JEE Chemistry Chemical Kinetics MCQs Set 24

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Chapter-wise Objective Questions: Chemical Kinetics

Navigate directly to the 50 objective questions for Chemical Kinetics using the digital viewer below. Each practice set includes verified answer keys, allowing students to instantly cross-check their work and identify areas requiring further revision.

Multiple Choice Questions

Question 1: A first order reaction is 50% complete in 120 minutes. The time required for 90% decomposition is nearly
(a) 240 minutes
(b) 480 minutes
(c) 450 minutes
(d) 400 minutes
Show Answer & Explanation

Answer: (d) 400 minutes

Explanation:
1. t(1/2) = 120 min, so \( k=\frac{0.693}{120} \) per min.
2. t(90%) = \( \frac{2.303}{k}\log10 \) = 3.32 x 120 = 398.6, i.e. nearly 400 minutes.

Teacher's Note:
1. For a first order reaction use \( k=\frac{0.693}{t_{1/2}}=\frac{2.303}{t}\log\frac{a}{a-x} \).
2. The half-life of a first order reaction is independent of the initial concentration.

Question 2: The half-life of a first order reaction is 100 s. The rate constant of the reaction is
(a) 6.93 x 10\( ^{-3} \) s\( ^{-1} \)
(b) 6.93 x 10\( ^{-4} \) s\( ^{-1} \)
(c) 0.693 s\( ^{-1} \)
(d) 69.3 s\( ^{-1} \)
Show Answer & Explanation

Answer: (a) 6.93 x 10\( ^{-3} \) s\( ^{-1} \)

Explanation:
1. k = 0.693/t(1/2) = 0.693/100 = 6.93 x 10\( ^{-3} \) s\( ^{-1} \).

Teacher's Note:
1. For a first order reaction use \( k=\frac{0.693}{t_{1/2}}=\frac{2.303}{t}\log\frac{a}{a-x} \).
2. The half-life of a first order reaction is independent of the initial concentration.

Question 3: The rate constant of a first order reaction whose half-life is 480 seconds is
(a) 2.88 x 10\( ^{-3} \) s\( ^{-1} \)
(b) 1.44 x 10\( ^{-3} \) s\( ^{-1} \)
(c) 1.44 s\( ^{-1} \)
(d) 0.72 x 10\( ^{-3} \) s\( ^{-1} \)
Show Answer & Explanation

Answer: (b) 1.44 x 10\( ^{-3} \) s\( ^{-1} \)

Explanation:
1. k = 0.693/480 = 1.44 x 10\( ^{-3} \) s\( ^{-1} \).

Teacher's Note:
1. For a first order reaction use \( k=\frac{0.693}{t_{1/2}}=\frac{2.303}{t}\log\frac{a}{a-x} \).
2. The half-life of a first order reaction is independent of the initial concentration.

Question 4: The conversion of A to B follows second order kinetics. Doubling the concentration of A will increase the rate of formation of B by a factor of
(a) 1/4
(b) 2
(c) 1/2
(d) 4
Show Answer & Explanation

Answer: (d) 4

Explanation:
1. Rate = k[A]\( ^2 \).
2. Doubling [A] gives a rate \( 2^2=4 \) times larger.

Teacher's Note:
1. The order is the sum of the powers of concentration terms in the experimentally found rate law, not the coefficients of the equation.
2. Molecularity is always a whole number; order can be zero or fractional.

Question 5: If the reaction between A and B to give C shows first order kinetics in A and second order in B, the rate equation is
(a) \( Rate=k[A][B]^{1/2} \)
(b) \( Rate=k[A]^{1/2}[B] \)
(c) \( Rate=k[A][B]^2 \)
(d) \( Rate=k[A]^2[B] \)
Show Answer & Explanation

Answer: (c) \( Rate=k[A][B]^2 \)

Explanation:
1. Order 1 in A and order 2 in B give Rate = k[A]\( ^1 \)[B]\( ^2 \).

Teacher's Note:
1. The order is the sum of the powers of concentration terms in the experimentally found rate law, not the coefficients of the equation.
2. Molecularity is always a whole number; order can be zero or fractional.

