## Gibbs free question

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

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

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

Belle Calforda3f
Posts: 67
Joined: Fri Sep 29, 2017 7:07 am

### Gibbs free question

In class Lavelle went over a problem where we subbed 0 for delta g and solved for the other variables. I know it has something to do with equilibrium, but I am really confused about why and how we were just able to put in a 0. Can someone explain? thanks

Rachel Formaker 1E
Posts: 86
Joined: Fri Sep 29, 2017 7:04 am
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### Re: Gibbs free question

At equilibrium, the free energy of the products is equal to the free energy of the reactants.
Since ΔG = Gproducts - Greactants
When Greactants = Gproducts, ΔG = 0

Therefore, at equilibrium, ΔG = 0

Shane Simon 2K
Posts: 30
Joined: Fri Sep 29, 2017 7:05 am

### Re: Gibbs free question

I believe you would also set ΔG = 0 when you want to find the range of temperatures that a reaction needs to be spontaneous. Since when ΔG < 0 the reaction is spontaneous, solving for T in the equation ΔG = ΔH - TΔS = 0 would give you the temperature needed for the reaction to be in equilibrium. Depending on the sign of ΔH and ΔS, if the temperature is lower or higher than this equilibrium temperature, the reaction would always be spontaneous. An example of using this concept is problem 9.67.