JEE Chemistry Equilibrium MCQs Set 10

Chemistry Objective Questions and Answers: Equilibrium

Access targeted multiple-choice questions for 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.

Download Equilibrium MCQs with Answers

Access the complete set of multiple-choice questions for 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: For the reaction \( \ce{PCl3(g) + Cl2(g) <=> PCl5(g)} \) the position of equilibrium can be shifted to the right by
(a) increasing the temperature
(b) doubling the volume
(c) adding \( \ce{Cl2} \) at constant volume
(d) adding \( \ce{PCl5} \) at constant volume
Show Answer & Explanation

Answer: (c) adding \( \ce{Cl2} \) at constant volume

Explanation:
1. Adding a reactant raises its concentration, so the system moves forward to use it up (Le Chatelier).
2. Higher temperature favours the endothermic (reverse) reaction, a larger volume favours the side with more moles (reverse), and adding \( \ce{PCl5} \) shifts the equilibrium to the left.

Teacher's Note:
1. Write the balanced equation and the expression for the equilibrium constant first.
2. Use equilibrium concentrations (moles divided by volume) in K, not initial ones.

Question 2: For \( \ce{A(g) + 2B(g) <=> C(g)} \) the equilibrium concentrations are [A] = 0.06, [B] = 0.12 and [C] = 0.216 mol/L. The value of \( K_c \) is
(a) 250
(b) 416
(c) \( 4\times10^{-3} \)
(d) 125
Show Answer & Explanation

Answer: (a) 250

Explanation:
1. \( K_c=\frac{[C]}{[A][B]^2}=\frac{0.216}{0.06\times(0.12)^2} \) = 250.

Teacher's Note:
1. Write the balanced equation and the expression for the equilibrium constant first.
2. Use equilibrium concentrations (moles divided by volume) in K, not initial ones.

Question 3: For \( \ce{A + 2B <=> C + 3D} \) the partial pressures at equilibrium are A = 0.20, B = 0.10, C = 0.30 and D = 0.50 atm. The numerical value of the equilibrium constant is
(a) 11.25
(b) 18.75
(c) 5
(d) 3.75
Show Answer & Explanation

Answer: (b) 18.75

Explanation:
1. \( K_p=\frac{p_C\,p_D^3}{p_A\,p_B^2}=\frac{0.30\times(0.50)^3}{0.20\times(0.10)^2} \) = 0.0375/0.002 = 18.75.

Teacher's Note:
1. Write the balanced equation and the expression for the equilibrium constant first.
2. Use equilibrium concentrations (moles divided by volume) in K, not initial ones.

Question 4: 2 moles of \( \ce{PCl5} \) were heated in a closed vessel of 2 litre capacity. At equilibrium 40% of \( \ce{PCl5} \) is dissociated into \( \ce{PCl3} \) and \( \ce{Cl2} \). The value of \( K_c \) is
(a) 0.266
(b) 0.53
(c) 2.66
(d) 5.3
Show Answer & Explanation

Answer: (a) 0.266

Explanation:
1. Initial [PCl5] = 1 M; dissociated = 0.4 M.
2. At equilibrium [PCl5] = 0.6 M, [PCl3] = [Cl2] = 0.4 M.
3. \( K_c=0.4\times0.4/0.6 \) = 0.267.

Teacher's Note:
1. Write the balanced equation and the expression for the equilibrium constant first.
2. Use equilibrium concentrations (moles divided by volume) in K, not initial ones.

Question 5: 4 moles of A are mixed with 4 moles of B. At equilibrium of \( \ce{A + B <=> C + D} \), 2 moles of C and D are formed. The equilibrium constant is
(a) 1/4
(b) 1/2
(c) 1
(d) 4
Show Answer & Explanation

Answer: (c) 1

Explanation:
1. At equilibrium: A = 2, B = 2, C = 2, D = 2 mol (volume cancels).
2. K = (2 x 2)/(2 x 2) = 1.

Teacher's Note:
1. Write the balanced equation and the expression for the equilibrium constant first.
2. Use equilibrium concentrations (moles divided by volume) in K, not initial ones.

