NEET Chemical Kinetics PYQs & Practice Questions

10 free NEET Chemistry questions on Chemical Kinetics, starting with previous year questions (2016–2026). Tap “Show answer” to check each one. No login needed.

Topics: Half Life · First Order Reaction Half-Life · Half-life and First Order Reactions · First Order Reaction - Time Calculation · Role of a Catalyst · Effect of a Catalyst on Heat of Reaction · Determining Activation Energy · Misconception About Reaction Order · How Reaction Rate Changes Over Time · Rate Constant is Independent of Concentration · Time for 100% Completion of a First-Order Reaction · What a Pseudo First-Order Rate Constant Depends On

1. Q1NEET 2026Arrhenius Equation - Graphical Analysis

   For an elementary chemical reaction, the Arrhenius plot of ln k versus 1/T is a straight line with y-intercept (at 1/T=0) equal to 6 and negative slope. If the energy of activation is 6.64 kJ mol⁻¹ and R = 8.3 J K⁻¹ mol⁻¹, the temperature at which the rate constant becomes e² min⁻¹, is:

   • A. 150 K
   • B. 200 K
   • C. 250 K
   • D. 125 K
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   Answer: A — 150 K

   From the Arrhenius equation, ln k = ln A - (Ea/R)(1/T). From the graph, ln A = 6 (y-intercept at 1/T=0). Ea/R = 6640/8.3 = 800 K. We want k = e², i.e., ln k = 2: 2 = 6 - 800(1/T), so 800/T = 4, giving T = 200 K.

2. Q2NEET 2026Zero-Order Reaction Kinetics

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

   • A. 0.75 min
   • B. 1.0 min
   • C. 2.0 min
   • D. 1.5 min
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   Answer: —

   Rate of reaction = -(1/2)d[A]/dt = k, so d[A]/dt = -2k = -2.0 mol L⁻¹ min⁻¹ (rate of consumption of A). 75% completion means 1.5 M of A (out of initial 2 M) is consumed. Time = (amount consumed)/(rate of consumption) = 1.5/2.0 = 0.75 min.

3. Q3NEET 2026Reaction order from rate vs concentration plot

   For a certain reaction R → Product, the plot of concentration [R] vs time has a negative slope as shown, with k=-slope. The order of the reaction is:

   • A. 1
   • B. 2.5
   • C. 2
   • D. 0
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   Answer: C — 2

   A straight-line plot of [R] vs t with a constant slope (k=-slope) indicates a zero order reaction.

4. Q4NEET 2026Activation energy from Arrhenius plot

   Given below is an expression for the rate constant of a first order reaction occurring at a certain temperature, T(K): ln k=14.34-1.25×10⁴/T. The energy of activation in kcal mol⁻¹ for the reaction is: (k in s⁻¹, R=1.987 cal mol⁻¹K⁻¹)

   • A. 12.42
   • B. 18.63
   • C. 14.34
   • D. 24.84
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   Answer: C — 14.34

   Comparing with ln k = ln A - Ea/RT, Ea/R=1.25×10⁴ → Ea=1.25×10⁴×1.987 cal/mol ≈ 24840 cal/mol = 24.84 kcal/mol.

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5. Q5NEET 2026Matching order of reaction to unit of rate constant

   Match List I with List II: List I (Order of reaction) — A. Zero order, B. First order, C. Second order, D. Third order. List II (Unit of rate constant) — I. mol⁻¹ L s⁻¹, II. mol⁻¹L²s⁻¹, III. s⁻¹, IV. mol L⁻¹ s⁻¹.

   • A. A-IV, B-III, C-II, D-I
   • B. A-IV, B-III, C-I, D-II
   • C. A-I, B-II, C-III, D-IV
   • D. A-IV, B-I, C-II, D-III
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   Answer: B — A-IV, B-III, C-I, D-II

   Zero order: mol L⁻¹s⁻¹ (IV); first order: s⁻¹ (III); second order: mol⁻¹Ls⁻¹ (I); third order: mol⁻²L²s⁻¹ (II). A-IV, B-III, C-I, D-II.

6. Q6NEET 2025First Order Reaction Half-Life

   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:

   • A. 2 minutes
   • B. 4 minutes
   • C. 5 minutes
   • D. 10 minutes
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   Answer: C — 5 minutes

   t1/2 = (2.303/k)log(A0/At). For 50% completion: 1 = (2.303/k)log(100/50). For 99.9% completion: t = (2.303/k)log(100/0.1) = (2.303/k)log(1000). Ratio: t/1 = log(1000)/log(2) = 3/0.301 ≈ 10. So t ≈ 10 minutes.

7. Q7NEET 2025First Order Reaction - Time Calculation

   If the rate constant of a reaction is 0.03s⁻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)

   • A. 69.3 s
   • B. 23.1 s
   • C. 210 s
   • D. 21.0 s
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   Answer: —

   t = (2.303/k)log(A0/At) = (2.303/0.03)log(7.2/0.9) = (2.303/0.03)log(8) = (2.303/0.03)×3log2 = (2.303/0.03)×0.903 ≈ 69.3 s.

8. Q8NEET 2023Activation Energy

   Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R: Assertion A: A reaction can have zero activation energy. Reason R: The minimum extra amount of energy absorbed by reactant molecules so that their energy becomes equal to threshold value, is called activation energy. In the light of the above statements, choose the correct answer from the options given below.

   • A. Both A and R are true, and R is NOT the correct explanation of A.
   • B. A is true but R is false.
   • C. A is false but R is true.
   • D. Both A and R are true and R is the correct explanation of A.
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   Answer: B — A is true but R is false.

   Assertion A is false — a reaction cannot have zero activation energy; every reaction requires some minimum energy for the reactants to reach the transition state. Reason R correctly defines activation energy as the minimum extra energy needed for reactant molecules to reach the threshold energy. Hence A is false but R is true.

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9. Q9NEET 2023Rate Law and Order of Reaction

   For a certain reaction, the rate = K[A]^2[B]. When the initial concentration of A is tripled keeping concentration of B constant, the initial rate would:

   • A. increase by a factor of six
   • B. increase by a factor of nine
   • C. increase by a factor of three
   • D. increase by a factor of nine
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   Answer: A — increase by a factor of six

   Rate = K[A]^2[B]. Tripling [A] gives Rate' = K(3[A])^2[B] = 9K[A]^2[B] = 9 x Rate. So the rate increases by a factor of nine.

10. Q10NEET 2022Order of Reaction

    The given graph is a representation of kinetics of a reaction (constant temperature T, axes y vs x). The y and x axes for zero and first order reactions, respectively are:

    • A. zero order (y = concentration and x = time), first order (y = rate constant and x = concentration)
    • B. zero order (y = rate and x = concentration), first order (y = t1/2 and x = concentration)
    • C. zero order (y = rate and x = concentration), first order (y = rate and x = t1/2)
    • D. zero order (y = concentration and x = time), first order (y = t1/2 and x = concentration)
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    Answer: A — zero order (y = concentration and x = time), first order (y = rate constant and x = concentration)

    For a zero order reaction, rate is constant and independent of concentration, so plotting rate (y) vs concentration (x) gives a horizontal line. For a first order reaction, half-life (t1/2) is independent of initial concentration, so plotting t1/2 (y) vs concentration (x) also gives a horizontal line.