JEE Chemistry Chemical Kinetics MCQs Set 23

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

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

Practice Chemical Kinetics MCQs for JEE Chemistry

Access the complete set of multiple-choice questions for Chemical Kinetics below. This focused format allows students to isolate specific topics for thorough review and uninterrupted practice alongside official JEE textbooks.

Multiple Choice Questions

Question 1: The velocity constant of a reaction at 290 K is \( 3.2\times10^{-3} \). Assuming the rate doubles for every 10 K rise, its value at 310 K will be about
(a) \( 1.28\times10^{-2} \)
(b) \( 9.6\times10^{-3} \)
(c) \( 6.4\times10^{-3} \)
(d) \( 3.2\times10^{-4} \)
Show Answer & Explanation

Answer: (a) \( 1.28\times10^{-2} \)

Explanation:
1. A 20 K rise is two 10 K steps, so the rate constant increases by a factor of \( 2^2=4 \).
2. \( k=3.2\times10^{-3}\times4=1.28\times10^{-2} \).

Teacher's Note:
1. Learn the units of k: zero order mol/(L s), first order s⁻¹, second order L/(mol s).
2. Molecularity is always a whole number, but order can be zero or fractional.

Question 2: For \( 2N_2O_5(g)\rightarrow4NO_2(g)+O_2(g) \), the concentration of \( NO_2 \) increases by \( 5.2\times10^{-3} \) M in 100 seconds. The rate of the reaction is
(a) \( 1.3\times10^{-5}\ M\,s^{-1} \)
(b) \( 5\times10^{-4}\ M\,s^{-1} \)
(c) \( 7.6\times10^{-4}\ M\,s^{-1} \)
(d) \( 2\times10^{-3}\ M\,s^{-1} \)
Show Answer & Explanation

Answer: (a) \( 1.3\times10^{-5}\ M\,s^{-1} \)

Explanation:
1. Rate of reaction = (1/4) x rate of formation of \( NO_2 \).
2. \( \frac14\times\frac{5.2\times10^{-3}}{100}=1.3\times10^{-5} \) M/s.

Teacher's Note:
1. Learn the units of k: zero order mol/(L s), first order s⁻¹, second order L/(mol s).
2. Molecularity is always a whole number, but order can be zero or fractional.

Question 3: A first order reaction is 10% complete in 20 minutes. The time required to complete 19% is
(a) 30 minutes
(b) 40 minutes
(c) 50 minutes
(d) 38 minutes
Show Answer & Explanation

Answer: (b) 40 minutes

Explanation:
1. Remaining after 10% is 0.90 and after 19% is 0.81 = \( 0.9^2 \).
2. For first order reactions, equal time gives an equal fraction, so 19% completion takes 2 x 20 = 40 minutes.

Teacher's Note:
1. For a first order reaction use \( k=\frac{2.303}{t}\log\frac{a}{a-x} \) and \( t_{1/2}=\frac{0.693}{k} \).
2. Half-life of a first order reaction does not depend on the initial concentration.

Question 4: The data for the reaction A + B → C are: (1) [A] = 0.012, [B] = 0.035, rate = 0.10; (2) [A] = 0.024, [B] = 0.070, rate = 0.80; (3) [A] = 0.024, [B] = 0.035, rate = 0.10; (4) [A] = 0.012, [B] = 0.070, rate = 0.80. The rate law is
(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: (a) \( Rate=k[B]^3 \)

Explanation:
1. Runs 1 and 3: [A] doubles, rate unchanged, so the order in A is 0.
2. Runs 1 and 4: [B] doubles, rate increases 8 times, so the order in B is 3.
3. Rate = k[B]\( ^3 \).

Teacher's Note:
1. Write the rate law from the experimental data by comparing two runs in which only one concentration changes.
2. The order is the sum of the powers of concentration terms in the rate law, not the sum of the stoichiometric coefficients.

