JEE Chemistry Chemical Equilibrium MCQs Set 01

Download JEE MCQs for JEE Chemistry: Chemical Equilibrium

Explore reliable objective questions for Chemical Equilibrium tailored for JEE learners. Utilizing these Chemistry multiple-choice formats ensures thorough preparation and strengthens problem-solving speed for upcoming school assessments.

Chapter-wise Objective Questions: Chemical Equilibrium

Access the complete set of multiple-choice questions for Chemical Equilibrium below. This focused format allows students to isolate specific topics for thorough review and uninterrupted practice alongside official JEE textbooks.

Multiple Choice Questions

Question 1: In an elementary reaction A + 2B → 2C + D, the concentration of A is increased four times and that of B is halved. The rate becomes
(a) Twice
(b) Half
(c) Unchanged
(d) One fourth of the rate
Show Answer & Explanation

Answer: (c) Unchanged

Explanation:
1. Rate = k[A][B]² for the elementary step.
2. New rate = k(4[A])(½[B])² = k[A][B]², the same as before.

Teacher's Note:
1. Raise each concentration to its stoichiometric power.
2. Check units and the stoichiometric coefficients before substituting.

Question 2: A reaction x ⇌ y is said to be in equilibrium when
(a) only 10% conversion of x to y has taken place
(b) complete conversion of x to y has taken place
(c) conversion of x to y is only 50% complete
(d) the rate of change of x to y equals the rate of change of y to x
Show Answer & Explanation

Answer: (d) the rate of change of x to y equals the rate of change of y to x

Explanation:
1. Equilibrium is dynamic: forward and backward rates are equal.
2. The extent of conversion has no fixed value.

Teacher's Note:
1. Apply the law of mass action to the given species.
2. Check units and the stoichiometric coefficients before substituting.

Question 3: In the chemical reaction N₂ + 3H₂ ⇌ 2NH₃ at equilibrium
(a) equal volumes of N₂ and H₂ are reacting
(b) equal masses of N₂ and H₂ are reacting
(c) the reaction has stopped
(d) the same amount of ammonia is formed as is decomposed into N₂ and H₂
Show Answer & Explanation

Answer: (d) the same amount of ammonia is formed as is decomposed into N₂ and H₂

Explanation:
1. At equilibrium the rate of formation of NH₃ equals its rate of decomposition.
2. Reactions continue but there is no net change.

Teacher's Note:
1. Apply the law of mass action to the given species.
2. Check units and the stoichiometric coefficients before substituting.

Question 4: The ratio of active masses of 22 g CO₂, 3 g H₂ and 7 g N₂ in a gaseous mixture is
(a) 22 : 3 : 7
(b) 0.5 : 3 : 7
(c) 1 : 3 : 1
(d) 1 : 3 : 0.5
Show Answer & Explanation

Answer: (d) 1 : 3 : 0.5

Explanation:
1. Moles: CO₂ = 22/44 = 0.5, H₂ = 3/2 = 1.5, N₂ = 7/28 = 0.25.
2. Active mass is proportional to moles in the same volume: 0.5 : 1.5 : 0.25 = 1 : 3 : 0.5.

Teacher's Note:
1. Apply the law of mass action to the given species.
2. Check units and the stoichiometric coefficients before substituting.

Question 5: Which of the following shows the effect of a catalyst on a reversible reaction?
(a) It gives a new reaction path with lower activation energy
(b) It shifts the equilibrium to the right
(c) It decreases the kinetic energy of activated molecules
(d) It decreases the rate of the backward reaction
Show Answer & Explanation

Answer: (a) It gives a new reaction path with lower activation energy

Explanation:
1. A catalyst lowers the activation energy for both forward and backward reactions.
2. It does not shift equilibrium or change the energy of molecules.

Teacher's Note:
1. Apply the law of mass action to the given species.
2. Check units and the stoichiometric coefficients before substituting.

Question 6: In a reversible chemical reaction, equilibrium is established when
(a) the reactant is completely converted into product
(b) the rates of forward and backward reactions are equal
(c) the yield of product is minimum
(d) the concentrations of reactant and product are equal
Show Answer & Explanation

Answer: (b) the rates of forward and backward reactions are equal

Explanation:
1. Equilibrium is defined by equal forward and backward rates, not equal concentrations.

Teacher's Note:
1. Apply the law of mass action to the given species.
2. Check units and the stoichiometric coefficients before substituting.

