Question

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

Answer 1

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`.