JEE Chemistry Chemical Kinetics MCQs Set 09

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Multiple Choice Questions

Question 1: Select the rate law that corresponds to the data for A + B → C: (1) [A] = 0.012, [B] = 0.035, rate = 0.10; (2) 0.024, 0.070, 1.6; (3) 0.024, 0.035, 0.20; (4) 0.012, 0.070, 0.80.
(a) Rate = K[B]3
(b) Rate = K[B]4
(c) Rate = K[A][B]3
(d) Rate = K[A]2[B]2
Show Answer & Explanation

Answer: (c) Rate = K[A][B]3

Explanation:
1. Experiments 1 and 3: [A] doubles, rate doubles (0.10 → 0.20), so order in A = 1.
2. Experiments 1 and 4: [B] doubles, rate increases 8 times (0.10 → 0.80), so order in B = 3.
3. Rate = K[A][B]3; check expt 2: 2 × 8 = 16 times 0.10 = 1.6.

Teacher's Note:
1. Change one concentration at a time to find each order.
2. Verify with the experiment where both change.

Question 2: In the gaseous reaction X + 3Y → XY3, the initial rates were: [X] = 0.1 M, [Y] = 0.1 M, rate = 0.002; [X] = 0.2, [Y] = 0.1, rate = 0.002; [X] = 0.3, [Y] = 0.2, rate = 0.008; [X] = 0.4, [Y] = 0.3, rate = 0.018 (M s−1). The rate law is
(a) r = K[X][Y]3
(b) r = K[X]0[Y]2
(c) r = K[X][Y]
(d) r = [X]4[Y]3
Show Answer & Explanation

Answer: (b) r = K[X]0[Y]2

Explanation:
1. Doubling [X] with [Y] fixed leaves the rate unchanged, so order in X = 0.
2. [Y] from 0.1 to 0.2 changes the rate from 0.002 to 0.008 (×4), so order in Y = 2.
3. Check: [Y] = 0.3 gives 9 × 0.002 = 0.018.

Teacher's Note:
1. A zero order species does not appear in the rate law.
2. Always check the last data row.

Question 3: For the reaction 2A + B → C + D the initial rate data are: (0.1, 0.1) → 7.5 × 10−3; (0.3, 0.2) → 9.0 × 10−2; (0.3, 0.4) → 3.6 × 10−1; (0.4, 0.1) → 3.0 × 10−2 mol L−1 min−1 for ([A], [B]). The rate law is
(a) Rate = k[A]2[B]
(b) Rate = k[A][B]2
(c) Rate = k[A][B]3
(d) Rate = k[A][B]
Show Answer & Explanation

Answer: (b) Rate = k[A][B]2

Explanation:
1. Rows 2 and 3: [B] doubles, rate ×4, so order in B is 2.
2. Rows 1 and 4: [A] ×4, rate ×4, so order in A is 1.
3. Check row 2: 3 × 22 = 12, 7.5 × 10−3 × 12 = 0.09.

Teacher's Note:
1. Pick pairs of experiments where only one concentration changes.
2. Test the final law on the leftover pair.

Question 4: For a hypothetical reaction A + B → C, data: [A] = 0.01, [B] = 0.01, rate = 1.0 × 10−4; [A] = 0.01, [B] = 0.03, rate = 9.0 × 10−4; [A] = 0.03, [B] = 0.03, rate = 2.7 × 10−3. The rate law is
(a) r = K[A]2[B]
(b) r = K[A][B]2
(c) r = K[A][B]
(d) none of these
Show Answer & Explanation

Answer: (b) r = K[A][B]2

Explanation:
1. Expt 1 to 2: [B] ×3, rate ×9, so order in B = 2.
2. Expt 2 to 3: [A] ×3, rate ×3 (9 × 10−4 to 2.7 × 10−3), so order in A = 1.

Teacher's Note:
1. 9 = 32 means second order; 3 = 31 means first order.

Question 5: Calculate the order of the reaction with respect to A and B from the data: ([A], [B], rate) = (0.05, 0.05, 1.2 × 10−3); (0.10, 0.05, 2.4 × 10−3); (0.05, 0.10, 1.2 × 10−3).
(a) 1 and 0
(b) 1 and 1
(c) 0 and 1
(d) none of these
Show Answer & Explanation

Answer: (a) 1 and 0

Explanation:
1. Doubling [A] doubles the rate: order 1 in A.
2. Doubling [B] leaves the rate unchanged: order 0 in B.

