JEE Chemistry Chemical Kinetics MCQs Set 30

Practice 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.

Access Chemical Kinetics Questions and Solutions

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: One microcurie of a radioactive substance is the quantity that produces
(a) 6.022 × 1010 disintegrations per second
(b) 3.7 × 1010 disintegrations per second
(c) 3.7 × 104 disintegrations per second
(d) 3.7 × 107 disintegrations per second
Show Answer & Explanation

Answer: (c) 3.7 × 104 disintegrations per second

Explanation:
1. 1 curie = 3.7 × 1010 disintegrations per second.
2. 1 microcurie = 10-6 curie = 3.7 × 104 disintegrations per second.

Teacher's Note:
1. A micro prefix means 10-6.
2. Curie is the older unit of activity; the becquerel is 1 disintegration per second.

Question 2: Using the same two mechanisms (I: rate = 1 × 10-5[A][B]; II: A → E with k1 = 1 × 10-4 s-1 is rate determining, then E + B → C + D with k2 = 1 × 1010 M-1 s-1), the rate by mechanism II with [A] = [B] = 1 M is
(a) 1 × 10-10 M s-1
(b) 1 × 1010 M s-1
(c) 1 × 10-6 M s-1
(d) 1 × 10-4 M s-1
Show Answer & Explanation

Answer: (d) 1 × 10-4 M s-1

Explanation:
1. The second step is extremely fast, so the first step controls the rate.
2. Rate = k1[A] = 1 × 10-4 x 1 = 1 × 10-4 M s-1.

Teacher's Note:
1. The slow step is rate determining.
2. B does not appear in the rate.

Question 3: A catalyst lowers the activation energy by 8.314 kJ mol-1 at 300 K. Given e3.33 = 28, the catalysed reaction is faster by a factor of about
(a) 15
(b) 38
(c) 28
(d) 22
Show Answer & Explanation

Answer: (c) 28

Explanation:
1. kcat/kuncat = eDeltaE/RT = e8314/(8.314 × 300) = e3.33.
2. Ratio = 28.

Teacher's Note:
1. Assume the same pre-exponential factor.
2. 8314 J / (8.314 × 300) = 3.33.

Question 4: At some temperature the rate constant for decomposition of HI on a gold surface is 0.08 M s-1. The time for [HI] to fall from 1.50 M to 0.30 M, in seconds, is
(a) 30
(b) 12
(c) 15
(d) 7.5
Show Answer & Explanation

Answer: (c) 15

Explanation:
1. Zero order (unit M s-1).
2. t = (1.50 - 0.30)/0.08 = 15 s.

Teacher's Note:
1. Zero order: [A] = [A]0 - kt.
2. Units identify the order.

Question 5: A 1 mL bacterial culture is diluted and plated at time 0 (48 colonies) and 10 minutes later in the same way (72 colonies). Assuming growth is exponential and each colony arises from one bacterium, the generation time (time to double) is about
(a) 10 min
(b) 24 min
(c) 34 min
(d) 17 min
Show Answer & Explanation

Answer: (d) 17 min

Explanation:
1. N = N0 ekt: k = ln(72/48)/10 = 0.0405 min-1.
2. Doubling time = ln 2/k = 0.693/0.0405 = 17.1 minutes.

Teacher's Note:
1. Population growth follows first order kinetics.
2. 1.5 times in 10 min is slower than 2 times.

Question 6: A bottle of 20 volume H2O2 was left open for 6 hours. 10 mL of the solution was diluted to 100 mL, and 10 mL of the diluted solution needed 25 mL of 0.025 M KMnO4 (acidic). Assuming first order decomposition, 100 k (k in hr-1) is about
(a) 22
(b) 4.4
(c) 0.22
(d) 2.2
Show Answer & Explanation

Answer: (d) 2.2

Explanation:
1. KMnO4 used = 25 × 0.025 = 0.625 mmol; H2O2 = 2.5 × 0.625 = 1.5625 mmol in 10 mL diluted = 1.5625 M in the original.
2. Volume strength = 11.2 × 1.5625 = 17.5 volumes.
3. k = (1/6) ln(20/17.5) = 0.1335/6 = 0.02225 hr-1, so 100 k = 2.2.

Teacher's Note:
1. 5 H2O2 : 2 KMnO4 in acid.
2. 1 M H2O2 is 11.2 volume.

Question 7: The units of rate and of rate constant are identical for
(a) first order reactions
(b) fractional order reactions
(c) second order reactions
(d) zero order reactions
Show Answer & Explanation

Answer: (d) zero order reactions

Explanation:
1. Rate = k c0 = k for a zero order reaction.
2. So both have units mol L-1 s-1.

Teacher's Note:
1. Check which order gives k = rate.
2. Only c0 equals 1 for all concentrations.

