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

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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.

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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

use it to examine additivity of entropies.

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

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

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

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

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

entropies. Is equal to the entropy of the products,

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

the entropy of the reactant, so that pretty straightforward.

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

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

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

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

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

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

that we have for the entropy of a reaction.

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

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

water. So that requires that we go and look up

the standard entropies for hydrogen, oxygen and water.

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

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

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

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

term appearing with a positive symbol in front of it.

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

absolute zero where a perfect crystal can be zero.

And now we subtract the entropy of the two reactants.

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

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

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

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

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

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

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

I get minus 163.3 joules per kelvin per mole.

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

increased. Order increased, disorder decreased.

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

gas. And gas is very disordered.

Molecules flying around from a big volume.

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

something, now we only have one mole of something.

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

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

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

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

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

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

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

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

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

monoxide. Both of these gasses are very useful

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

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

monoxide. And they are 130.7.

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

Once again your intuition should have said yes of course.

Bigger. Bigger mass than H2.

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

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

here 5.7. Again that came from a prior video.

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

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

So this is a very large positive entropy of reaction.

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

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

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

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

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

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

gas. So the entropy change makes intuitive

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

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

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

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

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