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Equilibrium Chapter - Chemistry JEE Main

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Concepts of Equilibrium for JEE Main Chemistry

Chemical equilibrium is defined as the state in which both products and reactants are present in the present concentrations which have no further tendency to change with time so that there is no observable change in the properties of the system which is present. This results in the state when the forward reaction proceeds at the same rate as the reverse reaction. The reaction rates of the backward and forward reactions are generally not zero, but equal. Thus, no net changes are seen in the concentrations of reactants and products. Such a state is called a dynamic equilibrium state.


Reactions do occur at the molecular level, Although the macroscopic equilibrium concentrations are constant in time. For example, when the acetic acid dissolved in water it forms acetate and hydronium ions,

$CH_3CO_2H + H_2O \rightleftharpoons CH_3CO^{-2} + H_3O^+$

Types of Chemical Equilibrium

There are two types of chemical equilibrium:

  • Homogeneous Equilibrium

  • Heterogeneous Equilibrium

Homogeneous Chemical Equilibrium

In Homogeneous Equilibrium, the reactants and the products of chemical equilibrium are all in the same phase. It can be further divided into two types: 

a) Reactions in which the number of molecules of the products is equal to the number of molecules of the reactants. For example,

$H_2 (g) + I_2 (g) \rightleftharpoons 2HI (g)$

$N_2 (g) + O_2 (g) \rightleftharpoons 2NO (g)$

b) Reactions in which the number of molecules of the products is not equal to the total number of reactant molecules. For example,

$2SO_2 (g) + O_2 (g) \rightleftharpoons 2SO_3 (g)$

$COCl_2 (g) \rightleftharpoons CO (g) + Cl_2 (g)$

Heterogeneous Chemical Equilibrium

In, Heterogeneous Equilibrium the reactants and the products of chemical equilibrium are present in different phases. Examples of heterogeneous equilibrium are listed below.

$CO_2 (g) + C (s) \rightleftharpoons 2CO (g)$

$CaCO_3 (s) \rightleftharpoons CaO (s) + CO_2 (g)$

Thus, the different types of chemical equilibrium are based on the phase of the reactants and products.

Because the opposing mechanisms only demand physical alterations, the equilibrium is called Physical Equilibrium. Chemical Equilibrium occurs when chemical reactions are in opposition to one another.

JEE Main Chemistry Chapters 2024 

Important Topics for Equilibrium Chapter

  • Equilibrium

  • Law of Mass Action

  • Equilibrium Constant

  • Le-Chatelier's Principle

  • Free Energy

  • Ionic Equilibrium

Equilibrium Important Concept for JEE Main

Equilibria Involving Physical Processes: Solid-Liquid, Liquid-Gas, and Solid-Gas Equilibria

Equilibria in chemistry describe the balance between opposing processes, and they manifest in various physical states: solid-liquid, liquid-gas, and solid-gas. These equilibria play a fundamental role in understanding phase transitions and gas solubility, and they adhere to specific principles and laws.

Solid-Liquid Equilibrium:

In the solid-liquid equilibrium, a dynamic balance exists between a solid and its corresponding liquid phase. For example, in the melting or dissolving process, the rate of solid-to-liquid conversion equals the rate of liquid-to-solid conversion.

Liquid-Gas Equilibrium:

Liquid-gas equilibrium is observed during processes like vaporization and condensation. Here, the rates of molecules escaping the liquid phase (vaporization) and returning to the liquid phase (condensation) are equal. It's the basis for phenomena like evaporation and boiling.

Solid-Gas Equilibrium:

In solid-gas equilibrium, substances in solid form can sublime or desublime, transitioning between solid and gas phases. The rate of sublimation is balanced by the rate of desublimation.

Henry's Law:

Henry's Law describes the solubility of a gas in a liquid. It states that the amount of gas dissolved in a liquid is directly proportional to the partial pressure of that gas above the liquid. Mathematically, it is represented as C = k * P, where C is the concentration of the gas in the liquid, P is the partial pressure of the gas, and k is Henry's law constant.

