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 8

This last module rounds out the course with the introduction of new state functions, namely, the Helmholtz and Gibbs free energies. The relevance of these state functions for predicting the direction of chemical processes in isothermal-isochoric and isothermal-isobaric ensembles, respectively, is derived. With the various state functions in hand, and with their respective definitions and knowledge of their so-called natural independent variables, Maxwell relations between different thermochemical properties are determined and employed to determine thermochemical quantities not readily subject to direct measurement (such as internal energy). Armed with a full thermochemical toolbox, we will explain the behavior of an elastomer (a rubber band, in this instance) as a function of temperature. Homework problems will provide you the opportunity to demonstrate mastery in the application of the above concepts. The final exam will offer you a chance to demonstrate your mastery of the entirety of the course material.

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

Chemistry

Let's wrap up this week's new material with an exploration of the pressure and

Â temperature dependence of the Gibbs free energy.

Â So I'll remind you and you can go look back at video 8.4 if you want to see the

Â derivation. That the partial derivative of the Gibbs

Â free energy with respect to pressure at constant temperature is the volume.

Â And so if I integrate delta G over a pressure change, I'll get, it's the

Â integral from P1 to P2 of the volume dP. So if I do this for a mole of an ideal

Â gas, then I'm going to get delta G over bar now to indicate a molar quantity, at

Â a given temperature T which I should specify.

Â It's going to be equal to RT where T is the temperature I'm working at, integral

Â from P1 to P2, dP over P. So that's a trivial integral to do.

Â It is RT log P2 over P1. So now, let me take the initial pressure

Â to be 1 bar. That defines P1.

Â And instead of writing P2, I'll just write P.

Â P could be any pressure. So in that case delta G over bar, so the

Â molar change in the free energy is equal to G bar at the new pressure, which I'm

Â just writing P now, not P2, minus G bar at the original pressure which we're

Â choosing to be 1 bar. And that's equal to RT log P over 1.

Â So, I've replaced P2 and P1 with the quantities I say I'm going to use.

Â So, of course one, that's a convenient thing to divide by.

Â That means I, I'm going to rewrite it looking slightly differently.

Â So I'm going to say then, that this quantity G bar at a given temperature and

Â pressure. Is equal to, so I'm going to move this

Â term over to the right-hand side, is equal to G superscript 0, where this is

Â the standard molar Gibbs free energy. That is, it is defined to be the free

Â energy of an ideal gas, this gas, behaving ideally, at one bar pressure.

Â And that depends only on the temperature. And then the other term is RT log P over

Â 1. P over 1 is just P, of course.

Â So I get plus RT log P. So if you like, this is a, this is a

Â reasonably important equation in some sense.

Â It says if I'd like to know the free energy of my gas at a given temperature,

Â and a given pressure, I might be able to go look up in a table the standard molar

Â Gibbs free energy at that temperature. And then all I have to do is know the

Â temperature and the pressure I want to take it to.

Â I just add RT log P to some number, this number G superscript 0.

Â So a very useful thing that doesn't require me to do a whole lot of

Â complicated measurements. It's, it's really quite straightforward.

Â So, let me pause here for a moment, and I'll let you work a bit with that

Â expression and think about the implications for an ideal gas.

Â Well now that we've seen the particularly simple dependence of the free energy on

Â pressure for a gas, let's take a look at the temperature dependence of the free

Â energy. So we start from G equals H minus TS.

Â And now I'd like to divide both sides by T.

Â So G over T is equal to H over T minus S. Now if I differentiate with respect to

Â temperature at constant P. I'll get, so the differentiation here is

Â pretty easy. I've got 1 over T, so when I take the

Â differential, I'll get minus 1 over T squared.

Â So minus H over T squared, plus, now I've got to differentiate enthalpy, so this is

Â a chain rule sort of differentiation, so the 1 over T stays there, but I have to

Â consider partial H partial T, the constant pressure.

Â Minus partial S partial T at constant pressure.

Â And now I've got some more things on the second line here.

Â Where did they come from? Well, if you go back and review video

Â 7.1, you'll discover that we determined that partial S partial T at constant

Â pressure is minus the heat capacity at constant pressure over T.

Â In fact, that's how we measure entropies basically, is we measure heat capacities

Â over temperature ranges, and use them to assemble entropy's third law, entropies,

Â and of course, where did this one come from?

Â Well, partial H, partial T, that's the definition of a constant pressure heat

Â capacity, and so I've got a heat capacity over T minus heat capacity over T.

Â These terms drop out. The second and third terms cancel.

Â I'm left only with this first term. That's a pretty important equation, it's

Â known as the Gibbs-Helmholtz equation. And it says, that the change in the free

Â energy divided by the temperature. With respect to a change in temperature

Â at constant pressure, will be equal to minus the enthalpy divided by T squared.

