Delta G=0 at equilibrium

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

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

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

Samantha Man 1L
Posts: 63
Joined: Fri Sep 28, 2018 12:22 am

Delta G=0 at equilibrium

I understand that at equilibrium, Gibbs free energy change is zero but I'm confused on the conceptual reasoning behind this. Could someone explain this to me?

Desiree1G
Posts: 62
Joined: Fri Sep 28, 2018 12:16 am
Been upvoted: 2 times

Re: Delta G=0 at equilibrium

When Gibbs free energy is equal to 0 that means that the reaction is at equilibrium meaning that no more work can be done.

Ethan Baurle 1A
Posts: 33
Joined: Fri Sep 28, 2018 12:19 am

Re: Delta G=0 at equilibrium

Gibbs free energy is the energy free to do work. At equilibrium, no net energy is being gained or lost (nothing is really changing). Thus, at equilibrium, Delta G is 0.

Bruce Chen 2H
Posts: 60
Joined: Fri Sep 28, 2018 12:26 am

Re: Delta G=0 at equilibrium

Gibbs free energy is the free energy in a system. When it is 0, there can be no gain or loss of energy.

michelle
Posts: 62
Joined: Fri Sep 28, 2018 12:23 am

Re: Delta G=0 at equilibrium

The total delta G of a reaction equals the total delta G of products minus that of all the reactants. At equilibrium, there is no difference between the delta G of the reactants and the products, so that total delta G equals 0.