How does change in concentration at equilibrium affects the value of K_C (equilibrium constant)?

1 Answer
May 1, 2018

Changes in concentration do not affect the value of K_c.

Explanation:

Let's say that this chemical reaction is at equilibrium, and that all reactants and products are either gaseous or aqueous:

A + B rightleftharpoons C + D

The equilibrium constant, K_c, would be represented as this:

K_c = ([C][D])/([A][B])

If the concentration of a reactant, such as A, increases, then we'd expect the value of K_c to decrease (because the denominator would increase).

However, this does not happen! When the concentration of a reactant, A, increases, what happens is that the forward reaction (which increases concentrations of C and D and decreases concentrations of A and B) will be favoured.

A + B -> C + D

So, the concentrations of C and D will increase, the concentration of A will decrease, and K_c will be kept constant.

The same thing happens when the concentration of a product is increased. Let's say that the concentration of C increased—we'd expect the value of K_c to increase (because the numerator would increase).

Again, that doesn't happen!
This is because, when the concentration of a product is increased, the backward reaction (which decreases concentrations of C and D and increases concentrations of A and B) will be favoured.

C + D -> A + B

This results in an increase in the concentration of A and B and a decrease in the concentration of C and D.
K_c is kept constant again.

This idea is reflected in Le Châtelier's Principle, which basically states that, when an equilibrium is disrupted, the position of the equilibrium (which is not K_c) will shift in the direction which reduces the effect of the disruption.

When we disrupted our equilibrium in the form of a change in concentration, either the forward or backward reactions are favoured in order to reduce the effect of that disruption and maintain K_c.