JEE Chemistry Chemical Kinetics MCQs Set 04

Multiple Choice Questions (MCQs) for JEE Chemistry: Chemical Kinetics

Review structured MCQ sets for JEE Chemistry Chemical Kinetics. Built according to official JEE guidelines, these downloadable questions support daily revision and core concept reinforcement.

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Multiple Choice Questions

Question 1: For NO₂ + CO → CO₂ + NO, rate = k[NO₂]². The number of CO molecules 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.
2. So it is not part of the slow (rate-determining) step; the slow step is 2NO₂ → NO₃ + NO.

Teacher's Note:
1. Only species in the slow step (and prior equilibria) appear in the rate law.
2. Zero order in CO means no CO in the slow step.

Question 2: The rate law of A + 2B → product is \( \frac{d[P]}{dt} = k[A]^2[B] \). 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. With A in large excess, [A] stays practically constant and merges into k.
2. Rate = k′[B], so the order is 1 (pseudo-first-order).

Teacher's Note:
1. Excess reactant gives a pseudo order.
2. This is why ester hydrolysis in water appears first order.

Question 3: The acid hydrolysis of an ester is
(a) A pseudo first-order reaction
(b) A bimolecular reaction
(c) A pseudo unimolecular reaction
(d) All of these
Show Answer & Explanation

Answer: (d) All of these

Explanation:
1. Ester + water → acid + alcohol is bimolecular.
2. Water is in large excess so its concentration is constant: the rate depends on [ester] only.
3. Hence it is pseudo first order and pseudo unimolecular, and bimolecular too.

Teacher's Note:
1. Order is experimental, molecularity is theoretical.
2. Water in excess makes the hydrolysis appear first order.

Question 4: For P → R the mechanism is: P ⇌ 2Q (fast, equilibrium constant K₁); 2Q + P → R (slow, constant k₂). The rate law is
(a) k₁k₂[P][Q]
(b) k₁k₂[P]
(c) k₁k₂[P]²
(d) k₁k₂[Q]
Show Answer & Explanation

Answer: (c) k₁k₂[P]²

Explanation:
1. Slow step: rate = k₂[Q]²[P].
2. From equilibrium, [Q]² = K₁[P].
3. Rate = k₂K₁[P]² = K₁k₂[P]².

Teacher's Note:
1. Eliminate the intermediate Q using the equilibrium.
2. The total order is 2.

Question 5: 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 the order is 2 + 1 = 3.

Teacher's Note:
1. Treat the fast step as an equilibrium.
2. Sum the powers after substitution.

Question 6: The rate constant is numerically the same for three reactions of first, second and third order. For concentrations of reactants all equal to and greater than 1 M, which is true for the rates r₁, r₂, r₃?
(a) r₁ = r₂ = r₃
(b) r₁ > r₂ > r₃
(c) r₁ < r₂ < r₃
(d) All of these
Show Answer & Explanation

Answer: (c) r₁ < r₂ < r₃

Explanation:
1. r₁ = kc, r₂ = kc², r₃ = kc³.
2. For c > 1, c < c² < c³, so r₁ < r₂ < r₃.

Teacher's Note:
1. If c were less than 1 the order of rates would reverse.
2. Higher order means stronger dependence on concentration.

Question 7: The rate constant of a zero-order reaction is 2 × 10⁻² mol L⁻¹ s⁻¹. If the concentration of the reactant after 25 s is 0.5 M, the initial concentration must have been
(a) 0.5 M
(b) 1.25 M
(c) 12.5 M
(d) 1.0 M
Show Answer & Explanation

Answer: (d) 1.0 M

Explanation:
1. For zero order, [A]₀ − [A] = kt = 2 × 10⁻² × 25 = 0.5 M.
2. [A]₀ = 0.5 + 0.5 = 1.0 M.

Teacher's Note:
1. Zero-order rate constants have units of concentration per time.
2. Concentration falls linearly with time.

Question 8: The decomposition 2N₂O₅ → 4NO₂ + O₂ follows first-order kinetics. Hence
(a) The reaction is bimolecular
(b) The reaction is unimolecular
(c) t₁/₂ ∝ a⁰
(d) t₁/₂ ∝ a²
Show Answer & Explanation

Answer: (c) t₁/₂ ∝ a⁰

Explanation:
1. For a first-order reaction t₁/₂ = 0.693/k, independent of initial concentration a.
2. So t₁/₂ ∝ a⁰.

Teacher's Note:
1. Molecularity cannot be inferred from the order.
2. Half-life independent of concentration is the signature of first order.

Question 9: A plot of log(a − x) against time t is a straight line. This indicates that the reaction is of
(a) Second order
(b) First order
(c) Zero order
(d) Third order
Show Answer & Explanation

Answer: (b) First order

Explanation:
1. For first order, \( \log(a-x) = \log a - \frac{kt}{2.303} \).
2. This is linear in t with slope −k/2.303.

Teacher's Note:
1. Linear plot of log concentration versus time identifies first order.
2. Zero order: [A] versus t; second order: 1/[A] versus t.