Question 5: For the reaction 2A + B\( _2 \) → 2AB the experimental data are: (1) [A] = 0.50, [B\( _2 \)] = 0.50, rate = \( 1.6\times10^{-4} \); (2) [A] = 0.50, [B\( _2 \)] = 1.00, rate = \( 3.2\times10^{-4} \); (3) [A] = 1.00, [B\( _2 \)] = 1.00, rate = \( 3.2\times10^{-4} \). The rate equation is
(a) \( Rate=k[B_2] \)
(b) \( Rate=k[B_2]^2 \)
(c) \( Rate=k[A]^2[B_2]^2 \)
(d) \( Rate=k[A]^2[B_2] \)
Show Answer & Explanation

Answer: (a) \( Rate=k[B_2] \)

Explanation:
1. Runs 1 and 2: [B2] doubles and the rate doubles, so the order in B2 is 1.
2. Runs 2 and 3: [A] doubles, rate unchanged, so the order in A is 0.
3. Rate = k[B2].

Teacher's Note:
1. Write the rate law from the experimental data by comparing two runs in which only one concentration changes.
2. The order is the sum of the powers of concentration terms in the rate law, not the sum of the stoichiometric coefficients.

Question 6: For a reaction 2A + B → products, doubling the initial concentration of both reactants increases the rate by a factor of 8, and doubling the concentration of B alone doubles the rate. The rate law is
(a) \( r=k[A][B]^2 \)
(b) \( r=k[A]^2[B] \)
(c) \( r=k[A][B] \)
(d) \( r=k[A]^2[B]^2 \)
Show Answer & Explanation

Answer: (b) \( r=k[A]^2[B] \)

Explanation:
1. Doubling B alone doubles the rate, so the order in B is 1.
2. Doubling both gives \( 2^{a+b}=8 \), so a + b = 3 and the order in A is 2.
3. r = k[A]\( ^2 \)[B].

Teacher's Note:
1. Write the rate law from the experimental data by comparing two runs in which only one concentration changes.
2. The order is the sum of the powers of concentration terms in the rate law, not the sum of the stoichiometric coefficients.

Question 7: For a reaction A + B → product, the rate increases four times when the concentration of A is doubled, while the rate is unaffected when B is doubled. The rate law is
(a) \( rate=k[A][B] \)
(b) \( rate=k[A]^2 \)
(c) \( rate=k[A]^2[B]^1 \)
(d) \( rate=k[A]^2[B]^2 \)
Show Answer & Explanation

Answer: (b) \( rate=k[A]^2 \)

Explanation:
1. Rate quadruples when [A] is doubled: order in A = 2.
2. No change with [B]: order in B = 0.
3. Rate = k[A]\( ^2 \).

Teacher's Note:
1. Write the rate law from the experimental data by comparing two runs in which only one concentration changes.
2. The order is the sum of the powers of concentration terms in the rate law, not the sum of the stoichiometric coefficients.

Question 8: In a first order reaction the concentration of the reactant decreases from 800 mol/dm\( ^3 \) to 50 mol/dm\( ^3 \) in \( 2\times10^2 \) s. The rate constant is
(a) \( 2\times10^4\ s^{-1} \)
(b) \( 3.45\times10^{-5}\ s^{-1} \)
(c) \( 1.386\times10^{-2}\ s^{-1} \)
(d) \( 2\times10^{-4}\ s^{-1} \)
Show Answer & Explanation

Answer: (c) \( 1.386\times10^{-2}\ s^{-1} \)

Explanation:
1. 800 to 50 is a fall by a factor of 16 = \( 2^4 \), i.e. 4 half-lives.
2. t\( _{1/2} \) = 200/4 = 50 s, so \( k=0.693/50=1.386\times10^{-2}\ s^{-1} \).

Teacher's Note:
1. For a first order reaction use \( k=\frac{2.303}{t}\log\frac{a}{a-x} \) and \( t_{1/2}=\frac{0.693}{k} \).
2. Half-life of a first order reaction does not depend on the initial concentration.

Question 9: For a reaction A → B, the rate of reaction quadruples when the concentration of A is doubled. For the rate expression \( r=k(A)^n \), the value of n is
(a) 1
(b) 0
(c) 3
(d) 2
Show Answer & Explanation

Answer: (d) 2

Explanation:
1. \( 2^n=4 \), so n = 2.

Teacher's Note:
1. Write the rate law from the experimental data by comparing two runs in which only one concentration changes.
2. The order is the sum of the powers of concentration terms in the rate law, not the sum of the stoichiometric coefficients.