Question 6: For a reaction A to B, the rate increases by a factor of 2.25 when the concentration of A is increased by 1.5 times. The order of the reaction is
(a) 3
(b) 0
(c) 2
(d) 1
Show Answer & Explanation

Answer: (c) 2

Explanation:
1. \( (1.5)^n=2.25 \) gives n = 2 since \( 1.5^2=2.25 \).

Teacher's Note:
1. The order is the sum of the powers of concentration terms in the experimentally found rate law, not the coefficients of the equation.
2. Molecularity is always a whole number; order can be zero or fractional.

Question 7: For the reaction A + 2B to C, the rate is given by R = [A][B]\( ^2 \). The order of the reaction is
(a) 3
(b) 6
(c) 5
(d) 7
Show Answer & Explanation

Answer: (a) 3

Explanation:
1. Order = 1 + 2 = 3.

Teacher's Note:
1. The order is the sum of the powers of concentration terms in the experimentally found rate law, not the coefficients of the equation.
2. Molecularity is always a whole number; order can be zero or fractional.

Question 8: Units of the rate constant of first order and zero order reactions in terms of molarity M are respectively
(a) \( s^{-1} \), \( M\,s^{-1} \)
(b) \( M\,s^{-1} \), \( s^{-1} \)
(c) \( s^{-1} \), M
(d) \( M \), \( s^{-1} \)
Show Answer & Explanation

Answer: (a) \( s^{-1} \), \( M\,s^{-1} \)

Explanation:
1. First order: rate/concentration = (M/s)/M = \( s^{-1} \).
2. Zero order: rate = k, so the unit is \( M\,s^{-1} \).

Teacher's Note:
1. The units of k are (concentration)^(1-n) time^(-1) for a reaction of order n.
2. Zero order: mol/(L s); first order: per second; second order: L/(mol s).

Question 9: The half-life of a first order reaction is 693 seconds. The rate constant of the reaction is
(a) 0.1 s\( ^{-1} \)
(b) 0.01 s\( ^{-1} \)
(c) 0.001 s\( ^{-1} \)
(d) 0.0001 s\( ^{-1} \)
Show Answer & Explanation

Answer: (c) 0.001 s\( ^{-1} \)

Explanation:
1. k = 0.693/693 = 0.001 s\( ^{-1} \).

Teacher's Note:
1. For a first order reaction use \( k=\frac{0.693}{t_{1/2}}=\frac{2.303}{t}\log\frac{a}{a-x} \).
2. The half-life of a first order reaction is independent of the initial concentration.

Question 10: For a first order reaction with rate constant \( k=10^{-3}\ s^{-1} \), the time required for two-thirds of the reaction to complete is nearly
(a) 1100 s
(b) 2200 s
(c) 3300 s
(d) 4400 s
Show Answer & Explanation

Answer: (a) 1100 s

Explanation:
1. Remaining fraction = 1/3, so \( t=\frac{2.303}{k}\log3 \).
2. t = 2303 x 0.477 = 1099 s, i.e. about 1100 s.

Teacher's Note:
1. For a first order reaction use \( k=\frac{0.693}{t_{1/2}}=\frac{2.303}{t}\log\frac{a}{a-x} \).
2. The half-life of a first order reaction is independent of the initial concentration.

Question 11: In a reaction the concentration of the reactant is increased two times and then three times, and the rate increases four times and nine times respectively. The order of the reaction is
(a) 0
(b) 1
(c) 2
(d) 3
Show Answer & Explanation

Answer: (c) 2

Explanation:
1. \( 2^n=4 \) and \( 3^n=9 \), both give n = 2.

Teacher's Note:
1. The order is the sum of the powers of concentration terms in the experimentally found rate law, not the coefficients of the equation.
2. Molecularity is always a whole number; order can be zero or fractional.

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

Answer: (a) 16 min

Explanation:
1. 75% completion is two half-lives (100 to 50 to 25).
2. So 2 t(1/2) = 32 min and t(1/2) = 16 min.

Teacher's Note:
1. For a first order reaction use \( k=\frac{0.693}{t_{1/2}}=\frac{2.303}{t}\log\frac{a}{a-x} \).
2. The half-life of a first order reaction is independent of the initial concentration.