General Characteristics of Equilibrium Involving Physical Processes:

Dynamic Nature

Equilibria are dynamic; although macroscopic properties remain constant, at the molecular level, there's continuous motion and interaction.

To know more about the Dynamic equilibrium and Henry’s law, you can cover Vedantu’s page on Dynamic Equilibrium.

Equilibrium Constants

Equilibrium constants (Kc, Kp) quantitatively describe the extent of a reaction at equilibrium, helping predict the direction in which a reaction will proceed.

Reversible and Irreversible Reactions

  • Reversible Reactions: Reversible reactions are ones in which the reactants are not completely transformed into products.

A reversible reaction is one in which an acid and a base, one or both of which are weak, neutralise each other.

This is the reaction that occurs when a weak acid (CH3COOH) reacts with a strong base (NaOH).

  • Irreversible Reactions: Irreversible reactions are ones in which the entire quantity of reactants is transformed into products.

Irreversible reactions include the neutralisation reaction between a strong acid (HCl) and a strong base (NaOH).
NaOH + HCl → NaCl + H2O.

Equilibrium and Its Dynamic Character

  • The state of equilibrium is when the concentrations of reactants and products do not change over time i.e. the reactant and product concentrations become constant.

  • A chemical equilibrium is dynamic in nature, which implies that reactions continue to occur at the same rate in both the forward and backward directions. 

  • As a result, the amount of product created reacts backwards to give reactants, leaving no change in reactant or product concentration with the passage of time.

Law of Mass Action and Equilibrium Constant

  • “At any given moment, the product of the molar concentrations of the reactants at a fixed temperature determines the pace of a chemical reaction."

  • Take a look at a simple reversible reaction, at a particular temperature.
    aA + bB ⇌ cC + dD.

  • Therefore at equilibrium ,
    Forward reaction rate = Backward reaction rate
    $\dfrac{k_{f}}{k_{b}}= K_{c} = \dfrac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}}$; Kc stands for equilibrium constant.

Relation Between the Three Constants - Kp, Kc, and Kx

  • The values of Kp can be obtained as follows:
    Kp = Kc (RT)Δn
    Kp = Kc (P)Δn
    Δn = In a chemical equation, the number of moles of gaseous products equals the number of moles of gaseous reactants.

Relationship Between Equilibrium Constant and G°

  • The relation relates ΔG for a reaction under any situation to ΔG°,
    ΔG = ΔG° + 2.303 RT log(Q)

  • At temperature T, the standard free energy change of a reaction and its equilibrium constant are related by the relation,
    ΔG° = - 2.303 RT logK

  • For any general reaction,
    aA + bB ⇌ cC + dD;
    The rate constant is given as,
    K = ((aC)c(aD)d)/((aA)a(aB)b); The activity of the reactants and products is represented by the letter a. It has a lower unit count.

  • For pure solids and liquids, use the following formula:
    a = 1

  • For gases, use:
    a = gas pressure measured in atm.

  • For components in solution: a = molar concentration.

Le-Chatelier's Principle

  • Changes in any of the components that determine a system's equilibrium conditions will cause the equilibrium to shift in such a way that the effect of the change is reduced or neutralised.

  • Application of Le-Chatelier's Principle:
    The Le-Chatelier's concept is extremely important in chemical, physical, and everyday life systems that are in a state of equilibrium.

Let’s dive into Le Chatelier’s - Principle and Its Examples for JEE from Vedantu’s page for more information. 

Applications to the Chemical Equilibrium




Synthesis of Ammonia i.e. Haber’s Process

N2 + 3H2 ⇌ 2NH3 + 23 kcal

- High-pressure situations (Δn<0).
- Low temperature.

- Excess amounts of N2 and H2.

- Removal of NH3 favours forward reaction.

