## 5G. 13

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

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

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

JamieVu_2C
Posts: 108
Joined: Thu Jul 25, 2019 12:16 am

### 5G. 13

(a) Calculate the reaction Gibbs free energy of I2(g) -->2I(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) Indicate whether this reaction mixture is likely to form reactants, is likely to form products, or is at equilibrium.

In the solutions manual, instead of finding $\Delta G^{o}$ using the free energies of the reactants and products, it just substituted in -RTlnK for it. So if the problem gives us the value of K (K=6.8), then can we just automatically assume that it wants us to calculate the reaction Gibbs free energy when the reaction is at equilibrium?

Goyama_2A
Posts: 107
Joined: Sat Aug 24, 2019 12:17 am

### Re: 5G. 13

The standard Gibbs of formation should only be used to calculate the delta G standard of a reaction at standard conditions (25C and 1 atm). If equilibrium isn't at those conditions, you should be using the given K to calculate the standard delta G of the reaction. In this case, they give you K so that you can calculate the standard delta Gibbs free energy and then use that to calculate the current delta G.

jisulee1C
Posts: 149
Joined: Thu Jul 25, 2019 12:17 am

### Re: 5G. 13

Since the problem is asking to calculate the reaction Gibbs Free energy at 1200K and standard conditions are at 298K the given K is at standard condition. Since the reaction is not at standard condition to calculate the total Gibbs Free energy of the reaction you have to calculate both the standard Gibbs free energy using the given K value and -RTlnK. Plug into the equation Gibbs free energy (reaction) = standard Gibbs free energy + RTlnQ to find the total Gibbs free energy of the reaction that is not under standard conditions.