Question 10: The rate constant of a first-order reaction is 4 × 10⁻³ s⁻¹. At a reactant concentration of 0.02 M, the rate is
(a) 8 × 10⁻⁵ M s⁻¹
(b) 4 × 10⁻³ M s⁻¹
(c) 2 × 10⁻¹ M s⁻¹
(d) 4 × 10⁻¹ M s⁻¹
Show Answer & Explanation

Answer: (a) 8 × 10⁻⁵ M s⁻¹

Explanation:
1. Rate = k[A] = 4 × 10⁻³ × 0.02 = 8 × 10⁻⁵ M s⁻¹.

Teacher's Note:
1. Multiply k by concentration for first order.
2. Check the units: s⁻¹ × M = M s⁻¹.

Question 11: In a first-order reaction a/(a − x) was found to be 8 after 10 minutes. The rate constant is
(a) \( \frac{2.303\times3\log2}{10} \)
(b) \( \frac{2.303\times2\log3}{10} \)
(c) \( 10\times2.303\times2\log3 \)
(d) \( 10\times2.303\times3\log2 \)
Show Answer & Explanation

Answer: (a) \( \frac{2.303\times3\log2}{10} \)

Explanation:
1. k = (2.303/t) log[a/(a−x)] = (2.303/10) log 8.
2. log 8 = 3 log 2, so k = 2.303 × 3 log 2 / 10.

Teacher's Note:
1. Substitute the ratio directly into the integrated rate law.
2. Express log 8 as 3 log 2 to match the form.

Question 12: 75% of a first-order reaction was found to complete in 32 min. When will 50% of the same reaction complete?
(a) 24 min
(b) 16 min
(c) 8 min
(d) 4 min
Show Answer & Explanation

Answer: (b) 16 min

Explanation:
1. 75% completion = 2 half-lives, so 2t₁/₂ = 32 min.
2. t₁/₂ = 16 min, which is the time for 50% completion.

Teacher's Note:
1. 50% completion is one half-life by definition.
2. 75% = two half-lives, 87.5% = three.

Question 13: A first-order reaction has a half-life of 69.3 s. At 0.10 mol L⁻¹ reactant concentration the rate will be
(a) 10⁻⁴ M s⁻¹
(b) 10⁻³ M s⁻¹
(c) 10⁻¹ M s⁻¹
(d) 6.93 × 10⁻¹ M s⁻¹
Show Answer & Explanation

Answer: (b) 10⁻³ M s⁻¹

Explanation:
1. k = 0.693/69.3 = 10⁻² s⁻¹.
2. Rate = k[A] = 10⁻² × 0.10 = 10⁻³ M s⁻¹.

Teacher's Note:
1. Find k from the half-life, then the rate from the rate law.
2. Units of rate are M s⁻¹.

Question 14: 99% of a first-order reaction was completed in 32 min. When will 99.9% of the reaction complete?
(a) 50 min
(b) 46 min
(c) 49 min
(d) 48 min
Show Answer & Explanation

Answer: (d) 48 min

Explanation:
1. t ∝ log[a/(a−x)]. For 99%, log 100 = 2; for 99.9%, log 1000 = 3.
2. t = 32 × 3/2 = 48 min.

Teacher's Note:
1. The time is proportional to the log of the reciprocal of the fraction left.
2. Ratio 3/2 gives the answer without needing k.

Question 15: The half-life of the first-order reaction N₂O₅ → 2NO₂ + ½O₂ is 24 h at 30 °C. Starting with 10 g of N₂O₅, how many grams remain after 96 h?
(a) 1.25 g
(b) 0.63 g
(c) 1.77 g
(d) 0.5 g
Show Answer & Explanation

Answer: (b) 0.63 g

Explanation:
1. 96 h = 4 half-lives.
2. Remaining = 10 × (½)⁴ = 10/16 = 0.625 ≈ 0.63 g.

Teacher's Note:
1. Number of half-lives n = t/t₁/₂.
2. Amount left = initial × (½)ⁿ.

Question 16: For a first-order reaction it takes 20 minutes for the concentration to drop from 1 M to 0.6 M. The time required for the concentration to drop from 0.6 M to 0.36 M will be
(a) More than 20 min
(b) Less than 20 min
(c) Equal to 20 min
(d) Infinity
Show Answer & Explanation

Answer: (c) Equal to 20 min

Explanation:
1. 0.6/1 = 0.6 and 0.36/0.6 = 0.6, the same fractional decrease.
2. In first-order reactions equal fractional changes take equal times, so 20 min.

Teacher's Note:
1. First-order kinetics: constant fraction per unit time.
2. Compare ratios, not differences.

Question 17: A first-order reaction is carried out with an initial concentration of 10 mol L⁻¹ and 80% of the reactant changes into product. If the same reaction is carried out with an initial concentration of 5 mol L⁻¹ for the same period, the percentage of reactant changing to product is
(a) 40
(b) 80
(c) 160
(d) Cannot be calculated
Show Answer & Explanation

Answer: (b) 80

Explanation:
1. For a first-order reaction the fraction reacted in a given time is independent of initial concentration.
2. So again 80%.

