The Equilibrium Constant

The equilibirium constant, K

The size of the equilibrium constant, K,tells us how the equilibrium mixture is made up with respect to reactants and products

$K = \frac{{[products]}_{eqm}}{{[reactants]}_{eqm}}$

If K > 1, the concentration of products is greater than the concentration of reactants and we say that the equilibrium lies to the right hand side
- When K >> 1, equilibrium lies far over to the right hand side and the reaction almost goes to completion
If K < 1, then the concentration of reactants is greater than the concentration of products and we say that the equilibrium lies to the left hand side
- When K << 1, equilibrium lies far over to the left hand side and the reaction hardly proceeds
When K = 1, at equilibrium, there are significant amounts of both reactants and products and equilibrium does not lie in favour of either the reactants or products
K is a constant at a specified temperature
Since temperature can affect the position of equilibrium, it follows that K is dependent on temperature

Worked example

When the following reactions reach equilibrium, state whether the equilibrium mixture contains mostly reactants or products. Assume the value of Kcorresponds to the temperature of the reaction mixture

1.	Ag⁺ (aq) + Fe²⁺ (aq) ⇌ Ag (s) + Fe³⁺ (aq)	K = 7.3 x 10^-26
2.	N₂ (g) + 3H₂ (g) ⇌ 2NH₃ (g)	K = 2.6 x 10^-18
3.	2SO₂ (g) + O₂ (g) ⇌ 2SO₃ (g)	K = 5.0 x 10¹³

Answer:

Reactions 1 and 2:
- Kis very much smaller than 1
- So, the denominator in the equilibrium constant expression must be much larger than the numerator
- This means that the concentration of the reactants is much larger than the concentration of products
- Therefore, the equilibrium lies far to the left and the equilibrium mixture contains mostly reactants
Reaction 3:
- Kis very much larger than 1
- So, the numerator in the equilibrium constant expression must be much larger than the denominator
- This means that the concentration of the products is much larger than the concentration of reactants
- Therefore, the equilibrium lies to the right-hand side and the reaction mixture contains mostly products

Examiner Tip

Stronger acids dissociate more than weaker acids in solution, meaning that equilibrium lies towards the products
So, stronger acids will have a higher value of K than weaker acids.

The relationship between K values for reactions that are the reverse of each other

The equilibrium constant expression is dependent on a specific reaction
For example, take the reaction between nitrogen and hydrogen to make ammonia:

N_2(g) + 3H_2(g) ⇌ 2NH_3(g)

The equilibrium constant expression for this reaction is:

$K = \frac{{[{NH}_{3}]}^{2}}{[N_{2}] {[H_{2}]}^{3}}$

If we reverse the equation:

2NH_3(g) ⇌ N_2(g) + 3H_2(g)

The equilibrium constant expression for the reverse of this reaction, K', is:

$K' = \frac{[N_{2}] {[H_{2}]}^{3}}{{[{NH}_{3}]}^{2}}$

What is the relationship between these two K values? At the same temperature, K' becomes the reciprocal of the original Kvalue:

$K' = \frac{1}{K}$ or $K' = K^{- 1}$

Worked example

The equilibrium constant for the following reaction is 7.1 × 10³².

2NO₂ (g) + F₂ (g) ⇌ 2NO₂F (g)

What is the equilibrium constant for the reverse at the same temperature?

Answer:

$K_{(reverse)} = \frac{1}{K_{(forward)}} = \frac{1}{7.1 \times 10^{32}} = 1.41 \times 10^{- 33}$

The Equilibrium Constant (SL IB Chemistry)

Revision Note