Question 8: 75% of a radioactive element disintegrates in 2 hours. Its half-life is
(a) two hours
(b) three hours
(c) one hour
(d) half an hour
Show Answer & Explanation

Answer: (c) one hour

Explanation:
1. 75% decayed means 25% = (1/2)2 is left, i.e. 2 half-lives.
2. So 2 h = 2 t1/2 and t1/2 = 1 h.

Teacher's Note:
1. Remaining fraction (1/2)^n gives n.
2. 75% decay always takes two half-lives.

Question 9: For A → B → C (first order steps with rate constants k1 and k2), the time at which the concentration of B is maximum is
(a) ln(k1/k2)/(k1 - k2)
(b) ln(k1 k2)/(k1 + k2)
(c) (1/k1) ln(k1/k2)
(d) 1/(k1 - k2)
Show Answer & Explanation

Answer: (a) ln(k1/k2)/(k1 - k2)

Explanation:
1. [B] = k1 a/(k2 - k1) (e-k1 t - e-k2 t).
2. d[B]/dt = 0 gives k1 e-k1 t = k2 e-k2 t, so (k2 - k1) t = ln(k2/k1).
3. tmax = ln(k2/k1)/(k2 - k1) = ln(k1/k2)/(k1 - k2).

Teacher's Note:
1. Differentiate the expression for B.
2. The result is symmetric in k1 and k2.

Question 10: For a gaseous reaction A + B → products, the activation energy is 2.27 kJ mol-1 at 273 K. The ratio k/A at 273 K is
(a) 0.434
(b) 0.368
(c) 4.34
(d) 3.68
Show Answer & Explanation

Answer: (b) 0.368

Explanation:
1. k/A = e-E/RT = e-2270/(8.314 × 273) = e-1.0001.
2. k/A = 0.368.

Teacher's Note:
1. E/RT is almost exactly 1.
2. e-1 = 0.368.

Question 11: For a zero order reaction, the unit of the rate constant is
(a) mol dm-3 s-1
(b) dm3 mol-1 s-1
(c) s-1
(d) mol2 dm-6 s-1
Show Answer & Explanation

Answer: (a) mol dm-3 s-1

Explanation:
1. Rate = k, so k has the units of rate: mol dm-3 s-1.

Teacher's Note:
1. Zero order means no concentration dependence.
2. The unit pattern is (mol dm-3)1-n s-1.

Question 12: For a reaction taking place in three steps with rate constants k1, k2, k3 and overall k = k1 k3/k2, the activation energies of the steps are 40, 30 and 20 kJ mol-1. Which statements are correct? (i) The overall activation energy is 10 kJ; (ii) the overall activation energy is 30 kJ; (iii) the mechanism 2A ⇌ A* + A, A* → product fits with overall order one.
(a) (i) and (iii) only
(b) (ii) only
(c) (i) only
(d) (ii) and (iii) only
Show Answer & Explanation

Answer: (d) (ii) and (iii) only

Explanation:
1. E = E1 + E3 - E2 = 40 + 20 - 30 = 30 kJ, so (ii) is correct and (i) is not.
2. For 2A ⇌ A* + A (k1, k2) and A* → product (k3): [A*] = k1[A]/k2, rate = k3[A*] = (k1k3/k2)[A], first order, so (iii) is correct.

Teacher's Note:
1. Arrhenius: ln k = ln k1 + ln k3 - ln k2.
2. Apply steady state to A*.

Question 13: A fossil bone has 15.0% of the 14C content of living bone. The half-life of 14C is 5.73 × 103 years. The age of the bone is about
(a) 388 years
(b) 15688 years
(c) 25488 years
(d) 6818 years
Show Answer & Explanation

Answer: (b) 15688 years

Explanation:
1. t = (t1/2/0.693) ln(100/15) = 8267 × 1.897.
2. t = about 15,680 years, nearest to 15688 years.

Teacher's Note:
1. ln(100/15) = 1.897.
2. Mean life = 1.44 x half-life.

Question 14: The activity of 1 g of 226Ra (t1/2 = 1600 years) is about
(a) 4.96 × 1012 dpm
(b) 2.19 × 1016 dpm
(c) 2.19 × 1012 dpm
(d) 2.19 × 1014 dpm
Show Answer & Explanation

Answer: (c) 2.19 × 1012 dpm

Explanation:
1. N = (1/226) × 6.022 × 1023 = 2.66 × 1021 atoms.
2. λ = 0.693/(1600 × 365.25 × 24 × 60) = 8.24 × 10-10 min-1.
3. Activity = λ N = 2.19 × 1012 disintegrations per minute.

Teacher's Note:
1. Convert the half-life to minutes.
2. This is close to the 3.7 × 1010 dps of one curie (1 Ci = 2.2 × 1012 dpm).