Â And if I want to think of it as a process.

Â So maybe I've got some reaction that occurs with a certain delta G.

Â And I would like to know how does the delta G for that reaction change.

Â As I change the temperature, at constant pressure, that is, it's, it's on my lab

Â bench, for instance. So I'm running the reaction at 100

Â degrees C, and I'm thinking about running it at 200 degrees C, and I want to know,

Â will it be spontaneous? Will it stop being spontaneous, and what

Â I can do is ask about delta H for that reaction.

Â Divided by T squared, and that'll be the relationship for delta G relative to

Â delta H, and so I gave the example of a reaction.

Â That's something that would go from one endpoint to another.

Â It's not just a differential form, an infinitesimal form.

Â Now there's another way that we could actually think about the temperature

Â dependence of G, and that is to recognize that.

Â If G is equal to H minus TS, and now I'll emphasize temperature dependence, so G at

Â a given temperature, H at a given temperature, S at a given temperature,

Â and let's work with molar quantities for convenience.

Â So I've got an overbar on all these symbols.

Â Well, let's take. Enthalpy at zero degrees kelvin as the

Â reference for our free energy. And why, why am I taking an enthalpy as a

Â reference for a free energy? Well if, if I'm at 0 degrees kelvin H

Â minus TS is H minus 0 times entropy and so G is equal to H at 0 degrees Kelvin.

Â So I can write GT minus H of 0 is equal to HT minus H of 0 minus TS.

Â And I've already gone through, you can go review video 5.8 if you want to see the

Â details. How do we get at this enthalpy change,

Â going from zero kelvin to some non zero temperature?

Â And remember, it's the integral over the constant pressure heat capacity for the

Â solid phase, plus the phase-change enthalpy, and then the liquid phase, plus

Â the vaporization enthalpy, plus, if we're, if we make it all the way up to a

Â gas, plus heat capacity integrated up to the temperature of interest.

Â And we have also looked at how to get the absolute entropy, the third law entropy.

Â Similar procedure, we begin integrating from zero kelvin, heat capacity divided

Â by temperature up to a phase change, then the liquid up to a phase change.

Â There may be multiple solids, by the way, there could be other phase changes in

Â there. But in any case we have a way, and video

Â 7.3 was where we did this for entropy, we have a way to assemble the numbers needed

Â to add together to answer the question, how much free energy is there compared to

Â at zero kelvin? So, let's just look at an example, and in

Â particular, let's take benzene as an example.

Â And we'll look at the free energy at 1 bar pressure.

Â So remember that dG is equal to minus SdT plus VdP.

Â And I put this up here just because it's sort of a sanity check, what do we expect

Â to happen to the free energy? Well.

Â As the temperature increases, so dT, a positive value, it's getting larger,

Â multipliying a negative quantity, and the absolute entropy is a positive value, so

Â we get that G should be getting more and more negative as the temperature goes up,

Â from this term. We're not saying much about this yet, but

Â actually, let's not worry about that for a moment.

Â Okay, and so what do we see, we're at constant pressure that's why we don't

Â need to worry about that, we're at 1 bar so dP is zero.

Â And so here is, presented graphically, the free energy relative to.

Â Enthalpy at zero, which is also free energy at zero.

Â And this is the expression, and so what you see is we're not starting at absolute

Â zero. We're at about 200 and 5 kelvin or so.

Â And this is the solid phase and the free energy is going down, down, down, down,

Â down. It hits the phase transition.

Â And it's a continuous function, but it's got a different slope.

Â And so why should it be continuous at a phase transition?

Â Well, remember that a phase transition, it occurs at equilibrium.

Â Right? The temperature is staying the same, the

Â pressure is staying the same, it's a spontaneous process, so it's at

Â equilibrium. And so delta G for the transition is 0.

Â G is not changing, that's the definition of an equilibrium process that delta G is

Â equal to 0, spontaneous at equilibrium. No free energy change.

Â Goes down by more, it hits the vaporization point, continues going down

Â but with a steeper slope. What are these slopes?

Â Well, If dP is 0, I can rearrange this expression, and I get that partial G

Â partial T, holding pressure constant, is minus S.

Â So the slope here is really the entropy. And what do we expect for the entropy of

Â the gas compared to the entropy of the liquid, compared to the entropy of the

Â solid, well, there's much more disorder in a gas.

Â It ought to have a higher entropy. So the slope should be more negative, and

Â in a liquid there's less than the gas, but more than the solid.

Â And that's exactly what we see in this graph, got a very high negative slope for

Â the gas, less negative for the liquid, still less negative for the solid.

Â So this is a typical picture of free energy of a substance as it passes

Â through its phase transitions. Well, that completes the material that's

Â new for this week. The last thing we need to do before

Â getting to homework and a final exam is to review it.

Â So next video, we'll take a look at the key concepts in week 8.

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