## K and Gibbs Free Energy

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

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

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

Olivia Young 1A
Posts: 60
Joined: Fri Sep 28, 2018 12:24 am

### K and Gibbs Free Energy

In lecture we discussed how if K is less than 1, then deltaG is positive, and visa versa. When K is less than one, there are more reactants than products at equilibrium, so would the reaction always shift right/favor the forward reaction when deltaG is positive?

Searra Harding 4I
Posts: 68
Joined: Fri Sep 28, 2018 12:29 am

### Re: K and Gibbs Free Energy

Yes when delta G is positive it is an endergonic reaction that favors the reactants.

Jesse Kuehn 1B
Posts: 40
Joined: Wed Nov 14, 2018 12:23 am

### Re: K and Gibbs Free Energy

I believe so because they are directly correlated, and a positive delta G means that the products have a higher G which means they have more energy available to do work and the way reactions do work is through expansion (when that side of the equation has more moles of gas or is favored). I may have made too many jumps here but I'm pretty sure this is right