This introductory physical chemistry course examines the connections between molecular properties and the behavior of macroscopic chemical systems.

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From the course by University of Minnesota

Statistical Molecular Thermodynamics

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This introductory physical chemistry course examines the connections between molecular properties and the behavior of macroscopic chemical systems.

From the lesson

Module 4

This module connects specific molecular properties to associated molecular partition functions. In particular, we will derive partition functions for atomic, diatomic, and polyatomic ideal gases, exploring how their quantized energy levels, which depend on their masses, moments of inertia, vibrational frequencies, and electronic states, affect the partition function's value for given choices of temperature, volume, and number of gas particles. We will examine specific examples in order to see how individual molecular properties influence associated partition functions and, through that influence, thermodynamic properties. Homework problems will provide you the opportunity to demonstrate mastery in the application of the above concepts.

- Dr. Christopher J. CramerDistinguished McKnight and University Teaching Professor of Chemistry and Chemical Physics

Chemistry

Let's put together all the pieces for our ideal diatomic gas and look at a couple

Â of examples now. So we've looked at the four different

Â contributions in considerable detail. This is just a review slide.

Â That is, the translational component, the rotational component, the vibrational

Â component, and the electronic energy component that all contribute to the

Â total energy of a diatomic. I do want to qualify what we looked at in

Â the last lecture which is the rotational component ever so slightly so I'll

Â introduce a wrinkle. And that comes in from the way the

Â quantum mechanics actually handles a diatomic and makes a distinction between

Â a homonuclear diatomic, that's like molecular oxygen O2.

Â And a heteronuclear diatomic, like carbon monoxide, CO.

Â So the boundary conditions associated with the quantum mechanics are a little

Â different in those two cases because the two atoms of oxygen are indistinguishable

Â one from another, but the carbon obviously is distinguishable from oxygen.

Â It introduces something called the symmetry number into the rotational

Â partition function. So when we did our derivation last time,

Â we ended up with the rotational partition function when the rotational temperature

Â is much lower than the actual temperature, was just t over theta rote.

Â Now, in fact, this symmetry number should appear there sigma, and sigma takes on a

Â value of 1 for a heteronuclear diatomic, but it takes on a value of 2, the

Â partition function is only half as large when it's a homonuclear diatomic.

Â So in fact the derivation we did in the last lecture was valid for the

Â hetero-nuclear, not for the homo-nuclear, which just has different levels that

Â contribute to that different sigma value. So we'll just take that as a given and

Â use it in the future. So, the full diatomic partition function

Â in that case, which remember is the product of all the individual partition

Â functions. So when you have an energy that's a sum

Â of energies, the corresponding partition function is a product of partition

Â functions. And so I've just expanded it down below.

Â It depends on volume and it depends on temperature.

Â The volume dependence comes in in the translational term.

Â Remember that's from particle in a box quantum mechanics.

Â And then there's the rotational term we looked at most recently, now with the

Â symmetry number appearing. The vibrational term, which we did before

Â we did rotation, and it involves the vibrational temperature.

Â And, finally, our assumption that the electronic component involves only the

Â ground state, and the ground state energy is defined relative to the dissociation

Â energy of the diatomic. And it gives rise to this, partition

Â function where the degeneracy of the ground state is included as well.

Â So if we now want to form from that full molecular partition function, the

Â ensemble partition function which now, in addition to depending on volume and

Â temperature. Depends on number of particles.

Â Recall that for an ideal gas we take the molecular partition function to the nth

Â power and we divide by n factorial. So this is all still sort of a review of

Â things we've done in the past, but what I want to do now is use that ensemble

Â partition function and look at The internal energy and the heat capacity.

Â So we've really done all the individual terms.

Â Remember we got this from the partial derivative of the log of the partition

Â function with respect to temperature multiplied by MKT squared, so I won't go

Â through all the derivations again. You can review the older videos if you'd

Â like to see them, but I'll just remind you of the results.

