Question 11.111 (Sixth Edition)

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Steve Magana 2I
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Joined: Fri Sep 28, 2018 12:24 am

Question 11.111 (Sixth Edition)

Postby Steve Magana 2I » Sun Mar 03, 2019 1:23 am

Question: A certain enzyme-catalyzed reaction in a biochemical cycle has an equilibrium constant that is 10 times the equilibrium constant of the next step in the cycle. If the standard Gibbs free energy of the first reaction is -200. kJ.mol^-1, what is the standard Gibbs free energy of the second reaction?

Can someone help me with this question, I keep getting a weird answer... Thank you!

Chris Freking 2G
Posts: 60
Joined: Fri Sep 28, 2018 12:29 am

Re: Question 11.111 (Sixth Edition)

Postby Chris Freking 2G » Sun Mar 03, 2019 9:12 am

Use the formula deltaGr⁰ = -(RT)lnK; solving for K gives K = e^-(deltaGr⁰/RT)

Solve for the first step reaction: K1 = 10K2 = e^-(G1⁰/RT) = e^(2.00*10^5/RT)

For the second step reaction: K2 = e^-(G2⁰/RT)

Combining the two steps gives the reaction 10 = e^(deltaG1⁰+deltaG2⁰/RT) = e^(2.00*10^5+deltaG2⁰/RT)

Since the reaction occurred at standard conditions (25C): 10 = e^[(2.00*10^5)+deltaG2⁰]/[(8.3145JK^-1mol^-1)(298.15K)

Solving for deltaG2⁰ gives deltaG2⁰ = -1.94 * 10^5 J = -194. kJ

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