delta G at equilibrium

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

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

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

Anushi Patel 1J
Posts: 61
Joined: Fri Sep 28, 2018 12:19 am
Been upvoted: 1 time

delta G at equilibrium

Why is delta G always 0 at equilibrium? I think I vaguely understand this but I would appreciate if someone explained it.

Marcela Udave 1F
Posts: 66
Joined: Fri Sep 28, 2018 12:16 am

Re: delta G at equilibrium

Gibbs free energy measures how much potential a reaction has left. Hence if the free energy is zero, then the reaction is at equilibrium, and no more work can be done.

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