## Circumstances of delta G

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

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

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

Emily Glaser 1F
Posts: 156
Joined: Thu Jul 27, 2017 3:01 am

### Circumstances of delta G

In what circumstance will delta G be zero?

When will delta S be zero?

(Please do not speak in terms of the values of enthalpy and entropy. I want a conceptual explanation, not a numerical one)

Kayla Danesh 1F
Posts: 38
Joined: Thu Jul 13, 2017 3:00 am

### Re: Circumstances of delta G

When deltaG=0, K is equal to one. This means that the reaction is at equilibrium, and the products/reactants are equally favored.

When deltaS=0, we would be talking about a perfect crystal, where there is only one possible state that each particle can be in.

Vasiliki G Dis1C
Posts: 53
Joined: Fri Sep 29, 2017 7:04 am

### Re: Circumstances of delta G

ΔG=0 when the system is at equilibrium. For instance, if ice and water in equilibrium then the Gibbs free energy of 1 mol H2O(l) will be the same as the Gibbs free energy of 1 mol H2O(s).
ΔS=0 is also equilibrium. If ΔS=0, it means that neither the forward nor the reverse reaction/process is spontaneous. It is important to remember that this is ΔStotal which is composed of the ΔSsystem and ΔSsurroundings. When S=0, we have a perfect crystal, not ΔS=0. Hope this helps!

TarynLane2J
Posts: 30
Joined: Sat Jul 22, 2017 3:00 am

### Re: Circumstances of delta G

In addition, Gibbs free energy represents energy that is free to do useful work, so when delta G is zero i guess that means that there is no change in the amount of energy that is free to do work