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 7

This module is relatively light, so if you've fallen a bit behind, you will possibly have the opportunity to catch up again. We examine the concept of the standard entropy made possible by the Third Law of Thermodynamics. The measurement of Third Law entropies from constant pressure heat capacities is explained and is compared for gases to values computed directly from molecular partition functions. The additivity of standard entropies is exploited to compute entropic changes for general chemical changes. 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

Alright, let's take advantage of the state function character of entropy and

Â use it to examine additivity of entropies.

Â So, entropy, much like entalpy, where we studied this property in some detail, is

Â additive when it comes to reactions. That is, the entropy of reaction can be

Â defined as, if I have a moles of reactant A plus little b moles of reactant B,

Â going to y moles of product Y plus z moles of product Z, then the entropy of

Â reactions, so an r subscript. Is equal to, and I'll use standard

Â entropies. Is equal to the entropy of the products,

Â which is going to be a molar times a kind of, number of moles for instance, minus

Â the entropy of the reactant, so that pretty straightforward.

Â lets actually do a couple of examples before we really do anything with

Â numbers. But let me give you a chance to take a

Â look at the first example and make a prediction about sign.

Â Okay. So, we're going to burn some hydrogen.

Â We're going to make a little water. One of the greener combustion reactions.

Â We make nice green water. So, here's the general expression then

Â that we have for the entropy of a reaction.

Â And here's the reaction we're interested in, gaseous hydrogen plus a half a mole

Â of gaseous oxygen, or half an equivalent I guess we could say, goes to liquid

Â water. So that requires that we go and look up

Â the standard entropies for hydrogen, oxygen and water.

Â And I've actually had those on the well we certainly had water in the last video

Â I, we didn't have hydrogen and oxygen but in any case, one can look them up in

Â tables and here they are. So the entropy of the product water is

Â 70.0. It's the only product, so it's the only

Â term appearing with a positive symbol in front of it.

Â Remember of course, the entropy itself must be positive, unless we're at

Â absolute zero where a perfect crystal can be zero.

Â And now we subtract the entropy of the two reactants.

Â So, minus 130.7, that's the entropy of molecular Hydrogen.

Â And minus 1 half, 205.2. And it's always good, every time you look

Â at equations like this to exercise your intuition and say, yes, the entropy of

Â oxygen is higher than that of hydrogen. They're both diatomics, but the mass of

Â oxygen is much greater than hydrogen, and it's moment of inertia is greater.

Â So, it oughtta have more entropy, and it does.

Â Well if I do the arithmetic, 70 minus 137 minus a half of this.

Â I get minus 163.3 joules per kelvin per mole.

Â So that is a very negative entropy change, which is to say that order

Â increased. Order increased, disorder decreased.

Â And does that make sense? Well yes, we took a mole and a half of

Â gas. And gas is very disordered.

Â Molecules flying around from a big volume.

Â And we condensed it into a single mole. So we had a one and half moles of

Â something, now we only have one mole of something.

Â And moreover we made it a liquid. So it's interacting with itself, it's

Â compressed, there's been a, a huge, loss of disorder, or increase in order.

Â Okay, well that was one example. Let's again do a little exercise here.

Â I'll offer you another example. You think about the sign of the entropy

Â change, and then we'll actually do the quantitative calculation.

Â So the reaction we're looking at this time has a name.

Â It's called the Water Shift Reaction and it involves taking extremely hot carbon,

Â so it could be graphite, for example, or coal, and treating it with water.

Â And at very high temperatures that makes gaseous hydrogen and gaseous carbon

Â monoxide. Both of these gasses are very useful

Â industrially. So, once again, to compute the entropy of

Â a reaction, we want to take the entropy of the products, hydrogen gas and carbon

Â monoxide. And they are 130.7.

Â We used that one when we burned it. 197.7 for carbon monoxide.

Â Once again your intuition should have said yes of course.

Â Bigger. Bigger mass than H2.

Â They're both diatomics but more mass for CO, more moment of inertia.

Â And then we subtract the entropy of carbon as, I've use the graphite value

Â here 5.7. Again that came from a prior video.

Â So that's a very small entropy. And gaseous water minus 188.8.

Â When you do the arithmetic, you get 133.9 Joules per kelvin per mol.

Â So this is a very large positive entropy of reaction.

Â And that's because, we've basically taken one mol of a gas, and a very ordered

Â solid. And we've made two moles of products.

Â So, two moles of, original material. Two moles of products.

Â But the products are both gases where one of the reactants was a solid.

Â So, in a sense, most of the entropy gained is achieved by taking a highly

Â ordered solid, and giving its atoms to something that became a very disordered

Â gas. So the entropy change makes intuitive

Â sense. Well, we've come to the end of the new

Â material associated with the third law of thermodynamics in this seventh week of

Â the course. All that's left is to do a quick recap of

Â the most important concepts and so let's do that next.

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