JEE Chemistry Chemical Kinetics MCQs Set 10

Chemistry Objective Questions and Answers: 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.

Download Chemical Kinetics MCQs with Answers

View or download the dedicated Chemical Kinetics 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: The rate constants of first, second and third order reactions are numerically equal. At a moment when the concentration of every reactant is the same and greater than 1 M, which is true?
(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. The order reverses if c < 1 M.
2. Compare using the powers of concentration.

Question 2: 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. Zero order: [A] = [A]0 − kt.
2. [A]0 = 0.5 + 2 × 10−2 × 25 = 1.0 M.

Teacher's Note:
1. Concentration falls linearly with time in a zero order reaction.

Question 3: The decomposition of N2O5 (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 first order, t1/2 = 0.693/k, independent of initial concentration a, so t1/2 ∝ a0.
2. Order is experimental and does not decide molecularity.

Teacher's Note:
1. First order does not necessarily mean unimolecular for the overall equation.

Question 4: In the graph of concentrations of X and Y versus time for X → Y, the point of intersection of the two curves represents
(a) t1/2
(b) t3/4
(c) t2/3
(d) data are insufficient to predict
Show Answer & Explanation

Answer: (a) t1/2

Explanation:
1. At the intersection [X] = [Y], and [X] + [Y] = initial concentration.
2. So half of X has reacted: the time is t1/2.

Teacher's Note:
1. Equal concentrations of reactant and product means 50% conversion.

Question 5: If a first order reaction involves gaseous reactants and gaseous products, the unit of its rate is
(a) atm
(b) atm s
(c) atm s−1
(d) atm2 s2
Show Answer & Explanation

Answer: (c) atm s−1

Explanation:
1. Rate = change of (partial) pressure per unit time = atm s−1.

Teacher's Note:
1. The unit of k for first order remains s−1.

Question 6: A plot of log(a − x) versus time t is a straight line. This indicates 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 − kt/2.303.
2. This is linear in t with slope −k/2.303.

Teacher's Note:
1. Linear graphs: zero order [A] vs t; first order ln[A] vs t; second order 1/[A] vs t.

Question 7: The rate constant of a first order reaction is 4 × 10−3 s−1. At a reactant concentration of 0.02 M, the rate of the reaction would be
(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 the rate constant by the concentration.

Question 8: In a first order reaction the concentration of the reactant is decreased from 1.0 M to 0.25 M in 20 min. The rate constant of the reaction would be
(a) 10 min−1
(b) 6.931 min−1
(c) 0.6931 min−1
(d) 0.06931 min−1
Show Answer & Explanation

Answer: (d) 0.06931 min−1

Explanation:
1. \( k = \dfrac{2.303}{t}\log\dfrac{a}{a-x} = \dfrac{2.303}{20}\log 4 = 0.0693 \) min−1.

Teacher's Note:
1. A fall to 1/4 takes two half-lives, so t1/2 = 10 min and k = 0.693/10.

Question 9: In a first order reaction a/(a − x) was found to be 8 after 10 minutes. The rate constant is
(a) \( \dfrac{2.303\times3\log2}{10} \)
(b) \( \dfrac{2.303\times2\log3}{10} \)
(c) \( 10\times2.303\times2\log3 \)
(d) \( 10\times2.303\times3\log2 \)
Show Answer & Explanation

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

Explanation:
1. \( k = \dfrac{2.303}{t}\log\dfrac{a}{a-x} = \dfrac{2.303}{10}\log 8 = \dfrac{2.303\times 3\log 2}{10} \).

Teacher's Note:
1. log 8 = 3 log 2.
2. Place the time in the denominator.

Question 10: 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 is two half-lives, so t1/2 = 32/2 = 16 min.
2. 50% completion = one half-life = 16 min.

Teacher's Note:
1. For first order, 75% = 2 t1/2 and 87.5% = 3 t1/2.

Question 11: 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 = 0.01 s−1.
2. Rate = 0.01 × 0.10 = 10−3 M s−1.

Teacher's Note:
1. Get k from the half-life first.

