Gibbs Energy Change and Equilibrium

Gibbs Energy Change and Equilibrium

We have already seen that the sign and magnitude of the standard Gibbs energy change, $\Delta_r G^\circ$, tells us two things: whether a reaction is spontaneous, and how much useful work it can provide. But all the reactions we considered so far were irreversible — they went to completion in one direction. Real chemical reactions often reach a point where they stop going further, because the forward and backward reactions are both happening at the same rate. That is equilibrium, and it requires a different way of thinking about Gibbs energy.

Reversible Reactions and the Criterion for Equilibrium

In thermodynamics, the word "reversible" has a very precise meaning. A reversible process is one carried out so slowly that the system is always in perfect equilibrium with its surroundings — every infinitesimal step could be reversed by an infinitesimal change in conditions. When we apply this idea to a chemical reaction like

the term "reversible" means the reaction can proceed in either direction simultaneously. A dynamic equilibrium is set up.

This creates an apparent puzzle. If the forward reaction is spontaneous, its Gibbs energy change must be negative. But the backward reaction is also happening, and if it too is spontaneous, its Gibbs energy change must also be negative. How can both be negative at the same time?

The answer is that at equilibrium, neither direction has a negative Gibbs energy change. Instead, the system has reached the lowest possible Gibbs energy it can have. If the Gibbs energy were not at a minimum, the system would spontaneously change to a configuration of lower Gibbs energy. So the condition for equilibrium is:

This is the fundamental thermodynamic criterion for equilibrium. The reaction has not stopped — forward and backward reactions are both occurring — but there is no net change in the composition of the system because the Gibbs energy is at its minimum.

Relating Standard Gibbs Energy Change to the Equilibrium Constant

We know that for a reaction, the Gibbs energy change under any conditions is related to the standard Gibbs energy change by

where $Q$ is the reaction quotient. At equilibrium, $\Delta_r G = 0$ and $Q$ becomes the equilibrium constant $K$. Substituting:

This gives the central relationship:

or equivalently, using base-10 logarithms:

This equation connects thermodynamics directly to the equilibrium constant — a quantity we can measure experimentally. It tells us that the standard Gibbs energy change determines the position of equilibrium.

Interpreting the Magnitude of K

The value of $K$ tells us how far a reaction proceeds before reaching equilibrium. Since $\Delta_r G^\circ = -RT \ln K$, a large negative $\Delta_r G^\circ$ gives a large positive $K$, meaning the reaction goes nearly to completion. A large positive $\Delta_r G^\circ$ gives a $K$ much smaller than 1, meaning very little product forms.

For strongly endothermic reactions, $\Delta_r H^\circ$ is large and positive. In such cases, $\Delta_r G^\circ$ is likely to be positive (unless the entropy change is large enough to overcome it), so $K$ will be much smaller than 1. The reaction will not produce much product.

For exothermic reactions, $\Delta_r H^\circ$ is large and negative. Here $\Delta_r G^\circ$ is likely to be large and negative too, so $K$ will be much larger than 1. Strongly exothermic reactions tend to have large equilibrium constants and can go to near completion.

But $\Delta_r G^\circ$ also depends on $\Delta_r S^\circ$. If the entropy change of the reaction is taken into account, the value of $K$ — and therefore the extent of the reaction — will be affected. A positive $\Delta_r S^\circ$ makes $\Delta_r G^\circ$ more negative (favouring products), while a negative $\Delta_r S^\circ$ makes $\Delta_r G^\circ$ less negative or even positive (favouring reactants).

Using the Relationship Between $\Delta_r G^\circ$ and $K$

The equation $\Delta_r G^\circ = -RT \ln K$ can be used in two complementary ways. …