# What happens when chemical equilibrium is restored after being disturbed?

I'm aware that, according to Le Chatelier's principle, a reaction will shift its equilibrium position in order to counteract the disturbance at equilibrium. However, I don't know what that means after equilibrium is re-established.

For example, take the generic reaction

$$\ce{a A + b B <=> cC}$$

At some point, equilibrium is disturbed, causing the forward reaction to be favored. Once equilibrium is re-established, how do the concentrations of the species compare to before equilibrium was disturbed?

• $$K_{eq} = \dfrac{[C]_0^c}{[A]_0^a[B]_0^b} = \dfrac{[C]_1^c}{[A]_1^a[B]_1^b}$$ – MaxW Feb 10 '19 at 23:29

Once equilibrium is re-established, how do the concentrations of the species compare to before equilibrium was disturbed?

Let's say it is a homogeneous equilibrium. You have to consider different cases:

1. The equilibrium was disturbed by changing one of the concentrations (removing or adding a substance). You can't get back to the same set of concentrations because there are too few or too many atoms.

2. The equilibrium was disturbed by adding solvent. Again, you can't get back to the same set of concentrations (all are lower than at equilibrium, and a net reaction in one or the other direction can't increase all of them at once).

3. You disturbed the equilibrium by changing the temperature. The equilibrium constant is at a different temperature will be different, so the old set of concentrations will not satisfy the new equilibrium constant.

4. You disturbed the equilibrium by intermittently changing the temperature (or - more elaborate - by attaching a power source to an electrochemical cell, and then replacing it by a wire shortcutting the cell). In this case, the concentrations change exclusively because of the chemical reaction (no "outside" disturbance), and they can return back to the original concentrations once the disturbance is removed.

In cases 1. and 2., however, the reaction quotient Q will go back to the original one (=K). In case 3., Q will be different.

Equilibrium for a reaction means that $$\sum \nu_i \mu_i = 0$$

where $$\nu_i$$ is the stoichiometric coefficient for species "i" and $$\mu_i$$ is the chemical potential, the full (total) chemical potential, of species "i".

So, the concentrations will change until this summation equals zero. It is that simple.

There is no need to try to explain this with equilibrium constants, but if you wish to, do make sure you do it with proper ratios of activity coefficients, there is little to be gained by approximating activity coefficients with concentrations, as is often done. It leads only to headaches later in life when you don't get thrown simple ivory tower mixtures :)