## 5G.17 HW

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

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

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

Hannah Lee 2F
Posts: 117
Joined: Thu Jul 11, 2019 12:15 am

### 5G.17 HW

Depict the progress of the reaction graphically (as in Fig. 5G.1) for the reaction in Exercise 5G.13.

(The question for 5G.13 is:(a) Calculate the reaction Gibbs free energy of I2(g) S 2 I(g) at 1200. K (K 5 6.8) when the partial pressures of I2 and I are 0.13 bar and 0.98 bar, respectively.)

Can someone explain the graph to me? Why is I listed as the reactant and I2 as the product in the solutions manual - is it the reverse rxn?

Alex Chen 2L
Posts: 86
Joined: Wed Sep 18, 2019 12:21 am

### Re: 5G.17 HW

Yes, the reaction is proceeding in the reverse direction. The Gibbs free energy change for the forward reaction is positive, so the reverse reaction will proceed spontaneously to form more I2 in order to reach equilibrium. In the reverse direction, I2 is the product and I is the reactant.

705302428
Posts: 50
Joined: Sat Aug 24, 2019 12:16 am

### Re: 5G.17 HW

The reaction is going in the reverse direction.