Formation of Sulphur Trioxide

2SO2 + O2 ⇌ 2SO3 + 45 kcal (exothermic)

- High-pressure situations (Δn<0).

- Low temperature.

- Excess amounts of SO2 and O2 favour forward reaction.

Nitric Oxide Synthesis

N2 + O2 ⇌ 2NO - 43.2 kcal

- Low temperature.

- Excess amounts of N2 and O2 favour forward reaction.

- Because there is no change in volume during the reaction, i.e. Δn = 0,  pressure has no effect on the equilibrium.

Nitrogen Dioxide Formation

2NO + O2 ⇌ 2NO2 + 27.8 kcal. 

- High pressure

- Low temperature

- Excess amounts of NO and O2 favour the reaction in the forward direction.

Phosphorus Pentachloride Dissociation

PCl5 ⇌ PCl3 + Cl2 - 15 kcal.

- Low pressure or a high volume of the container,
Δn > 0,

- High temperature

-  Excess of PCl5.

Ionic Equilibrium: Understanding the Key Concepts

Ionic equilibrium is a fundamental topic in chemistry that delves into the behavior of electrolytes, acids, and bases. It encompasses various concepts, from the ionization of electrolytes to the definitions of acids and bases as proposed by Arrhenius, Bronsted-Lowry, and Lewis. Let's explore these concepts and their applications:

1. Weak and Strong Electrolytes:

Weak Electrolytes: These substances partially ionize in solution. Examples include weak acids like acetic acid and weak bases like ammonia.

Strong Electrolytes: They completely ionize in solution, producing a high concentration of ions. Common examples are strong acids (HCl, H2SO4) and strong bases (NaOH, KOH).

2. Ionization of Electrolytes:

Electrolytes, when dissolved in water, dissociate into ions. This ionization process plays a critical role in understanding the behavior of solutes in solution.

3. Concepts of Acids and Bases:

Arrhenius Theory: Acids release H+ ions in aqueous solutions, and bases release OH- ions. This theory is limited to aqueous solutions.

Bronsted-Lowry Theory: Acids donate protons (H+) and bases accept protons. This theory is more comprehensive as it extends beyond aqueous solutions.

Lewis Theory: Acids are electron pair acceptors, and bases are electron pair donors. It provides a broad perspective on acid-base reactions.

4. Acid-Base Equilibria:

Acid-base equilibria describe the balance between the concentration of the ions involved in acid-base reactions.

Multistage ionization involves acids or bases with multiple dissociation steps (polyprotic), leading to stepwise equilibrium expressions.

Ionization constants (Ka for acids, Kb for bases) quantify the extent of ionization in these equilibria.

5. Ionization of Water:

Water itself undergoes ionization to form H+ and OH- ions, maintaining neutrality. The equilibrium constant for this process is the ion product of water (Kw).

6. pH Scale:

The pH scale quantifies the acidity or basicity of a solution. It's defined as the negative logarithm of the H+ ion concentration. pH = -log[H+].

7. Common Ion Effect:

When a solution already contains one of the ions of a sparingly soluble salt, the solubility of that salt decreases. This is known as the common ion effect.

8. Hydrolysis of Salts:

Some salts undergo hydrolysis when dissolved in water, leading to the formation of acidic or basic solutions.

9. Solubility of Sparing Salts and Solubility Products:

Solubility products (Ksp) describe the equilibrium constant for the dissolution of sparingly soluble salts. They help predict the solubility of such compounds in solution.

10. Buffer Solutions:

Buffers are solutions that resist changes in pH when an acid or base is added. They consist of a weak acid and its conjugate base or a weak base and its conjugate acid.

JEE Main Equilibrium Solved Examples

Question 1: A critical stage in the industrial manufacture of sulfuric acid is the reaction between gaseous sulphur dioxide and oxygen:

2SO2(g) + O2(g) → 2SO3(g)

The system was kept at 800 K with a mixture of SO2 and O2 until it reached equilibrium. The equilibrium mixture was made up of 5.0 × 10−2 M SO3, 3.5 × 10−3 M O2, and 3.0 × 10−3 M SO2. At this temperature, calculate K and Kp.