Teacher's Note:
1. Rate = k[A].
2. Zero order species have no effect on the rate.

Question 6: For A + B → product, the order is one with respect to each of A and B. Find x and y: (rate, [A], [B]) = (0.10, 0.20 M, 0.05 M); (0.40, x, 0.05 M); (0.80, 0.40 M, y).
(a) 0.20, 0.80
(b) 0.80, 0.40
(c) 0.80, 0.20
(d) 0.40, 0.20
Show Answer & Explanation

Answer: (c) 0.80, 0.20

Explanation:
1. Rate = k[A][B], so k = 0.10/(0.20 × 0.05) = 10.
2. Row 2: 0.40 = 10 × x × 0.05, so x = 0.80.
3. Row 3: 0.80 = 10 × 0.40 × y, so y = 0.20.

Teacher's Note:
1. Find k from the complete row first.
2. Then use it to fill the missing entries.

Question 7: For a reaction A + B → products, doubling the concentration of A quadruples the rate, but doubling the amount of B does not affect it. The rate equation is
(a) Rate = K[A][B]
(b) Rate = K[A]2
(c) Rate = K[A]2[B]
(d) Rate = K[A]2[B]2
Show Answer & Explanation

Answer: (b) Rate = K[A]2

Explanation:
1. Rate ∝ [A]n with 2n = 4, so n = 2.
2. Rate is independent of [B], so the order in B is 0.

Teacher's Note:
1. Orders come from the experiment, not from the equation.
2. Write only species that affect the rate.

Question 8: Which statement is incorrect?
(a) The rate law is an experimental value
(b) The law of mass action is a theoretical proposal
(c) The rate law is more informative than the law of mass action for developing a mechanism
(d) The rate law is always different from the expression of the law of mass action
Show Answer & Explanation

Answer: (d) The rate law is always different from the expression of the law of mass action

Explanation:
1. For elementary reactions the rate law and the mass-action expression are identical.
2. Statement (d) says always different, which is wrong.

Teacher's Note:
1. Rate laws are determined by experiment.
2. They agree with the stoichiometric expression only for single-step reactions.

Question 9: For A + B → products, the rate doubles when [A] is doubled. The rate is again doubled when [A] and [B] are both doubled. The orders with respect to A and B are
(a) 1, 1
(b) 2, 0
(c) 1, 0
(d) 0, 1
Show Answer & Explanation

Answer: (c) 1, 0

Explanation:
1. Doubling [A] doubles the rate: first order in A.
2. Doubling both gives only a further factor of 1, so doubling [B] has no effect: order in B = 0.

Teacher's Note:
1. Isolate one change at a time.

Question 10: For A + B → products, the order in A is 1 and in B is 1/2. When the concentrations of both A and B are increased four times, the rate increases by a factor of
(a) 16
(b) 8
(c) 6
(d) 4
Show Answer & Explanation

Answer: (b) 8

Explanation:
1. Rate ∝ [A][B]1/2.
2. New rate = 4 × 41/2 = 4 × 2 = 8 times.

Teacher's Note:
1. Fractional powers: 41/2 = 2.

Question 11: The rate law for A + B → product is Rate = k[A][B]2. Which statement is false?
(a) If [B] is constant and [A] is doubled, the reaction proceeds twice as fast
(b) If [A] is constant and [B] is reduced to one quarter, the rate will be halved
(c) If [A] and [B] are both doubled, the reaction proceeds 8 times as fast
(d) The reaction is of third order
Show Answer & Explanation

Answer: (b) If [A] is constant and [B] is reduced to one quarter, the rate will be halved

Explanation:
1. [B] → 1/4 gives (1/4)2 = 1/16 of the rate, not 1/2.
2. The other statements are true: 2, 2 × 4 = 8, and order 1 + 2 = 3.

Teacher's Note:
1. Apply the power to the factor by which concentration changes.

