## Equilibrium

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

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

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

Eileen Si 1G
Posts: 120
Joined: Fri Aug 30, 2019 12:17 am

### Equilibrium

Do delta G, delta H, and delta S have nonzero values at equilibrium, or are they equal to 0 since there is no net change occurring due to the system being at equilibrium?

Hope Hyland 2D
Posts: 50
Joined: Wed Feb 20, 2019 12:16 am

### Re: Equilibrium

I know delta G is for sure 0 at equilibrium, since there's no change in the free energy, but I'm not sure about delta H or delta S. I don't think they would be at zero, since the reaction is still occurring (because equilibrium is dynamic).

Lauren Tanaka 1A
Posts: 109
Joined: Sat Aug 17, 2019 12:18 am

### Re: Equilibrium

At equilibrium the delta G is 0 but delta H and delta S could still be nonzero numbers.

WYacob_2C
Posts: 102
Joined: Sat Jul 20, 2019 12:16 am

### Re: Equilibrium

At equilibrium, delta G will be zero, but I don't believe that always mean that delta H and delta S will also be zero.

Jacob Motawakel
Posts: 103
Joined: Wed Sep 18, 2019 12:20 am
Been upvoted: 1 time

### Re: Equilibrium

delta G is zero at equilibrium, but delta H and delta S do not have to be zero at equilibrium.

Andres Merlos 2L
Posts: 46
Joined: Wed Sep 18, 2019 12:17 am

### Re: Equilibrium

Delta G is 0 at equilibrium, but delta S and delta H could be nonzero in order for delta G to be 0. This is as long as the equation results in 0.