2008 Final Question 2B






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704578485
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2008 Final Question 2B

Postby 704578485 » Tue Mar 08, 2016 3:07 pm

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?

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Re: 2008 Final Question 2B

Postby Chem_Mod » Wed Mar 09, 2016 12:03 pm

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
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Re: 2008 Final Question 2B

Postby 704578485 » Wed Mar 09, 2016 2:08 pm

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