  • In the equilibrium constant expression, substitute the relevant equilibrium concentrations,

  • K = [SO3]2/([SO2]2[O2])

K = (5.0 × 10−2)2/(3.0 × 10−3)2.(3.5 × 10−3

∴ K = 7.9 × 104.

  • We use the relationship between K and Kp to solve for Kp, where
    Kp = K(RT)Δn

Kp = 7.9 × 104((0.08206 L⋅atm/mol⋅K)(800 K))−1 

∴ Kp = 1.2 × 103

Key Points to Remember: The relationship between K and Kp is Kp = K(RT)Δn.

Question 2: When 0.1 M CH3COOH (50 ml) and 0.1 M NaOH (50 ml) are mixed, calculate the pH of the solution [Ka (CH3COOH) = 10-5].


  • The three dissociation into ions in given chemical reactions are given as:
    CH3COOH ⇌ CH3COO- + H+,
    NaOH → Na+ + OH-, and H+ + OH- ⇌ H2O.

  • For the reaction,
    CH3COOH + OH- ⇌ CH3COO- + H2O
    Keq = Ka/Kw.

  • Concentration of H2O thus remains constant,
    109 = x/(0.05-x)2

  • Because the value of equilibrium constant is very high, x >> 0.05

  • ∴ Let 0.05 - x = a

109 = 0.05/a2

a = 7.07 x 10-6

pOH = 6 - log(7.07)

pOH = 6 - 0.85

  • pH = 14 - (6 - 0.85) = 8.85

Hence, the pH is 8.85 which is basic.

Key Points to Remember: The relation between equilibrium constant and acid dissociation constant is Keq = Ka/Kw.

Solved Examples from Previous Year Question Papers

Question 1: For the reaction, 2SO2(g) + O2(g) ⇌ 2SO3(g), ∆H = - 57.2 kJ mol–1 and Kc = 1.7 × 1016. Which of the following statements is incorrect?

(a) The equilibrium will shift in the forward direction as the pressure increases.

(b) The addition of inert gas at constant volume will not affect the equilibrium constant.

(c) The equilibrium constant is large, suggestive of reaction going to completion and so no catalyst is required.

(d) The equilibrium constant decreases as the temperature increases.


  • The big value of Kc indicates that the reaction is nearly complete. SO2 oxidation to SO3 is a slow process. 

  • As a result, adding a catalyst speeds up the process. 

  • As a result, option (c) is the correct answer.

Question 2: 20 mL of 0.1 M H2SO4 solution is added to 30 mL of 0.2 M NH4OH solution. The pH of the resultant mixture is [pKb of NH4OH = 4.7]

(a) 9.4

(b) 9.0

(c) 5.0

(d) 5.2


  • Given that NH4OH has a pKb of 4.7,

  • 0.1 M H2SO4 (20 ml) ⇒ nH+ = 4.

  • nNH4OH = 6 from 30 ml 0.2 M NH4OH.

  • The chemical reaction is given as:
    NH4OH + H+ ⇌ NH4+ + H2O
        6           4            0          0
        2           0            4          4  

  • Basic buffer is the solution.

  • pOH = pKb + log [NH4+]/[NH4OH] = 4.7 + log (4/2) pOH = pKb + log [NH4+]/[NH4OH] = 4.7 + log (4/2) pOH = pK

  • pOH = log 2 + 4.7

pOH = 4.7 + 0.3

 pOH = 5

  • 14 - pOH, is the pH value.

14 - 5 = 9

  • As a result, option (b) is the correct answer.

Question 3: ​​The increase of pressure on the ice water system at constant temperature will lead to

(a) no effect on that equilibrium.

(b) a decrease in the entropy of the system.

(c) a shift of the equilibrium in the forward direction.

