JEE Chemistry Chemical Equilibrium MCQs Set 03

Practice MCQs for JEE Chemistry Chemical Equilibrium

Access targeted multiple-choice questions for Chemical Equilibrium designed to align with the latest JEE academic syllabus for JEE Chemistry. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.

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View or download the dedicated Chemical Equilibrium MCQ resource below. Practicing these 50 objective questions regularly builds familiarity with standard exam patterns and helps secure higher marks in final Chemistry evaluations.

Multiple Choice Questions

Question 1: On cooling the system CO₂(s) ⇌ CO₂(g) at equilibrium
(a) there is no effect on the equilibrium state
(b) more gas is formed
(c) more gas solidifies
(d) none of the above
Show Answer & Explanation

Answer: (c) more gas solidifies

Explanation:
1. Sublimation is endothermic, so cooling favours the solid: more gas solidifies.

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

Question 2: On heating PCl₅, PCl₃ and Cl₂ form. The vapour density of the gas mixture is 70.2 at 200°C and 57.9 at 250°C. The degrees of dissociation at 200°C and 250°C are
(a) 48.50% and 80%
(b) 60% and 70%
(c) 70% and 80%
(d) 80% and 90%
Show Answer & Explanation

Answer: (a) 48.50% and 80%

Explanation:
1. D = 104.25 (theoretical); α = (D − d)/d.
2. At 200°C: (104.25 − 70.2)/70.2 = 0.485. At 250°C: (104.25 − 57.9)/57.9 = 0.80.

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: The vapour density of PCl₅ is 104.16, but when heated to 230°C it falls to 62. The degree of dissociation is
(a) 6.8%
(b) 68%
(c) 46%
(d) 64%
Show Answer & Explanation

Answer: (b) 68%

Explanation:
1. α = (D − d)/d = (104.16 − 62)/62 = 0.68, i.e. 68%.

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: For CH₄(g) + 2O₂(g) ⇌ CO₂(g) + 2H₂O(l), ΔH = −170.8 kJ mol⁻¹, which statement is not true?
(a) At equilibrium the concentrations of CO₂(g) and H₂O(l) are not equal
(b) The equilibrium constant is Kp = [CO₂]/([CH₄][O₂])
(c) Addition of CH₄(g) or O₂(g) at equilibrium causes a shift to the right
(d) The reaction is exothermic
Show Answer & Explanation

Answer: (b) The equilibrium constant is Kp = [CO₂]/([CH₄][O₂])

Explanation:
1. The correct expression is Kp = [CO₂]/([CH₄][O₂]²) since O₂ has coefficient 2, and liquid water is omitted.

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: Given N₂ + 3H₂ ⇌ 2NH₃ (K₁), N₂ + O₂ ⇌ 2NO (K₂), H₂ + ½O₂ ⇌ H₂O (K₃), the equilibrium constant of 2NH₃ + (5/2)O₂ ⇌ 2NO + 3H₂O is
(a) K₁K₂/K₃
(b) K₂K₃²/K₁
(c) K₂K₃³/K₁
(d) K₁K₂K₃
Show Answer & Explanation

Answer: (c) K₂K₃³/K₁

Explanation:
1. Take N₂ + O₂ ⇌ 2NO (K₂), 3 × [H₂ + ½O₂ ⇌ H₂O] (K₃³), and reverse the ammonia equation (1/K₁).
2. K = K₂K₃³/K₁.

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: The equilibrium constant of HI(g) ⇌ ½H₂(g) + ½I₂(g) is 8.0. The equilibrium constant of H₂(g) + I₂(g) ⇌ 2HI(g) is
(a) 1/64
(b) 16
(c) 1/8
(d) 1/16
Show Answer & Explanation

Answer: (a) 1/64

Explanation:
1. Reversing gives K = 1/8 for ½H₂ + ½I₂ ⇌ HI; doubling the equation squares it: (1/8)² = 1/64.

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: The Kp values for X ⇌ Y + Z (1) and A ⇌ 2B (2) are in the ratio 9 : 1. If the degrees of dissociation of X and A are equal, the ratio of the total pressures at equilibrium for (1) and (2) is
(a) 1 : 9
(b) 36 : 1
(c) 1 : 1
(d) 3 : 1
Show Answer & Explanation

Answer: (b) 36 : 1

Explanation:
1. Kp1 = α²P₁/(1 − α²) and Kp2 = 4α²P₂/(1 − α²).
2. Kp1/Kp2 = P₁/(4P₂) = 9, so P₁/P₂ = 36.

