Practice Midterm Winter 2013: Q5B

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William_Lee_Dis_3H
Posts: 15
Joined: Wed Sep 21, 2016 2:57 pm

Practice Midterm Winter 2013: Q5B

Postby William_Lee_Dis_3H » Sun Feb 12, 2017 3:42 pm

There was a previous response that stated that the degeneracy of configurations takes priority over masses. I am confused as to the answer for question 5b. Shouldn't CF4 have the lowest molar entropy because it only has 1 possible orientation and CH2F2 should have the highest because it has 6 orientations. I know they are all tetrahedral structures but textbook problem 9.25 had a similar molecule that is tetrahedral, SO2F2, and it had 6 orientations as well.

Xinyi_Yan_3L
Posts: 21
Joined: Wed Sep 21, 2016 2:56 pm

Re: Practice Midterm Winter 2013: Q5B

Postby Xinyi_Yan_3L » Sun Feb 12, 2017 5:21 pm

I think here "configuration" infers electron configuration rather than geometry orientation.
Clearly F has a more complex electron configuration than H. A more complex electron configuration means more possible microstates. Therefore the more F the molecule has, the larger the molar entropy.

April Z 1K
Posts: 10
Joined: Fri Jul 22, 2016 3:00 am

Re: Practice Midterm Winter 2013: Q5B

Postby April Z 1K » Sun Feb 12, 2017 5:36 pm

Another way you can look at it is the amount of "disorder," I think. CF bonds are more polar than CH bonds, so having 4 CF bonds vs. 2 makes for more "disorder" and thus larger molar entropy.

MayaKhalil_1L
Posts: 34
Joined: Wed Sep 21, 2016 2:59 pm

Re: Practice Midterm Winter 2013: Q5B

Postby MayaKhalil_1L » Sun Feb 12, 2017 7:17 pm

Ok but also there was one like this on the quiz, comparing HOCl and H2O, and HOCl had greater entropy bc you can make more arrangements (saying it had more mass was wrong). These questions are very contradictory.


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