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Topic: Molar fraction and equilibrium constant - simple problem  (Read 1722 times)

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Offline Peder

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Molar fraction and equilibrium constant - simple problem
« on: September 06, 2014, 08:43:24 AM »
Hello, first time poster here. Hope I didn't mess up with the code.

I ran into this problem when working on a lab project concerning the differences in energy for different cyclohexane substituent conformations.

I need to find out the molar fractions for axial and equatorial conformations. The equation used is

[tex]ΔG°=-RTlnK[/tex] where [tex]K=X(ax)/X(eq)[/tex]   ax is for the axial confromation and eq for the equatorial.

To find out the X(ax) fraction, I need to bring the [tex]K=x(ax)/x(eq)[/tex] in to this form: [tex] x(ax)= K/1+K [/tex]

I'm sure that the mathematics involved are rather easy, but I just can't seem to figure it out. Any help would be appreciated.
« Last Edit: September 06, 2014, 08:56:56 AM by Peder »

Offline Borek

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Re: Molar fraction and equilibrium constant - simple problem
« Reply #1 on: September 06, 2014, 04:20:30 PM »
xax + xeq = ...
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