9.25 6th edition

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Kathryn 1F
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Joined: Fri Sep 28, 2018 12:19 am

9.25 6th edition

Postby Kathryn 1F » Tue Feb 12, 2019 9:54 am

The question is:
If SO2F2 adopts a positionally disordered arrangement in its crystal form, what might its residual molar entropy be?

How would you so this? I assume use the formula S=kln(W) but I don't know how to find W in this case

kamalkolluri
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Re: 9.25 6th edition

Postby kamalkolluri » Tue Feb 12, 2019 10:23 am

To find W, you would draw out possible configurations for SO2F2 and get a value of 6. You then simply plug it into the equation getting S = kb*ln(6)

Grace Kim 1J
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Re: 9.25 6th edition

Postby Grace Kim 1J » Tue Feb 12, 2019 11:26 am

SO2F2 has six possible configurations, which means that the degeneracy of this molecule is W=6. The change in entropy in this instance is equal to Kb(ln(W^Na)), where Kb=Boltzmann's constant, W= degeneracy, and Na=the number of particles within the molecule. To find Na, you take the one mole of SO2F2 and multiply that by Avogadro's number (6.02*10^23), to calculate how many particles are present. After plugging everything in, you should get 14.9 J/K.

Hope this helped!

Kathryn 1F
Posts: 66
Joined: Fri Sep 28, 2018 12:19 am

Re: 9.25 6th edition

Postby Kathryn 1F » Tue Feb 12, 2019 5:24 pm

I just worked through this one and thought to note,
because of log rules S=kbln(W^Na) is the same as S=Na*kbln(W)
It made calculating it a lot easier!

Stevin1H
Posts: 89
Joined: Fri Sep 28, 2018 12:17 am

Re: 9.25 6th edition

Postby Stevin1H » Fri Feb 15, 2019 3:33 pm

To solve this problem you would have to know that W = (# of positions)^#of molecules. And since the molecule has 6 different positions for the atoms and 1 mol for SO2F2, W would be 6^6.022x10^-23. Then you substitute this W into the equation S=KblnW and find the entropy.


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