Page 1 of 1

### 11.15

Posted: Tue Feb 13, 2018 11:15 am
(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?

### Re: 11.15

Posted: Tue Feb 13, 2018 11:19 am
the question states that the reaction is at 1200K

### Re: 11.15

Posted: Tue Feb 13, 2018 11:19 am
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.