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Just imagine that water is flowing from the upper floor to the lower floor. Will it flow forever? Of course not. The flow will stop when the two levels become equal. This state can be considered as a state of equilibrium. This idea also applies to many chemical reactions and can be explained by the Law of Mass Action.
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Law of Mass Action
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The Law of Mass Action states that the rate of a chemical reaction is proportional to the product of the active masses of the reactants.
Consider a reaction-
A + B ⇔ C + D
Let the concentration of reactions be [A] and [B]
Rate of reaction ∝ [A] [B]
R ∝ [A] [B]
R = K [A] [B]
Here, K is the rate constant.
Active Mass
Active Mass means molar concentration. The active mass refers to the number of moles dissolved per litre of the solution.
This law explains the behavior exhibited by solutions in dynamic equilibrium. The law of mass action also states that the ratio between the concentrations of reactants and concentrations of products is constant at the state of equilibrium.
Kc = [C][D] / [A][B]
Also Read: Law of Mass Action in Detail
Law of Mass Action: Equilibrium Constant
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At the state of chemical equilibrium, the ratio of reactant concentration and product concentration is constant, which can be explained with the help of the equilibrium constant (Kc).
Consider a simple reversible reaction where A & B are the reactants and C & D are the products.
A + B ⇔ C + D
According to the Law of Mass Action,
Kc = [C][D] / [A][B]
Here,
Kc is called the equilibrium constant.
In this equation, the concentration of A at equilibrium is represented as [A] (similarly for B, C, and D), and the stoichiometric coefficients of the reactants and products are 1. It has been experimentally observed that the equilibrium constant is also dependant on the stoichiometric coefficients of the reactants and products.
Therefore, the law of mass action explains that the equilibrium constant, at a constant temperature, is equal to the product of the concentration of products divided by the product of the concentration o reactant, each raised to the corresponding stoichiometric coefficient.
This is also known as the equilibrium law or the law of chemical equilibrium.
Law of Mass Action: Representation of Equilibrium Constant
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Equilibrium Constant for the balanced reaction of the following type:
aA + bB ⇔ cC + dD
According to the Law of Mass Action, the equilibrium constant is the value obtained by relating the equilibrium concentration of the products and the reactants.
For forward reaction, the equilibrium constant can be given by:
Kc = [C]c[D]d / [A]a[B]b
For reverse reaction, the equilibrium constant can be expressed as the inverse of the forward reaction and can be expressed by:
K’c = 1/Kc = [A]a[B]b / [C]c[D]d
If the coefficient of the chemical equations is multiplied by a factor ‘n’ then the equilibrium constant becomes Kcn.
| Equilibrium Constant Representation | Expressed in terms of | Expressed as |
|---|---|---|
| Kc | Concentrations of reactants and products | [C]c[D]d/[A]a[B]b |
| Kp | Partial pressures of reactants and products. | PcC PdD / PaAPbB |
| Kx | Mole fractions of reactants and products | [XC]c[XD]d / [XA]a[XB]b |
Law of Mass Action: Relation between Kc, Kp and Kx
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Kp = Kc(RT)Δng
Kp = Kx. PΔng
Where,
Δng = moles of gaseous products - moles of gaseous reactants
Kc = Equilibrium constant expressed in terms of the concentration of the reactants/products
Kp = Constant of partial pressures of reactants and products.
Kx = Mole fraction of reactants and products
Applications of the Law of Mass Action
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The Law of Mass action applies to semiconductors and, therefore, it has a number of important applications in the fields of electronics and semiconductor physics. It also depicts a relationship between the concentrations of electron holes and free electrons when the semiconductor system is in a state of thermal equilibrium.
The law of mass action also applies to the following fields:
- In semiconductor physics and Sociophysics.
- In Mathematical ecology
- For the explanation of diffusion in condensed matter
- For solving the model of the spread of disease in mathematical epidemiology
- Ostwald’s Dilution Law for determining the dissociation equilibrium of weak electrolytes
Law of Mass Action: Things to Remember
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- According to the Law of Mass Action, the rate of a chemical reaction is proportional to the product of the active masses of the reactants.
- Law of Mass Action also states that the ratio between the concentrations of reactants and concentrations of products is constant at the state of equilibrium.
Kc = [C][D] / [A][B]
- The active mass of a substance can be defined as the number of moles dissolved per litre of the solution.
- The relation between Kp, Kc, and Kx can be expressed as:
Kp = Kc(RT)Δng and Kp = Kx. PΔng
- The Law of Mass action has various applications in the field of semiconductor physics, Sociophysics, Mathematical ecology and mathematical epidemiology.
Law of Mass Action: Important Questions
Ques.1: (a) What is meant by equilibrium?
(b) State Law of Mass Action. (2 Marks)
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Answer: (a) In a chemical reaction, the state at which the concentration of both the reactants and products is constant and will not change further with time is termed as the state of equilibrium.
(b) The Law of Mass Action states that the rate of a chemical reaction is proportional to the product of the active masses of the reactants. It also states that the ratio between the concentrations of reactants and concentrations of products is constant at the state of equilibrium.
Ques.2: The degree of dissociation of N2O4 is α according to the reaction- N2O4 (g) ⇔ 2NO2 (g) at temperature T and total pressure P. Find the expression for the equilibrium constant of this reaction. (2 Marks)
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Answer: At Equilibrium-

Ques.3: The equilibrium constants for the reactions- N2 + O2 ⇔ 2NO and 2NO + O2 ⇔ 2NO2 are K1 and K2 respectively, then what would be the equilibrium constant for the reactions. (3 Marks)
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Answer: Given that:
The equilibrium constant for the reaction- N2 + O2 ⇔ 2NO is K1
The equilibrium constant for the reaction- 2NO + O2 ⇔ 2NO2 is K2
Then,

Ques.4: (a) At 773 K, the equilibrium constant Kc for the reaction N2 (g) + 3H2 (g) ⇔ 2NH3 (g) is 6.02 × 10-2 L2 mol-2. Calculate the value of Kp at the same temperature.
(b) If Kp for the reaction N2 (g) + 3H2 (g) ⇔ 2NH3 (g) is 49 at a certain temperature. Calculate the value of Kp at the same temperature for the reaction
NH3 ⇔ 1/2 N2 + 3/2 H2 (g). (5 Marks)
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Answer: (a) We know that:
Kp = Kc(RT)Δng
Given that:
Kc = 6.02 x 10-2

(b) Given for the reaction N2 (g) + 3H2 (g) ⇔ 2NH3 (g),
Kp = 49
∴ For the reverse reaction- 2NH3 (g) ⇔ N2 (g) + 3H2 (g)

Ques.5: (a) In the equilibrium CaCO3 (s) ⇔ CaO (s) + CO2 (g) at 1073 K, the pressure of CO2 is formed to be 2.5 × 104 Pa. What is the equilibrium constant for this reaction at 1073 K.
(b) Determine the concentration of CO2 which will be in equilibrium with 2.5 × 10-2 mol L-1 of CO at 100oC for the reaction FeO (s) + CO (g) ⇔ Fe (s) + CO2 (g); Kc= 5.0.
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Answer: (a) With reference to the standard state pressure of 1 bar which is = 105 Pa

(b) Given that: Kc = 5, Concentration of CO = 2.5 × 10-2, mol L-







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