(d) an increase in the Gibbs energy of the system.

Solution: When the pressure on this system in equilibrium is increased, the equilibrium tends to shift in the direction of volume reduction, i.e., forward.

As a result, option (c) is the correct answer.

Practice Questions

Question 1: The reaction between hydrogen gas and iodine produces hydrogen iodide:
H2(g)+ I2(g) → 2HI(g)

At 740 K, a combination of H2 and I2 was kept until the system reached equilibrium. The equilibrium mixture was made up of
1.37 × 10−2 M HI, 6.47 × 10−3 M H2, and 5.94 × 10−4 M I2.
For this reaction, calculate K and Kp.

Answer: K = 4.88, and Kp = 48.8.

Question 2: For a chemical reaction in equilibrium, which of the following is true?

(a) The forward reaction is the only thing that comes to a halt.

(b) Only the opposite reaction is halted.

(c) The forward and reverse reactions have the same rate constants.

(d) The forward and reverse reactions have the same rates.

Answer: (d) The forward and reverse reactions have the same rates. 

JEE Main Chemistry Equilibrium Study Materials

Here, you'll find a comprehensive collection of study resources for Equilibrium designed to help you excel in your JEE Main preparation. These materials cover various topics, providing you with a range of valuable content to support your studies. Simply click on the links below to access the study materials of Equilibrium and enhance your preparation for this challenging exam.

JEE Main Chemistry Study and Practice Materials

Explore an array of resources in the JEE Main Chemistry Study and Practice Materials section. Our practice materials offer a wide variety of questions, comprehensive solutions, and a realistic test experience to elevate your preparation for the JEE Main exam. These tools are indispensable for self-assessment, boosting confidence, and refining problem-solving abilities, guaranteeing your readiness for the test. Explore the links below to enrich your Chemistry preparation.


When the observable attributes of a process, such as colour, temperature, pressure, concentration, and so on, do not vary, it is said to be in equilibrium.

The word equilibrium implies 'balance,' implying that the reactants and products involved in a chemical reaction are in a state of equilibrium. The equilibrium condition can also be seen in physical processes like the melting point of ice at 0°C, when both ice and water are present at the same time.

FAQs on Equilibrium Chapter - Chemistry JEE Main

1. What does it mean to be in a state of equilibrium?

Equilibrium is a condition of equilibrium or a stable situation in which opposing forces cancel each other out and no changes occur. In Economics, equilibrium occurs when supply and demand are equal. When you are peaceful and stable, this is an example of balance.

2. Which of the following is an example of equilibrium?

A book kept at rest on a table is an illustration of equilibrium. A car that moves at a steady speed. A chemical process in which the forward and backward reaction rates are equal.

3. What is the difference between equilibrium and balance?

Balance is (uncountable) a state in which opposing forces harmonise; equilibrium is the condition of a system in which competing influences are balanced, resulting in no net change.

4. What are the different types of equilibria discussed in this chapter?

The chapter covers different types of equilibria, including chemical equilibria, ionic equilibria, and solubility equilibria. Each of these has its unique principles and applications, making this chapter quite comprehensive.

5. How is Le Chatelier's Principle applied in this chapter, and why is it important?

Le Chatelier's Principle is vital in understanding how changes in concentration, pressure, and temperature affect the position of equilibrium in chemical reactions. It helps predict how systems respond to external perturbations, a key concept in this chapter.

6. What are some practical applications of the concepts learned in the Equilibrium chapter?

Equilibrium concepts find applications in various fields, including chemical engineering, environmental science (acid rain), pharmaceuticals (drug reactions), and industrial processes. Understanding equilibrium is essential for solving real-world problems.

7. How can I effectively prepare for the Equilibrium chapter for JEE exams?

To excel in this chapter, thoroughly understand the underlying concepts, practice problem-solving, use textbooks and online resources, and focus on the principles of chemical and ionic equilibria. Also, familiarize yourself with common equilibrium constants and their significance.