JEE Chemistry Chemical Kinetics MCQs Set 06

Chemistry Objective Questions and Answers: 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.

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: In the Haber process \( N_2 + 3H_2 \rightarrow 2NH_3 \), the rate of reaction measured as \( \dfrac{\Delta[NH_3]}{\Delta t} \) is \( 2\times10^{-4} \) mol L\(^{-1}\) s\(^{-1}\). The rate of disappearance of \( H_2 \) is
(a) \( 1\times10^{-4} \) mol L\(^{-1}\) s\(^{-1}\)
(b) \( 3\times10^{-4} \) mol L\(^{-1}\) s\(^{-1}\)
(c) \( 2\times10^{-4} \) mol L\(^{-1}\) s\(^{-1}\)
(d) \( 6\times10^{-4} \) mol L\(^{-1}\) s\(^{-1}\)
Show Answer & Explanation

Answer: (b) \( 3\times10^{-4} \) mol L\(^{-1}\) s\(^{-1}\)

Explanation:
1. Rate of reaction \( = \dfrac{1}{2}\dfrac{d[NH_3]}{dt} = -\dfrac{1}{3}\dfrac{d[H_2]}{dt} = -\dfrac{d[N_2]}{dt} \).
2. Rate of reaction \( = 1\times10^{-4} \) mol L\(^{-1}\) s\(^{-1}\), so rate of disappearance of \( H_2 = 3\times1\times10^{-4} = 3\times10^{-4} \).

Teacher's Note:
1. Divide the rate of change of each species by its stoichiometric coefficient to get the rate of reaction. For N2 the answer would be 1 x 10-4.

Question 2: For \( 2A + B + C \rightarrow D + E \), the reaction is first order in A, second order in B and zero order in C. If the concentrations of A, B and C are all doubled, the rate becomes
(a) 8 times
(b) 4 times
(c) 6 times
(d) 16 times
Show Answer & Explanation

Answer: (a) 8 times

Explanation:
1. Rate \( = k[A][B]^2[C]^0 \).
2. Doubling all concentrations gives rate \( = k(2A)(2B)^2 = 8\,k[A][B]^2 \).
3. The rate increases 8 times.

Teacher's Note:
1. Zero-order components do not change the rate. Multiply 2 to the power of each order for the others.

Question 3: For the elementary reaction \( 2A + B_2 \rightarrow 2AB \), the volume of the vessel is reduced to one third of its original volume. The rate of reaction becomes
(a) 9 times
(b) 3 times
(c) 81 times
(d) 27 times
Show Answer & Explanation

Answer: (d) 27 times

Explanation:
1. Rate \( = k[A]^2[B_2] \) for an elementary reaction.
2. Reducing the volume to one third triples every concentration.
3. New rate \( = k(3A)^2(3B_2) = 27\,k[A]^2[B_2] \).

Teacher's Note:
1. For an elementary reaction the orders equal the coefficients, so the total order is 3 and the factor is 3 cubed.

Question 4: For \( 4NH_3 + 5O_2 \rightarrow 4NO + 6H_2O \), the concentration of NO increases by \( 1.08\times10^{-2} \) mol L\(^{-1}\) in 3 s. The rate of formation of water is
(a) \( 3.6\times10^{-3} \) mol L\(^{-1}\) s\(^{-1}\)
(b) \( 9\times10^{-4} \) mol L\(^{-1}\) s\(^{-1}\)
(c) \( 5.4\times10^{-3} \) mol L\(^{-1}\) s\(^{-1}\)
(d) \( 1.08\times10^{-2} \) mol L\(^{-1}\) s\(^{-1}\)
Show Answer & Explanation

Answer: (c) \( 5.4\times10^{-3} \) mol L\(^{-1}\) s\(^{-1}\)

Explanation:
1. Rate of formation of NO \( = 1.08\times10^{-2}/3 = 3.6\times10^{-3} \).
2. Rate of reaction \( = 3.6\times10^{-3}/4 = 9\times10^{-4} \).
3. Rate of formation of \( H_2O = 6\times9\times10^{-4} = 5.4\times10^{-3} \) mol L\(^{-1}\) s\(^{-1}\).

