CBSE Class 12 Chemistry Chemical Kinetics MCQs Set 16

Multiple Choice Questions (MCQs) for Class 12 Chemistry: Unit 03 Chemical Kinetics

Access targeted multiple-choice questions for Unit 03 Chemical Kinetics designed to align with the latest CBSE academic syllabus for Class 12 Chemistry. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.

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Multiple Choice Questions

Question 1: For the reaction 2A + B → 3C + D, which of the following does not express the rate of the reaction?
(a) \( -\frac{1}{3}\frac{d[C]}{dt} \)
(b) \( -\frac{d[B]}{dt} \)
(c) \( \frac{d[D]}{dt} \)
(d) \( -\frac{1}{2}\frac{d[A]}{dt} \)
Show Answer & Explanation

Answer: (a) \( -\frac{1}{3}\frac{d[C]}{dt} \)

Explanation:
1. The rate of reaction is \( -\frac{1}{2}\frac{d[A]}{dt} = -\frac{d[B]}{dt} = \frac{1}{3}\frac{d[C]}{dt} = \frac{d[D]}{dt} \).
2. C is a product, so its term must carry a positive sign: \( +\frac{1}{3}\frac{d[C]}{dt} \).
3. The expression with a negative sign for C is therefore wrong; the other three are correct.

Teacher's Note:
1. Reactants take a minus sign and products a plus sign.
2. Divide each concentration change by its stoichiometric coefficient.

Question 2: For the reaction 2A + 3B → products, the rates of disappearance of A and B are r₁ and r₂ respectively. They are related as:
(a) r₁ = r₂
(b) 2r₁ = 3r₂
(c) 3r₁ = 2r₂
(d) r₁² = 2r₂²
Show Answer & Explanation

Answer: (c) 3r₁ = 2r₂

Explanation:
1. Rate of reaction = r₁/2 = r₂/3, because the rate of disappearance of each reactant is divided by its coefficient.
2. Cross-multiplying gives 3r₁ = 2r₂.

Teacher's Note:
1. Write the rate of reaction using both reactants first.
2. The species with the larger coefficient disappears faster.

Question 3: In the reaction BrO3⁻ + 5Br⁻ + 6H⁺ → 3Br2 + 3H2O, how is the rate of appearance of Br2 related to the rate of disappearance of Br⁻?
(a) d[Br2]/dt = −d[Br⁻]/dt
(b) d[Br2]/dt = +(3/5) d[Br⁻]/dt
(c) d[Br2]/dt = −(5/3) d[Br⁻]/dt
(d) d[Br2]/dt = −(3/5) d[Br⁻]/dt
Show Answer & Explanation

Answer: (d) d[Br2]/dt = −(3/5) d[Br⁻]/dt

Explanation:
1. Rate of reaction = \( -\frac{1}{5}\frac{d[Br⁻]}{dt} = \frac{1}{3}\frac{d[Br_2]}{dt} \).
2. Therefore d[Br2]/dt = −(3/5) d[Br⁻]/dt. The minus sign appears because [Br⁻] decreases while [Br2] increases.

Teacher's Note:
1. Equate the two rate expressions and solve for the required ratio.
2. Check the sign: one concentration falls while the other rises.

Question 4: For the reaction N2O5(g) → 2NO2(g) + 1/2 O2(g), the rate of disappearance of N2O5 is 6.25 × 10⁻³ mol L⁻¹ s⁻¹. The rates of formation of NO2 and O2 are respectively:
(a) 1.25 × 10⁻² and 3.125 × 10⁻³ mol L⁻¹ s⁻¹
(b) 6.25 × 10⁻³ and 6.25 × 10⁻³ mol L⁻¹ s⁻¹
(c) 1.25 × 10⁻² and 6.25 × 10⁻³ mol L⁻¹ s⁻¹
(d) 6.25 × 10⁻³ and 3.125 × 10⁻³ mol L⁻¹ s⁻¹
Show Answer & Explanation

Answer: (a) 1.25 × 10⁻² and 3.125 × 10⁻³ mol L⁻¹ s⁻¹

Explanation:
1. Rate = \( -\frac{d[N_2O_5]}{dt} = \frac{1}{2}\frac{d[NO_2]}{dt} = 2\frac{d[O_2]}{dt} \).
2. d[NO2]/dt = 2 × 6.25 × 10⁻³ = 1.25 × 10⁻² mol L⁻¹ s⁻¹.
3. d[O2]/dt = ½ × 6.25 × 10⁻³ = 3.125 × 10⁻³ mol L⁻¹ s⁻¹.

