## Molar Entropy

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Joined: Wed Sep 21, 2016 2:58 pm

### Molar Entropy

Why is it that on Quiz 1 question 1 CO has a greater Molar entropy than N2. Is it based on the structure of the molecule or is it based on the molar mass of the molecule. Since in the 2013 Winter Midterm question 5B where it also asks about the molar entropy of CHF3(g), CF4(g), CH3F(g), CH2F2(g) the order was CH3F(g)<CH2F2(g)<CHF3(g)<CF4(g). Is it really based on Molar mass even though CO and N2 both have a molar mass of 28 g/mol. So if the molecules were CO and S2 would S2 have a greater molar entropy.

Christopher Reed 1H
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Joined: Wed Sep 21, 2016 2:57 pm
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### Re: Molar Entropy

Hi!

I have looked into your question as I was wondering the same thing as well.

For starters, in the quiz it mentions that T=0K so we know that nothing is moving/vibrating. When T=0K I believe (someone correct me if I'm wrong) that we are only to look at the structure because nothing is moving. A way of thinking about it is if our two molecules were happily bouncing about and then they were to suddenly freeze, how many different positions could we catch them in? For N2, they would all look the same which is why it has a lower molar entropy.

I think that when we are looking at crystals at T=0K we only consider S=KblnW for determining molar entropy.

When looking at standard molar entropy at 25C we look at molecular weight because particles start to move around and the more particles you have the more "randomness" one has.

Applying this to your example CO > S2 in terms of molar entropy at T=0K (crystal), but CO < S2 in terms of STANDARD molar entropy.

I hope this helps.

Posts: 14
Joined: Wed Sep 21, 2016 2:58 pm

### Re: Molar Entropy

Oh okay thank you that definitely clarifies it. So follow up question, you know how you said that was based on the structure of it, does having the same geometry not mean they have the same structure, therefore wouldn't that be the same? Or am i going off somewhere else here?

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### Re: Molar Entropy

Christopher is right about the residual entropy vs. the standard molar entropy. Since the question mentioned 0 K, we know that we have to use the S = Kb * lnW equation because at this temperature, we are looking at the residual entropy, which has to do with degeneracy (W). Degeneracy is essentially the number of microstates that a molecule or compound can exist in; in other words, the number of different shapes/configurations that the molecule can have. N2 does not have as many different configurations has CO because N2 is made up of the same atoms, so there's really only one microstate/configuration to exist in. Since S is directly proportional to W at 0 K, when W is higher, S is higher; thus making CO have the higher entropy.