Â 3 halves RT for translation. RT for rotation.

Â Something involving the vibrational temperatures.

Â So this one varies a bit because excited vibrational states get picked up as

Â temperature goes up. So r times theta viable over 2.

Â an additional term involving vibrational temperature And finally the term that

Â comes from the dissociation energy. And so that's just what I said,

Â translations, rotations. this is actually the zero point component

Â of the vibrational energy. That's the vibrational energy beyond the

Â zero point because of thermal effects, and finally, there's the electronic.

Â And so if you also review what we determined for heat capacities we get 5

Â halves R from the translations and rotations.

Â Remember just the derivative of the internal energy with respect temperature.

Â So we get 3 halves R, R. T does not appear in here.

Â Remember the vibrational temperature is it has temperature units.

Â But it doesn't involve T so there's no T dependent so it's zero.

Â There is a t hiding in here, and if I carry out this differentiation i get this

Â term. And there's not a T here, either, so that

Â does not contribute to the heat capacity. So here's the molar heat capacity for an

Â ideal diatomic gas. So given those expressions for internal

Â energy and for heat capacity, let's pause for a moment, and I'll let you see if you

Â can work with them to work out a problem. Alright.

Â So the only term in the heat capacity that is dependent on the temperature is

Â the term that comes from the vibrations. Let, let's take a closer look at that for

Â a moment. And so I'll remind you that the

Â vibrational molar heat capacity has this form.

Â And let's just take a look for two different diatomics of some interest.

Â So let's do molecular nitrogen. We could call that, reagent grade air if

Â you like, 78% that we're all breathing in.

Â And then we'll also take a look at molecular bromine as a gas.

Â that would not be pleasant to breath in, but we're just working with it on a

Â computer, happily. and because bromine has two isotopes that

Â are both quite abundant at at natural abundance on earth.

Â And we do need to specify the moment of inertia specifically, let's work with

Â double bromine 79, bromine gas. So, if you go and you measure with an

Â infrared spectrometer. Actually, you can't use infrared, you'd

Â have to use ROM in here, but that's the subject of a different class.

Â In any case, the vibrational frequencies can be determined.

Â And they are 2230 wavenumbers for molecular nitrogen, and only 323

Â wavenumbers for molecular bromine. So, a much higher energy stretch for N2,

Â with its triple bond, than for bromine with its single bond.

Â And if you remember that the vibrational temperature is Planck's constant, times

Â the frequency, divided by Boltzmann's constant, and then noting that h nu,

Â Planck's constant times vibrational frequency, that has units of energy.

Â And wavenumbers is already in energy. We're expressing the vibrational

Â frequency not in per second, but in energy units.

Â And so we can just look up Boltzmann's constant in wave number units, and you'll

Â discover that it's 0.695 wavenumbers per kelvin.

Â And as a result, when I plug that value in, 2230 divided by 0.695 is about 3353.

Â So, that's the vibrational temperature in Kelvin for nitrogen.

Â And 465 Kelvin for double 79 bromine. So, let's take our gases up to 500

Â Kelvin. Reasonably warm gases actually.

Â And ask the question, what's the vibrational heat capacity for nitrogen?

Â So I'm just going to substitute in then, all the things I need here.

Â R is 8.314 Joules per Kelvin per mol. Theta vib I just worked, it is 3353, so

Â here it is, divided by our temperature of 500 squared And then I also need to plug

Â it into this exponential here and this exponential here.

Â So, I've just done this; 3353 and 500 appearing in the right places.

Â And if you type this into your favorite spreadsheet, or punch it into your

Â favorite calculator, you'll find that you end up with 0.459 joules per kelvin per

Â mole. And of course that's just a number, it

Â probably doesn't look terribly meaningful necessarily.