Question 12: For a first order reaction A → products, the rate of reaction at [A] = 0.2 M is 1 × 10−2 mol L−1 min−1. The half life period of the reaction is
(a) 832 min
(b) 440 min
(c) 416 min
(d) 14 min
Show Answer & Explanation

Answer: (d) 14 min

Explanation:
1. k = rate/[A] = 0.01/0.2 = 0.05 min−1.
2. t1/2 = 0.693/0.05 = 13.9 min, about 14 min.

Teacher's Note:
1. Rate divided by concentration gives k for a first order reaction.

Question 13: 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 = (2.303/k) log (100/(100 − %)): 99% gives log 100 = 2; 99.9% gives log 1000 = 3.
2. t99.9/t99 = 3/2, so t = 1.5 × 32 = 48 min.

Teacher's Note:
1. Completion times scale with log of 100/(100 − percentage).

Question 14: The half life for the first order reaction N2O5 → 2NO2 + 1/2 O2 is 24 hours at 30°C. Starting with 10 g of N2O5, how many grams will remain after 96 hours?
(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 × (1/2)4 = 0.625 g, about 0.63 g.

Teacher's Note:
1. Amount left = initial × (1/2)n, n = t/t1/2.

Question 15: What is the half life of a radioactive substance if 87.5% of any given amount of the substance disintegrates in 40 minutes?
(a) 160 min
(b) 10 min
(c) 20 min
(d) 13 min 20 s
Show Answer & Explanation

Answer: (d) 13 min 20 s

Explanation:
1. 87.5% decayed leaves 12.5% = (1/2)3, so three half lives.
2. t1/2 = 40/3 = 13.33 min = 13 min 20 s.

Teacher's Note:
1. Radioactive decay is first order.

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. For first order, equal fractional change takes equal time: 20 min.

Teacher's Note:
1. First order: the time depends only on the ratio, not on the actual concentrations.

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 time, 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 first order, the fraction reacted in a given time is independent of the initial concentration.
2. So 80% again.

Teacher's Note:
1. Option (c) is impossible: more than 100%.

Question 18: For A + Reagent → Product and B + Reagent → Product (both first order), calculate k1/k2 if 50% of B and 94% of A have reacted in the 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) = 2.814/t.
2. k2 = (2.303/t) log 2 = 0.693/t.
3. k1/k2 = 2.814/0.693 = 4.06.

Teacher's Note:
1. For the same time, the ratio of k equals the ratio of ln terms.

Question 19: The reaction L → M is started with 10 g/L. After 30 min and 90 min, 5 g/L and 1.25 g/L are left respectively. The order of the reaction is
(a) 0
(b) 2
(c) 1
(d) 3
Show Answer & Explanation

Answer: (c) 1

Explanation:
1. 10 → 5 in 30 min: t1/2 = 30 min.
2. From 5 at 30 min to 1.25 at 90 min is two half lives in 60 min, again 30 min each.
3. Constant half-life means first order.

Teacher's Note:
1. A constant half life is the signature of first order.

Question 20: Doubling the initial concentration of a reactant doubles the t1/2 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 ∝ a1−n.
2. Doubling t1/2 with doubled a means 1 − n = 1, n = 0.

Teacher's Note:
1. Zero order half life: t1/2 = a/2k.

Question 21: The half life period for the catalytic decomposition of AB3 at 50 mm is 4 hours and at 100 mm it is 2 hours. The order of reaction is
(a) 3
(b) 1
(c) 2
(d) 0
Show Answer & Explanation

Answer: (c) 2

Explanation:
1. Doubling the pressure halves t1/2, so t1/2 ∝ 1/a.
2. 1 − n = −1 gives n = 2.

Teacher's Note:
1. t1/2 ∝ a1−n.

Question 22: A substance A decomposes in solution following 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 will be the time 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. 1 M → 0.25 M is two half lives, so t1/2 = 4 h.
2. Flask II: 0.6 M → 0.3 M is one half life = 4.0 h.

Teacher's Note:
1. The volume of solution does not matter for first order.

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