## 11.83

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

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

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

Virpal Gill 1B
Posts: 32
Joined: Thu Jul 27, 2017 3:00 am

### 11.83

The problem states to calculate the equilibrium constant at 25 C and at 150 C. The solutions manual states that because the equilibrium constant is at 2 different temperatures we need to find the standard enthalpy of the reaction and the standard entropy of the reaction so the standard Gibbs free energy can be calculated. Can we just find the standard Gibbs free energy using the standard Gibbs free energies of formation or can that not be done because of the temperature change?

Lena Nguyen 2H
Posts: 51
Joined: Fri Sep 29, 2017 7:06 am
Been upvoted: 1 time

### Re: 11.83

It's correct that using the standard Gibbs free energy of formations wouldn't work for 150 C because Gibbs free energy is sensitive to temperature changes. Since the standard enthalpy of reaction and standard entropy of reaction are mostly independent of temperature, they are the same at both temperatures.

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