Problem 11.111-Sixth Edition






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Kate_Santoso_4F
Posts: 72
Joined: Fri Sep 28, 2018 12:29 am

Problem 11.111-Sixth Edition

Postby Kate_Santoso_4F » Fri Feb 22, 2019 9:02 am

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/mole, what is the standard Gibbs free energy of the second reaction?
Can someone please correctly explain how to set up this problem? I keep getting different answers

Griffin Carter 2I
Posts: 16
Joined: Wed Nov 14, 2018 12:23 am

Re: Problem 11.111-Sixth Edition

Postby Griffin Carter 2I » Fri Feb 22, 2019 11:43 am

So from the problem we can see that K1 = 10*K2. We use the equation Delta Gr = -RT lnK (i will now refer to Delta Gr as G1 or G2 for convenience purposes), so K = e^-(G/RT).
K1 = 10*K2 = e^-(G1/RT) and K2 = e^-(G2/RT). From here you can divide the first equation (K1) by the second equation (K2) to isolate the unknown G2 variable and solve from there.


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