Question 13: For the reaction \( 2NO(g)+O_2(g)\rightarrow2NO_2(g) \), the volume is suddenly made half by raising the pressure. If the reaction is of first order in O2 and second order in NO, the rate will
(a) diminish to one fourth of its initial value
(b) diminish to one eighth of its initial value
(c) increase to eight times its initial value
(d) increase to four times its initial value
Show Answer & Explanation

Answer: (c) increase to eight times its initial value

Explanation:
1. Halving the volume doubles every concentration.
2. Rate = k[NO]\( ^2 \)[O\( _2 \)]; new rate = \( 2^2\times2=8 \) times the initial rate.

Teacher's Note:
1. The order is the sum of the powers of concentration terms in the experimentally found rate law, not the coefficients of the equation.
2. Molecularity is always a whole number; order can be zero or fractional.

Question 14: A to B follows first order kinetics. The time taken for 0.8 mol of A to produce 0.6 mol of B is 1 hour. What is the time taken for conversion of 0.9 mol of A to produce 0.675 mol of B?
(a) 2 hours
(b) 1 hour
(c) 0.5 hour
(d) 0.25 hour
Show Answer & Explanation

Answer: (b) 1 hour

Explanation:
1. In both cases 75% of A is converted (0.6/0.8 = 0.675/0.9 = 0.75).
2. For a first order reaction the time for a given fraction is independent of the initial amount, so the time is again 1 hour.

Teacher's Note:
1. For a first order reaction use \( k=\frac{0.693}{t_{1/2}}=\frac{2.303}{t}\log\frac{a}{a-x} \).
2. The half-life of a first order reaction is independent of the initial concentration.

Question 15: Decomposition of nitrogen pentoxide is a first order reaction. 75% of the oxide decomposed in the first 24 minutes. At the end of one hour from the start, the amount of oxide left is about
(a) nil
(b) 1%
(c) 2%
(d) 3%
Show Answer & Explanation

Answer: (d) 3%

Explanation:
1. 75% decomposed in 24 min means two half-lives, so t(1/2) = 12 min.
2. 60 min = 5 half-lives, so the fraction left = \( (1/2)^5 \) = 1/32 = 3.1%, about 3%.

Teacher's Note:
1. For a first order reaction use \( k=\frac{0.693}{t_{1/2}}=\frac{2.303}{t}\log\frac{a}{a-x} \).
2. The half-life of a first order reaction is independent of the initial concentration.

Question 16: The alkaline hydrolysis of ethyl acetate, \( CH_3COOC_2H_5+NaOH\rightarrow CH_3COONa+C_2H_5OH \), follows the rate law \( \frac{dx}{dt}=k[CH_3COOC_2H_5][NaOH] \). The reaction is
(a) bimolecular and of first order
(b) bimolecular and of second order
(c) pseudo-bimolecular
(d) pseudo-unimolecular
Show Answer & Explanation

Answer: (b) bimolecular and of second order

Explanation:
1. Rate depends on the first power of each of two reactants: order = 2.
2. Two molecules take part in the elementary step, so it is bimolecular (and the order equals the molecularity).

Teacher's Note:
1. The order is the sum of the powers of concentration terms in the experimentally found rate law, not the coefficients of the equation.
2. Molecularity is always a whole number; order can be zero or fractional.

Question 17: An endothermic reaction A to B has an activation energy of 15 kcal/mol and the energy of reaction is 5 kcal/mol. The activation energy for the reverse reaction B to A is
(a) 20 kcal/mol
(b) 15 kcal/mol
(c) 10 kcal/mol
(d) none of these
Show Answer & Explanation

Answer: (c) 10 kcal/mol

Explanation:
1. For an endothermic reaction \( \Delta H=E_f-E_b \).
2. E(b) = 15 - 5 = 10 kcal/mol.

Teacher's Note:
1. Arrhenius equation: \( k=Ae^{-E_a/RT} \) and \( \log\frac{k_2}{k_1}=\frac{E_a}{2.303R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right) \).
2. If the activation energy is zero the rate constant does not depend on temperature.