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: The reaction 2A(g) + B(g) ⇌ 3C(g) + D(g) starts with [A] = [B] = 1.00 M. At equilibrium [D] = 0.25 M. The equilibrium constant is given by
(a) [(0.75)³(0.25)]/[(0.50)²(0.75)]
(b) [(0.75)³(0.25)]/[(0.50)²(0.25)]
(c) [(0.75)³(0.25)]/[(0.75)²(0.25)]
(d) [(0.75)³(0.25)]/[(1.00)²(1.00)]
Show Answer & Explanation

Answer: (a) [(0.75)³(0.25)]/[(0.50)²(0.75)]

Explanation:
1. D = 0.25 means B used = 0.25, A used = 0.50 and C formed = 0.75.
2. [A] = 0.50, [B] = 0.75, so K = (0.75)³(0.25)/((0.50)²(0.75)).

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: In which equilibrium are Kc and Kp not equal?
(a) 2C(s) + O₂(g) ⇌ 2CO₂(g)
(b) 2NO(g) ⇌ N₂(g) + O₂(g)
(c) SO₂(g) + NO₂(g) ⇌ SO₃(g) + NO(g)
(d) H₂(g) + I₂(g) ⇌ 2HI(g)
Show Answer & Explanation

Answer: (a) 2C(s) + O₂(g) ⇌ 2CO₂(g)

Explanation:
1. Kp = Kc only when Δn(gas) = 0.
2. In 2C(s) + O₂(g) ⇌ 2CO₂(g), Δn(gas) = 2 − 1 = 1, so Kp ≠ Kc. The other three reactions have Δn = 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 10: For N₂(g) + O₂(g) ⇌ 2NO(g), K = K₁; for 2NO(g) + O₂(g) ⇌ 2NO₂(g), K = K₂. The K for NO₂(g) ⇌ ½N₂(g) + O₂(g) is
(a) 1/(2K₁K₂)
(b) 1/(4K₁K₂)
(c) [1/(K₁K₂)]^(1/2)
(d) 1/(K₁K₂)
Show Answer & Explanation

Answer: (c) [1/(K₁K₂)]^(1/2)

Explanation:
1. Adding the equations gives N₂ + 2O₂ ⇌ 2NO₂ with K₁K₂.
2. Reversing and halving: K = (1/(K₁K₂))^(1/2).

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: Which of the following is not a characteristic of equilibrium?
(a) The rate is equal in both directions
(b) Measurable quantities are constant at equilibrium
(c) Equilibrium occurs in a reversible condition
(d) Equilibrium occurs only in an open vessel at constant temperature
Show Answer & Explanation

Answer: (d) Equilibrium occurs only in an open vessel at constant temperature

Explanation:
1. Chemical equilibrium needs a closed system at constant temperature, not an open 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 12: For A₂(g) + B₂(g) ⇌ 2AB(g) the equilibrium concentrations are A₂ = 3.0 × 10⁻³ M, B₂ = 4.2 × 10⁻³ M and AB = 2.8 × 10⁻³ M. If the reaction is in a sealed vessel at 527°C, Kc is
(a) 0.62
(b) 4.5
(c) 2.0
(d) 1.9
Show Answer & Explanation

Answer: (a) 0.62

Explanation:
1. Kc = [AB]²/([A₂][B₂]) = (2.8)²/(3.0 × 4.2) = 7.84/12.6 = 0.62.

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: Kp/Kc for CO(g) + ½O₂(g) → CO₂(g) is
(a) RT
(b) 1/RT
(c) 1/√(RT)
(d) RT/2
Show Answer & Explanation

Answer: (c) 1/√(RT)

Explanation:
1. Δn = 1 − 1.5 = −½, so Kp/Kc = (RT)^(−1/2) = 1/√(RT).

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 the reversible reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g) + heat, the equilibrium shifts in the forward direction
(a) by increasing the concentration of NH₃
(b) by decreasing the pressure
(c) by decreasing the concentrations of N₂ and H₂
(d) by increasing the pressure and decreasing the temperature
Show Answer & Explanation

Answer: (d) by increasing the pressure and decreasing the temperature

Explanation:
1. High pressure favours fewer gas moles and low temperature favours the exothermic forward reaction.

