## 11.111

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

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

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

Lucia H 2L
Posts: 43
Joined: Mon Nov 14, 2016 3:00 am
Been upvoted: 1 time

### 11.111

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 rst reaction is 200. kJ mol 1, what is the standard Gibbs free energy of the second reaction?

So k2 = k1/10
using delta G = -RTlnk, k1 = e^(200/RT)
plugging in the value for k2, delta G = -RTln(1/10(e^200/RT))
How do you calculate this without a given temp?

Andres Reynoso 1J
Posts: 30
Joined: Fri Sep 29, 2017 7:06 am

### Re: 11.111

Because the question asks for the standard Gibbs free energy of the second reaction, we assume that the temperature the reaction occurred at is 25°C since that is the accepted temperature for standard conditions.