9.75  [ENDORSED]

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Rakhi Ratanjee 1D
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Joined: Fri Sep 29, 2017 7:04 am

9.75

Postby Rakhi Ratanjee 1D » Fri Feb 09, 2018 11:56 pm

Question 9.75 on the homework asks to explain whether a crystal of octahedral cis-MX2Y4 would have a higher molar entropy from the trans isomer and then explain why or why not.The answer calculated the actual molar entropies for both isomers and then compared them to find out that the cis isomer had a higher molar entropy. It says that the cis compound has 12 different orientations and the trans compound only has 3 orientations. How do you find these values?

Joanne Guan 1B
Posts: 30
Joined: Sat Jul 22, 2017 3:01 am

Re: 9.75  [ENDORSED]

Postby Joanne Guan 1B » Sat Feb 10, 2018 11:13 am

I'm not sure if there's any mathematical way of discerning the number of microstates there are, but I usually just visualize and count the different orientations a molecule can have. To explain, imagine cis-MX2Y4 replacing all the X atoms with Y atoms. There's only one microstate because M must remain in the center and even if you shuffle the Y atoms around, the orientation is still the same. Now, imagine 1 X atom with 5 Y atoms. There are 6 microstates because the X atom can be positioned at any of the Y atom spots. Now, to find the microstates for cis-MX2Y4, the XMX section of the molecule must remain in a right angle shape. If you can imagine the XYZ axes, there are four possibilities on the XZ plane, four on the XY plane, and four on the ZY plane. Therefore, there are 12 orientations.

For trans-MX2Y4, the first possibility is already given to you, and there are two on the XZ plane.


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