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Class XII · Chemistry

Chemical Kinetics: NEET MCQ practice

223 practice questions on Chemical Kinetics, each with an explanation. Practise them chapter by chapter in the GyaanLoop app.

223

Practice questions

99

Easy

109

Medium

15

Hard

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What Chemical Kinetics covers

Topics in the NEET (UG) 2026–27 syllabus for this chapter.

  • Rate of a chemical reaction, factors affecting rate- concentration, temperature, pressure, catalyst
  • Elementary and complex reactions, order and molecularity of reactions
  • Rate law, rate constant, differential and integral forms of zero and first order reactions, half-lives
  • Effect of temperature on rate of reactions- Arrhenius theory, activation energy
  • Collision theory of bimolecular gaseous reactions (no derivation)

Sample questions on Chemical Kinetics

Try each one, then open the answer and explanation.

Question 1 · Practice question

Q. For zeroth order reaction, A → B [A] 0 = 0.01 M, [A] t = 0.008Matter 10 min. Thus, half-life is

  1. A.5 min
  2. B.12 min
  3. C.16 min
  4. D.25 min
Show answer

Answer: D. 25 min

For a zeroth order reaction, the half-life is given by t½ = [A]₀/(2k). First calculate the rate constant k from the given data: k = ([A]₀ - [A]ₜ)/t = (0.01 M - 0.008 M)/10 min = 0.0002 M/min. Then t½ = 0.01 M / (2 × 0.0002 M/min) = 25 min. The tempting quick wrong answer 5 min might come from mistakenly using first-order half-life (t½ = 0.693/k), which would require calculating k differently.

Question 2 · Practice question

A Certain Zeroth Order reaction has k = 0.025 Ms -1 for the disappearance of A. What will be the concentration of A after 15s, if the intial concentration is 0.50 M?

  1. A.0.50 M
  2. B.0.375 M
  3. C.0.125 M
  4. D.0.060 M
Show answer

Answer: C. 0.125 M

For a zeroth order reaction, concentration decreases linearly with time: [A]t = [A]₀ - kt. Substituting given values: 0.50 - (0.025 × 15) = 0.125 M. Option (b) 0.375 M is tempting if you mistakenly multiply k and t but forget to subtract from initial concentration.

Question 3 · Practice question

In a first order reaction, the concentration of the reactant decreases from 0.8 M to 0.4 M is 15 min. The time taken for the concentration to change from 0.1 M to 0.025 M is

  1. A.30 min
  2. B.15 min
  3. C.7.5 min
  4. D.60 min
Show answer

Answer: A. 30 min

For a first-order reaction, half-life is constant. The first part tells you that half-life is 15 min. Going from 0.1 M to 0.025 M is two half-lives (because 0.1 → 0.05 → 0.025), so total time = 2 × 15 = 30 min. Students might mistakenly think the time is the same 15 min by comparing the same concentration difference, but here it involves two half-lives, not one.

Question 4 · Practice question

The half-life period of a first order chemical reaction is 6.93 min. The time required for the completion of 99% of the chemical reaction will be (log2 = 0.3010) [AIEEE 2009]

  1. A.230.3 min
  2. B.23.03 min
  3. C.46.06 min
  4. D.460.6 min
Show answer

Answer: C. 46.06 min

For a first-order reaction, k = 0.693 / t₁/₂ = 0.1 min⁻¹. Time for 99% completion: t = (2.303/k) log(100) = 23.03 × 2 = 46.06 min. Option (a) 230.3 min comes from forgetting to take log(100) = 2, while option (b) 23.03 min comes from using log(10) = 1 instead of log(100) = 2.

Question 5 · Practice question

For the first order reaction, A → Product Q. The concentration of A changes from 0.1 M to 0.025 M in 40 min. The rate of the reaction when the concentration of A is 0.01 M is [AIEEE 2012]

  1. A.1.73 x 10 -6 Mmin -1
  2. B.3.47 x 10 -4 M min -1
  3. C.3.47 x 10 -5 M min -1
  4. D.1.73 x 10 -6 M min -1
Show answer

Answer: B. 3.47 x 10 -4 M min -1

For a first‑order reaction, rate = k[A]. First find k from the concentration change: k = (2.303/t) log([A]₀/[A]ₜ) ≈ 0.0347 min⁻¹. Then at [A] = 0.01 M, rate = 0.0347 × 0.01 = 3.47×10⁻⁴ M min⁻¹. Option (a) and (d) are much smaller values (1.73×10⁻⁶), likely arising from misplacing the decimal or using the half‑life formula incorrectly.

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