Question 10: The reaction \( N_2O_5 \) (in \( CCl_4 \)) → \( 2NO_2 \) + 1/2 \( O_2 \) is first order in \( N_2O_5 \) with rate constant \( 6.2\times10^{-1}\ s^{-1} \). The rate of reaction when [N\( _2 \)O\( _5 \)] = 1.25 mol/L is
(a) \( 7.75\times10^{-1}\ mol\,L^{-1}s^{-1} \)
(b) \( 6.35\times10^{-3}\ mol\,L^{-1}s^{-1} \)
(c) \( 5.15\times10^{-5}\ mol\,L^{-1}s^{-1} \)
(d) \( 3.85\times10^{-1}\ mol\,L^{-1}s^{-1} \)
Show Answer & Explanation

Answer: (a) \( 7.75\times10^{-1}\ mol\,L^{-1}s^{-1} \)

Explanation:
1. Rate = k[N2O5] = 0.62 x 1.25 = 0.775 mol L\( ^{-1} \) s\( ^{-1} \).

Teacher's Note:
1. For a first order reaction use \( k=\frac{2.303}{t}\log\frac{a}{a-x} \) and \( t_{1/2}=\frac{0.693}{k} \).
2. Half-life of a first order reaction does not depend on the initial concentration.

Question 11: A reaction that is first order in reactant A has a rate constant of 6 \( min^{-1} \). Starting with [A] = 0.5 mol/L, the time for [A] to reach 0.05 mol/L is
(a) 0.384 min
(b) 0.15 min
(c) 3 min
(d) 3.84 min
Show Answer & Explanation

Answer: (a) 0.384 min

Explanation:
1. \( t=\frac{2.303}{k}\log\frac{0.5}{0.05}=\frac{2.303}{6}\times1 \) = 0.384 min.

Teacher's Note:
1. For a first order reaction use \( k=\frac{2.303}{t}\log\frac{a}{a-x} \) and \( t_{1/2}=\frac{0.693}{k} \).
2. Half-life of a first order reaction does not depend on the initial concentration.

Question 12: For the first order decomposition of \( N_2O_5 \), the rate of reaction and the rate constant are \( 1.02\times10^{-4} \) and \( 3.4\times10^{-5}\ s^{-1} \) respectively at a certain instant. The concentration of \( N_2O_5 \) at that time is (mol/L)
(a) 1.732
(b) 3
(c) \( 1.02\times10^{-4} \)
(d) \( 3.4\times10^{5} \)
Show Answer & Explanation

Answer: (b) 3

Explanation:
1. Rate = k[N2O5].
2. [N2O5] = \( 1.02\times10^{-4}/3.4\times10^{-5} \) = 3 mol/L.

Teacher's Note:
1. For a first order reaction use \( k=\frac{2.303}{t}\log\frac{a}{a-x} \) and \( t_{1/2}=\frac{0.693}{k} \).
2. Half-life of a first order reaction does not depend on the initial concentration.

Question 13: The rate equation for 2A + B → C is found to be rate = k[A][B]. The correct statement about this reaction is that
(a) the rate of formation of C is twice the rate of disappearance of A
(b) the half-life is a constant
(c) the unit of k must be \( s^{-1} \)
(d) the value of k is independent of the initial concentrations of A and B
Show Answer & Explanation

Answer: (d) the value of k is independent of the initial concentrations of A and B

Explanation:
1. The rate constant depends only on temperature (and catalyst), not on concentrations.
2. Rate of formation of C is half the rate of disappearance of A, and the reaction is second order, so t\( _{1/2} \) is not constant and k is not in s\( ^{-1} \).

Teacher's Note:
1. Learn the units of k: zero order mol/(L s), first order s⁻¹, second order L/(mol s).
2. Molecularity is always a whole number, but order can be zero or fractional.

Question 14: A first order reaction which is 30% complete in 30 minutes has a half-life of
(a) 24.2 min
(b) 58.2 min
(c) 102.2 min
(d) 120.2 min
Show Answer & Explanation

Answer: (b) 58.2 min

Explanation:
1. \( k=\frac{2.303}{30}\log\frac{100}{70}=0.0119\ min^{-1} \).
2. t\( _{1/2} \) = 0.693/0.0119 = 58.3 min, i.e. about 58.2 min.

Teacher's Note:
1. For a first order reaction use \( k=\frac{2.303}{t}\log\frac{a}{a-x} \) and \( t_{1/2}=\frac{0.693}{k} \).
2. Half-life of a first order reaction does not depend on the initial concentration.

