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Topic: Entropy of Formation Question  (Read 2742 times)

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Offline Cooper

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Entropy of Formation Question
« on: October 23, 2014, 11:05:26 AM »
Hi,

I am having conceptual difficulty with finding the entropy of formation of urea.

See the standard molar entropies below...

[tex]C(s)+\frac{1}{2}O_2(g)+N_2(g)+2H_2(g)\rightarrow CO(NH_2)_2(s)\\\Delta S^{\phi}_m=\underset{p}\Sigma-\underset{r}\Sigma\\\Delta S^{\phi}_m=[650.0-5.74-\frac{1}{2}(205.138)-191.61-2(130.684)]JK^{-1}mol^{-1}\\\Delta S^{\phi}_m=+88.71JK^{-1}mol^{-1}[/tex]

I don't see how the entropy can be positive when you are going from three moles of gas to one mole of solid. Am I doing the math wrong?

Thanks
« Last Edit: October 23, 2014, 11:20:05 AM by Cooper »
~Cooper :)

Offline Corribus

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Re: Entropy of Formation Question
« Reply #1 on: October 23, 2014, 11:55:48 AM »
I think your intuition is right. According to NIST, the standard molar entropy of solid urea is ~104 J/K mol, not 650 J/K mol. Where did you get your value? 

http://webbook.nist.gov/cgi/cbook.cgi?ID=C57136&Mask=2#Thermo-Condensed
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Offline Cooper

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Re: Entropy of Formation Question
« Reply #2 on: October 23, 2014, 12:22:26 PM »
I think your intuition is right. According to NIST, the standard molar entropy of solid urea is ~104 J/K mol, not 650 J/K mol. Where did you get your value? 

http://webbook.nist.gov/cgi/cbook.cgi?ID=C57136&Mask=2#Thermo-Condensed

Thanks! We were just given the values for a homework problem, no idea where they came from.
~Cooper :)

Offline Corribus

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Re: Entropy of Formation Question
« Reply #3 on: October 23, 2014, 01:18:40 PM »
It's good to see you're thinking about your answers rather than chugging out numbers. This is the mark of a true budding scientist!
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

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