Question 6: The equilibrium concentrations of HI, \( \ce{H2} \) and \( \ce{I2} \) are 0.80, 0.10 and 0.10 mol/L. The equilibrium constant for \( \ce{H2 + I2 <=> 2HI} \) is
(a) 64
(b) 12
(c) 8
(d) 0.8
Show Answer & Explanation

Answer: (a) 64

Explanation:
1. \( K_c=\frac{(0.80)^2}{0.10\times0.10} \) = 64.

Teacher's Note:
1. Write the balanced equation and the expression for the equilibrium constant first.
2. Use equilibrium concentrations (moles divided by volume) in K, not initial ones.

Question 7: In \( \ce{A + B <=> C + D} \), one mole each of the two reactants is mixed and 0.6 mole each of the products is formed. The equilibrium constant is
(a) 1
(b) 0.36
(c) 2.25
(d) 4/9
Show Answer & Explanation

Answer: (c) 2.25

Explanation:
1. Reactants left = 1 - 0.6 = 0.4 mol each.
2. K = (0.6 x 0.6)/(0.4 x 0.4) = 2.25.

Teacher's Note:
1. Write the balanced equation and the expression for the equilibrium constant first.
2. Use equilibrium concentrations (moles divided by volume) in K, not initial ones.

Question 8: The decomposition of \( \ce{N2O4} \) to \( \ce{NO2} \) is carried out at 280 K in chloroform. At equilibrium 0.2 mol of \( \ce{N2O4} \) and \( 2\times10^{-3} \) mol of \( \ce{NO2} \) are present in 2 litre of solution. The equilibrium constant for \( \ce{N2O4 <=> 2NO2} \) is
(a) \( 1\times10^{-2} \)
(b) \( 2\times10^{-3} \)
(c) \( 1\times10^{-5} \)
(d) \( 2\times10^{-5} \)
Show Answer & Explanation

Answer: (c) \( 1\times10^{-5} \)

Explanation:
1. [N2O4] = 0.1 M; [NO2] = \( 1\times10^{-3} \) M.
2. \( K_c=\frac{(10^{-3})^2}{0.1}=1\times10^{-5} \).

Teacher's Note:
1. Write the balanced equation and the expression for the equilibrium constant first.
2. Use equilibrium concentrations (moles divided by volume) in K, not initial ones.

Question 9: A quantity of \( \ce{PCl5} \) was heated in a 10 litre vessel at 250 C. At equilibrium the vessel contains 0.1 mol \( \ce{PCl5} \), 0.20 mol \( \ce{PCl3} \) and 0.2 mol \( \ce{Cl2} \). The equilibrium constant is
(a) 0.02
(b) 0.05
(c) 0.04
(d) 0.025
Show Answer & Explanation

Answer: (c) 0.04

Explanation:
1. Concentrations: 0.01, 0.02 and 0.02 mol/L.
2. \( K_c=\frac{0.02\times0.02}{0.01} \) = 0.04.

Teacher's Note:
1. Write the balanced equation and the expression for the equilibrium constant first.
2. Use equilibrium concentrations (moles divided by volume) in K, not initial ones.

Question 10: 0.3 mole of \( \ce{H2} \) and 0.3 mole of \( \ce{I2} \) are allowed to react in a 10 litre flask at 500 C. If \( K_c \) for \( \ce{H2 + I2 <=> 2HI} \) is 64, the amount of unreacted \( \ce{I2} \) at equilibrium is
(a) 0.15 mole
(b) 0.06 mole
(c) 0.03 mole
(d) 0.2 mole
Show Answer & Explanation

Answer: (b) 0.06 mole

Explanation:
1. Let x mol each of H2 and I2 react: HI = 2x. Volume cancels since the moles of products and reactants are equal.
2. \( \frac{(2x)^2}{(0.3-x)^2}=64 \) so 2x/(0.3 - x) = 8 and x = 0.24.
3. Unreacted I2 = 0.3 - 0.24 = 0.06 mol.

