## 2008 Final Question 2B

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

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

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

704578485
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### 2008 Final Question 2B

This question asks if a reaction given reaction has multiple steps and the ratio of step 1's equilibrium constant to step 2's is 10/1, what is deltaG(step 2), given that deltaG(step 1) is -200kJ. The final answer is -194kJ. Since deltaG=-RTlnk and the only thing changing is lnk, I thought you could find the ratio of lnk/ln10k and just multiple -200 by this value. Why doesn't this work?

Chem_Mod
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### Re: 2008 Final Question 2B

Please include a photo of the question since most students will only have the previous final exams starting at 2011 in their course readers.

704578485
Posts: 62
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### Re: 2008 Final Question 2B

Attached is a photo of the problem
Attachments
Chem Final problem.pdf