Application of Equilibrium Constant: Definition, Formula, Factors, Applications & Class 11 Notes

Chemistry Chemical Equilibrium 2025

Vikash Kumar Vishwakarma
Updated on Sep 12, 2025 17:35 IST

By Vikash Kumar Vishwakarma

In Chemistry, it is noted that not every reaction reaches completion, but they reach a state of balance, which is known as chemical equilibrium. In equilibrium, the rate of forward and reverse reactions is equal. The concentration of reactants and products remains constant; however, the reaction is still in process. To understand the relationship between the concentration of reactants and products at equilibrium, we use the chemical equilibrium constant (Kc or Kp). Understanding of the equilibrium constant will build a foundation for topics such as dynamic equilibrium, reaction rates, and Le Chatelier's principle.

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Table of content
  • What is Equilibrium Constant?
  • Unit of Equilibrium Constant
  • Equilibrium Constant Formula (Kc and Kp)
  • Equilibrium Concentration for Predicting the Direction of a Reaction
  • Factors Affecting Equilibrium Constant
  • Application of Equilibrium Constants
  • Application of Equilibrium Constant: Key Features
  • Sample JEE Main Problems
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What is Equilibrium Constant?

The equilibrium constant expresses the relationship between the concentrations of products and reactants in a reversible chemical reaction at equilibrium. It is denoted by K. 
For a reaction a A + b B < = > c C + d D ,

The concentration-based constant is: K c = [ C ] c [ D ] d [ A ] a [ B ] b

Where 
[A],[B],[C],[D] is are molar concentrations
a, b, c, and d are stoichiometric coefficients

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Unit of Equilibrium Constant

The ratio of product concentration to reactant concentration, each raised to its respective stoichiometric coefficients, is called the equilibrium constant. The unit if equilibrium constant Kc is  [ Mole L - 1 ] Δ n .
Where,

∆n = sum of stoichiometric coefficients of products – sum of stoichiometric coefficients of reactants.

Related Links: NCERT Solution | Class 11 Chemistry NCERT Solution

 

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Equilibrium Constant Formula (Kc and Kp)

In equilibrium, the equilibrium constant is used to calculate the ratio of concentrations of products to reactants, each raised to the power of their stoichiometric coefficients.

Reaction in terms of concentration (Kc):

General reversible reaction: a A + b B c C + d D

Equilibrium constant (Kc) is: K c = [ C ] c [ D ] d [ A ] a [ B ] b Where 
[A],[B],[C],[D] is are molar concentrations
a, b, c, and d are stoichiometric coefficients

For Gaseous Reactions in term of Pressure (Kp)

K p = ( P C ) c ( P D ) d ( P A ) a ( P B ) b

Where,
P A, P B, P C, P D are the partial pressures of gases.

Relationship between Kp and Kc

K p = K c ( R T ) Δ n Where, 
R-gas constant
T- temperature 
Δn - sum of stoichiometric coefficients of products – sum of stoichiometric coefficients of reactants.

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Equilibrium Concentration for Predicting the Direction of a Reaction

The direction of reaction can be predicted using the equilibrium constant. The reaction quotient (Qc calculated using concentration, or Qp calculated using partial pressures) is required, which is similar to the equilibrium constant, but the conditions are not in equilibrium.

Also Read: NCERT Solution for Class 11 Chemistry Chapter 6

A balanced reaction equation is: a A + b B c C + d D

Reaction quotient (Qc or Qp) is given as: Q c = [ C ] c [ D ] d [ A ] a [ B ] b and Q p = ( P C ) c ( P D ) d ( P A ) a ( P B ) b

Where,

While comparing to Kc and the direction of reaction

  • Case 1: Q = Kc, reaction is in equilibrium [ where Kc is the equilibrium constant]
  • Case 2: Q greater than Kc, Q tends to decrease until equal to K. The reaction will proceed in the reverse direction.
  • Case 3: Q less than Kc, Q tends to increase until equal to K. Reaction proceeds in the forward direction.

