Equilibrium constant (Kc) and its characteristics

Some main characteristics of equilibrium constant are described below:

• The value of the equilibrium constant of a reaction is the same, at constant 'T' and 'P'. However, if either temperature 'T' or pressure 'P' or both are changed, the value of the equilibrium constant may also change. For example, the value of K_p (equilibrium constant in terms of partial pressures) for the reaction,

at 700 K and 800 K are 1.5 x 10^-4 and 1.4 x 10^-5 respectively.

• The value of equilibrium constant of a reaction does not depend upon the concentrations of the various species involved in the reaction. For example, the value of the equilibrium constant for the reaction

is 48 at 720 K. Whatever may be the concentrations of H₂(g), I₂(g) and HI(g), the value of the equilibrium constant remains the same provided temperature 'T' and pressure 'P' are held constant.

• For a reversible reaction, the equilibrium constant for the backward reaction is inverse (reciprocal) of the equilibrium constant for the forward reaction. For example, if the equilibrium constant for the forward reaction,

is K, then the equilibrium constant for the backward reaction,

is 1/K. From these two relationships, we can show,

• The value of the equilibrium constant of a reversible reaction gives an idea about the more favoured direction of the reaction. For example, when, (i) K>> 1, the forward direction is more favoured. As a result, at equilibrium, the concentration of the products will be much higher than the concentration of the reactants. Very large value of K, e.g.,

K = 10¹⁰ means that the reaction proceeds in the forward direction to almost completion.

(ii) When, K = 1, both directions of a reversible reaction are equally favoured and the concentration of the reactants and products at equilibrium will be comparable (nearly equal).

(iii) When, K <<>

Effect of catalyst

A catalyst has no effect on the value of the equilibrium constant of a reaction. This is because, a catalyst affects the rates of forward and backward reactions to the same extent. A catalyst however, helps in attaining the state of equilibrium faster.

Stoichiometry of a chemical equation

The numerical value of equilibrium constant depends upon the stoichiometry of the chemical equation used in representing the reaction. For example, the reaction

can also be expressed as

Then, the corresponding equilibrium constants are

When the coefficients are doubled, then the K for the new reaction is square of the original 'K'. In general when we multiply the terms in an equation by a certain value, we must change the 'K' for the equation to a power equal to that value.

Thus, if n= ½, K_new = (K)^1/2 or

Concentration units

The value of the equilibrium constant also depends upon the units in which concentrations is expressed.

Problems

8. The equilibrium constant for the reaction

at 298 K is 138. What is the value of equilibrium constant for the reaction,

at the same temperature. Comment upon the values of the two equilibrium constants.

Solution

Equilibrium constant for the two reactions may be written as follows

and

From above, we have

The values of K₁ and K₂ show that the value of equilibrium constant for a reaction depends upon the stoichiometry of the chemical equation for the reaction.

9. At 700 K, the equilibrium constant for the reaction

is 54.8. If 0.5 mole/litre of HI_(g) is present at equilibrium at 700 K, what are the concentrations of H_2(g)and I_2(g), assuming that we initially started with HI_(g) and allowed it to reach equilibrium at 700 K?

Solution

Let the concentrations of H_2(g) and I_2(g) at equilibrium be 'x' mol L^-1. Then for the reaction,

The equilibrium constant, is,

But, K = 54.8.

So,

Thus, at equilibrium, H_2(g) = 0.067 mol/L and I_2(g) = 0.067 mol/L.

10. 1 mole of H₂O and 1 mole of CO are taken in a 10 litre vessel and heated to 725 K. 40 per cent of water (by mass) reacts with carbon monoxide at equilibrium according to the equation:

Calculate the equilibrium constant for the reaction.

Solution

The reaction is,

Initial amount: 1 mol 1 mol

= 0.4 mol = 0.4 mol

Amounts at equilibrium:(1-0.4)mol (1-0.4)mol 0.4 mol 0.4 mol

= 0.6 mol = 0.6 mol 0.4 mol 0.4 mol

Volume of the reaction vessel is 10 L.

The equilibrium constant of this reaction is given by,

11. One mole of oxygen and 2 moles of SO₂ are heated in a closed vessel of 1 dm³ capacity at 1098 K. At equilibrium 1.6 moles of sulphur trioxide are formed. Calculate the equilibrium constant for the reaction.

Solution

According to the equation, if 2 moles of SO₃ are formed, 2 moles of SO₂ and 1 mole of O₂ are used up. Hence 1.6 moles of SO3 are formed from 1.6 moles of SO₂ and 0.8 mole of O₂. Therefore,

SO₂ left over at equilibrium = 2 - 1.6 = 0.4 mole

O₂ left over at equilibrium = 1 - 0.8 = 0.2 mole

At equilibrium [SO₂] = 0.4 mole dm^-3

[O₂] = 0.2 mole dm^-3

[SO³] = 1.6 mole dm^-3