CBSE Class 11 Chemistry Chemical Equilibrium MCQs Set 08

Practice MCQs for Class 11 Chemistry Chapter 06 Equilibrium

Review structured MCQ sets for Class 11 Chemistry Chapter 06 Equilibrium. Built according to official CBSE guidelines, these downloadable questions support daily revision and core concept reinforcement.

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View or download the dedicated Chapter 06 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: In the reversible reaction A ⇌ B, the system is at equilibrium when:
(a) A has been completely changed into B
(b) half of A has changed into B
(c) only 10% of A has changed into B
(d) the rate of change of A into B equals the rate of change of B into A
Show Answer & Explanation

Answer: (d) the rate of change of A into B equals the rate of change of B into A

Explanation:
1. Equilibrium is a dynamic state: the forward and backward reactions both continue.
2. It is reached when the two rates become equal, and not when the amounts of A and B are equal or when A is used up.
3. The composition at equilibrium can be anything, depending on the equilibrium constant.

Teacher's Note:
1. Equal rates, not equal amounts, define chemical equilibrium.
2. The reactions do not stop at equilibrium; they go on at the same rate in both directions.

Question 2: According to the law of mass action, the rate of a chemical reaction is proportional to:
(a) the concentration of the products
(b) the molar concentrations of the reactants, each raised to the power of its coefficient
(c) the molar concentration of the products only
(d) the temperature of the reaction only
Show Answer & Explanation

Answer: (b) the molar concentrations of the reactants, each raised to the power of its coefficient

Explanation:
1. Guldberg and Waage stated that the rate depends on the active masses (molar concentrations) of the reactants.
2. Each concentration is raised to a power equal to its stoichiometric coefficient in the balanced equation.

Teacher's Note:
1. For aA + bB → products, rate ∝ [A]^a[B]^b.
2. Apply the law to the forward and backward reactions to get the equilibrium constant.

Question 3: What is the active mass (molar concentration) of 64 g of HI in a 2 litre flask? (H = 1, I = 127)
(a) 2 mol L⁻¹
(b) 1 mol L⁻¹
(c) 5 mol L⁻¹
(d) 0.25 mol L⁻¹
Show Answer & Explanation

Answer: (d) 0.25 mol L⁻¹

Explanation:
1. Molar mass of HI = 1 + 127 = 128 g mol⁻¹, so 64 g = 0.5 mol.
2. Active mass = moles ÷ volume in litres = 0.5 ÷ 2 = 0.25 mol L⁻¹.

Teacher's Note:
1. Active mass means molar concentration.
2. Always change grams to moles before dividing by the volume.

Question 4: The rate constants for the forward and backward reactions of the hydrolysis of an ester are 1.1 × 10⁻² and 1.5 × 10⁻³ per minute respectively. The equilibrium constant for CH3COOC2H5 + H2O ⇌ CH3COOH + C2H5OH is:
(a) 4.33
(b) 5.33
(c) 7.33
(d) 6.33
Show Answer & Explanation

Answer: (c) 7.33

Explanation:
1. Kc = kf ÷ kb.
2. Kc = (1.1 × 10⁻²) ÷ (1.5 × 10⁻³) = 7.33.

Teacher's Note:
1. At equilibrium the forward and backward rates are equal, so Kc is the ratio of the two rate constants.
2. Divide forward by backward.

Question 5: For the reaction 3A + 2B ⇌ C, the expression for the equilibrium constant is:
(a) \( \frac{[3A][2B]}{[C]} \)
(b) \( \frac{[C]}{[A]^3[B]^2} \)
(c) \( \frac{[A]^3[B]^2}{[C]} \)
(d) \( \frac{[C]}{[3A][2B]} \)
Show Answer & Explanation

Answer: (b) \( \frac{[C]}{[A]^3[B]^2} \)

Explanation:
1. Products go in the numerator and reactants in the denominator.
2. Each concentration is raised to the power of its coefficient, so [A] is cubed and [B] is squared.
3. Kc = [C] ÷ ([A]³[B]²).