 

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Factors Affecting Equilibrium Constant

The only factor that affects the equilibrium constant is temperature, and the nature of the reactants and products. 

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Application of Equilibrium Constants

There are various applications of the equilibrium constant; let's discuss them below. 

  • Direction of Reaction:
    • When the equilibrium constant is greater than 1 ( Kc > 1), products are favoured and the reaction moves forward.
    • When the equilibrium constant is smaller than 1 ( Kc < 1 ), reactants are favoured and the reaction moves in reverse.
    • When Kc ≈ 1: A Significant amount of both product and reactants are present. 
  • Calculating Equilibrium Concentrations

K c = [ Products ] coefficients [ Reactants ] coefficients Using the formula, you can calculate the concentrations of species at equilibrium.

  • Determining Feasibility of Reactions: The feasibility of a reaction means that the product formed in a chemical reaction is economically viable to use in an industrial process. 
    • The greater the value of the equilibrium constant, the more feasible the chemical reaction will be. 
  • Understanding Biological Systems
    • Maintaining homeostasis in living organisms, enzyme reactions, buffer systems, and metabolic pathways depends on equilibrium constants. 
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Application of Equilibrium Constant: Key Features

Applications of equilibrium constants include:

  • Predicting Direction: Compare Q c with K c . If Q c < K c , reaction proceeds forward; if Q c > K c , backwards; if Q c = K c , equilibrium.
  • Extent of Reaction: Large K c (> 10 3 ) indicates near-complete reaction; small K c < 10 - 3  suggests minimal products; K c 1 means comparable concentrations.
  • Equilibrium Concentrations: Use K c and ICE tables to find concentrations at equilibrium.
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Sample JEE Main Problems

Problem 1: For 2 S O 2 ( g ) + O 2 ( g ) < = > 2 S O 3 ( g ) , K c = 2.8 × 10 2 at 1000 K . Initial concentrations: [ S O 2 ] = 0.1 M , [ O 2 ] = 0.05 M , [ S O 3 ] = 0.01 M . Predict reaction direction.

Q c = ( 0.01 ) 2 ( 0.1 ) 2 ( 0.05 ) = 0.2

Since Q c < K c , the reaction proceeds toward S O 3 .

Problem 2: For H 2 ( g ) + I 2 ( g ) < = > 2 H I ( g ) , K c = 49 at 730 K . If 1 mol each of H 2 and I 2 are in a 2 L vessel, find [HI] at equilibrium. Initial: [ H 2 ] = [ I 2 ] = 0.5 M . Let [ H I ] = 2 x , [ H 2 ] = [ I 2 ] = 0.5 - x . Then:

K c = ( 2 x ) 2 ( 0.5 - x ) 2 = 49 , 2 x 0.5 - x = 7 , x = 0.35 , [ H I ] = 0.7 M

Problem 3: For C O ( g ) + H 2 O ( g ) < = > C O 2 ( g ) + H 2 ( g ) , K c = 4.24 at 800 K.

Initial: [ C O ] = 0.1 M , [ H 2 O ] = 0.2 M . Find [CO2] at equilibrium.

Let [ C O 2 ] = x , so [ H 2 ] = x , [ C O ] = 0.1 - x , [ H 2 O ] = 0.2 - x . Then: K c = x 2 ( 0.1 - x ) ( 0.2 - x ) = 4.24

Solving the quadratic: 4.24 x 2 = 0.02 - 0.3 x + x 2 , 3.24 x 2 + 0.3 x - 0.02 = 0 , x 0.048 M .
Problem 4: For N 2 ( g ) + O 2 ( g ) < = > 2 N O ( g ) , K p = 0.05  at 2000 K . Initial pressures: p N 2 = p O 2 = 1  bar. Find p N O  at equilibrium.

Let p N O = 2 x , p N 2 = p O 2 = 1 - x . Then: K p = ( 2 x ) 2 ( 1 - x ) 2 = 0.05 , 2 x 1 - x = 0.05 0.224 , x 0.092 , p N O = 0.184 b a r

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