Teacher's Note:
1. k and t fix the fraction reacted, not the amount.
2. This property is not true for zero or second order.

Question 18: A + Reagent → Product and B + Reagent → Product are both first order. Calculate k₁/k₂ if 50% of B has reacted when 94% of A has reacted (same time).
(a) 4.06
(b) 0.246
(c) 2.06
(d) 0.06
Show Answer & Explanation

Answer: (a) 4.06

Explanation:
1. k₁ = (2.303/t) log(100/6) and k₂ = (2.303/t) log(100/50).
2. k₁/k₂ = log 16.67 / log 2 = 1.222/0.301 = 4.06.

Teacher's Note:
1. Same time t means the ratio of rate constants equals the ratio of logs.
2. Use log(100/(100−%)).

Question 19: Doubling the initial concentration of a reactant doubles the half-life of the reaction. The order of the reaction is
(a) 3
(b) 2
(c) 1
(d) 0
Show Answer & Explanation

Answer: (d) 0

Explanation:
1. t₁/₂ ∝ a^{1−n}.
2. Doubling a doubles t₁/₂, so 1 − n = 1 and n = 0.

Teacher's Note:
1. Zero order: t₁/₂ = a/(2k).
2. Direct proportionality to a means zero order.

Question 20: The half-life for the catalytic decomposition of AB₃ at 50 mm is 4 h and at 100 mm it is 2 h. The order of the reaction is
(a) 3
(b) 1
(c) 2
(d) 0
Show Answer & Explanation

Answer: (c) 2

Explanation:
1. Doubling the pressure halves the half-life, so t₁/₂ ∝ 1/a.
2. That corresponds to second order.

Teacher's Note:
1. Inverse proportionality of t₁/₂ to initial concentration means second order.
2. Pressure plays the role of concentration.

Question 21: A substance A decomposes in solution by first-order kinetics. Flask I contains 1 L of 1 M solution of A and Flask II contains 100 mL of 0.6 M solution. After 8 hours the concentration of A in Flask I becomes 0.25 M. What time is needed for the concentration of A in Flask II to become 0.3 M?
(a) 0.4 hr
(b) 2.4 hr
(c) 4.0 hr
(d) Unpredictable as the rate constant is not given
Show Answer & Explanation

Answer: (c) 4.0 hr

Explanation:
1. In Flask I, 1 → 0.25 M is two half-lives, so t₁/₂ = 4 h.
2. In Flask II, 0.6 → 0.3 M is one half-life = 4.0 h.

Teacher's Note:
1. Half-life of a first-order reaction does not depend on volume or initial concentration.
2. Count half-lives rather than computing k.

Question 22: The rate constant k for 2A + B → product was 2.5 × 10⁻⁵ L mol⁻¹ s⁻¹ after 15 s, 2.60 × 10⁻⁵ after 30 s and 2.55 × 10⁻⁵ after 50 s. The order of the reaction is
(a) 2
(b) 3
(c) 0
(d) 1
Show Answer & Explanation

Answer: (a) 2

Explanation:
1. k is practically constant over time, with the unit L mol⁻¹ s⁻¹.
2. That unit corresponds to second order.

Teacher's Note:
1. Small scatter in k is experimental error.
2. Unit L mol⁻¹ s⁻¹ means second order.

Question 23: The rate constant of a second-order reaction is 8 × 10⁻⁵ M⁻¹ min⁻¹. How long will it take for a 1 M solution to be reduced to 0.5 M?
(a) 8.665 × 10³ min
(b) 8 × 10⁻³ min
(c) 1.25 × 10⁴ min
(d) 4 × 10⁻⁵ min
Show Answer & Explanation

Answer: (c) 1.25 × 10⁴ min

Explanation:
1. For second order, \( kt = \frac{1}{[A]}-\frac{1}{[A]_0} \).
2. t = (2 − 1)/(8 × 10⁻⁵) = 1.25 × 10⁴ min.

Teacher's Note:
1. For second order t₁/₂ = 1/(k a).
2. Here a = 1 M, so t₁/₂ = 1/k.

Question 24: For a first-order reaction, the ratio of the time required for 99.9% completion to that for 50% completion is
(a) 2
(b) 5
(c) 10
(d) None of these
Show Answer & Explanation

Answer: (c) 10

Explanation:
1. t₉₉.₉ = (2.303/k) log 1000 = 6.909/k and t₅₀ = 0.693/k.
2. Ratio = 6.909/0.693 ≈ 10.

Teacher's Note:
1. t₉₉.₉ ≈ 10 t₁/₂ is a useful rule of thumb.
2. log 1000 = 3 and log 2 = 0.301.

Question 25: From different sets of data of t₁/₂ at different initial concentrations a for a given reaction, the product t₁/₂ × a is found to be constant. The order of the reaction is
(a) 0
(b) 1
(c) 2
(d) 3
Show Answer & Explanation

Answer: (c) 2

Explanation:
1. t₁/₂ × a = constant means t₁/₂ ∝ 1/a.
2. That is second order.

Teacher's Note:
1. Relate t₁/₂ ∝ a^{1−n} to the observation.
2. Here 1 − n = −1.

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