## Hmwk 9.83

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

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

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

kshalbi
Posts: 68
Joined: Fri Sep 25, 2015 3:00 am

### Hmwk 9.83

Hi. So, this question asks you to find thr standard Gibbs free energy of H2 (g) +Br2 (g)--> 2HBr (g) at 298K. Why can't you just answer it by giving the Gibbs free energy of formation for HBr (g)? Instead you have to subtract the energy of the products from that of the reactants.

kshalbi
Posts: 68
Joined: Fri Sep 25, 2015 3:00 am

### Re: Hmwk 9.83

Jk. Br(g) isnt the pure state. The liquid form is pure.