Question 15: The rate constant of a reaction increases by 5% when the temperature rises from 27 °C to 28 °C. The activation energy is about
(a) 46.6 kJ mol-1
(b) 16.6 kJ mol-1
(c) 26.6 kJ mol-1
(d) 36.6 kJ mol-1
Show Answer & Explanation

Answer: (d) 36.6 kJ mol-1

Explanation:
1. ln(k2/k1) = (E/R)(1/T1 - 1/T2) with k2/k1 = 1.05.
2. E = 8.314 x ln 1.05 × 300 × 301/1 = about 36.6 kJ mol-1 (using T = 300 K and 301 K).

Teacher's Note:
1. ln 1.05 = 0.0488.
2. Use absolute temperatures.

Question 16: The order of a reaction whose rate constant is k = 2.3 × 10-5 L mol-1 s-1 is
(a) one
(b) three
(c) zero
(d) two
Show Answer & Explanation

Answer: (d) two

Explanation:
1. Units of k are (mol L-1)1-n s-1.
2. L mol-1 s-1 corresponds to 1 - n = -1, so n = 2.

Teacher's Note:
1. Match units to the pattern.
2. Zero order has mol L-1 s-1.

Question 17: A radioactive element has a half-life of 14 hours. The fraction that disintegrates in 56 hours is
(a) 0.75
(b) 0.875
(c) 0.9375
(d) 0.60
Show Answer & Explanation

Answer: (c) 0.9375

Explanation:
1. 56/14 = 4 half-lives, so (1/2)4 = 1/16 remains.
2. Fraction disintegrated = 1 - 1/16 = 0.9375.

Teacher's Note:
1. Distinguish fraction remaining from fraction decayed.
2. (1/2)4 = 0.0625.

Question 18: The number of α particles emitted per second by a radioactive element falls to 6.25% of its original value in 48 days. The half-life in days is
(a) 12
(b) 16
(c) 8
(d) 3
Show Answer & Explanation

Answer: (a) 12

Explanation:
1. 6.25% = 1/16 = (1/2)4, so 4 half-lives have elapsed.
2. t1/2 = 48/4 = 12 days.

Teacher's Note:
1. Emission rate is proportional to the number of nuclei.
2. 100 / 16 = 6.25.

Question 19: For NH4+ + NO2- → N2 + 2H2O: Set 1 [NH4+] = 0.010, [NO2-] = 0.020, rate 0.020; Set 2 [NH4+] = 0.015, [NO2-] = 0.020, rate 0.030; Set 3 [NH4+] = 0.010, [NO2-] = 0.010, rate 0.005. The rate law is
(a) rate = k[NH4+][NO2-]
(b) rate = k[NH4+]2[NO2-]
(c) rate = k[NH4+]2[NO2-]2
(d) rate = k[NH4+][NO2-]2
Show Answer & Explanation

Answer: (d) rate = k[NH4+][NO2-]2

Explanation:
1. Set 1 to 2: [NH4+] × 1.5, rate x 1.5, so first order in NH4+.
2. Set 1 to 3: [NO2-] halved, rate becomes 1/4, so second order in NO2-.

Teacher's Note:
1. Use ratios of rates.
2. (1/2)^y = 1/4 gives y = 2.

Question 20: Two mechanisms are proposed for A + B → C + D. Mechanism I: A + B → AB* → C + D, k = 1 × 10-5 M-1 s-1. Mechanism II: A → A* → E (k1 = 1 × 10-4 s-1) followed by E + B → C + D (k2 = 1 × 1010 M-1 s-1). The rate by mechanism I with [A] = [B] = 0.1 M is
(a) 1 × 10-4 M s-1
(b) 1 × 10-5 M s-1
(c) 1 × 10-6 M s-1
(d) 1 × 10-7 M s-1
Show Answer & Explanation

Answer: (d) 1 × 10-7 M s-1

Explanation:
1. Rate = k[A][B] = 10-5 x 0.1 × 0.1 = 1 × 10-7 M s-1.

Teacher's Note:
1. Mechanism I is bimolecular.
2. Substitute directly.

Question 21: The inversion of cane sugar in excess water with an acid catalyst is a reaction that is
(a) second order
(b) first order with respect to cane sugar
(c) zero order with respect to cane sugar
(d) third order
Show Answer & Explanation

Answer: (b) first order with respect to cane sugar

Explanation:
1. Water is present in large excess, so its concentration is effectively constant.
2. Rate = k[sucrose][H2O] reduces to k'[sucrose], a pseudo first order law.

Teacher's Note:
1. Such reactions are called pseudo first order.
2. Molecularity is two but the order is one.