Teacher's Note:
1. Use the coefficients as ratios: N2O5 : NO2 : O2 = 1 : 2 : ½.
2. NO2 forms twice as fast as N2O5 is used up.

Question 5: For the reaction P(g) + 3Q(g) → 2R(g), the rate of appearance of R is 2.5 × 10⁻⁴ mol L⁻¹ s⁻¹. The rate of reaction and the rate of disappearance of Q (in mol L⁻¹ s⁻¹) are respectively:
(a) 3.75 × 10⁻⁴ and 1.25 × 10⁻⁴
(b) 1.25 × 10⁻⁴ and 3.75 × 10⁻⁴
(c) 2.5 × 10⁻⁴ and 3.75 × 10⁻⁴
(d) 1.25 × 10⁻⁴ and 2.5 × 10⁻⁴
Show Answer & Explanation

Answer: (b) 1.25 × 10⁻⁴ and 3.75 × 10⁻⁴

Explanation:
1. Rate of reaction = ½ × rate of appearance of R = ½ × 2.5 × 10⁻⁴ = 1.25 × 10⁻⁴ mol L⁻¹ s⁻¹.
2. Rate of disappearance of Q = 3 × rate of reaction = 3 × 1.25 × 10⁻⁴ = 3.75 × 10⁻⁴ mol L⁻¹ s⁻¹.

Teacher's Note:
1. Find the rate of reaction first, then scale it by each coefficient.
2. The rate of reaction is not the same as the rate for one species unless its coefficient is 1.

Question 6: In the reaction 2MnO4⁻ + 10I⁻ + 16H⁺ → 2Mn²⁺ + 5I2 + 8H2O, MnO4⁻ disappears at 4.56 × 10⁻³ mol L⁻¹ s⁻¹. The rate of appearance of I2 is:
(a) 1.14 × 10⁻³ mol L⁻¹ s⁻¹
(b) 5.7 × 10⁻³ mol L⁻¹ s⁻¹
(c) 1.14 × 10⁻² mol L⁻¹ s⁻¹
(d) 4.56 × 10⁻⁴ mol L⁻¹ s⁻¹
Show Answer & Explanation

Answer: (c) 1.14 × 10⁻² mol L⁻¹ s⁻¹

Explanation:
1. From the equation, 2 mol of MnO4⁻ disappear for every 5 mol of I2 formed.
2. d[I2]/dt = (5/2) × 4.56 × 10⁻³ = 1.14 × 10⁻² mol L⁻¹ s⁻¹.

Teacher's Note:
1. Use the ratio of coefficients 5 : 2 for I2 : MnO4⁻.
2. I2 is formed faster than MnO4⁻ is consumed.

Question 7: For SO3 → SO2 + 1/2 O2, the rates are given by −d[SO3]/dt = k₁[SO3], d[SO2]/dt = k₂[SO3] and d[O2]/dt = k₃[SO3]. The relation between k₁, k₂ and k₃ is:
(a) 2k₁ = k₂ = 4k₃
(b) k₁ = 2k₂ = k₃
(c) k₁ = k₂ = k₃
(d) k₁ = k₂ = 2k₃
Show Answer & Explanation

Answer: (d) k₁ = k₂ = 2k₃

Explanation:
1. SO3 disappears at the same rate at which SO2 appears, so k₁ = k₂.
2. O2 appears at half that rate (coefficient ½), so k₃ = k₁/2, that is k₁ = 2k₃.
3. Hence k₁ = k₂ = 2k₃.

Teacher's Note:
1. Compare each rate with the rate of SO3 using the coefficients.
2. O2 has coefficient ½, so its rate constant is half.

Question 8: For the reaction 4A + B → 2C + 2D, the incorrect statement is:
(a) The rate of disappearance of B is one-fourth the rate of disappearance of A
(b) The rate of appearance of C is half the rate of disappearance of B
(c) The rate of formation of D is half the rate of consumption of A
(d) The rates of formation of C and D are equal
Show Answer & Explanation

Answer: (b) The rate of appearance of C is half the rate of disappearance of B

Explanation:
1. Rate of reaction = \( -\frac{1}{4}\frac{d[A]}{dt} = -\frac{d[B]}{dt} = \frac{1}{2}\frac{d[C]}{dt} = \frac{1}{2}\frac{d[D]}{dt} \).
2. So d[C]/dt = 2 × (−d[B]/dt): C appears twice, not half, as fast as B disappears.
3. The other three statements follow directly from the same relation.