Teacher's Note:
1. The coefficient becomes a power, not a multiplier.
2. Write products over reactants.

Question 6: 4 mol of A and 4 mol of B are mixed in a closed vessel. At equilibrium for A + B ⇌ C + D, 2 mol each of C and D are present. The equilibrium constant is:
(a) 1
(b) 4
(c) 0.25
(d) 2
Show Answer & Explanation

Answer: (a) 1

Explanation:
1. Moles at equilibrium: A = 4 − 2 = 2, B = 2, C = 2, D = 2.
2. Because Δn = 0, the volume cancels: Kc = (2 × 2) ÷ (2 × 2) = 1.

Teacher's Note:
1. If the number of moles of gas is the same on both sides, you can use moles in place of concentrations.
2. Find the moles of A and B that remain.

Question 7: The equilibrium concentrations of HI, H2 and I2 are 0.80, 0.10 and 0.10 mol L⁻¹ respectively. The equilibrium constant for H2 + I2 ⇌ 2HI is:
(a) 12
(b) 8
(c) 64
(d) 0.8
Show Answer & Explanation

Answer: (c) 64

Explanation:
1. Kc = [HI]² ÷ ([H2][I2]).
2. Kc = (0.80)² ÷ (0.10 × 0.10) = 0.64 ÷ 0.01 = 64.

Teacher's Note:
1. Square [HI] because its coefficient is 2.
2. Do not divide by [HI] alone.

Question 8: For A(g) + 2B(g) ⇌ C(g), the equilibrium concentrations are [A] = 0.06 mol L⁻¹, [B] = 0.12 mol L⁻¹ and [C] = 0.216 mol L⁻¹. The equilibrium constant is:
(a) 250
(b) 416
(c) 4 × 10⁻³
(d) 125
Show Answer & Explanation

Answer: (a) 250

Explanation:
1. Kc = [C] ÷ ([A][B]²).
2. Kc = 0.216 ÷ (0.06 × 0.0144) = 0.216 ÷ 0.000864 = 250.

Teacher's Note:
1. B appears with coefficient 2, so square it.
2. Keep the units out of the working and add them at the end.

Question 9: The unit of the equilibrium constant Kc for the reaction 2SO2(g) + O2(g) ⇌ 2SO3(g) is:
(a) mol L⁻¹
(b) L mol⁻¹
(c) mol² L⁻²
(d) L² mol⁻²
Show Answer & Explanation

Answer: (b) L mol⁻¹

Explanation:
1. Δn = 2 − 3 = −1.
2. The unit of Kc is (mol L⁻¹)Δn = (mol L⁻¹)⁻¹ = L mol⁻¹.

Teacher's Note:
1. Unit of Kc = (mol L⁻¹)^Δn.
2. K has no unit when Δn = 0.

Question 10: For the reaction 2X + Y ⇌ YX2, the equilibrium concentrations of X, Y and YX2 are 4, 2 and 2 mol L⁻¹ respectively. The value of Kc is:
(a) 0.0625
(b) 0.625
(c) 6.25
(d) 0.00625
Show Answer & Explanation

Answer: (a) 0.0625

Explanation:
1. Kc = [YX2] ÷ ([X]²[Y]).
2. Kc = 2 ÷ (16 × 2) = 2 ÷ 32 = 0.0625.

Teacher's Note:
1. Square the concentration of X.
2. Check your answer by recalculating the ratio.

Question 11: CO and Cl2 are mixed in a 500 mL vessel to form COCl2. At equilibrium the vessel contains 0.2 mol of COCl2 and 0.1 mol each of CO and Cl2. The equilibrium constant for CO + Cl2 ⇌ COCl2 is:
(a) 5
(b) 10
(c) 20
(d) 15
Show Answer & Explanation

Answer: (b) 10

Explanation:
1. Concentrations: [COCl2] = 0.2 ÷ 0.5 = 0.4 M; [CO] = [Cl2] = 0.1 ÷ 0.5 = 0.2 M.
2. Kc = 0.4 ÷ (0.2 × 0.2) = 10.

