# Why does the position of equilibrium depend only on the relative Gibbs energies of the products and the reactants?

The position of equilibrium, I read, is dependent only on the relative Gibbs energies of the products and the reactants. Why is this so? Surely one must take into account the entropy of mixing of the products and the reactants, which wouldn't be dependent on the Gibbs energies, to work out the lowest point (energy wise) and find the position of equilibrium. Why, therefore, is the position of equilibrium between products and reactants only dependent on the relative gibbs energies of the products and reactants. Thanks!

• The entropy of mixing of what exactly? Commented Feb 5, 2014 at 11:16
• If $\Delta G = \Delta H - T\Delta S$, then there is a Gibbs energy of mixing that could in theory be factored into the maths. However, see tschoppi's answer. Commented Feb 5, 2014 at 12:45

## 2 Answers

Why does the position of equilibrium depend only on the relative Gibbs energies of the products and the reactants?

The short answer is that the approximate entropy of mixing is a simple function of the reaction quotient Q, so it is implicitly considered when calculating the equilibrium constant from $$\Delta_\mathrm{r} G$$.

Surely one must take into account the entropy of mixing of the products and the reactants, which wouldn't be dependent on the Gibbs energies, to work out the lowest point (energy wise) and find the position of equilibrium.

Yes, the entropy of mixing is important in describing an equilibrium and figuring out the equilibrium constant. Without the entropy of mixing, all reactions would go to completion, see Why does equilibrium exist? and Why can't a reaction go to completion?.

The Gibbs energy of reaction has an enthalpy and an entropy component:

$$\Delta_\mathrm{r} G = \Delta_\mathrm{r} H - T \Delta_\mathrm{r} S$$

The enthalpy does not depend on concentrations (or reaction quotient Q), but the entropy does:

$$\Delta_\mathrm{r} S = \Delta_\mathrm{r} S^\circ - R \ln Q$$

The term $$R \ln Q$$ is the entropy of mixing (relative to everything being at standard state, i.e. $$Q = 1$$). Having both reactants and products present results in a higher entropy than having just reactants or just products.

Where is the mixing?

There is no actual mixing, just making reactants or products, but the expression is the same. You could do a thought experiment where the reaction proceeds via an enzyme that takes up reactants from one container and puts them into another (so reactants and products would never be in the same container). In a second step, you would mix the two containers so that reactants and products are in the same container.

Equilibrium and the entropy of mixing

As a result of the concentration-dependent entropy term, the Gibbs energy of reaction changes (increases) as the reaction goes towards equilibrium. When it reaches equilibrium, it is zero. The expression for $$\Delta_\mathrm{r} G$$ is:

$$\tag{1}\Delta_\mathrm{r} G = \Delta_\mathrm{r} H - T \Delta_\mathrm{r} S$$ $$= \Delta_\mathrm{r} H^\circ - T (\Delta_\mathrm{r} S^\circ - R \ln Q)$$ $$= \Delta_\mathrm{r} G^\circ + R T \ln Q$$

At equilibrium, $$Q = K$$ (that's the special name we give to $$Q$$ for a set of concentrations that is at equilibrium) and $$\Delta_\mathrm{r} G = 0$$. If we make these substitutions in equation (1), we get:

$$\tag{2}0 = \Delta_\mathrm{r} G^\circ + R T \ln K$$

Solving for $$\Delta_\mathrm{r} G^\circ$$, we get the well-known relationship between standard Gibbs energy of reaction and the equilibrium constant:

$$\Delta_\mathrm{r} G^\circ = - R T \ln K$$

The left-hand side does not make any reference to concentration dependence (it refers to a single state, the standard state), whereas the right-hand side contains the concentration-dependent term (entropy of mixing times temperature, again compared to the standard state where Q is one).

I quote Wikipedia:

In thermodynamics the entropy of mixing is the increase in the total entropy when several initially separate systems of different composition, each in a thermodynamic state of internal equilibrium, are mixed without chemical reaction by the thermodynamic operation of removal of impermeable partition(s) between them, followed by a time for establishment of a new thermodynamic state of internal equilibrium in the new unpartitioned closed system.

It seems to me that the scenario which you describe is an already mixed solution of reactants. So you have already maximized the entropy, because for every molecule that changes according to the reaction, you get the new molecule at the same place. The overall entropy with respect to the mixing doesn't change.

Note that the position of the equilibrium is only then dependent on the relative gibbs free energies when the reaction is not kinetically hindered (when the activation energy barrier is low enough to be overcome).

• If I have reactant A and product B. At 100% A the Gibbs Energy is the gibbs energy of A. If at 100% B gibbs Energy of B. But in the middle, we have entropy of mixing A and B, causing more "disorderliness", and therefore more entropy. Also the relative enthalpies of each substance has to be taken into account. Why, therefore, does it depend only on the relative Gibbs Energies, when the entropy of mixing is not dependent on them?? Thanks for the additional detail at the bottom! :) Commented Feb 6, 2014 at 20:38
• The last paragraph does not make sense to me. When the reaction is kinetically hindered, it does not attain equilibrium, but the position of equilibrium is still defined by $\Delta_\mathrm{r} G = 0$. If you can find the appropriate catalyst, the reaction will proceed to equilibrium.
– Karsten
Commented Apr 25, 2019 at 22:06
• In this answer, there is a good explanation how the entropy of mixing relates to reactions not going to completion.
– Karsten
Commented Apr 25, 2019 at 22:07