Question 7: In a chemical equilibrium the rate constant of the backward reaction is 7.5 × 10⁻⁴ and the equilibrium constant is 1.5. The rate constant of the forward reaction is
(a) 2 × 10⁻³
(b) 5 × 10⁻⁴
(c) 1.12 × 10⁻³
(d) 9.0 × 10⁻⁴
Show Answer & Explanation

Answer: (c) 1.12 × 10⁻³

Explanation:
1. K = kf/kb, so kf = K × kb.
2. kf = 1.5 × 7.5 × 10⁻⁴ = 1.125 × 10⁻³ ≈ 1.12 × 10⁻³.

Teacher's Note:
1. Apply the law of mass action to the given species.
2. Check units and the stoichiometric coefficients before substituting.

Question 8: For Ag⁺ + 2NH₃ ⇌ Ag(NH₃)₂⁺ at 298 K, the equilibrium molar concentrations of Ag⁺, Ag(NH₃)₂⁺ and NH₃ are 10⁻¹, 10⁻¹ and 10³ respectively. The value of Kc is
(a) 10⁻⁶
(b) 10⁶
(c) 2 × 10⁻³
(d) 2 × 10⁶
Show Answer & Explanation

Answer: (a) 10⁻⁶

Explanation:
1. Kc = [Ag(NH₃)₂⁺]/([Ag⁺][NH₃]²).
2. = 10⁻¹/(10⁻¹ × (10³)²) = 10⁻⁶.

Teacher's Note:
1. Apply the law of mass action to the given species.
2. Check units and the stoichiometric coefficients before substituting.

Question 9: Equilibrium constants are: (a) x ⇌ y, K = 10⁻¹; (b) y ⇌ z, K = 2 × 10⁻²; (c) P ⇌ Q, K = 3 × 10⁻⁴; (d) R ⇌ S, K = 2 × 10⁻³. With equal initial reactant concentrations, in which reactions are the reactants and the products respectively present at the highest concentration?
(a) d, c
(b) c, a
(c) a, d
(d) b, c
Show Answer & Explanation

Answer: (b) c, a

Explanation:
1. A small K means the reactant is favoured; a large K means the product is favoured.
2. The smallest K is (c) 3 × 10⁻⁴ and the largest is (a) 10⁻¹, so the answer is c, a.

Teacher's Note:
1. Apply the law of mass action to the given species.
2. Check units and the stoichiometric coefficients before substituting.

Question 10: For C(s) + CO₂(g) ⇌ 2CO(g), the partial pressures of CO and CO₂ at equilibrium are 2.0 and 4.0 atm. Kp is
(a) 0.5
(b) 4.0
(c) 8.0
(d) 1
Show Answer & Explanation

Answer: (d) 1

Explanation:
1. The solid is omitted: Kp = p²(CO)/p(CO₂).
2. = (2.0)²/4.0 = 1.

Teacher's Note:
1. Apply the law of mass action to the given species.
2. Check units and the stoichiometric coefficients before substituting.

Question 11: For which reaction is Kp = Kc?
(a) 2NOCl(g) ⇌ 2NO(g) + Cl₂(g)
(b) N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
(c) H₂(g) + Cl₂(g) ⇌ 2HCl(g)
(d) 2SO₂(g) + O₂(g) ⇌ 2SO₃(g)
Show Answer & Explanation

Answer: (c) H₂(g) + Cl₂(g) ⇌ 2HCl(g)

Explanation:
1. Kp = Kc(RT)^Δn, so they are equal when Δn = 0.
2. Only H₂ + Cl₂ ⇌ 2HCl has Δn = 2 − 2 = 0.

Teacher's Note:
1. Apply the law of mass action to the given species.
2. Check units and the stoichiometric coefficients before substituting.

Question 12: log(Kp/Kc) + log RT = 0 holds for which reaction?
(a) PCl₅ ⇌ PCl₃ + Cl₂
(b) 2SO₂ + O₂ ⇌ 2SO₃
(c) N₂ + 3H₂ ⇌ 2NH₃
(d) (2) and (3) both
Show Answer & Explanation

Answer: (b) 2SO₂ + O₂ ⇌ 2SO₃

Explanation:
1. The condition gives Kp/Kc = 1/RT = (RT)⁻¹, so Δn = −1.
2. Δn is +1, −1 and −2 for the three reactions; only 2SO₂ + O₂ ⇌ 2SO₃ has −1.

Teacher's Note:
1. Apply the law of mass action to the given species.
2. Check units and the stoichiometric coefficients before substituting.