Teacher's Note:
1. Here Δn = −1, so the volume does not cancel.
2. Convert moles to concentrations first.

Question 12: An equilibrium mixture of 2H2S(g) ⇌ 2H2(g) + S2(g) contains 0.5 mol H2S, 0.10 mol H2 and 0.4 mol S2 in a one litre vessel. The value of Kc in mol L⁻¹ is:
(a) 0.016
(b) 0.008
(c) 0.004
(d) 0.160
Show Answer & Explanation

Answer: (a) 0.016

Explanation:
1. Kc = [H2]²[S2] ÷ [H2S]².
2. Kc = (0.10)² × 0.4 ÷ (0.5)² = 0.004 ÷ 0.25 = 0.016 mol L⁻¹.

Teacher's Note:
1. Square H2 and H2S, but not S2, which has a coefficient of 1.
2. Write each concentration in mol L⁻¹ before substituting.

Question 13: For N2 + 3H2 ⇌ 2NH3, Kc at 1000 K is 2.37 × 10⁻³. If [N2] = 2 M and [H2] = 3 M at equilibrium, the concentration of NH3 is:
(a) 0.00358 M
(b) 0.358 M
(c) 0.0358 M
(d) 3.58 M
Show Answer & Explanation

Answer: (b) 0.358 M

Explanation:
1. [NH3]² = Kc × [N2][H2]³ = 2.37 × 10⁻³ × 2 × 27 = 0.128.
2. [NH3] = √0.128 = 0.358 M.

Teacher's Note:
1. Rearrange Kc for the unknown and then take the square root.
2. (3)³ = 27.

Question 14: Two moles of NH3 are placed in a previously evacuated 1 litre vessel. At equilibrium 1 mole of NH3 is present. The equilibrium constant for 2NH3 ⇌ N2 + 3H2 is:
(a) 3/4 mol² L⁻²
(b) 27/64 mol² L⁻²
(c) 27/16 mol² L⁻²
(d) 27/32 mol² L⁻²
Show Answer & Explanation

Answer: (c) 27/16 mol² L⁻²

Explanation:
1. 1 mol of NH3 has decomposed, giving 0.5 mol N2 and 1.5 mol H2.
2. Kc = [N2][H2]³ ÷ [NH3]² = 0.5 × (1.5)³ ÷ 1² = 0.5 × 3.375 = 1.6875 = 27/16.

Teacher's Note:
1. Use the mole ratio 2 : 1 : 3 from the equation.
2. Δn = 2, so Kc has the unit mol² L⁻².

Question 15: The equilibrium constants K1 and K2 for NO(g) + ½O2(g) ⇌ NO2(g) and 2NO2(g) ⇌ 2NO(g) + O2(g) respectively are related as:
(a) K2 = 1/K1
(b) K2 = K1²
(c) K2 = K1/2
(d) K2 = 1/K1²
Show Answer & Explanation

Answer: (d) K2 = 1/K1²

Explanation:
1. The second reaction is the reverse of the first, multiplied by 2.
2. Reversing gives 1/K1; multiplying the equation by 2 squares it, so K2 = 1/K1².

Teacher's Note:
1. Reverse: take the reciprocal. Multiply by n: raise to the power n.
2. Do both changes.

Question 16: At a given temperature, Kc for PCl5(g) ⇌ PCl3(g) + Cl2(g) is 2.4 × 10⁻³. At the same temperature the equilibrium constant for PCl3(g) + Cl2(g) ⇌ PCl5(g) is:
(a) 2.4 × 10⁻³
(b) −2.4 × 10⁻³
(c) 4.2 × 10²
(d) 4.8 × 10⁻²
Show Answer & Explanation

Answer: (c) 4.2 × 10²

Explanation:
1. The new reaction is the reverse of the old one, so K′ = 1 ÷ K.
2. K′ = 1 ÷ (2.4 × 10⁻³) = 4.2 × 10².