Question 18: For a reaction the activation energy is zero and the rate constant at 300 K is 3.2 x 10\( ^6 \) s\( ^{-1} \). The rate constant at 310 K is
(a) 3.2 x 10\( ^{-12} \) s\( ^{-1} \)
(b) 3.2 x 10\( ^6 \) s\( ^{-1} \)
(c) 6.4 x 10\( ^{12} \) s\( ^{-1} \)
(d) 6.4 x 10\( ^6 \) s\( ^{-1} \)
Show Answer & Explanation

Answer: (b) 3.2 x 10\( ^6 \) s\( ^{-1} \)

Explanation:
1. If \( E_a=0 \), \( k=Ae^{0}=A \), so k does not change with temperature.
2. k at 310 K = 3.2 x 10\( ^6 \) s\( ^{-1} \).

Teacher's Note:
1. Arrhenius equation: \( k=Ae^{-E_a/RT} \) and \( \log\frac{k_2}{k_1}=\frac{E_a}{2.303R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right) \).
2. If the activation energy is zero the rate constant does not depend on temperature.

Question 19: The rate constant of a reaction at 200 K is 10 times less than the rate constant at 400 K. The activation energy of the reaction is (R = gas constant)
(a) 1842.4 R
(b) 921.2 R
(c) 230.3 R
(d) 460.6 R
Show Answer & Explanation

Answer: (b) 921.2 R

Explanation:
1. \( \log\frac{k_{400}}{k_{200}}=\frac{E_a}{2.303R}\left(\frac{1}{200}-\frac{1}{400}\right) \).
2. 1 = \( \frac{E_a}{2.303R}\times\frac{1}{400} \), so \( E_a=921.2R \).

Teacher's Note:
1. Arrhenius equation: \( k=Ae^{-E_a/RT} \) and \( \log\frac{k_2}{k_1}=\frac{E_a}{2.303R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right) \).
2. If the activation energy is zero the rate constant does not depend on temperature.

Question 20: The rate constant is doubled when the temperature increases from 27 °C to 37 °C. The activation energy in kJ/mol is nearly (R = 8.314 J/(mol K))
(a) 34
(b) 54
(c) 100
(d) 50
Show Answer & Explanation

Answer: (b) 54

Explanation:
1. \( E_a=2.303R\log2\times\frac{T_1T_2}{T_2-T_1} \) = 2.303 x 8.314 x 0.301 x (300 x 310/10).
2. E(a) = 53.6 kJ/mol, about 54.

Teacher's Note:
1. Arrhenius equation: \( k=Ae^{-E_a/RT} \) and \( \log\frac{k_2}{k_1}=\frac{E_a}{2.303R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right) \).
2. If the activation energy is zero the rate constant does not depend on temperature.

Question 21: The half-life of a reaction is found to be inversely proportional to the cube of its initial concentration. The order of the reaction is
(a) 2
(b) 5
(c) 3
(d) 4
Show Answer & Explanation

Answer: (d) 4

Explanation:
1. For order n, \( t_{1/2}\propto a^{1-n} \).
2. Inverse cube means \( 1-n=-3 \), so n = 4.

Teacher's Note:
1. For a first order reaction use \( k=\frac{0.693}{t_{1/2}}=\frac{2.303}{t}\log\frac{a}{a-x} \).
2. The half-life of a first order reaction is independent of the initial concentration.

Question 22: The activation energy of a reaction is 9.0 kcal/mol. The percentage increase in the rate constant when the temperature is raised from 298 K to 308 K is nearly (R = 1.987 cal/(mol K))
(a) 63%
(b) 50%
(c) 100%
(d) 10%
Show Answer & Explanation

Answer: (a) 63%

Explanation:
1. \( \ln\frac{k_2}{k_1}=\frac{9000}{1.987}\times\frac{10}{298\times308} \) = 0.4935.
2. k2/k1 = 1.64, an increase of about 63%.

Teacher's Note:
1. Arrhenius equation: \( k=Ae^{-E_a/RT} \) and \( \log\frac{k_2}{k_1}=\frac{E_a}{2.303R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right) \).
2. If the activation energy is zero the rate constant does not depend on temperature.

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Multiple Choice Questions (MCQs) for JEE Chemistry Chemical Kinetics

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