JEE Chemistry Chemical Equilibrium MCQs Set 02

Multiple Choice Questions (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.

Practice Chemical Equilibrium MCQs for JEE Chemistry

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: Match List X with List Y: (A) Active mass, (B) Dynamic nature, (C) A + heat ⇌ B, (D) log(Kp₂/Kp₁) = (ΔH/2.303R)[1/T₁ − 1/T₂], (E) 2A(g) + B(g) ⇌ 3C(g) with (I) Δn = 0, (II) molar concentration, (III) van't Hoff equation, (IV) adaptation if temperature increases, (V) chemical equilibrium
(a) A–V, B–II, C–III, D–I, E–IV
(b) A–V, B–IV, C–III, D–II, E–I
(c) A–II, B–V, C–IV, D–III, E–I
(d) none of these
Show Answer & Explanation

Answer: (c) A–II, B–V, C–IV, D–III, E–I

Explanation:
1. Active mass is the molar concentration (II); dynamic nature belongs to chemical equilibrium (V).
2. A + heat ⇌ B shifts forward when temperature rises (IV); the K–T relation is van't Hoff's equation (III); 2A + B ⇌ 3C has Δn = 0 (I).

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: 2CO₂ ⇌ 2CO + O₂. If 2 mol of CO₂ is taken and 40% dissociates, the total number of moles at equilibrium is
(a) 2.4
(b) 2.0
(c) 1.2
(d) 5
Show Answer & Explanation

Answer: (a) 2.4

Explanation:
1. CO₂ left = 1.2 mol; CO formed = 0.8 mol; O₂ formed = 0.4 mol.
2. Total = 1.2 + 0.8 + 0.4 = 2.4 mol.

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 A₃(g) ⇌ 3A(g) the initial concentration of A₃ is a mol L⁻¹ and x is the degree of dissociation. The total concentration at equilibrium is
(a) a − ax/3
(b) a/3 − x
(c) (a − ax)/2
(d) a + 2ax
Show Answer & Explanation

Answer: (d) a + 2ax

Explanation:
1. A₃ left = a(1 − x); A formed = 3ax.
2. Total = a − ax + 3ax = a + 2ax.

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: A 13 L vessel initially contains 12 g C, 64 g S₂ and 76 g CS₂ for C(s) + S₂(g) ⇌ CS₂(g) at 1027°C. The total pressure is
(a) 200R
(b) 158R
(c) 100R
(d) 79R
Show Answer & Explanation

Answer: (a) 200R

Explanation:
1. Moles: C = 1 (solid), S₂ = 1, CS₂ = 1. Gas moles = 2.
2. P = nRT/V = 2 × R × 1300/13 = 200R.

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: A + B ⇌ C + D in a 1 L vessel at 250°C. Initial [A] = 3n and [B] = n. At equilibrium [C] equals [B]. The equilibrium concentration of D is
(a) n/2
(b) 3n − n/2
(c) n + n/2
(d) n
Show Answer & Explanation

Answer: (a) n/2

Explanation:
1. If x of each is formed, [C] = x and [B] = n − x.
2. [C] = [B] gives x = n/2, so [D] = n/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 6: The initial pressure of an N₂ and H₂ mixture in a closed apparatus is 100 atm. At equilibrium 20% of the mixture reacts. The total pressure at the same temperature is
(a) 100
(b) 90
(c) 85
(d) 80
Show Answer & Explanation

Answer: (b) 90

Explanation:
1. N₂ + 3H₂ → 2NH₃: 4 volumes become 2, a decrease of half the reacted amount.
2. Reacted = 20 atm, decrease = 10 atm, so the total pressure is 90 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 7: 4 mol of PCl₅ are heated at constant temperature in a closed container. If the degree of dissociation is 0.5, the total number of moles at equilibrium is
(a) 4.5
(b) 6
(c) 3
(d) 4
Show Answer & Explanation

Answer: (b) 6

Explanation:
1. PCl₅ = 4(1 − 0.5) = 2; PCl₃ = 2; Cl₂ = 2.
2. Total = 6 mol.