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: If the equilibrium constant for a particular reaction is 1.6 × 10¹², then at equilibrium the system will contain
(a) mostly reactants
(b) mostly products
(c) similar amounts of reactants and products
(d) all reactants
Show Answer & Explanation

Answer: (b) mostly products

Explanation:
1. A very large K means the products dominate at 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 16: If K is the equilibrium constant for N₂(g) + O₂(g) ⇌ 2NO(g), the equilibrium constant for ½N₂(g) + ½O₂(g) ⇌ NO(g) is
(a) K
(b) K²
(c) K^(1/2)
(d) K/2
Show Answer & Explanation

Answer: (c) K^(1/2)

Explanation:
1. Halving an equation takes the square root of K.

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: For A(g) ⇌ B(g), the initial pressure at 300°C is 1 atm and Kp = 4. The pressure of B at equilibrium is
(a) 1/5 atm
(b) 2/5 atm
(c) 4/5 atm
(d) 3/5 atm
Show Answer & Explanation

Answer: (c) 4/5 atm

Explanation:
1. If p(B) = x, then p(A) = 1 − x and x/(1 − x) = 4.
2. x = 4/5 atm.

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: A 20 L container at 400 K holds CO₂(g) at 0.4 atm with excess SrO (volume of solid neglected). The piston is moved to decrease the volume. The maximum volume of the container when the pressure of CO₂ reaches its maximum value is (Kp for SrCO₃(s) ⇌ SrO(s) + CO₂(g) is 1.6 atm)
(a) 10 litre
(b) 4 litre
(c) 2 litre
(d) 5 litre
Show Answer & Explanation

Answer: (d) 5 litre

Explanation:
1. The maximum CO₂ pressure is Kp = 1.6 atm.
2. Boyle's law: 0.4 × 20 = 1.6 × V gives V = 5 L.

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: At a certain temperature and total pressure of 1 atm, an equilibrium mixture contains 20% by volume of A atoms for A₂(g) ⇌ 2A(g). Kp is
(a) 5 × 10⁻²
(b) 2 × 10⁻³
(c) 5 × 10⁻³
(d) 4 × 10⁻⁴
Show Answer & Explanation

Answer: (a) 5 × 10⁻²

Explanation:
1. p(A) = 0.2 atm and p(A₂) = 0.8 atm.
2. Kp = (0.2)²/0.8 = 0.05 = 5 × 10⁻².

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(g) + B(g) ⇌ C(g), 1 mol of each reactant is taken in a 2 L container. At equilibrium 20% of A has reacted. Kc is
(a) 0.225
(b) 0.412
(c) 0.625
(d) 0.925
Show Answer & Explanation

Answer: (c) 0.625

Explanation:
1. A = B = 0.8 mol/2 L = 0.4 M; C = 0.2/2 = 0.1 M.
2. Kc = 0.1/(0.4 × 0.4) = 0.625.

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: Which condition will favour the maximum formation of product in A₂(g) + B₂(g) ⇌ X₂(g), ΔH = −X kJ?
(a) Low temperature and high pressure
(b) Low temperature and low pressure
(c) High temperature and high pressure
(d) High temperature and low pressure
Show Answer & Explanation

Answer: (a) Low temperature and high pressure

Explanation:
1. Exothermic: low temperature favours the product; fewer moles on the right: high pressure favours it.

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: At constant temperature, on increasing pressure which is correct?
(a) Solid ⇌ liquid: more liquid is formed
(b) Formation of ammonia from N₂ and H₂ decreases
(c) CO₂(g) + C(s) ⇌ 2CO(g) shifts in the forward direction
(d) Solubility of a gas in a liquid increases
Show Answer & Explanation

Answer: (d) Solubility of a gas in a liquid increases

Explanation:
1. Higher pressure increases the solubility of a gas in a liquid (Henry's law).

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: A + 2B ⇌ 2C. 2 moles each of A and B are taken in a 10 L flask and 1 mole of C is formed at equilibrium. Kc is
(a) 6.66
(b) 3.33
(c) 2.22
(d) 1.11
Show Answer & Explanation

Answer: (a) 6.66

Explanation:
1. A = 1.5 mol, B = 1 mol, C = 1 mol in 10 L: 0.15, 0.1, 0.1 M.
2. Kc = (0.1)²/(0.15 × (0.1)²) = 6.67.

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: At constant temperature in a closed container CaCO₃(s) ⇌ CaO(s) + CO₂(g). Which of the following changes the pressure?
(a) Size of the container
(b) Temperature
(c) Addition of CaO(s)
(d) Amount of CaCO₃(s)
Show Answer & Explanation

Answer: (b) Temperature

Explanation:
1. The pressure equals Kp, which depends only on temperature.

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

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