## Problem 11.111

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

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

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

Varsha Swamy 2J
Posts: 14
Joined: Sat Jul 22, 2017 3:00 am

### Problem 11.111

Problem 11.111 is as follows:

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, what is the standard Gibbs free energy of the second reaction?

I set it up using the equation for standard Gibbs free energy and using properties of logs and substituting 10K2 for K1 and ended up here:

standard Gibbs free energy of second reaction = -RT ln(10K1) = -200 + RT ln(10)

However, I don't know what to do from here as the problem does not give the temperature, but the solutions at the back of the book have the answer -194. kJ.

Andrea Grigsby 1I
Posts: 60
Joined: Fri Sep 29, 2017 7:03 am

### Re: Problem 11.111

use the equation deltaG = -RTlnK
and rearrange
lnK= - 200000/(8.314*298)
= 80.7
K = 1.14*10^35
divide by 10 so K2 = 1.14*10^34
and then use it in the same equation as above
deltaG2 = -8.314*298*ln1.14*10^34
= -194 kJ

Varsha Swamy 2J
Posts: 14
Joined: Sat Jul 22, 2017 3:00 am

### Re: Problem 11.111

Got it, thanks for the explanation! I didn't know to use 298K for the temperature when doing the homework problem. Plugging that in for T in the expression I got leads to the same answer of -194 kJ/mol.