Question 13: For which reactions at 298 K is Kp/Kc maximum and minimum respectively? (a) N₂O₄ ⇌ 2NO₂; (b) 2SO₂ + O₂ ⇌ 2SO₃; (c) X + Y ⇌ 4Z; (d) A + 3B ⇌ 7C
(a) d, c
(b) d, b
(c) c, b
(d) d, a
Show Answer & Explanation

Answer: (b) d, b

Explanation:
1. Kp/Kc = (RT)^Δn, with Δn = +1, −1, +2 and +3 for (a), (b), (c), (d).
2. The largest Δn is in (d) and the smallest in (b), so d, b.

Teacher's Note:
1. Apply the law of mass action to the given species.
2. Check units and the stoichiometric coefficients before substituting.

Question 14: For A ⇌ B, Kc = 2; B ⇌ C, Kc = 4; C ⇌ D, Kc = 6. Kc for A ⇌ D is
(a) 12
(b) 4/3
(c) 24
(d) 48
Show Answer & Explanation

Answer: (d) 48

Explanation:
1. Adding the reactions multiplies the constants.
2. K = 2 × 4 × 6 = 48.

Teacher's Note:
1. Apply the law of mass action to the given species.
2. Check units and the stoichiometric coefficients before substituting.

Question 15: For A ⇌ B (Kc = 3), B ⇌ C (Kc = 5) and C ⇌ D (Kc = 2), the equilibrium constant for D ⇌ A is
(a) 15
(b) 0.3
(c) 30
(d) 0.03
Show Answer & Explanation

Answer: (d) 0.03

Explanation:
1. A ⇌ D has K = 3 × 5 × 2 = 30.
2. The reverse reaction has K = 1/30 ≈ 0.03.

Teacher's Note:
1. Apply the law of mass action to the given species.
2. Check units and the stoichiometric coefficients before substituting.

Question 16: The equilibrium constant of a reversible reaction at a given temperature
(a) depends on the initial concentration of the reactants
(b) depends on the concentration of the products at equilibrium
(c) does not depend on the initial concentrations
(d) is not characteristic of the reaction
Show Answer & Explanation

Answer: (c) does not depend on the initial concentrations

Explanation:
1. K depends only on temperature (and the reaction), not on the starting concentrations.

Teacher's Note:
1. Apply the law of mass action to the given species.
2. Check units and the stoichiometric coefficients before substituting.

Question 17: Which statement about the equilibrium constant is correct?
(a) The equilibrium constant of a reaction changes with temperature
(b) It depends on the concentrations of reactants with which we start
(c) The constant for 3Fe(s) + 4H₂O(g) ⇌ Fe₃O₄(s) + 4H₂(g) is the same in an open or closed vessel
(d) It becomes double if the reaction is multiplied by 2 throughout
Show Answer & Explanation

Answer: (a) The equilibrium constant of a reaction changes with temperature

Explanation:
1. K is a function of temperature only.
2. Multiplying the equation by 2 squares K, and a gas-phase equilibrium needs a closed vessel.

Teacher's Note:
1. Apply the law of mass action to the given species.
2. Check units and the stoichiometric coefficients before substituting.

Question 18: For any reversible reaction, if the concentration of reactants is increased, the effect on the equilibrium constant is
(a) depends on the amount of concentration
(b) unchanged
(c) decreases
(d) increases
Show Answer & Explanation

Answer: (b) unchanged

Explanation:
1. The equilibrium constant depends only on temperature, so it is unchanged.

Teacher's Note:
1. Apply the law of mass action to the given species.
2. Check units and the stoichiometric coefficients before substituting.

Question 19: The effect of temperature on K is log K₂ − log K₁ = (−ΔH/2.303R)[1/T₂ − 1/T₁]. For an endothermic reaction (T₂ > T₁) the false statement is
(a) [1/T₂ − 1/T₁] = positive
(b) log K₂ > log K₁
(c) ΔH = positive
(d) K₂ > K₁
Show Answer & Explanation

Answer: (a) [1/T₂ − 1/T₁] = positive

Explanation:
1. T₂ > T₁ gives 1/T₂ < 1/T₁, so the bracket is negative.
2. With ΔH positive, log K₂ − log K₁ is positive, so K₂ > K₁. Only statement 1 is false.

Teacher's Note:
1. Apply the law of mass action to the given species.
2. Check units and the stoichiometric coefficients before substituting.

Question 20: For A + B ⇌ C + D the equilibrium constant is K when the initial concentrations of A and B are each 0.1 mol L⁻¹. If the experiment is repeated with initial concentrations of 2 and 3 mol L⁻¹, the equilibrium constant is
(a) K/2
(b) K
(c) K²
(d) 1/K
Show Answer & Explanation

Answer: (b) K

Explanation:
1. K is independent of the initial concentrations at a fixed temperature.

