Chemical Kinetics NEET PYQs (2005–2026) with Solutions

10 solved Chemical Kinetics PYQs

Real NEET and AIPMT questions from this chapter, newest first. Try each one, then open the answer and the step-by-step explanation.

1. Question 1 (NEET 2026)

   NEET 2026MediumOrder & Molecularity

   Match List I with List II: Choose the correct answer from the options given below :

   1. AOption A: A–IV, B–II, C–I, D–III
   2. BOption B: A–IV, B–III, C–I, D–II
   3. COption C: A–IV, B–III, C–II, D–I
   4. DOption D: A–I, B–II, C–III, D–IV

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   Correct answer: (B) A–IV, B–III, C–I, D–II

   Explanation

   Unit of rate constant for nth order reaction = (mol L⁻¹)^(1−n)s⁻¹. Thus: zero order → mol L⁻¹ s⁻¹, first order → s⁻¹, second order → mol⁻¹ L s⁻¹, third order → mol⁻² L² s⁻¹. Hence option (B) is correct.

2. Question 2 (NEET 2026)

   NEET 2026MediumIntegrated rate law for zero-order reactions

   $2\mathrm{A} \xrightarrow{k} \mathrm{B}$ is a zero-order reaction, where $k = 1.0\ \mathrm{mol\ L^{-1}\ min^{-1}}$. If the initial concentration of A is 2 M, then the time taken to complete $75\%$ of the reaction will be

   1. AOption A: $1.5\ \mathrm{min}$
   2. BOption B: $0.75\ \mathrm{min}$
   3. COption C: $1.0\ \mathrm{min}$
   4. DOption D: $2.0\ \mathrm{min}$

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   Correct answer: (B) $0.75\ \mathrm{min}$

   Explanation

   For $2\mathrm{A} \rightarrow \mathrm{B}$, the rate of reaction is $-\frac{1}{2}\frac{d[\mathrm{A}]}{dt} = k$ (zero order), so A is used up at $2k$: $$[\mathrm{A}]_0 - [\mathrm{A}] = 2kt$$ For $75\%$completion, the amount of A consumed$= 0.75 \times 2\ \mathrm{M} = 1.5\ \mathrm{M}$.$t = \frac{1.5}{2 \times 1.0} = 0.75\ \mathrm{min}$, option (B).

3. Question 3 (NEET 2026)

   NEET 2026MediumArrhenius equation and its plot

   For an elementary chemical reaction, the Arrhenius plot is given below. If the energy of activation is $6.64\ \mathrm{kJ\ mol^{-1}}$ and $R = 8.3\ \mathrm{J\ K^{-1}\ mol^{-1}}$, the temperature at which the rate constant becomes $e^2\ \mathrm{min^{-1}}$, is

   1. AOption A: $125\ \mathrm{K}$
   2. BOption B: $150\ \mathrm{K}$
   3. COption C: $200\ \mathrm{K}$
   4. DOption D: $250\ \mathrm{K}$

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   Correct answer: (C) $200\ \mathrm{K}$

   Explanation

   Arrhenius equation: $\ln k = \ln A - \frac{E_a}{RT}$. A plot of $\ln k$ against $\frac{1}{T}$ is a straight line with slope $-\frac{E_a}{R}$ and intercept $\ln A$. From the graph, the intercept on the $\ln k$ axis is 6, so $\ln A = 6$. $\frac{E_a}{R} = \frac{6640\ \mathrm{J\ mol^{-1}}}{8.3\ \mathrm{J\ K^{-1}\ mol^{-1}}} = 800\ \mathrm{K}$. For $k = e^2$, $\ln k = 2$: $$2 = 6 - \frac{800}{T} \Rightarrow \frac{800}{T} = 4 \Rightarrow T = 200\ \mathrm{K}$$ So the answer is option (C).

4. Question 4 (NEET 2026)

   NEET 2026EasyIntegrated Rate Equations

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

   1. AOption A: 0
   2. BOption B: 1
   3. COption C: 2
   4. DOption D: 2.5

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   Correct answer: (A) 0

   Explanation

   For a zero-order reaction: [R]=[R₀]−kt, which gives a straight line with negative slope (−k) when concentration is plotted against time. Hence option (A) is correct.