Teacher's Note:
1. Write the full rate expression before judging the statements.
2. A quick check: C has coefficient 2 and B has coefficient 1.

Question 9: In the reaction 2A → products, the concentration of A falls from 0.50 mol L⁻¹ to 0.40 mol L⁻¹ in 10 minutes. The average rate of the reaction, defined as \( -\frac{1}{2}\frac{\Delta[A]}{\Delta t} \), is:
(a) 5.0 × 10⁻³ mol L⁻¹ min⁻¹
(b) 1.0 × 10⁻² mol L⁻¹ min⁻¹
(c) 2.0 × 10⁻² mol L⁻¹ min⁻¹
(d) 1.3 mol L⁻¹ min⁻¹
Show Answer & Explanation

Answer: (a) 5.0 × 10⁻³ mol L⁻¹ min⁻¹

Explanation:
1. Δ[A] = 0.40 − 0.50 = −0.10 mol L⁻¹ and Δt = 10 min.
2. Rate = ½ × 0.10 ÷ 10 = 5.0 × 10⁻³ mol L⁻¹ min⁻¹.

Teacher's Note:
1. Take the positive value after dividing by the coefficient 2.
2. Always check that the unit is mol L⁻¹ per unit time.

Question 10: The rate law of a reaction is rate = k[A][B]2/3[C]⁰. The order of the reaction is:
(a) 2
(b) 1
(c) 5/3
(d) zero
Show Answer & Explanation

Answer: (c) 5/3

Explanation:
1. Add the powers of the concentration terms: 1 + 2/3 + 0 = 5/3.
2. Order can be a fraction because it is found from experiment.

Teacher's Note:
1. Any term with power zero does not contribute to the order.
2. Order need not be a whole number.

Question 11: The rate constant of a reaction is 10.8 × 10⁻⁵ mol L⁻¹ s⁻¹. The reaction is of:
(a) first order
(b) zero order
(c) second order
(d) half order
Show Answer & Explanation

Answer: (b) zero order

Explanation:
1. The unit of k is mol L⁻¹ s⁻¹, which is the same as the unit of rate.
2. This is the unit of k for a zero order reaction (unit = (mol L⁻¹)1−n s⁻¹ with n = 0).

Teacher's Note:
1. Match the unit of k with the order: s⁻¹ for first order, mol⁻¹ L s⁻¹ for second order.
2. For zero order, k has the same unit as rate.

Question 12: Which of the following statements about the order of a reaction is correct?
(a) The order of a reaction must be a positive integer
(b) A second order reaction is always bimolecular
(c) The order of a reaction increases with temperature
(d) The order of a reaction can be found only by experiment
Show Answer & Explanation

Answer: (d) The order of a reaction can be found only by experiment

Explanation:
1. Order depends on how the rate varies with concentrations, and this has to be measured.
2. Order can be zero or fractional, so it need not be a positive integer.
3. A second order reaction need not be bimolecular, and the order does not change with temperature.

Teacher's Note:
1. Order is experimental; molecularity is theoretical.
2. Temperature changes k, not the order.

Question 13: For the reaction A + B → products, the rate law is rate = k[A][B]². Which of the following statements is false?
(a) If [B] is kept constant and [A] is doubled, the rate doubles
(b) If [A] and [B] are both doubled, the rate increases eight times
(c) If [A] is kept constant and [B] is reduced to one-quarter, the rate is halved
(d) The reaction is of third order
Show Answer & Explanation

Answer: (c) If [A] is kept constant and [B] is reduced to one-quarter, the rate is halved

Explanation:
1. If [B] falls to ¼, the rate becomes (¼)² = 1/16 of its value, not ½.
2. Doubling [A] doubles the rate, and doubling both gives 2 × 2² = 8 times; the order is 1 + 2 = 3.

Teacher's Note:
1. Raise the change in concentration to the power of its order.
2. Test each statement with a quick calculation.

Question 14: The elementary gaseous reaction 2NO + O2 → 2NO2 takes place in a closed vessel. If the volume of the vessel is doubled, the rate of the reaction will:
(a) increase eight times
(b) decrease to one-eighth of its initial value
(c) decrease to one-fourth
(d) increase four times
Show Answer & Explanation

Answer: (b) decrease to one-eighth of its initial value

Explanation:
1. For an elementary reaction rate = k[NO]²[O2], which is of third order.
2. Doubling the volume halves every concentration, so the rate becomes (½)³ = 1/8 of the initial rate.

Teacher's Note:
1. Doubling the volume halves all concentrations.
2. Order three means the factor is (½)³.