Â But let's think of that as a fraction of R, because recall that we get one half R

Â from, say, a translation or one half R from one of the, degrees of freedom of a

Â rotation. So this says we get about 0.06 R, that is

Â this is about six percent of R. And so we're only getting six percent,

Â instead of the fifty percent we'd be getting from a rotation or a

Â translational degree of freedom. So really, even at 500 Kelvin, not much

Â heat capacity associated with putting energy into higher vibrational states

Â form electrical nitrogen. So I'm going to let you play again with a

Â problem. And I'm going to ask you, in fact, to

Â take a look at double 79 bromine. And see what fraction of r it contributes

Â to the total heat capacity. Alright, hopefully that was straight

Â forward. And you determined, we've already started

Â with these data. We took the Bi rational frequencies.

Â We transformed them to vibrational frequencies.

Â We transformed them to vibrational temperatures.

Â And doing the calculations we found that molecular nitrogen has a vibrational

Â component to the molar heat capacity of 0.06 R, but molecular bromine 0.93 R, so

Â almost an entire factor of R. And as the temperature of a diatomic

Â ideal gas goes from below it's vibrational temperature to above it's

Â vibrational temperature The molar heat capacity will go from 5 halves r, so

Â that's just translations and rotations, up to 7 halves r.

Â As you get a full factor of r from that one vibrational degree or freedom.

Â So keep in mind that those are different. Vibrations can contribute up to r,

Â translations and rotations only one half r.

Â The last thing I want to do, you might recall in the very beginning video I

Â talked about how, we would need to build some tools, before we could build a

Â house. But then I showed you a few houses and we

Â did some experiments and we talked about for example, the hydrogen chloride

Â cannon. And at some point I just plugged in a

Â value of 7 halves R, for the molar heat capacity of hydrogen chloride.

Â And I said at some point, maybe we'd figure out why.

Â So now you finally see why. So because that hydrogen chloride went to

Â very high temperature as it reacted, it was a high temperature diatomic and it

Â should have had a heat capacity of close to 7 halves R.

Â And so we've got at least one hammered in our toolkit that we know how to use and

Â hopefully we're going to enjoy building some more as we go along.

Â So we are going to keep going along. And we've wrapped up monatomic ideal

Â gases and diatomic ideal gases. So perhaps you won't be surprised to hear

Â that next we're going to take a look at ideal polyatomic gases.

Â And that's going to occur in a couple of different lectures.

Â The first one will come next. Before we do that though, I think it's

Â time to take a look at another demonstration.

Â This one designed, hopefully, to cement these ideas of where the energy flows and

Â how it flows as you look at the heat capacity of diatomics.

Â This is a very quick and simple demonstration designed to give you a

Â visual memory to reinforce the concepts of energy distribution.

Â In translational, rotational, and vibrational molecular energy levels.

Â The tubing on this large board is connected to a reservoir full of a

Â colored liquid that we may think of as representing thermal energy.

Â As we pour liquid into the system. That is, as we provide energy which

Â raises the temperature. That energy fills the translational and

Â rotational energy levels in a continuous way.

Â Now, we do know that the translational and rotational energy levels are

Â quantized, but the separations between the allowed levels are so tiny that for

Â practical purposes we can treat them as continuous.

Â And for a linear molecule, they fill up equally fast, with one half RT worth of

Â energy for each translational and rotational degree of freedom, which is a

Â total of five halves RT for a linear molecule.

Â On the right-hand side of the board, we have a schematic for vibrational energy

Â levels. Our visual aid breaks down slightly here.

Â Because really, we do not suddenly feel a vibrational energy level when the total

Â added energy hits some characteristic temperature.

Â But the take home message is to remember that the separation between vibrational

Â levels. Is such that usually we have population

Â of only the lowest, or at most lowest few levels.

Â Unlike the rotational and translational states, which are so much more dense that

Â they are nearly continuous. Our work with molecular partition

Â functions will provide the quantitative companion to this demonstration.

Â But hopefully, you've found the visual analogy, helpful.

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