## 11.15

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

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

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

Ryan Fang 1D
Posts: 53
Joined: Sat Jul 22, 2017 3:00 am

### 11.15

(a) Calculate the reaction Gibbs free energy of I2(g) -> 2 I(g) at 1200. K (K =6.8) when the partial pressures of I2 and I are 0.13 bar and 0.98 bar, respectively. (b) What is the spontaneous direction of the reaction? Explain briefly.

For this question, why does the solutions manual calculate ∆G° using 1200. K instead of 298 K? Since ∆G° is the Gibbs Free Energy of the reaction at standard state, shouldn't we use 298 K?
Last edited by Ryan Fang 1D on Tue Feb 13, 2018 11:31 am, edited 1 time in total.

Kelly Kiremidjian 1C
Posts: 62
Joined: Fri Sep 29, 2017 7:04 am

### Re: 11.15

the question states that the reaction is at 1200K

Britney Alvey 1B
Posts: 31
Joined: Fri Jun 23, 2017 11:39 am

### Re: 11.15

The question states that the reaction is happening at 1200K so you would use this value to calculate deltaG. The solutions manual calculates delta G not $\Delta G^{\circ}$. This is an important distinction because $\Delta G^{\circ}$ would be calculated at 298K by definition.