Equilibrium constant (Keq)

What is equilibrium constant (K_eq)?

The equilibrium constant may be defined as the ratio between the product of the molar concentrations of the products to that of the product of the molar concentrations of the reactants with each concentration term raised to a power equal to the stoichiometric coefficient in the balanced chemical reaction.

The equilibrium constant at a given temperature is the ratio of the rate constant of forward and backward reactions.

Table of contents

• Determination of equilibrium constant(K_eq)
• Equilibrium constant(K_eq) depends on the mode of reaction
• Equilibrium constant(K_eq) on stoichiometric representation
• Equilibrium constant(K_eq) in terms of pressure
• Equilibrium constant(K_eq) in terms of mole fraction
• Unit of equilibrium constant -K_eq
• Frequently Asked Questions-FAQs

Determination of equilibrium constant(K_eq)

Let us consider the reversible reaction:

A + B ⇋ C +D

The equilibrium constant for the reaction is

Equilibrium constant(K_eq) depends on the mode of reaction

The equilibrium constant of the reaction depends on the mode of reaction, let us consider a reversible reaction.

A + B ⇋ C + D

The equilibrium constant for the reaction is   K_c = [C] [D] /[A] [B]

Now if we start from the product mode then the reaction will be,

C + D ⇋ A + B

The equilibrium constant for the reaction is  K_c’ = [A] [B] /[C] [D]

∴ The equilibrium constant K_c’, is actually the reciprocal of K_c.

K_c’ = 1/K_c

Equilibrium constant(K_eq) on stoichiometric representation

(a) When a reversible reaction is written with the help of two or more stoichiometric equations, the value of equilibrium constant will be numerically different from each other.

Let us consider a general reversible reaction:

A + B ⇋ C + D

The equilibrium constant for the reaction is K_c = [C] [D] /[A] [B] ………..(1)

If we multiply a certain value n, the equilibrium constant for the new equation will be K_c’.

nA + nB ⇋ nC + nD

K_c’ = [C]ⁿ [D]ⁿ /[A]ⁿ [B]ⁿ ………..(2)

From equation (1) and (2),

K_c’ = (K_c)ⁿ

Example:

N₂ + 2O₂ ⇋ 2NO₂ ; K_c = [NO₂]² /{[N₂] [O₂ ]²}

½ N₂ + O₂ ⇋ NO₂ ; K_c’ = [NO₂] /{[N₂]^½ [O₂ ]}

Thus the two constants are related to each other as: K_c’ = (K_c)^½ .

(b) If the two reversible reactions are added together:

Example: N₂ + O₂ ⇋ 2NO ; K₁ = [NO]² /{[N₂] [O₂ ]} …….. (1)

2NO + Cl₂ ⇋ 2NOCl ; K₂ = [NOCl]² /{[NO]² [Cl₂ ]} …….. (2)

Combining above two equations (1) and (2)

N₂ + O₂ + Cl₂ ⇋ 2NOCl ; K_c = [NOCl]² /{[N₂] [O₂ ] [Cl₂ ]}

Hence K₁ x K₂ = [NO]² /{[N₂] [O₂ ]} x [NOCl]² /{[NO]² [Cl₂ ]} = [NOCl]² /{[N₂] [O₂ ] [Cl₂ ]}

∴ K₁ x K₂ = K_c

Equilibrium constant(K_eq) in terms of pressure

When the reactants and products are in a gaseous state, the partial pressure can be used instead of concentration and the equilibrium constant is known as K_p.

Consider a general reversible gaseous reaction:

aA + bB ⇋ cC + dD

For gases, PV = nRT ⇒ P = (n/V) RT = CRT

∴ K_p = K_c (RT)^Δng

Where Δng = (c + d) – (a + b)

Equilibrium constant(K_eq) in terms of mole fraction

Consider a general reversible reaction: aA + bB ⇋ cC + dD

Equilibrium constant in terms of mole fraction ‘χ’ may be given as,

According to Dalton’s law of partial pressures:

Partial pressure of a gas = mole fraction of the gas x Total pressure

∴ K_P = K_X P^Δng

Unit of equilibrium constant -K_eq

• The unit constant of K_c = [M]^Δn ,
• where M = mol. Lit^-1 and Δn = (no. of moles of product – no. of moles of reactant)
• The unit constant of K_p = [atm]^Δn