equilibrium - Gibbs free energy-minimum or zero?

A reaction proceeds towards the direction of lesser Gibbs free energy (at constant $T$ (temperature) and $P$ (pressure)). So, we could say that Gibbs free energy at equilibrium is minimum.

On the other hand, we have

$$\Delta G=\Delta G^\circ + RT\ln Q$$ , where $Q$ is the reaction quotient. At equilibrium, $Q=K_\text{eq}$, and we already know that $\Delta G^\circ =-RT\ln K_\text{eq}$.

Substituting, we get $\Delta G=0$ at equilibrium. But, we know that $G$ minimized itself—thus there was a change in $G$ and $\Delta G < 0$.

What am I missing here?

Answer

I think your question really arises from some confusion about what $\Delta G$ represents. In general, $\Delta X$ for a thermodynamic quantity $X$ is the change of $X$ along some process. You could make it clear by actually writing $\Delta G(\text{A}\rightarrow\text{B})$ where A and B are before and after states. (We'll note that, in the general case, $\Delta X$ depends on the path take from A to B, making this notation improper. If $X$ is a function of state, though, you're good to go.)

However, in the equation you quote:

$$\Delta G = \Delta G^0 + RT \ln Q$$

the $\Delta G$ is a free energy of reaction and should thus be denoted $\Delta_r G$, with the correct equation being:

$$\Delta_r G = \Delta_r G^0 + RT \ln Q$$

The free energy of reaction is defined as $\Delta_r G = G_{\text{products}} - G_{\text{reactants}}$.

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Thus, this $\Delta_r G$ is not the variation of $G$ over the entire reaction, which would be the $\Delta G$ of the system between the start of the reaction and the equilibrium.

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PS: I think this link is the online resource I found with the clearer use and explanation of notations. Notations are important in thermodynamics!