Boltzmann Equation for Entropy:

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Emilie Hoffman 1E
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Joined: Fri Sep 29, 2017 7:04 am


Postby Emilie Hoffman 1E » Sat Feb 10, 2018 9:37 pm

Can anyone explain how they got to the conclusion that the cis molecule has more available states than the trans version of the molecule?

Considering positional disorder, would you expect a crystal of octahedral cis-MX2Y4 to have the same, higher, or lower residual entropy than the corresponding trans isomer? Explain your conclusion.

Jakob von Morgenland 2C
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Re: 9.75

Postby Jakob von Morgenland 2C » Sat Feb 10, 2018 10:31 pm

It's based on the degeneracy of the two respective molecules. The trans molecule has less possible equivalent energy states/positions of its elements than the cis model. If you draw out the two molecules and move the elements around while preserving their cis/trans conditions, you can see the cis has many more states than the trans.

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