## Change in free energy for a reaction at equilibrium

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

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

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

Sam Joslyn 1G
Posts: 34
Joined: Fri Sep 28, 2018 12:20 am

### Change in free energy for a reaction at equilibrium

I am a little confused by the notes I took during Friday week 6's lecture, and was wondering if anyone could help clarify. I am confused as to whether the change in Gibbs free energy for a reaction at equilibrium is 0 or -RTln(K). And if it is the latter, how can a reaction at equilibrium have a change in energy?

Jayasuriya Senthilvelan 4I
Posts: 31
Joined: Thu Jan 10, 2019 12:17 am

### Re: Change in free energy for a reaction at equilibrium

The change in Gibbs free energy for a reaction at equilibrium is 0. When delta G is zero, it means neither the forward reaction nor the reverse reaction is more spontaneous or more favored than the other. That is, both directions are equally favored. This matches with the definition of equilibrium (both forward and reverse reactions occur at equal rates).

Sydney Tay 2B
Posts: 64
Joined: Fri Sep 28, 2018 12:20 am

### Re: Change in free energy for a reaction at equilibrium

At equilibrium deltaG is 0 when at constant temperature and pressure. The example they give in the textbook is that when the temperature is 0 C, solid water and liquid water are at an equilibrium (as seen in the heating curve because they are both present), so the deltaG is 0.

Brian Hom 2F
Posts: 60
Joined: Fri Sep 28, 2018 12:24 am

### Re: Change in free energy for a reaction at equilibrium

The change in Gibbs free energy for a reaction at equilibrium is 0 J because at equilibrium neither the forward reaction nor the reverse reaction is favored so the change in Gibbs free energy is 0. Both reactions are equally spontaneous.