## Standard ∆G for a system at equilibrium

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

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

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

Xihui Yin 1I
Posts: 40
Joined: Fri Sep 29, 2017 7:06 am

### Standard ∆G for a system at equilibrium

It says that standard ∆G for a system at equilibrium=0, but why for varying values of K, representing a system at equilibrium as well, not equal to ∆G=0 itself? I.e. K>1, ∆G<0 as opposed to ∆G=0.

Xihui Yin 1I
Posts: 40
Joined: Fri Sep 29, 2017 7:06 am

### Re: Standard ∆G for a system at equilibrium

Nvm, I figured out that ∆G=0 for equilibirum but ∆G° varies for different magnitudes of K.

Ramya Lakkaraju 1B
Posts: 67
Joined: Fri Sep 29, 2017 7:03 am

### Re: Standard ∆G for a system at equilibrium

Make sure to distinguish between delta G and delta G knot since I think that is what is confusing.

Harrison Wang 1H
Posts: 50
Joined: Fri Sep 29, 2017 7:07 am

### Re: Standard ∆G for a system at equilibrium

Standard change in free energy of a system at equilibrium is not 0, but rather related to the equilibrium constant K at that temperature. If K is >1, then standard change in free energy will be -, and if K<1, then standard change in free energy is +. It is delta g or change in energy that is 0 for a system at equilibrium