## 11.111 [ENDORSED]

$\Delta G^{\circ}= \Delta H^{\circ} - T \Delta S^{\circ}$

$\Delta G^{\circ}= -RT\ln K$

$\Delta G^{\circ}= \sum \Delta G_{f}^{\circ}(products) - \sum \Delta G_{f}^{\circ}(reactants)$

ZoeHahn1J
Posts: 63
Joined: Sat Jul 22, 2017 3:01 am
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### 11.111

When doing 11.111, I used dG of rxn 1 to find K of reaction one, multiplied K by 10 to get K of rxn 2, and then used that to find dG of reaction 2 (this gave the incorrect answer). However, I'm not sure of how to solve this even looking at the solutions manual; why would we add the two delta G values together in e^(deltaG1 + deltaG2 /RT) then solve for dG?
I hope this is clear! Thank you so much!

Jessica Lutz 2E
Posts: 56
Joined: Fri Sep 29, 2017 7:04 am

### Re: 11.111

I thought about it the same way that you did but, because the first reaction is ten times the second reaction, you would actually divide (not multiply) the first K value by ten to find the second K value. Then you can solve for dG and you should get the right answer.

ZoeHahn1J
Posts: 63
Joined: Sat Jul 22, 2017 3:01 am
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### Re: 11.111

Oh! Thank you so much!

Nhan Nguyen 2F
Posts: 52
Joined: Thu Jul 13, 2017 3:00 am
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### Re: 11.111  [ENDORSED]

It's actually 10= e^-(deltaG1)/RT divided by e^- (deltaG2 /RT) because 10=k1/k2
When e^x/ e^y, it becomes e^(x-y)

so then it's e^(-deltaG1 + deltaG2 /RT)