## Gibb's Free Energy

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

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

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

Audrey Goodman 1F
Posts: 31
Joined: Fri Sep 29, 2017 7:06 am

### Gibb's Free Energy

What is the main difference between Gibbs free energy of reaction and the standard Gibbs free energy of reaction? I get that the second is in terms of standard molar Gibbs energies, but in lecture, we said that when Gibbs free energy of reaction = 0 the reaction is at equilibrium, but when the standard free energy = 0, it's not. Why is this?

Varsha Sivaganesh 1A
Posts: 51
Joined: Thu Jul 13, 2017 3:00 am

### Re: Gibb's Free Energy

$\Delta$G° is the change in Gibbs free energy at the standard state, meaning at either 1 atm or 1 M. Also, I think that when $\Delta$G° = 0, the reaction is still at equilibrium... In the example we did in lecture with the Br2(l) turning into Br2(g), we assumed that equilibrium was when $\Delta$G° = 0, and used that to solve for temperature.

Aijun Zhang 1D
Posts: 53
Joined: Tue Oct 10, 2017 7:13 am

### Re: Gibb's Free Energy

Gibb's free energy is just "G" which is a state function, but standard free energy change is $\Delta$G°, indicating substance in a standard state.

$\Delta$G is the G at equilibrium since G is a quadratic function. Its tangent line at local minimum is $\Delta$G. And at that point, $\Delta$G = 0. But $\Delta$G° is calculated through $\Delta$G. At equlibrium, $\Delta$G°= -RTlnK. If K is equal to 1, then $\Delta$G°=0. If K is not equal to one, then $\Delta$G° will be either positive or negative. But still, $\Delta$G is 0.