Teacher's Note:
1. An equilibrium constant is never negative.
2. Reciprocal for the reverse reaction.

Question 17: In which of the following reactions is Kp equal to Kc?
(a) PCl5 ⇌ PCl3 + Cl2
(b) 2NH3 ⇌ N2 + 3H2
(c) 2SO2 + O2 ⇌ 2SO3
(d) H2 + I2 ⇌ 2HI
Show Answer & Explanation

Answer: (d) H2 + I2 ⇌ 2HI

Explanation:
1. Kp = Kc(RT)Δn, so Kp = Kc only when Δn = 0.
2. For H2 + I2 ⇌ 2HI, Δn = 2 − 2 = 0. The others have Δn = +1, +2 and −1.

Teacher's Note:
1. Count the moles of gas on each side.
2. No change in the number of gas moles means Δn = 0.

Question 18: For the reaction N2(g) + 3H2(g) ⇌ 2NH3(g) the relation between Kp and Kc is:
(a) Kp = Kc(RT)⁻²
(b) Kp = Kc(RT)⁻¹
(c) Kp = Kc(RT)²
(d) Kp = Kc(RT)
Show Answer & Explanation

Answer: (a) Kp = Kc(RT)⁻²

Explanation:
1. Δn = 2 − (1 + 3) = −2.
2. Kp = Kc(RT)Δn = Kc(RT)⁻².

Teacher's Note:
1. Δn = moles of gaseous products − moles of gaseous reactants.
2. Here Kp is smaller than Kc.

Question 19: For N2(g) + 3H2(g) ⇌ 2NH3(g), Kc = 6.0 × 10⁻² at 500 °C. The numerical value of Kp for this reaction is about: (R = 0.0821 L atm K⁻¹ mol⁻¹)
(a) 1.5 × 10⁶
(b) 1.5 × 10⁵
(c) 1.5 × 10⁻⁶
(d) 1.5 × 10⁻⁵
Show Answer & Explanation

Answer: (d) 1.5 × 10⁻⁵

Explanation:
1. T = 500 + 273 = 773 K and Δn = −2.
2. RT = 0.0821 × 773 = 63.5, so Kp = 6.0 × 10⁻² ÷ (63.5)² = 6.0 × 10⁻² ÷ 4028 = 1.5 × 10⁻⁵.

Teacher's Note:
1. Change °C to kelvin first.
2. Because Δn is negative, Kp comes out smaller than Kc.

Question 20: Which of the following statements about chemical equilibrium is wrong?
(a) The state of equilibrium can be changed by altering temperature or pressure.
(b) Equilibrium is dynamic in nature.
(c) The forward reaction is favoured by the addition of a catalyst.
(d) The same equilibrium state is reached whether we start from reactants or from products.
Show Answer & Explanation

Answer: (c) The forward reaction is favoured by the addition of a catalyst.

Explanation:
1. A catalyst speeds up the forward and backward reactions equally, so it does not favour either direction or change the equilibrium composition.
2. The other three statements are correct: equilibrium shifts with temperature (and with pressure for gases), is dynamic, and can be approached from both sides.

Teacher's Note:
1. A catalyst only shortens the time to reach equilibrium.
2. It also leaves the value of K unchanged.

Free study material for Chemistry

Practice MCQs for Class 11 Chemistry Chapter 06 Equilibrium

Class 11 Chemistry Chapter 06 Equilibrium Objective Test Questions

Review structured objective questions for Class 11 Chemistry Chapter 06 Equilibrium. Built according to official CBSE guidelines, these MCQ sets support daily revision and core concept reinforcement.

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