Teacher's Note:
1. Find the rate of reaction first, then multiply by the coefficient of the species asked.

Question 5: For the elementary reaction \( A(g) + 2B(g) \rightarrow C(g) + D(g) \), the initial partial pressures are \( P_A = 0.6 \) atm and \( P_B = 0.8 \) atm. When \( P_C = 0.2 \) atm, the rate relative to the initial rate is
(a) \( \dfrac{1}{3} \)
(b) \( \dfrac{1}{6} \)
(c) \( \dfrac{1}{4} \)
(d) \( \dfrac{1}{8} \)
Show Answer & Explanation

Answer: (b) \( \dfrac{1}{6} \)

Explanation:
1. When 0.2 atm of C forms, A used = 0.2 atm and B used = 0.4 atm.
2. So \( P_A = 0.4 \) and \( P_B = 0.4 \) atm.
3. Rate ratio \( = \dfrac{0.4\times(0.4)^2}{0.6\times(0.8)^2} = \dfrac{0.064}{0.384} = \dfrac{1}{6} \).

Teacher's Note:
1. Use stoichiometry to update the partial pressures and then substitute in the rate law of the elementary step.

Question 6: For a zero order reaction \( A \rightarrow B \) with \( k = 1.2\times10^{-2} \) M s\(^{-1}\), starting with 10 M of A, the concentration of B after 10 minutes is
(a) 7.2 M
(b) 0.72 M
(c) 12 M
(d) 3.6 M
Show Answer & Explanation

Answer: (a) 7.2 M

Explanation:
1. For zero order, \( [B] = kt = 1.2\times10^{-2}\times600 = 7.2 \) M.
2. This is less than 10 M, so A is not exhausted.

Teacher's Note:
1. Convert minutes to seconds before using k in M per second. After 20 minutes the reaction would be complete, since 14.4 M cannot form from 10 M.

Question 7: The rate constant of a zero order reaction is \( 2\times10^{-2} \) mol L\(^{-1}\) s\(^{-1}\). If the concentration of the reactant after 25 s is 0.25 M, the initial concentration is
(a) 0.50 M
(b) 0.25 M
(c) 1.00 M
(d) 0.75 M
Show Answer & Explanation

Answer: (d) 0.75 M

Explanation:
1. \( [A]_0 - [A] = kt = 2\times10^{-2}\times25 = 0.5 \) M.
2. \( [A]_0 = 0.25 + 0.50 = 0.75 \) M.

Teacher's Note:
1. For zero order the concentration falls linearly with time.

Question 8: A zero order reaction of A with an equimolar quantity of B is 75% complete in one hour. The time at which A is 10% unreacted is
(a) 1.0 hour
(b) 1.5 hours
(c) 1.2 hours
(d) 0.9 hours
Show Answer & Explanation

Answer: (c) 1.2 hours

Explanation:
1. Zero order: amount reacted is proportional to time, so 75% in 1 h gives 75 units per hour.
2. 10% unreacted means 90% reacted, \( t = 90/75 = 1.2 \) h.

Teacher's Note:
1. For zero order the percentage reacted is linear in t, unlike first order.

Question 9: A first order reaction is 75% complete in 72 minutes. The time required for 87.5% completion is
(a) 144 min
(b) 108 min
(c) 90 min
(d) 96 min
Show Answer & Explanation

Answer: (b) 108 min

Explanation:
1. 75% completion takes two half-lives, so \( t_{1/2} = 36 \) min.
2. 87.5% completion takes three half-lives \( = 3\times36 = 108 \) min.

Teacher's Note:
1. Count half-lives: 50% is 1, 75% is 2, 87.5% is 3, 93.75% is 4.

Question 10: A first order reaction is 20% complete in 10 minutes. The time taken to reach 75% completion is nearly
(a) 62 min
(b) 31 min
(c) 46 min
(d) 100 min
Show Answer & Explanation

Answer: (a) 62 min

Explanation:
1. \( k = \dfrac{2.303}{10}\log\dfrac{100}{80} = 0.0223 \) min\(^{-1}\).
2. \( t = \dfrac{2.303}{0.0223}\log\dfrac{100}{25} = \dfrac{1.386}{0.0223} \approx 62 \) min.

