To complete our understanding of systems that have achieved equilibrium, we need to look at the position of equilibrium through the lens of Gibbs energy.
A reaction is spontaneous or feasible if the total entropy of the universe decreases or there is a decrease in Gibbs energy (Gibbs energy takes into account both the enthalpy and entropy changes of a reaction which simplifies everything neatly).
But if this is the case, how do we explain a reversible reaction?
2NO2(g) ⇌ N2O4(g)
Nitrogen(IV) oxide exists in an equilibrium mixture with dinitrogen tetroxide. The reaction comes to a position of equilibrium in which there are significant amounts of both NO2 and N2O4 present, whether we start with a gas syringe full of NO2 or N2O4.
Moving to the equilibrium position in both cases must reflect a decrease in Gibbs energy.
Once we’ve reached the position of equilibrium and the minimum Gibbs energy, moving towards completion or back to a syringe of pure reactants is not spontaneous as it would involve an increase in Gibbs energy.
The actual position of equilibrium (remember that this simply describes a set of equilibrium concentrations of reactants and products in the mixture) reflects the system with the highest entropy. If we start with a gas syringe of pure NO2 and only a very small amount converts to N2O4, the fact that we now have 2 different molecules present leads to a big increase in entropy because there will be a big increase in the degree of disorder or chaos in the syringe.
And since …
ΔG = ΔH – TΔS
an increase in entropy (ΔS) leads to a decrease in Gibbs energy (note the negative TΔS term in the equation).
It turns out that the position of equilibrium actually depends only on the difference between the molar Gibbs energy of the pure reactants and products.
Imagine a reversible reaction in which reactant A forms product B:
A ⇌ B
At equilibrium, the concentration of A and B present in the mixture is described by the equilibrium constant, K
It therefore follows, with a bit of maths and a sprinkle of magic, that
If Gm(B) is less than Gm(A) then the position of equilibrium lies to the right, the product side. This means we have a greater concentration of product B in the equilibrium mixture than reactant A, so K will be greater than 1, ΔG is negative and lnK is positive.
If the position of equilibrium lies to the left, we will have a greater concentration of A in the equilibrium mixture and K will be less than 1. ΔG will be positive and lnK will be negative.
We can summarise all this very simply 😊.
For many reactions, moving to an equilibrium mixture of products and reactants leads to a lower minimum Gibbs energy than would be the case if the reaction proceeded to completion.