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: 8 mol of PCl₅ are heated in a 10 L closed vessel and 25% dissociates into PCl₃ and Cl₂. At equilibrium Kp equals (P = total pressure)
(a) P/30
(b) P/15
(c) 2/3P
(d) 3/2P
Show Answer & Explanation

Answer: (b) P/15

Explanation:
1. Moles: PCl₅ = 6, PCl₃ = 2, Cl₂ = 2; total = 10.
2. Kp = (0.2P)(0.2P)/(0.6P) = P/15.

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: For PCl₅ ⇌ PCl₃ + Cl₂ the equilibrium partial pressures of PCl₃, Cl₂ and PCl₅ are 0.3, 0.2 and 0.6 atm. If the partial pressures of PCl₃ and Cl₂ are doubled, the partial pressure of PCl₅ at the new equilibrium is
(a) 0.3
(b) 1.2
(c) 2.4
(d) 0.15
Show Answer & Explanation

Answer: (c) 2.4

Explanation:
1. Kp = 0.3 × 0.2/0.6 = 0.1 atm.
2. New: 0.6 × 0.4/p = 0.1 gives p = 2.4 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 10: 4 mol of NO in a 0.25 L tube is 10% dissociated. Kp for 2NO ⇌ N₂ + O₂ is
(a) 1/(18)²
(b) 1/(8)²
(c) 1/16
(d) 1/32
Show Answer & Explanation

Answer: (a) 1/(18)²

Explanation:
1. NO = 3.6 mol, N₂ = 0.2 mol, O₂ = 0.2 mol; Δn = 0 so Kp = Kc.
2. K = (0.2)(0.2)/(3.6)² = 1/(18)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: In A + B ⇌ C + D, one mole each of the reactants are mixed and 0.4 mol each of the products are formed. The equilibrium constant is
(a) 1
(b) 0.36
(c) 2.25
(d) 4/9
Show Answer & Explanation

Answer: (d) 4/9

Explanation:
1. Equilibrium amounts: A = B = 0.6, C = D = 0.4.
2. K = (0.4 × 0.4)/(0.6 × 0.6) = 4/9.

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: Kp for H₂ + I₂ ⇌ 2HI. 2 moles each of H₂ and I₂ are taken initially and 2 moles of HI are present at equilibrium. Kp is
(a) 2.5
(b) 4
(c) 0.25
(d) 1.0
Show Answer & Explanation

Answer: (b) 4

Explanation:
1. HI = 2 means 1 mol of each reactant is used; H₂ = I₂ = 1.
2. K = 2²/(1 × 1) = 4.

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: 4 mol of A are mixed with 4 mol of B. 2 mol of C are formed at equilibrium for A + B ⇌ C + D. The equilibrium constant is
(a) 4
(b) 1
(c) √2
(d) √4
Show Answer & Explanation

Answer: (b) 1

Explanation:
1. A = B = 2, C = D = 2 mol; Δn = 0.
2. K = (2 × 2)/(2 × 2) = 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 14: 4.5 mol each of hydrogen and iodine are heated in a sealed 10 L vessel. At equilibrium 3 mol of HI are found. K for H₂ + I₂ ⇌ 2HI is
(a) 1
(b) 10
(c) 5
(d) 0.33
Show Answer & Explanation

Answer: (a) 1

Explanation:
1. HI = 3 mol means 1.5 mol of each reactant used; H₂ = I₂ = 3 mol.
2. K = 3²/(3 × 3) = 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 15: 1.50 mol each of H₂ and I₂ are placed in a sealed 10 L container at 717 K. At equilibrium 1.25 mol of each are left. Kc for H₂ + I₂ ⇌ 2HI is
(a) 0.4
(b) 0.16
(c) 25
(d) 50
Show Answer & Explanation

Answer: (b) 0.16

Explanation:
1. HI formed = 2 × 0.25 = 0.5 mol.
2. Kc = (0.5)²/(1.25)² = 0.16.

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: For cis-2-pentene ⇌ trans-2-pentene, ΔG° at 400 K is −3.67 kJ mol⁻¹. If excess trans-2-pentene is added to the system
(a) additional trans-2-pentene will form
(b) excess of cis-2-pentene will form
(c) equilibrium will proceed in the forward direction
(d) equilibrium will remain unaffected
Show Answer & Explanation

Answer: (b) excess of cis-2-pentene will form

Explanation:
1. Adding product makes Q > K, so the reaction shifts backward and cis-2-pentene forms.