Teacher's Note:
1. Always use remaining amount in the log: 80 for the first part and 25 for the second.

Question 11: A first order reaction has \( k = 1.5\times10^{-3} \) s\(^{-1}\). The time taken for 5.0 g of the reactant to reduce to 1.25 g is nearly
(a) 462 s
(b) 1848 s
(c) 693 s
(d) 924 s
Show Answer & Explanation

Answer: (d) 924 s

Explanation:
1. 5.0 g to 1.25 g is two half-lives (5.0 to 2.5 to 1.25).
2. \( t_{1/2} = 0.693/1.5\times10^{-3} = 462 \) s, so \( t = 2\times462 = 924 \) s.

Teacher's Note:
1. Spot the number of half-lives before doing any calculation.

Question 12: A drug is ineffective after 30% decomposition. A sample of 500 units/ml analysed after 20 months is 420 units/ml. Assuming first order decomposition, the expiry time of the drug is nearly
(a) 20 months
(b) 30 months
(c) 41 months
(d) 60 months
Show Answer & Explanation

Answer: (c) 41 months

Explanation:
1. \( k = \dfrac{2.303}{20}\log\dfrac{500}{420} = 8.72\times10^{-3} \) month\(^{-1}\).
2. Expiry when 70% remains: \( t = \dfrac{2.303}{k}\log\dfrac{100}{70} = \dfrac{0.3567}{8.72\times10^{-3}} \approx 41 \) months.

Teacher's Note:
1. Expiry means 30% decomposed, which leaves 70% of the original concentration.

Question 13: The first order reaction \( SO_2Cl_2(g) \rightarrow SO_2(g) + Cl_2(g) \) has \( k = 2.2\times10^{-5} \) s\(^{-1}\) at 320 \(^\circ\)C. The percentage of \( SO_2Cl_2 \) decomposed on heating for 90 minutes is nearly
(a) 22.4%
(b) 11.2%
(c) 5.6%
(d) 88.8%
Show Answer & Explanation

Answer: (b) 11.2%

Explanation:
1. \( t = 90\times60 = 5400 \) s, so \( kt = 0.1188 \).
2. Fraction left \( = e^{-0.1188} = 0.888 \).
3. Decomposed = 11.2%.

Teacher's Note:
1. Convert minutes to seconds. For small kt the decomposed fraction is close to kt.

Question 14: For the reaction \( 2NO + H_2 \rightarrow N_2 + H_2O \) at 800 \(^\circ\)C, \( -\dfrac12\dfrac{d[NO]}{dt} \) is \( 4.4\times10^{-4} \) M s\(^{-1}\) for [NO] = \( 1.5\times10^{-4} \) M and [H\(_2\)] = \( 4\times10^{-3} \) M; \( 2.2\times10^{-4} \) for [NO] = \( 1.5\times10^{-4} \), [H\(_2\)] = \( 2\times10^{-3} \); and \( 8.8\times10^{-4} \) for [NO] = \( 3.0\times10^{-4} \), [H\(_2\)] = \( 2\times10^{-3} \). The rate law and overall order are
(a) rate \( = k[NO]^2[H_2] \), order 3
(b) rate \( = k[NO][H_2] \), order 2
(c) rate \( = k[NO][H_2]^2 \), order 3
(d) rate \( = k[NO]^2[H_2]^2 \), order 4
Show Answer & Explanation

Answer: (a) rate \( = k[NO]^2[H_2] \), order 3

Explanation:
1. Rows 1 and 2: [H\(_2\)] halves, rate halves, so order in \( H_2 \) is 1.
2. Rows 2 and 3: [NO] doubles, rate becomes 4 times, so order in NO is 2.
3. Rate \( = k[NO]^2[H_2] \), overall order 3.

Teacher's Note:
1. Change one concentration at a time and compare the rates.