Question 15: The first order rate constant for the decomposition of \( N_2O_5 \) is \( 6.2\times10^{-4}\ s^{-1} \). Its half-life in seconds is
(a) 1117.7
(b) 111.7
(c) 223.4
(d) 160.9
Show Answer & Explanation

Answer: (a) 1117.7

Explanation:
1. \( t_{1/2}=\frac{0.693}{6.2\times10^{-4}} \) = 1117.7 s.

Teacher's Note:
1. For a first order reaction use \( k=\frac{2.303}{t}\log\frac{a}{a-x} \) and \( t_{1/2}=\frac{0.693}{k} \).
2. Half-life of a first order reaction does not depend on the initial concentration.

Question 16: The decomposition of \( N_2O_5 \) is a first order reaction: \( N_2O_5\rightarrow N_2O_4+\tfrac12O_2 \). After 15 minutes the volume of \( O_2 \) produced is 9 mL and at the end of the reaction 35 mL. The rate constant is
(a) \( \frac1{15}\ln\frac{35}{44} \)
(b) \( \frac1{15}\ln\frac{44}{35} \)
(c) \( \frac1{15}\ln\frac{35}{9} \)
(d) \( \frac1{15}\ln\frac{35}{26} \)
Show Answer & Explanation

Answer: (d) \( \frac1{15}\ln\frac{35}{26} \)

Explanation:
1. Volume of O2 is proportional to the amount decomposed: a = 35, x = 9, a - x = 26.
2. \( k=\frac1t\ln\frac{a}{a-x}=\frac1{15}\ln\frac{35}{26} \).

Teacher's Note:
1. For a first order reaction use \( k=\frac{2.303}{t}\log\frac{a}{a-x} \) and \( t_{1/2}=\frac{0.693}{k} \).
2. Half-life of a first order reaction does not depend on the initial concentration.

Question 17: The decay constant of a reaction is \( 1.1\times10^{-9}\ s^{-1} \). Its half-life (in seconds) is
(a) \( 1.2\times10^8 \)
(b) \( 6.3\times10^8 \)
(c) \( 3.3\times10^8 \)
(d) \( 2.1\times10^8 \)
Show Answer & Explanation

Answer: (b) \( 6.3\times10^8 \)

Explanation:
1. \( t_{1/2}=\frac{0.693}{1.1\times10^{-9}}=6.3\times10^{8} \) s.

Teacher's Note:
1. For a first order reaction use \( k=\frac{2.303}{t}\log\frac{a}{a-x} \) and \( t_{1/2}=\frac{0.693}{k} \).
2. Half-life of a first order reaction does not depend on the initial concentration.

Question 18: Decomposition of \( N_2O_5 \) is a first order reaction. In the first 24 minutes 75% of the oxide had decomposed. At the end of one hour after the start, the amount of oxide left will be
(a) nil
(b) about 1%
(c) about 2%
(d) about 3%
Show Answer & Explanation

Answer: (d) about 3%

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

Teacher's Note:
1. For a first order reaction use \( k=\frac{2.303}{t}\log\frac{a}{a-x} \) and \( t_{1/2}=\frac{0.693}{k} \).
2. Half-life of a first order reaction does not depend on the initial concentration.

Question 19: In presence of HCl, sucrose is hydrolysed into glucose and fructose. The concentration of sucrose was found to fall from 0.4 M to 0.2 M in 1 hour and to 0.1 M in 2 hours. The order of the reaction is
(a) zero
(b) one
(c) two
(d) none of these
Show Answer & Explanation

Answer: (b) one

Explanation:
1. The concentration halves in each hour: the half-life (1 h) is constant.
2. A constant half-life means a first order reaction.

Teacher's Note:
1. For a first order reaction use \( k=\frac{2.303}{t}\log\frac{a}{a-x} \) and \( t_{1/2}=\frac{0.693}{k} \).
2. Half-life of a first order reaction does not depend on the initial concentration.

Question 20: The rates of a certain reaction at different times are: 0 min, \( 2.8\times10^{-2} \); 10 min, \( 2.78\times10^{-2} \); 20 min, \( 2.81\times10^{-2} \); 30 min, \( 2.79\times10^{-2} \) mol/(L s). The reaction is
(a) zero order
(b) first order
(c) second order
(d) third order
Show Answer & Explanation

Answer: (a) zero order

Explanation:
1. The rate stays constant (within experimental error) although the reactant is being consumed.
2. A rate independent of concentration indicates zero order.