Question 15: For a reaction A + B → products, doubling the concentration of A quadruples the rate, but doubling the concentration of B has no effect on the rate. The rate law is:
(a) rate = k[A]²
(b) rate = k[A][B]
(c) rate = k[A]²[B]
(d) rate = k[A][B]²
Show Answer & Explanation

Answer: (a) rate = k[A]²

Explanation:
1. Doubling [A] gives 2ⁿ = 4, so the order in A is 2.
2. Doubling [B] changes nothing, so the order in B is zero and [B] does not appear in the rate law.

Teacher's Note:
1. Use 2ⁿ = (factor by which the rate changes).
2. Zero order in a reactant means it drops out of the rate law.

Question 16: The rate of a reaction increases by a factor of 2 when the concentration of the reactant is increased 8 times. The order of the reaction with respect to that reactant is:
(a) 3
(b) 1/2
(c) zero
(d) 1/3
Show Answer & Explanation

Answer: (d) 1/3

Explanation:
1. 8ⁿ = 2 gives n = 1/3 (since the cube root of 8 is 2).
2. A fractional order is possible and shows that the rate depends weakly on concentration.

Teacher's Note:
1. Write the change as (concentration factor)ⁿ = (rate factor).
2. Cube root of 8 is 2.

Question 17: For a reaction A + B → products, the order is 1 with respect to A and ½ with respect to B. If the concentrations of both A and B are increased four times, the rate increases by a factor of:
(a) 8
(b) 16
(c) 6
(d) 4
Show Answer & Explanation

Answer: (a) 8

Explanation:
1. Rate = k[A][B]½, so the factor is 4 × √4 = 4 × 2 = 8.
2. The total order is 3/2, and 4 raised to the power 3/2 is 8.

Teacher's Note:
1. Apply the factor to each term and multiply.
2. The square root of 4 is 2.

Question 18: The following initial rate data were obtained for X + 3Y → XY₃:
[X] (M)[Y] (M)Rate (M s⁻¹)
0.10.10.002
0.20.10.002
0.30.20.008
0.40.30.018
The rate law is:
(a) rate = k[X][Y]³
(b) rate = k[X][Y]²
(c) rate = k[Y]²
(d) rate = k[X]²[Y]
Show Answer & Explanation

Answer: (c) rate = k[Y]²

Explanation:
1. Rows 1 and 2: [X] is doubled with no change in rate, so the order in X is zero.
2. Rows 1 and 3 (using the zero order in X): [Y] doubles from 0.1 to 0.2 and the rate becomes 4 times, so the order in Y is 2.
3. Check with row 4: [Y] is 3 times, so the rate is 9 × 0.002 = 0.018. It agrees.

Teacher's Note:
1. Compare two rows in which only one concentration changes.
2. Verify the rate law with an unused row.

Question 19: The initial rates for the reaction 2A + B → C + D are:
Expt.[A] (mol L⁻¹)[B] (mol L⁻¹)Initial rate (mol L⁻¹ min⁻¹)
10.10.17.5 × 10⁻³
20.30.29.0 × 10⁻²
30.30.43.6 × 10⁻¹
40.40.13.0 × 10⁻²
The rate law is:
(a) rate = k[A]²[B]
(b) rate = k[A][B]²
(c) rate = k[A][B]
(d) rate = k[A][B]³
Show Answer & Explanation

Answer: (b) rate = k[A][B]²

Explanation:
1. Experiments 1 and 4: [A] becomes 4 times with [B] constant, and the rate becomes 4 times. The order in A is 1.
2. Experiments 2 and 3: [B] doubles with [A] constant and the rate becomes 4 times, so the order in B is 2.
3. Experiment 2 from 1: 3 × 2² = 12 and 7.5 × 10⁻³ × 12 = 9.0 × 10⁻², which agrees.

Teacher's Note:
1. Always choose pairs of experiments where only one concentration changes.
2. Check the answer with the remaining experiment.

Question 20: For an elementary reaction 2A + B → 2C, the molecularity is:
(a) 2
(b) 1
(c) cannot be predicted
(d) 3
Show Answer & Explanation

Answer: (d) 3

Explanation:
1. In an elementary reaction the molecularity is the number of reacting species colliding in that single step.
2. Two molecules of A and one of B collide, so the molecularity is 2 + 1 = 3 (termolecular).

Teacher's Note:
1. Molecularity is defined only for elementary reactions.
2. Count the species on the reactant side of the step.

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Unit 03 Chemical Kinetics Objective Questions & Solutions for Class 12 Chemistry

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