# Do maxima in Gibbs energy also correspond to equilibrium positions?

ΔrG=(∂G/∂ξ ) at constant p and constant T, where ξ is the extent of reaction.

By Second Law implication, ΔrG= 0= (∂G/∂ξ ) at equilibrium. If we consider a graph of G against ξ, this would mean we have a turning point (where first derivative=0) when ξ is that extent of reaction when equilibrium attained:

(∂G/∂ξ )= 0 when ξ= extent of reaction until equilibrium attained

So, ΔrG=0 when equilibrium attained.

However, first derivative=0 does not mean minimum point only, it can mean maximum point too! Let's say you do work until you reach Gibbs energy maxima. My question: Does a Gibbs energy maxima correspond to equilibrium state or not?

(a) If yes, doesn't this violate the Second Law which implies that Gibbs energy should be minimized whenever possible? No matter which direction you move, your Gibbs energy will always decrease. Just simply move from the maximum point and you will at least be better than your current state:

(b) If no, then it will mean that we have to move from the maximum point, but in which direction? In case the graph is a symmetrical parabola, where same dξ results in same dG, what will you observe in real life? Shift to reactants or products?

G vs. $$\xi$$ does not have a maximum

If all reactants and products are pure liquids and solids, G vs. $$\xi$$ is linear. If some of the species are in mixtures, the entropy of mixing is responsible for the "sagging" shape of the curve. If the curve has an extreme value, it will be a minimum.

Analogy to mechanics

In mechanics, this situation would be called metastability. Giving a nudge in either direction will allow the system to reach lower potential energy.

• So all G vs Xi graph do not have maximum point? I clicked the metastability hyperlink but saw there is a graph with local maximum point? – The99sLearner May 30 at 1:20
• That's a transition state, I guess. Here, you are considering the ground states of reactants and products, and your x-axis is the extent of reaction. In the graph you are referencing, the x-axis is the reaction coordinate for a single reaction. – Karsten Theis May 30 at 1:23

Does a Gibbs energy maxima correspond to equilibrium state or not?

(a) If yes, doesn't this violate the Second Law which implies that Gibbs energy should be minimized whenever possible? No matter which direction you move, your Gibbs energy will always decrease.

There are a number of central concepts in thermodynamics that are relevant here: equilibrium, spontaneity, reversibility, work, heat. The perfect example to illustrate how these concepts are related to each other is that of a voltaic (or galvanic) cell or battery. This example can also provide a good approximation to the situation you describe in your question.

When you charge a battery you are increasing its Gibbs free energy. If performed reversibly (usually this means sufficiently slowly), the electrical work done during charging equals the gain in Gibbs free energy.

Is this a violation of the second law? No, because you did work to charge the system. If you did so reversibly, work was efficiently converted into a gain in Gibbs free energy. Or, if you did not charge the battery reversibly, you did more work than was necessary and wasted some of the energy as heat.

Now what happens if you disconnect the charged battery, leaving the circuit open? Not much! The battery does not discharge$$^\dagger$$. Is the system at equilibrium at that point? Yes$$^\dagger$$, as long as the circuit remains open. You can think of a very large resistance keeping the system stable despite the higher free energy.

Finally, if you connect the leads again to close the cicruit, what will happen? The battery will discharge and return to its original state. This will occur spontaneously, driven by a decrease in the Gibbs free energy. And the battery will do work as electrons are transmitted from one to the other terminal.

$$\dagger$$ To a first approximation; batteries tend to self-discharge.