Question 15: For the rate law found in the previous reaction, \( -\frac12\frac{d[NO]}{dt} = k[NO]^2[H_2] \), the rate at [H\(_2\)] = \( 1.5\times10^{-3} \) M and [NO] = \( 1.1\times10^{-3} \) M is nearly
(a) \( 4.4\times10^{-4} \) M s\(^{-1}\)
(b) \( 8.9\times10^{-4} \) M s\(^{-1}\)
(c) \( 1.8\times10^{-2} \) M s\(^{-1}\)
(d) \( 8.9\times10^{-3} \) M s\(^{-1}\)
Show Answer & Explanation

Answer: (d) \( 8.9\times10^{-3} \) M s\(^{-1}\)

Explanation:
1. From the first row, \( k = \dfrac{4.4\times10^{-4}}{(1.5\times10^{-4})^2(4\times10^{-3})} = 4.89\times10^{6} \) M\(^{-2}\) s\(^{-1}\).
2. Rate \( = 4.89\times10^6\times(1.1\times10^{-3})^2\times(1.5\times10^{-3}) = 8.9\times10^{-3} \) M s\(^{-1}\).

Teacher's Note:
1. Find k from any data row, then substitute the new concentrations.

Question 16: The reaction of NO with \( Cl_2 \) at 295 K gives initial rates \( 1\times10^{-3} \) (\( [Cl_2] = 0.05 \), [NO] = 0.05), \( 3\times10^{-3} \) (0.15, 0.05) and \( 9\times10^{-3} \) M s\(^{-1}\) (0.05, 0.15). The rate when \( [Cl_2] = 0.2 \) M and [NO] = 0.4 M is
(a) 0.128 M s\(^{-1}\)
(b) 0.064 M s\(^{-1}\)
(c) 0.256 M s\(^{-1}\)
(d) 1.024 M s\(^{-1}\)
Show Answer & Explanation

Answer: (c) 0.256 M s\(^{-1}\)

Explanation:
1. Tripling \( [Cl_2] \) triples the rate, so first order in \( Cl_2 \). Tripling [NO] raises the rate 9 times, so second order in NO.
2. \( k = \dfrac{1\times10^{-3}}{(0.05)^2(0.05)} = 8 \) L\(^2\) mol\(^{-2}\) s\(^{-1}\).
3. Rate \( = 8\times(0.4)^2\times0.2 = 0.256 \) M s\(^{-1}\).

Teacher's Note:
1. Orders come from ratios of rates; k then comes from any single row.

Question 17: Catalytic decomposition of \( N_2O \) on gold at 900 \(^\circ\)C and initial pressure 200 mm is 50% complete in 53 min and 73% complete in 100 min. The reaction is first order. The value of k is nearly
(a) \( 2.62\times10^{-2} \) min\(^{-1}\)
(b) \( 1.31\times10^{-2} \) min\(^{-1}\)
(c) \( 6.5\times10^{-3} \) min\(^{-1}\)
(d) \( 1.31\times10^{-3} \) min\(^{-1}\)
Show Answer & Explanation

Answer: (b) \( 1.31\times10^{-2} \) min\(^{-1}\)

Explanation:
1. Half life is 53 min. If the reaction were zero order, 73% in 100 min would not fit; for first order 73% needs 1.9 half-lives \( = 100 \) min, which fits.
2. \( k = 0.693/53 = 1.31\times10^{-2} \) min\(^{-1}\).

Teacher's Note:
1. Check order by comparing times for 50% and 75% (for first order they differ by a factor of 2).

Question 18: For the reaction \( A + B \rightarrow \) products, the initial rates are \( 4.0\times10^{-4} \) (A = 0.1, B = 0.1), \( 1.6\times10^{-3} \) (0.2, 0.2), \( 2.0\times10^{-3} \) (0.5, 0.1) and \( 1.0\times10^{-2} \) mol L\(^{-1}\) s\(^{-1}\) (0.5, 0.5). The rate for A = 0.2 M and B = 0.35 M is
(a) \( 2.8\times10^{-3} \) M s\(^{-1}\)
(b) \( 1.4\times10^{-3} \) M s\(^{-1}\)
(c) \( 5.6\times10^{-3} \) M s\(^{-1}\)
(d) \( 2.8\times10^{-2} \) M s\(^{-1}\)
Show Answer & Explanation