Teacher's Note:
1. Write the balanced equation and the expression for the equilibrium constant first.
2. Use equilibrium concentrations (moles divided by volume) in K, not initial ones.

Question 11: At equilibrium the rate constant of the backward reaction is \( 7.5\times10^{-4} \) and the equilibrium constant is 1.5. The rate constant of the forward reaction is
(a) \( 5\times10^{-4} \)
(b) \( 2\times10^{-3} \)
(c) \( 1.125\times10^{-3} \)
(d) \( 9.0\times10^{-4} \)
Show Answer & Explanation

Answer: (c) \( 1.125\times10^{-3} \)

Explanation:
1. \( K_c=k_f/k_b \) so \( k_f=1.5\times7.5\times10^{-4} \) = \( 1.125\times10^{-3} \).

Teacher's Note:
1. Write the balanced equation and the expression for the equilibrium constant first.
2. Use equilibrium concentrations (moles divided by volume) in K, not initial ones.

Question 12: 28 g of \( \ce{N2} \) and 6 g of \( \ce{H2} \) were kept at 400 C in a 1 litre vessel. The equilibrium mixture contained 27.54 g of \( \ce{NH3} \). The approximate value of \( K_c \) (in \( mol^{-2}L^2 \)) is
(a) 75
(b) 50
(c) 25
(d) 100
Show Answer & Explanation

Answer: (a) 75

Explanation:
1. Initial: N2 = 1 mol, H2 = 3 mol. NH3 formed = 27.54/17 = 1.62 mol.
2. N2 left = 1 - 0.81 = 0.19; H2 left = 3 - 2.43 = 0.57.
3. \( K_c=\frac{(1.62)^2}{0.19\times(0.57)^3} \) = 75.

Teacher's Note:
1. Write the balanced equation and the expression for the equilibrium constant first.
2. Use equilibrium concentrations (moles divided by volume) in K, not initial ones.

Question 13: The equilibrium concentrations of X, Y and \( \ce{YX2} \) are 4, 2 and 2 mol/L for \( \ce{2X + Y <=> YX2} \). The value of \( K_c \) is
(a) 0.625
(b) 0.0625
(c) 6.25
(d) 0.00625
Show Answer & Explanation

Answer: (b) 0.0625

Explanation:
1. \( K_c=\frac{[YX_2]}{[X]^2[Y]}=\frac{2}{16\times2} \) = 0.0625.

Teacher's Note:
1. Write the balanced equation and the expression for the equilibrium constant first.
2. Use equilibrium concentrations (moles divided by volume) in K, not initial ones.

Question 14: Solid \( \ce{NH4HS} \) is placed in a flask already containing ammonia at 0.50 atm. It decomposes to \( \ce{NH3} \) and \( \ce{H2S} \), and at equilibrium the total pressure is 0.84 atm. The equilibrium constant \( K_p \) is
(a) 0.30
(b) 0.18
(c) 0.17
(d) 0.11
Show Answer & Explanation

Answer: (d) 0.11

Explanation:
1. Let each gas formed have partial pressure x: total = 0.50 + x + x = 0.84, so x = 0.17 atm.
2. p(NH3) = 0.67 atm, p(H2S) = 0.17 atm.
3. \( K_p=0.67\times0.17 \) = 0.11.

Teacher's Note:
1. Write the balanced equation and the expression for the equilibrium constant first.
2. Use equilibrium concentrations (moles divided by volume) in K, not initial ones.

Question 15: In \( \ce{A + 2B <=> 2C} \), 2.0 mol of A, 3.0 mol of B and 2.0 mol of C are placed in a 2.0 L flask and the equilibrium concentration of C is 0.5 mol/L. The value of \( K_c \) is
(a) 0.073
(b) 0.147
(c) 0.05
(d) 0.026
Show Answer & Explanation

Answer: (c) 0.05

Explanation:
1. Initial concentrations: A = 1.0, B = 1.5, C = 1.0 mol/L.
2. C falls to 0.5 M, so the reaction runs backwards by 0.5 M of C: A = 1.0 + 0.25 = 1.25 M and B = 1.5 + 0.5 = 2.0 M.
3. \( K_c=\frac{(0.5)^2}{1.25\times(2.0)^2} \) = 0.05.

