Why is Gibbs free energy zero at equilibrium?

Jul 24, 2016

Because at equilibrium, everything that $G$ depends on is constant.

The Gibbs' free energy may have been given to you without really emphasizing that it is actually a function of temperature, pressure, and the $\text{mol}$s of involved reactants and products:

$\setminus m a t h b f \left(G = G \left(T , P , {n}_{1} , {n}_{2} , . . . , {n}_{N}\right)\right)$

When we are at equilibrium, we really mean all of the following are occurring:

• thermal equilibrium - the temperature is constant.
• pressure equilibrium - the pressure is constant.
• dynamic equilibrium - the concentration, and hence the $\text{mol}$s, of reactants and products is steady/constant (but may react if disturbed).

But when all those things are constant, the Gibbs' free energy simply isn't changing at all. If the variables on which $G$ is dependent are constant, $G$ (for a single state) is constant.

Obviously, $\Delta G = {G}_{f} - {G}_{i}$, but when $G$ is constant, ${G}_{f} = {G}_{i}$.

Therefore, $\setminus m a t h b f \left(\Delta {G}_{\text{eq}} = 0\right)$.