Question 22: The half-life of a radionuclide is 2.0 × 105 s. Its decay constant is
(a) 2.89 × 105 s-1
(b) 5.0 × 10-6 s-1
(c) 3.465 s-1
(d) 3.465 × 10-6 s-1
Show Answer & Explanation

Answer: (d) 3.465 × 10-6 s-1

Explanation:
1. λ = 0.693 / t1/2 = 0.693 / (2.0 × 105).
2. λ = 3.465 × 10-6 s-1.

Teacher's Note:
1. Decay constant has units of reciprocal time.
2. Do not confuse it with the mean life 1/λ.

Question 23: In a second order reaction, 20% of a substance dissociates in 40 minutes. The time for 80% dissociation is
(a) 200 min
(b) 320 min
(c) 640 min
(d) 160 min
Show Answer & Explanation

Answer: (c) 640 min

Explanation:
1. ka t = x/(1-x): at 20%, ka x 40 = 0.2/0.8 = 0.25, so ka = 0.00625 min-1.
2. At 80%: ka t = 0.8/0.2 = 4, t = 4/0.00625 = 640 min.

Teacher's Note:
1. Use kat = x/(1-x) for second order.
2. The ratio of (x/(1-x)) values is 16.

Question 24: For a gaseous first order decomposition at 400 K the half-life is 150 minutes and the activation energy is 65.0 kJ mol-1. The fraction of molecules having sufficient energy to react is about
(a) 4.6 × 10-12
(b) 6.5 × 10-5
(c) 1.0 × 10-3
(d) 3.2 × 10-9
Show Answer & Explanation

Answer: (d) 3.2 × 10-9

Explanation:
1. Fraction = e-Ea/RT = e-65000/(8.314 × 400).
2. Exponent = -19.55, so fraction = 3.2 × 10-9.

Teacher's Note:
1. The half-life is not needed.
2. Boltzmann factor.

Question 25: The units of the rate constant of a first order reaction, with time in seconds, are
(a) L mol-1 s-1
(b) s-1
(c) s
(d) mol L-1 s-1
Show Answer & Explanation

Answer: (b) s-1

Explanation:
1. Rate = k[A], so k = (mol L-1 s-1) / (mol L-1) = s-1.

Teacher's Note:
1. Concentration cancels in a first order law.
2. Compare with zero order: mol L-1 s-1.

Question 26: For H3AsO4 + 2H3O+ + 3I- → HAsO2 + I3- + 4H2O, the initial rates (x 10-5 M/s) are: 3.7 for (0.001 H3AsO4, 0.01 H3O+, 0.10 I-); 7.4 for (0.001, 0.01, 0.20); 7.4 for (0.002, 0.01, 0.10); 3.7 for (0.002, 0.005, 0.20). The overall order is
(a) 5
(b) 4
(c) 3
(d) 2
Show Answer & Explanation

Answer: (b) 4

Explanation:
1. Doubling I- doubles the rate: first order in I-. Doubling H3AsO4 doubles the rate: first order in H3AsO4.
2. Run 4 vs run 1: H3AsO4 × 2 and I- x 2 give x 4; the observed rate is x 1, so (1/2)^m = 1/4, m = 2 for H3O+.
3. Overall order = 1 + 1 + 2 = 4.

Teacher's Note:
1. Compare runs pairwise.
2. Account for the already known orders in run 4.

Question 27: The half-time of a second order reaction with equal reactant concentrations is 35 s. The time for 99% completion is
(a) 69 s
(b) 138 s
(c) 1733 s
(d) 3465 s
Show Answer & Explanation

Answer: (d) 3465 s

Explanation:
1. t1/2 = 1/(ka) = 35 s, so ka = 1/35.
2. For fraction x: kt a = x/(1-x) = 0.99/0.01 = 99.
3. t = 99 × 35 = 3465 s.

Teacher's Note:
1. For second order, kat = x/(1-x).
2. At x = 0.5 this gives t1/2.

Free study material for Chemistry

Practice MCQs for JEE Chemistry Chemical Kinetics

JEE Chemistry Chemical Kinetics Objective Test Questions

Explore reliable practice questions for Chemical Kinetics tailored for JEE Chemistry learners. Use these multiple-choice formats to evaluate preparedness and strengthen problem-solving skills.

NCERT-Aligned Objective Questions and Solutions

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.

Next Steps in Your Exam Preparation

Follow up your worksheet practice by attempting the interactive online Chemistry MCQ test for this chapter to evaluate your execution speed. All platform resources are free to access.

FAQs

Where can I access latest JEE Chemistry Chemical Kinetics MCQs Set 30?

You can get most exhaustive JEE Chemistry Chemical Kinetics MCQs Set 30 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 30 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 30, 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 30?

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 30 on StudiesToday.com as they provide instant answers and score to help you track your progress in Chemistry.