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: When NaNO₃ is heated in a closed vessel, O₂ is liberated and NaNO₂ is left behind. At equilibrium
(a) addition of NaNO₃ favours the forward reaction
(b) addition of NaNO₂ favours the reverse reaction
(c) increasing pressure favours the reverse reaction
(d) decreasing temperature favours the forward reaction
Show Answer & Explanation

Answer: (c) increasing pressure favours the reverse reaction

Explanation:
1. 2NaNO₃(s) ⇌ 2NaNO₂(s) + O₂(g): adding solids has no effect.
2. Raising the pressure shifts toward fewer gas moles, which is the reverse direction.

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: In the manufacture of NO, the reaction of N₂ and O₂ to form NO is favourable if
(a) pressure is increased
(b) pressure is decreased
(c) temperature is increased
(d) temperature is decreased
Show Answer & Explanation

Answer: (c) temperature is increased

Explanation:
1. N₂ + O₂ ⇌ 2NO is endothermic with Δn = 0, so only a higher temperature favours NO.

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: In which equilibrium would the position shift to the right if the total pressure is decreased?
(a) N₂ + 3H₂ ⇌ 2NH₃
(b) H₂ + I₂ ⇌ 2HI
(c) N₂O₄ ⇌ 2NO₂
(d) H₂ + Cl₂ ⇌ 2HCl
Show Answer & Explanation

Answer: (c) N₂O₄ ⇌ 2NO₂

Explanation:
1. Lower pressure favours the side with more gas moles.
2. Only N₂O₄ ⇌ 2NO₂ has more moles on the right (Δn = +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 20: For the manufacture of ammonia N₂ + 3H₂ ⇌ 2NH₃ + 21.9 kCal, the favourable conditions are
(a) low temperature, low pressure and a catalyst
(b) low temperature, high pressure and a catalyst
(c) high temperature, low pressure and a catalyst
(d) high temperature, high pressure and a catalyst
Show Answer & Explanation

Answer: (b) low temperature, high pressure and a catalyst

Explanation:
1. Exothermic: low temperature favours NH₃. Fewer gas moles: high pressure favours NH₃. A catalyst speeds up attainment.

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 2A(g) + B(g) ⇌ C(g) + 362 kCal, which combination of pressure and temperature gives the highest yield of C?
(a) 1000 atm and 500°C
(b) 500 atm and 500°C
(c) 1000 atm and 50°C
(d) 500 atm and 100°C
Show Answer & Explanation

Answer: (c) 1000 atm and 50°C

Explanation:
1. The reaction is exothermic and reduces gas moles, so the highest pressure and the lowest temperature are best.

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: For aA ⇌ bB + cC, ΔH = −x kcal, high pressure and low temperature favour the formation of the products. Hence
(a) a > b + c
(b) a < b + c
(c) a = b + c
(d) none of them
Show Answer & Explanation

Answer: (a) a > b + c

Explanation:
1. High pressure favours the side with fewer moles, so the product side has fewer moles: a > b + c.

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 reaction in which the yield of product cannot be increased by applying high pressure is
(a) PCl₃ + Cl₂ ⇌ PCl₅
(b) N₂ + O₂ ⇌ 2NO
(c) N₂ + 3H₂ ⇌ 2NH₃
(d) 2SO₂ + O₂ ⇌ 2SO₃
Show Answer & Explanation

Answer: (b) N₂ + O₂ ⇌ 2NO

Explanation:
1. Pressure has no effect when Δn = 0, as in N₂ + O₂ ⇌ 2NO.

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: For H₂O(l) ⇌ H₂O(g), if pressure is applied
(a) more water evaporates
(b) the boiling point of water is increased
(c) there is no effect on the boiling point
(d) none of the above
Show Answer & Explanation

Answer: (b) the boiling point of water is increased

Explanation:
1. A higher external pressure requires a higher vapour pressure to boil, so the boiling point rises.

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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Multiple Choice Questions (MCQs) for JEE Chemistry Chemical Equilibrium

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