Question 12: For A → B, the rate of reaction doubles when the concentration of A is increased 8 times. The order of reaction with respect to A is
(a) 3
(b) 1/2
(c) 1/3
(d) 0
Show Answer & Explanation

Answer: (c) 1/3

Explanation:
1. 8n = 2, so 23n = 21 and n = 1/3.

Teacher's Note:
1. Take logarithms: n = log 2 / log 8.

Question 13: The following mechanism is proposed for 2A + B → D + E: A + B → C + D (slow); A + C → E (fast). The rate law for the reaction is
(a) r = K[A]2[B]
(b) r = K[A][B]
(c) r = K[A]2
(d) r = K[A][C]
Show Answer & Explanation

Answer: (b) r = K[A][B]

Explanation:
1. The slow step is rate determining: rate = k[A][B].
2. C is an intermediate and does not appear in the final rate law.

Teacher's Note:
1. Slowest step determines the rate.
2. Its molecularity gives the rate law for the overall reaction.

Question 14: The reaction 2O3 → 3O2 proceeds as O3 ⇌ O2 + O (fast), O + O3 → 2O2 (slow). The rate law expression should be
(a) r = K[O3]2
(b) r = K[O3]2[O2]−1
(c) r = K[O3][O2]
(d) unpredictable
Show Answer & Explanation

Answer: (b) r = K[O3]2[O2]−1

Explanation:
1. Slow step: rate = k2[O][O3].
2. From the fast equilibrium [O] = K[O3]/[O2].
3. Rate = k2K[O3]2[O2]−1.

Teacher's Note:
1. An intermediate in the slow step is replaced using the earlier equilibrium.
2. The product O2 appears in the denominator and slows the reaction.

Question 15: The hypothetical reaction A2 + B2 → 2AB follows the mechanism A2 ⇌ A + A (fast); A + B2 → AB + B (slow); A + B → AB (fast). The order of the overall reaction is
(a) 2
(b) 1
(c) 1 1/2
(d) 0
Show Answer & Explanation

Answer: (c) 1 1/2

Explanation:
1. Slow step: rate = k[A][B2].
2. From the equilibrium, [A] = (K[A2])1/2.
3. Rate = kK1/2[A2]1/2[B2], order 1.5.

Teacher's Note:
1. Pre-equilibrium dissociation gives a square-root dependence.

Question 16: The rate for RCl + NaOH(aq) → ROH + NaCl is given by rate = k1[RCl]. The rate of the reaction is
(a) doubled on doubling the concentration of NaOH
(b) halved on reducing the concentration of RCl to half
(c) decreased on increasing the temperature
(d) unaffected by increasing the temperature
Show Answer & Explanation

Answer: (b) halved on reducing the concentration of RCl to half

Explanation:
1. Rate depends only on [RCl] (first order), so halving [RCl] halves the rate.
2. NaOH concentration has no effect, and temperature increases k.

Teacher's Note:
1. A reaction can have a rate independent of a reactant.

Question 17: For NO2 + CO → CO2 + NO, the rate expression is Rate = k[NO2]2. The number of molecules of CO involved in the slowest step is
(a) 0
(b) 1
(c) 2
(d) 3
Show Answer & Explanation

Answer: (a) 0

Explanation:
1. CO does not appear in the rate law, so it is not involved in the slow step.

Teacher's Note:
1. The slow step is 2NO2 → NO3 + NO; CO reacts later with NO3.

Question 18: The rate law of A + 2B → product is d[P]/dt = K[A]2[B]. If A is taken in large excess, the order of the reaction will be
(a) 0
(b) 1
(c) 2
(d) 3
Show Answer & Explanation

Answer: (b) 1

Explanation:
1. Large excess of A keeps [A] constant, so rate = k′[B].
2. The reaction is pseudo first order.

Teacher's Note:
1. Pseudo order counts only the reactant that changes.

Question 19: The acid hydrolysis of an ester is
(a) pseudo first order
(b) bimolecular
(c) pseudo unimolecular
(d) all of these
Show Answer & Explanation

Answer: (d) all of these

Explanation:
1. Rate = k[ester] because water is in large excess: pseudo first order.
2. It involves two molecules (ester and water) so is bimolecular, but behaves as unimolecular.

Teacher's Note:
1. Excess solvent keeps its concentration constant.

