3.2.3 (f,g) The Equilibrium Constant, Kc

(f) expressions for the equilibrium constant, K_c, for homogeneous reactions and calculations of the equilibrium constant, Kc , from provided equilibrium concentrations

{Learners will not need to determine the units for K_c.}

(g) estimation of the position of equilibrium from the magnitude of K_c.

{A qualitative estimation only is required.}

The equilibrium constant K_c is a way to quantify (put a number to) the position of an equilibrium rather than just saying "to the left" or "on the right".

For any equilibrium the Equilibrium Constant:

1. Will have the concentrations of products on the top of the expression and reactants on the bottom

2. Has the coefficients (big numbers) of each reactant/product from the equation as the power

3. The terms are multiplied by each other rather than added as in the chemical equation.

For example:

In the reaction,

N₂ + 3 H₂ ⇌ 2NH₃

K_c = [NH₃]²

[N₂][H₂]³

If equilibrium is reached when [N₂] = 2, [H₂] = 2, and [NH₃] = 4

K_c = [4]²

[2][2]³

K_c = 16 = 1

[2][8]

If we balance the equation differently

2 N₂ + 6 H₂ ⇌4 NH₃

K_c = [NH₃]⁴

[N₂]²[H₂]⁶

But the equilibrium concentrations are still [N₂] = 2, [H₂] = 2, and [NH₃] = 4

K_c = [4]⁴

[2]²[2]⁶

K_c = 256 = 256 = 1

[4][64] 256

Take the equilibrium;

NO + 0.5 O₂ ⇌ NO₂

If the equilibrium lay far to the left the concentration of reactants would be high and the concentration of products would be low.

eg [NO] = 4, [O₂] = 4, and [NO₂] = 1

This would generate a low K_c

K_c = [NO₂]

[NO][O₂]^0.5

K_c = 1

4 x [4]^0.5

K_c = 1

8

If the equilibrium moves to the right the concentration of reactants would be low and the concentration of products would be high.

eg [NO] = 1, [O₂] = 2.5, and [NO₂] = 4

This would generate a high K_c

K_c = [NO₂]

[NO][O₂]^0.5

K_c = 4

1 x [2.5]^0.5

K_c = 2.5

So, if an examiner gives you enough hints to suggest that the equilibrium moves right you can predict a higher K_c.

Or, if (s)he tells you the K_c is decreasing then you should know that the equilibrium is moving left.

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