of modes that is leftover for the internal degrees of freedom
for rotation and also for vibration.
And what we are going to see, next time, is that these
internal degrees of freedom, these internal modes are going
to be quantized. That is, this molecule is going
to be able to vibrate with only certain amounts of energy.
There is going to be a ground vibrational state.
There is going to be a first excited vibrational state.
A second excited vibrational state, just like we talked about
the energy levels of a hydrogen atom, where the electron was
bound with different amounts of energy dictated by a principle
quantum number. The vibrational degree of
freedom is also going to be dictated by a vibrational
quantum number. In addition,
the rotations of the molecules, they are also quantized.
A molecule is going to be able to rotate with only one energy,
or another energy, or another energy,
but not any energies in between.
There are discrete rotational states.
We are going to be talking about another kind of quantum
number. In that case,
a rotational quantum number. That is where we are going to
pick up on Monday. See you then.