Question 20: In the sequence A → B → C → D with rate constants K1, K2, K3 and K3 > K2 > K1, the rate determining step is
(a) A → B
(b) B → C
(c) C → D
(d) A → D
Show Answer & Explanation

Answer: (a) A → B

Explanation:
1. The step with the smallest rate constant is the slowest.
2. K1 is the smallest.

Teacher's Note:
1. In a sequence, the slowest step controls the overall rate.

Question 21: The mechanism for P → R is P ⇌ 2Q (fast, K1 = equilibrium constant) and 2Q + P → R (slow, k2). The rate law for the main reaction is
(a) k1[P][Q]
(b) k1k2[P]
(c) k1k2[P]2
(d) k1k2[a]
Show Answer & Explanation

Answer: (c) k1k2[P]2

Explanation:
1. Slow step: rate = k2[Q]2[P].
2. From the fast equilibrium, [Q]2 = K1[P].
3. Rate = k2K1[P]2.

Teacher's Note:
1. Eliminate the intermediate Q with the equilibrium constant.

Question 22: The reaction 2A + B → P follows the mechanism 2A ⇌ A2 (fast), A2 + B → P (slow). The order of the reaction is
(a) 1.5
(b) 3
(c) 1
(d) 2
Show Answer & Explanation

Answer: (b) 3

Explanation:
1. Rate = k[A2][B] and [A2] = K[A]2.
2. Rate = kK[A]2[B], so order = 3.

Teacher's Note:
1. Intermediate A2 is replaced by [A]2.

Question 23: For 2NO + Cl2 → 2NOCl, the mechanism is NO + Cl2 ⇌ NOCl2 (fast); NOCl2 + NO → 2NOCl (slow). The rate law is
(a) Rate = K[NO]2[Cl2]
(b) Rate = K[NO][Cl2]2
(c) Rate = K[NOCl2]
(d) Rate = K[NOCl]2
Show Answer & Explanation

Answer: (a) Rate = K[NO]2[Cl2]

Explanation:
1. Slow step: rate = k[NOCl2][NO].
2. [NOCl2] = K[NO][Cl2] from the fast equilibrium.
3. Rate = kK[NO]2[Cl2].

Teacher's Note:
1. Final rate laws contain only reactants, never intermediates.

Question 24: For the bromination of acetone CH3COCH3 + Br2 → CH3COCH2Br + H+ + Br− the data are: ([acetone], [Br2], [H+], rate) = (0.30, 0.05, 0.05, 5.7 × 10−5); (0.30, 0.10, 0.05, 5.7 × 10−5); (0.30, 0.10, 0.10, 1.2 × 10−4); (0.40, 0.05, 0.20, 3.1 × 10−4). The rate equation is
(a) Rate = k[CH3COCH3][Br2]
(b) Rate = k[CH3COCH3][Br2][H+]2
(c) Rate = k[CH3COCH3][Br2][H+]
(d) Rate = k[CH3COCH3][H+]
Show Answer & Explanation

Answer: (d) Rate = k[CH3COCH3][H+]

Explanation:
1. Rows 1 and 2: [Br2] doubles, rate unchanged, so zero order in Br2.
2. Rows 2 and 3: [H+] doubles, rate doubles (5.7 × 10−5 to 1.2 × 10−4), so first order in H+.
3. Row 4 check: (0.40/0.30) × (0.20/0.05) = 5.33, giving 3.04 × 10−4, which matches first order in acetone.

Teacher's Note:
1. Zero order in bromine is why bromination rate is independent of [Br2].
2. Acid catalysis appears in the rate law.

Question 25: For a reaction A + B → products, doubling the initial concentration of A alone doubles the rate, and doubling both A and B changes the rate by a factor of 8. The rate law is
(a) rate = k[A][B]
(b) rate = k[A]2[B]
(c) rate = k[A][B]2
(d) rate = k[A]2[B]2
Show Answer & Explanation

Answer: (c) rate = k[A][B]2

Explanation:
1. Order in A = 1.
2. Doubling both gives 2 × 2m = 8, so m = 2.
3. Rate = k[A][B]2.

Teacher's Note:
1. Separate the factor due to A from the combined factor.

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