Chemical Kinetics

13 NEET questions on Chemical Kinetics. A sample is shown below with full explanations; the rest are available free after signing up.

Chemical Kinetics2026hard

Q1.2A → B is a zero-order reaction, where k = 1.0 mol L⁻¹ min⁻¹. If the initial concentration of A is 2 M, then the time taken to complete 75% of the reaction will be

  • A1.5 min
  • B0.75 minAnswer
  • C1.0 min
  • D2.0 min

Explanation

For zero order the rate is constant, and with the stoichiometric coefficient 2 the rate law is −(1/2)d[A]/dt = k, so ([A]₀ − [A]t) = 2kt. Completing 75% of the reaction leaves [A]t = 0.5 M, giving t = (2 − 0.5)/(2 × 1.0) = 0.75 min. Forgetting the factor of 2 is the usual slip here.

Source: NEET 2026 re-exam (21 June), Code 50, Q60

Chemical Kinetics2025easy

Q2.If the half-life (t₁/₂) for a first order reaction is 1 minute, then the time required for 99.9% completion of the reaction is closest to

  • A2 minutes
  • B4 minutes
  • C5 minutes
  • D10 minutesAnswer

Explanation

99.9% completion leaves 0.1% of the reactant, so C₀/C = 1000. For a first order reaction the number of half-lives n satisfies 2ⁿ = 1000, and since 2¹⁰ = 1024 that is very close to ten half-lives. With t₁/₂ = 1 minute the time needed is about 10 minutes. The same result follows from t = (2.303/k)·log 1000 with k = 0.693/1.

Source: NEET 2025 (4 May), Code 45, Q56

Chemical Kinetics2025medium

Q3.If the rate constant of a reaction is 0.03 s⁻¹, how much time does it take for 7.2 mol L⁻¹ concentration of the reactant to get reduced to 0.9 mol L⁻¹? (Given: log 2 = 0.301)

  • A69.3 sAnswer
  • B23.1 s
  • C210 s
  • D21.0 s

Explanation

The units of the rate constant, s⁻¹, identify this as a first order reaction, for which t = (2.303/k)·log(C₀/C). Here C₀/C = 7.2/0.9 = 8, and log 8 = 3 log 2 = 3 × 0.301 = 0.903. Substituting, t = (2.303/0.03) × 0.903 = 76.77 × 0.903 ≈ 69.3 s. Equivalently the concentration falls by a factor of 8, which is exactly three half-lives of 23.1 s each.

Source: NEET 2025 (4 May), Code 45, Q89

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