## Gibbs Free Energy -/+

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

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

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

ClaireHW
Posts: 60
Joined: Fri Sep 29, 2017 7:07 am
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### Gibbs Free Energy -/+

Why it is more favorable for Gibbs Free Energy to be negative?
Thanks!
(Claire Woolson Dis 1K)

Nancy Le - 1F
Posts: 52
Joined: Fri Sep 29, 2017 7:07 am

### Re: Gibbs Free Energy -/+

A negative Gibbs Free Energy means that the energy of the products is less than the energy of the reactants. If it were the reverse, an input of energy would be needed, which is not favorable. Therefore, a negative Gibbs free energy is favored because it drives the reaction forward spontaneously.

Michael Cheng 1C
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Joined: Thu Jul 13, 2017 3:00 am

### Re: Gibbs Free Energy -/+

so spontaneous reaction is associated with - Gibbs Free Energy?

Anna Li 2E
Posts: 21
Joined: Sat Jul 22, 2017 3:00 am

### Re: Gibbs Free Energy -/+

For Michael Cheng, when Gibbs Free Energy is negative it is associated with a spontaneous reaction. When gibbs is negative, the system is releasing energy (exergonic), and the definition of spontaneous is when a reaction proceeds without any outside force.

Jessica_Singh_1J
Posts: 50
Joined: Fri Sep 29, 2017 7:03 am

### Re: Gibbs Free Energy -/+

A negative deltaG is favored because then the reaction will proceed spontaneously and exothermically, releasing heat and increasing the entropy of its surroundings. Reactions that increase entropy are usually more favorable than reactions that require energy and do the opposite.