Teacher's Note:
1. Write the balanced equation and the expression for the equilibrium constant first.
2. Use equilibrium concentrations (moles divided by volume) in K, not initial ones.

Question 16: In a 500 mL vessel CO and \( \ce{Cl2} \) are mixed to form \( \ce{COCl2} \). At equilibrium it contains 0.2 mol \( \ce{COCl2} \) and 0.1 mol each of CO and \( \ce{Cl2} \). The equilibrium constant \( K_c \) for \( \ce{CO + Cl2 <=> COCl2} \) is
(a) 5
(b) 10
(c) 15
(d) 20
Show Answer & Explanation

Answer: (b) 10

Explanation:
1. Concentrations: COCl2 = 0.4 M, CO = 0.2 M, Cl2 = 0.2 M.
2. \( K_c=\frac{0.4}{0.2\times0.2} \) = 10.

Teacher's Note:
1. Write the balanced equation and the expression for the equilibrium constant first.
2. Use equilibrium concentrations (moles divided by volume) in K, not initial ones.

Question 17: The equilibrium concentrations of HI, \( \ce{I2} \) and \( \ce{H2} \) are 0.7, 0.1 and 0.1 M. The equilibrium constant for \( \ce{I2 + H2 <=> 2HI} \) is
(a) 36
(b) 49
(c) 0.49
(d) 0.36
Show Answer & Explanation

Answer: (b) 49

Explanation:
1. \( K_c=\frac{(0.7)^2}{0.1\times0.1} \) = 49.

Teacher's Note:
1. Write the balanced equation and the expression for the equilibrium constant first.
2. Use equilibrium concentrations (moles divided by volume) in K, not initial ones.

Question 18: 4.5 moles each of hydrogen and iodine are heated in a sealed ten litre vessel. At equilibrium 3 moles of HI are found. The equilibrium constant for \( \ce{H2 + I2 <=> 2HI} \) is
(a) 1
(b) 10
(c) 5
(d) 0.33
Show Answer & Explanation

Answer: (a) 1

Explanation:
1. HI = 3 mol so 1.5 mol each of H2 and I2 reacted; 3.0 mol of each remain.
2. Volume cancels: K = 3^2/(3.0 x 3.0) = 1.

Teacher's Note:
1. Write the balanced equation and the expression for the equilibrium constant first.
2. Use equilibrium concentrations (moles divided by volume) in K, not initial ones.

Question 19: An equilibrium mixture of \( \ce{2H2S <=> 2H2 + S2} \) has 0.5 mol \( \ce{H2S} \), 0.10 mol \( \ce{H2} \) and 0.4 mol \( \ce{S2} \) in a one litre vessel. The value of K (mol \( L^{-1} \)) is
(a) 0.004
(b) 0.008
(c) 0.016
(d) 0.160
Show Answer & Explanation

Answer: (c) 0.016

Explanation:
1. \( K=\frac{[H_2]^2[S_2]}{[H_2S]^2}=\frac{(0.10)^2\times0.4}{(0.5)^2} \) = 0.016.

Teacher's Note:
1. Write the balanced equation and the expression for the equilibrium constant first.
2. Use equilibrium concentrations (moles divided by volume) in K, not initial ones.

Question 20: The rate constants for the forward and backward reactions of the hydrolysis of an ester are \( 1.1\times10^{-2} \) and \( 1.5\times10^{-3} \) per minute. The equilibrium constant is
(a) 4.33
(b) 5.33
(c) 6.33
(d) 7.33
Show Answer & Explanation

Answer: (d) 7.33

Explanation:
1. K = \( k_f/k_b=1.1\times10^{-2}/1.5\times10^{-3} \) = 7.33.