Answer: (a) \( 2.8\times10^{-3} \) M s\(^{-1}\)

Explanation:
1. Rows 1 and 3: A rises 5 times, rate rises 5 times, so first order in A.
2. Rows 3 and 4: B rises 5 times, rate rises 5 times, so first order in B.
3. \( k = 4\times10^{-4}/(0.1\times0.1) = 4\times10^{-2} \) M\(^{-1}\) s\(^{-1}\); rate \( = 4\times10^{-2}\times0.2\times0.35 = 2.8\times10^{-3} \).

Teacher's Note:
1. Check the answer against row 2 as well: 0.04 x 0.2 x 0.2 = 1.6 x 10-3.

Question 19: The pressure of a gas decomposing on a solid catalyst is \( 4\times10^3 \), \( 3.5\times10^3 \), \( 3\times10^3 \) and \( 2.5\times10^3 \) Pa at t = 0, 100, 200 and 300 s. The order and rate constant are
(a) first order, 5 Pa s\(^{-1}\)
(b) zero order, 50 Pa s\(^{-1}\)
(c) first order, 1.3 x 10\(^{-3}\) s\(^{-1}\)
(d) zero order, 5 Pa s\(^{-1}\)
Show Answer & Explanation

Answer: (d) zero order, 5 Pa s\(^{-1}\)

Explanation:
1. The pressure falls by 500 Pa in every 100 s, a constant amount per unit time, so the reaction is zero order.
2. \( k = 500/100 = 5 \) Pa s\(^{-1}\).

Teacher's Note:
1. A constant decrease per equal time interval is the signature of zero order.

Question 20: The half life of a compound is 50 minutes. When the initial concentration is halved the half life becomes 25 minutes. The order of the reaction is
(a) one
(b) two
(c) zero
(d) three
Show Answer & Explanation

Answer: (c) zero

Explanation:
1. For order n, \( t_{1/2}\propto a^{1-n} \).
2. Half life is proportional to initial concentration, so \( 1-n = 1 \) and \( n = 0 \).

Teacher's Note:
1. Zero order: half life proportional to a. First order: independent of a. Second order: inversely proportional to a.

Question 21: The initial rates of decomposition of acetone at 600 \(^\circ\)C are \( 5.2\times10^{-5} \) M s\(^{-1}\) at \( 6.0\times10^{-3} \) M and \( 7.8\times10^{-5} \) at \( 9.0\times10^{-3} \) M. The rate at \( 1.8\times10^{-3} \) M is
(a) \( 2.6\times10^{-5} \) M s\(^{-1}\)
(b) \( 1.56\times10^{-5} \) M s\(^{-1}\)
(c) \( 4.7\times10^{-6} \) M s\(^{-1}\)
(d) \( 5.2\times10^{-5} \) M s\(^{-1}\)
Show Answer & Explanation

Answer: (b) \( 1.56\times10^{-5} \) M s\(^{-1}\)

Explanation:
1. Concentration rises 1.5 times and the rate rises 1.5 times, so the reaction is first order.
2. \( k = 5.2\times10^{-5}/6.0\times10^{-3} = 8.67\times10^{-3} \) s\(^{-1}\).
3. Rate \( = 8.67\times10^{-3}\times1.8\times10^{-3} = 1.56\times10^{-5} \) M s\(^{-1}\).

Teacher's Note:
1. The rate is directly proportional to concentration for first order.

Question 22: The half life of a first order reaction is 50 minutes. The time for 90% completion is nearly
(a) 166 min
(b) 100 min
(c) 150 min
(d) 200 min
Show Answer & Explanation

Answer: (a) 166 min

Explanation:
1. \( k = 0.693/50 \) min\(^{-1}\).
2. \( t = \dfrac{2.303}{k}\log 10 = \dfrac{2.303\times50}{0.693} \approx 166 \) min.

Teacher's Note:
1. For 90% completion t is 3.32 times t half.