Teacher's Note:
1. Apply the law of mass action to the given species.
2. Check units and the stoichiometric coefficients before substituting.

Question 21: In A(s) ⇌ 2B(g) + 3C(g) at equilibrium, the concentration of C is doubled. The new equilibrium concentration of B is
(a) double its original value
(b) half its original value
(c) 2√2 times its original value
(d) 1/(2√2) times its original value
Show Answer & Explanation

Answer: (d) 1/(2√2) times its original value

Explanation:
1. K = [B]²[C]³ is constant.
2. Doubling [C] multiplies [C]³ by 8, so [B]² becomes 1/8 and [B] becomes 1/(2√2) of the original.

Teacher's Note:
1. Apply the law of mass action to the given species.
2. Check units and the stoichiometric coefficients before substituting.

Question 22: Kp for PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) is 16. If the volume of the container is halved, Kp at the same temperature is
(a) 32
(b) 64
(c) 16
(d) 4
Show Answer & Explanation

Answer: (c) 16

Explanation:
1. Kp depends only on temperature, so it stays 16 although the equilibrium shifts backward.

Teacher's Note:
1. Apply the law of mass action to the given species.
2. Check units and the stoichiometric coefficients before substituting.

Question 23: The equilibrium constant of N₂(g) + O₂(g) ⇌ 2NO(g) at 2000 K is 4 × 10⁴. In the presence of a catalyst the equilibrium is reached ten times faster at the same temperature. The equilibrium constant is now
(a) 40 × 10⁻⁴
(b) 4 × 10⁻⁴
(c) 4 × 10⁴
(d) none
Show Answer & Explanation

Answer: (c) 4 × 10⁴

Explanation:
1. A catalyst does not alter K; it only speeds up the approach to equilibrium.

Teacher's Note:
1. Apply the law of mass action to the given species.
2. Check units and the stoichiometric coefficients before substituting.

Question 24: Some helium is introduced into PCl₅ ⇌ PCl₃ + Cl₂ at constant pressure and temperature. The equilibrium constant
(a) increases
(b) decreases
(c) remains unchanged
(d) cannot be said
Show Answer & Explanation

Answer: (c) remains unchanged

Explanation:
1. An inert gas does not change K at constant temperature. It only shifts the position of equilibrium.

Teacher's Note:
1. Apply the law of mass action to the given species.
2. Check units and the stoichiometric coefficients before substituting.

Free study material for Chemistry

Chemical Equilibrium Objective Questions & Solutions for JEE Chemistry

Download Multiple Choice Questions: Chemical Equilibrium (JEE Chemistry)

Review structured objective questions for JEE Chemistry Chemical Equilibrium. Built according to official JEE guidelines, these MCQ sets support daily revision and core concept reinforcement.

Concept Clarification for Chemical Equilibrium

Each question includes structured solution keys mapped directly to standard JEE textbooks, helping students evaluate their reasoning and correct mistakes early in their revision.

Additional Study Resources for JEE Chemistry

Follow up your worksheet practice by attempting the interactive online Chemistry MCQ test for this chapter to evaluate your execution speed. All platform resources are free to access.

FAQs

Where can I access latest JEE Chemistry Chemical Equilibrium MCQs Set 01?

You can get most exhaustive JEE Chemistry Chemical Equilibrium MCQs Set 01 for free on StudiesToday.com. These MCQs for JEE Chemistry are updated for the 2026-27 academic session as per JEE examination standards.

Are Assertion-Reasoning and Case-Study MCQs included in the Chemistry JEE material?

Yes, our JEE Chemistry Chemical Equilibrium MCQs Set 01 include the latest type of questions, such as Assertion-Reasoning and Case-based MCQs. 50% of the JEE paper is now competency-based.

How do practicing Chemistry MCQs help in scoring full marks in JEE exams?

By solving our JEE Chemistry Chemical Equilibrium MCQs Set 01, JEE students can improve their accuracy and speed which is important as objective questions provide a chance to secure 100% marks in the Chemistry.

Do you provide answers and explanations for JEE Chemistry Chemical Equilibrium MCQs Set 01?

Yes, Chemistry MCQs for JEE have answer key and brief explanations to help students understand logic behind the correct option as its important for 2026 competency-focused JEE exams.

Can I practice these Chemistry JEE MCQs online?

Yes, you can also access online interactive tests for JEE Chemistry Chemical Equilibrium MCQs Set 01 on StudiesToday.com as they provide instant answers and score to help you track your progress in Chemistry.