Teacher's Note:
1. Write the balanced equation and the expression for the equilibrium constant first.
2. Use equilibrium concentrations (moles divided by volume) in K, not initial ones.

Question 21: 3 mol of A and 1 mol of B are mixed in a 1 L vessel: \( \ce{A(g) + B(g) <=> 2C(g)} \). 1.5 mol of C is formed. The equilibrium constant is
(a) 0.12
(b) 0.25
(c) 0.50
(d) 4.0
Show Answer & Explanation

Answer: (d) 4.0

Explanation:
1. C = 1.5 mol means 0.75 mol each of A and B reacted.
2. A = 2.25, B = 0.25, C = 1.5 mol/L.
3. \( K_c=\frac{(1.5)^2}{2.25\times0.25} \) = 4.0.

Teacher's Note:
1. Write the balanced equation and the expression for the equilibrium constant first.
2. Use equilibrium concentrations (moles divided by volume) in K, not initial ones.

Question 22: In \( \ce{A + B <=> 2C} \) the equilibrium concentrations of A and B are 0.20 mol/L each and that of C is 0.60 mol/L. The equilibrium constant is
(a) 2.4
(b) 18
(c) 4.8
(d) 9
Show Answer & Explanation

Answer: (d) 9

Explanation:
1. \( K_c=\frac{(0.60)^2}{0.20\times0.20} \) = 9.

Teacher's Note:
1. Write the balanced equation and the expression for the equilibrium constant first.
2. Use equilibrium concentrations (moles divided by volume) in K, not initial ones.

Question 23: 15 moles of \( \ce{H2} \) and 5.2 moles of \( \ce{I2} \) are mixed and allowed to attain equilibrium at 500 C. At equilibrium 10 moles of HI are found. The equilibrium constant for the formation of HI is
(a) 50
(b) 15
(c) 100
(d) 25
Show Answer & Explanation

Answer: (a) 50

Explanation:
1. 10 mol HI needs 5 mol each of H2 and I2.
2. H2 left = 10 mol, I2 left = 0.2 mol.
3. K = 10^2/(10 x 0.2) = 50 (volume cancels).

Teacher's Note:
1. Write the balanced equation and the expression for the equilibrium constant first.
2. Use equilibrium concentrations (moles divided by volume) in K, not initial ones.

Question 24: In the Haber process 30 L of dihydrogen and 30 L of dinitrogen are taken and only 50% of the expected product is obtained. The gaseous mixture at the end is
(a) 20 L ammonia, 25 L nitrogen, 15 L hydrogen
(b) 20 L ammonia, 20 L nitrogen, 20 L hydrogen
(c) 10 L ammonia, 25 L nitrogen, 15 L hydrogen
(d) 20 L ammonia, 10 L nitrogen, 30 L hydrogen
Show Answer & Explanation

Answer: (c) 10 L ammonia, 25 L nitrogen, 15 L hydrogen

Explanation:
1. H2 is limiting: 30 L H2 can give 20 L \( \ce{NH3} \) (3 : 1 : 2 by volume).
2. 50% yield gives 10 L \( \ce{NH3} \), using 15 L H2 and 5 L N2.
3. Left: H2 = 15 L, N2 = 25 L, NH3 = 10 L.

Teacher's Note:
1. Write the balanced equation and the expression for the equilibrium constant first.
2. Use equilibrium concentrations (moles divided by volume) in K, not initial ones.

Question 25: 3.2 moles of HI were heated in a sealed bulb at 444 C until equilibrium. Its degree of dissociation was 22%. The number of moles of HI at equilibrium is
(a) 2.496
(b) 1.87
(c) 2
(d) 4
Show Answer & Explanation

Answer: (a) 2.496

Explanation:
1. HI left = 3.2 x (1 - 0.22) = 3.2 x 0.78 = 2.496 mol.

Teacher's Note:
1. Write the balanced equation and the expression for the equilibrium constant first.
2. Use equilibrium concentrations (moles divided by volume) in K, not initial ones.

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Multiple Choice Questions (MCQs) for JEE Chemistry Equilibrium

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