5. Question 5 (NEET 2026)

   NEET 2026MediumArrhenius Equation & Collision Theory

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

   1. AOption A: 24.84
   2. BOption B: 14.34
   3. COption C: 18.63
   4. DOption D: 12.42

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   Correct answer: (A) 24.84

   Explanation

   Arrhenius equation: ln k = ln A − Ea/RT. Comparing, Ea/R=1.25×10⁴. Therefore Ea=(1.25×10⁴)(1.987)=24837 cal mol⁻¹≈24.84 kcal mol⁻¹. Hence option (A) is correct.

6. Question 6 (NEET 2025)

   NEET 2025MediumIntegrated Rate Equations

   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: (1) 2 minutes (2) 4 minutes (3) 5 minutes (4) 10 minutes

   1. AOption A: 2 minutes
   2. BOption B: 4 minutes
   3. COption C: 5 minutes
   4. DOption D: 10 minutes

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   Correct answer: (D) 10 minutes

   Explanation

   For a first order reaction: k = 0.693 / t₁/₂ = 0.693 min⁻¹ For 99.9% completion, 0.1% reactant remains: 0.001 = e^(−kt) t = 6.9 / 0.693 ≈ 10 minutes. Hence, the correct answer is 10 minutes.

7. Question 7 (NEET 2025)

   NEET 2025MediumArrhenius Equation & Collision Theory

   For the reaction A(g) ⇌ 2B(g), the backward reaction rate constant is higher than the forward reaction rate constant by a factor of 2500, at 1000 K. [Given : R = 0.0831 L atm mol⁻¹ K⁻¹] Kₚ for the reaction at 1000 K is

   1. AOption A: 83.1
   2. BOption B: 2.077 × 10⁵
   3. COption C: 0.033
   4. DOption D: 0.021

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   Correct answer: (C) 0.033

   Explanation

   For a reversible reaction: Kc = kf / kb Given that the backward rate constant is 2500 times the forward rate constant: Kc = 1 / 2500 = 0.0004 For the reaction: A(g) ⇌ 2B(g) Δn = 2 − 1 = 1 Using: Kp = Kc(RT)^Δn Kp = 0.0004 × (0.0831 × 1000) Kp = 0.0004 × 83.1 = 0.03324 ≈ 0.033 Hence, option (C) is correct.

8. Question 8 (NEET 2025)

   NEET 2025MediumIntegrated Rate Equations

   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 reactant to get reduced to $0.9\ mol\ L^{-1}$? (Given: $\log 2 = 0.301$)

   1. AOption A: 69.3 s
   2. BOption B: 23.1 s
   3. COption C: 210 s
   4. DOption D: 21.0 s

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   Correct answer: (A) 69.3 s

   Explanation

   For a first order reaction: $$t = \frac{2.303}{k} \log \frac{[R]_0}{[R]}$$ Substituting values: $$t = \frac{2.303}{0.03} \log \frac{7.2}{0.9}$$ $$= \frac{2.303}{0.03} \log 8$$ Since $$\log 8 = 3 \log 2 = 3 \times 0.301 = 0.903$$ $$t = \frac{2.303 \times 0.903}{0.03} \approx 69.3\ s$$ Hence the correct answer is 69.3 s.

9. Question 9 (NEET 2024)

   NEET 2024EasyArrhenius Equation & Collision Theory

   Activation energy of any chemical reaction can be calculated if one knows the value of:

   1. AOption A: orientation of reactant molecules during collision.
   2. BOption B: rate constant at two different temperatures.
   3. COption C: rate constant at standard temperature.
   4. DOption D: probability of collision.

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   Correct answer: (B) rate constant at two different temperatures.

   Explanation

   Using Arrhenius equation: $$k = Ae^{-E_a/RT}$$ Taking logarithm at two temperatures: $$\ln\frac{k_2}{k_1} = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$$ Thus activation energy can be calculated if rate constants at two different temperatures are known. Hence, option (B) is correct.

10. Question 10 (NEET 2023)

    NEET 2023EasyRate 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:

    1. AOption A: decrease by a factor of nine.
    2. BOption B: increase by a factor of six.
    3. COption C: increase by a factor of nine.
    4. DOption D: increase by a factor of three.

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    Correct answer: (C) increase by a factor of nine.

    Explanation

    Given: Rate = k[A]²[B] If concentration of A is tripled: New rate = k(3[A])²[B] = 9k[A]²[B] Thus, the rate increases by a factor of 9. Hence, the correct answer is option (C).