## Gibbs Free Energy

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

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

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

Heidi Ibarra Castillo 1D
Posts: 60
Joined: Fri Sep 28, 2018 12:19 am

### Gibbs Free Energy

When delta G is equal to 0 it means it is at equilibrium?

skyeblee2F
Posts: 62
Joined: Fri Sep 28, 2018 12:23 am

### Re: Gibbs Free Energy

haha yea

Jessica Castro 2H
Posts: 60
Joined: Fri Sep 28, 2018 12:29 am

### Re: Gibbs Free Energy

Yes, a equilibrium, since both the forward and reverse reactions are occurring at the same rate, K = 1. Therefore, looking at the equation ΔG°= =RTlnK, plugging in K=1 would result in ΔG°= 0. Thinking at it conceptually, since both the forward and reverse reactions are occurring at the same rate, Gr = Gp and therefore no work is involved.