Question 23: A first order reaction has \( k = 1.5\times10^{-6} \) s\(^{-1}\) at 200 \(^\circ\)C. The half life is
(a) \( 2.31\times10^{5} \) s
(b) \( 1.04\times10^{-6} \) s
(c) \( 9.24\times10^{5} \) s
(d) \( 4.62\times10^{5} \) s
Show Answer & Explanation

Answer: (d) \( 4.62\times10^{5} \) s

Explanation:
1. \( t_{1/2} = 0.693/k = 0.693/(1.5\times10^{-6}) = 4.62\times10^5 \) s.

Teacher's Note:
1. This is about 128 hours.

Question 24: The half life of the first order decomposition of nitramide \( NH_2NO_2 \to N_2O + H_2O \) is 2.1 h at 15 \(^\circ\)C. The time taken for 99% decomposition is nearly
(a) 7 h
(b) 21 h
(c) 14 h
(d) 4.2 h
Show Answer & Explanation

Answer: (c) 14 h

Explanation:
1. \( t = t_{1/2}\times\log_2 100 = 2.1\times6.64 \approx 13.95 \) h.

Teacher's Note:
1. 99% completion needs 6.64 half-lives.

Question 25: If 6.2 g of \( NH_2NO_2 \) (molar mass 62 g mol\(^{-1}\)) decomposes 99% by the reaction \( NH_2NO_2 \to N_2O(g) + H_2O(l) \), the volume of dry \( N_2O \) at STP is
(a) 0.224 L
(b) 2.2176 L
(c) 22.4 L
(d) 1.108 L
Show Answer & Explanation

Answer: (b) 2.2176 L

Explanation:
1. Moles of \( NH_2NO_2 = 6.2/62 = 0.1 \).
2. Moles decomposed \( = 0.099 \), so moles of \( N_2O = 0.099 \).
3. Volume \( = 0.099\times22.4 = 2.2176 \) L.

Teacher's Note:
1. One mole of N2O is formed per mole of nitramide, and water is a liquid so it adds no gas volume.

Free study material for Chemistry

Practice MCQs for JEE Chemistry Chemical Kinetics

Download Multiple Choice Questions: Chemical Kinetics (JEE Chemistry)

Test your conceptual understanding of Chemical Kinetics with these targeted multiple-choice questions. Designed in alignment with the latest JEE curriculum for JEE Chemistry, these problem sets build accuracy and prepare students for objective exams.

Concept Clarification for Chemical Kinetics

Built using the official NCERT book for JEE, these Chemistry objective sets provide reliable academic guidance. Pair your practice with our recommended NCERT solutions to master optimal problem-solving approaches.

Additional Study Resources for JEE Chemistry

Wrap up your chapter revision by testing your knowledge against standard question formats. Everything on our platform is provided free of charge.

FAQs

Where can I access latest JEE Chemistry Chemical Kinetics MCQs Set 06?

You can get most exhaustive JEE Chemistry Chemical Kinetics MCQs Set 06 for free on StudiesToday.com. These MCQs for JEE Chemistry are updated for the 2026-27 academic session as per JEE examination standards.

Are Assertion-Reasoning and Case-Study MCQs included in the Chemistry JEE material?

Yes, our JEE Chemistry Chemical Kinetics MCQs Set 06 include the latest type of questions, such as Assertion-Reasoning and Case-based MCQs. 50% of the JEE paper is now competency-based.

How do practicing Chemistry MCQs help in scoring full marks in JEE exams?

By solving our JEE Chemistry Chemical Kinetics MCQs Set 06, JEE students can improve their accuracy and speed which is important as objective questions provide a chance to secure 100% marks in the Chemistry.

Do you provide answers and explanations for JEE Chemistry Chemical Kinetics MCQs Set 06?

Yes, Chemistry MCQs for JEE have answer key and brief explanations to help students understand logic behind the correct option as its important for 2026 competency-focused JEE exams.

Can I practice these Chemistry JEE MCQs online?

Yes, you can also access online interactive tests for JEE Chemistry Chemical Kinetics MCQs Set 06 on StudiesToday.com as they provide instant answers and score to help you track your progress in Chemistry.