Teacher's Note:
1. Write the rate law from the experimental data by comparing two runs in which only one concentration changes.
2. The order is the sum of the powers of concentration terms in the rate law, not the sum of the stoichiometric coefficients.

Question 21: For a chemical reaction A → B the rate doubles when the concentration of A is increased four times. The order in A is
(a) two
(b) one
(c) half
(d) zero
Show Answer & Explanation

Answer: (c) half

Explanation:
1. \( 4^n=2 \), so n = 1/2.

Teacher's Note:
1. Write the rate law from the experimental data by comparing two runs in which only one concentration changes.
2. The order is the sum of the powers of concentration terms in the rate law, not the sum of the stoichiometric coefficients.

Question 22: The decomposition of ammonium nitrate in aqueous solution gave the following volumes of \( N_2 \) (cc): 6.25 at 10 min, 9.50 at 15 min, 11.42 at 20 min, 13.65 at 25 min and 35.05 finally. The order of the reaction is
(a) zero
(b) one
(c) two
(d) three
Show Answer & Explanation

Answer: (b) one

Explanation:
1. Take a = 35.05 cc (final volume). The value of \( \frac{2.303}{t}\log\frac{a}{a-x} \) is 0.0196, 0.0211, 0.0197 and 0.0197 min\( ^{-1} \) at the four times.
2. The nearly constant k confirms a first order reaction.

Teacher's Note:
1. For a first order reaction use \( k=\frac{2.303}{t}\log\frac{a}{a-x} \) and \( t_{1/2}=\frac{0.693}{k} \).
2. Half-life of a first order reaction does not depend on the initial concentration.

Question 23: If 'a' is the initial concentration and 'n' the order of the reaction, the half-life T is proportional to
(a) \( a^{n-1} \)
(b) \( a^{n} \)
(c) \( \frac1{a^{n}} \)
(d) \( \frac1{a^{n-1}} \)
Show Answer & Explanation

Answer: (d) \( \frac1{a^{n-1}} \)

Explanation:
1. For an nth order reaction \( t_{1/2}\propto\frac{1}{a^{n-1}} \).
2. Check: n = 1 gives no dependence on a; n = 2 gives 1/a.

Teacher's Note:
1. Learn the units of k: zero order mol/(L s), first order s⁻¹, second order L/(mol s).
2. Molecularity is always a whole number, but order can be zero or fractional.

Question 24: For the reaction A + 2B → C + D, if \( -\frac{d[A]}{dt}=5\times10^{-4}\ mol\,L^{-1}s^{-1} \), then \( -\frac{d[B]}{dt} \) is
(a) \( 2.5\times10^{-4}\ mol\,L^{-1}s^{-1} \)
(b) \( 5.0\times10^{-4}\ mol\,L^{-1}s^{-1} \)
(c) \( 2.5\times10^{-3}\ mol\,L^{-1}s^{-1} \)
(d) \( 1.0\times10^{-3}\ mol\,L^{-1}s^{-1} \)
Show Answer & Explanation

Answer: (d) \( 1.0\times10^{-3}\ mol\,L^{-1}s^{-1} \)

Explanation:
1. For every mole of A used, 2 mol of B is used.
2. B disappears twice as fast: 2 x \( 5\times10^{-4}=1.0\times10^{-3} \).

Teacher's Note:
1. Learn the units of k: zero order mol/(L s), first order s⁻¹, second order L/(mol s).
2. Molecularity is always a whole number, but order can be zero or fractional.

Question 25: The unit of the rate constant for a zero order reaction is
(a) \( s^{-1} \)
(b) \( L\ s^{-1} \)
(c) \( mol\,L^{-1}s^{-1} \)
(d) \( mol^{-1}L\,s^{-1} \)
Show Answer & Explanation

Answer: (c) \( mol\,L^{-1}s^{-1} \)

Explanation:
1. For zero order, rate = k, so k has the units of rate: mol L\( ^{-1} \) s\( ^{-1} \).

Teacher's Note:
1. Learn the units of k: zero order mol/(L s), first order s⁻¹, second order L/(mol s).
2. Molecularity is always a whole number, but order can be zero or fractional.

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