Chemical Kinetics JEE Main previous year questions with solutions

5 solved JEE Main questions on Chemical Kinetics, free to read — no sign-in needed. The full chapter has 235 questions; sign in to attempt the remaining 230 in the exam simulator.

Answers checked against the official answer keys.

1. Q1JEE Main 2026 (06 Apr, Shift 2)Arrhenius theory
   Decomposition of a hydrocarbon follows the equation $k = (5.5 \times 10^{11}\,\text{s}^{-1})\,e^{\frac{-28000\,\text{K}}{T}}$. The activation energy of reaction is __________ kJ mol$^{-1}$. (Nearest Integer) Given : R $= 8.3$ J K$^{-1}$ mol$^{-1}$
   Show answer & solution

   Answer: 232

   According to the Arrhenius equation, $k = A e^{-\dfrac{E_a}{RT}}$. The given equation is $k = (5.5 \times 10^{11} \text{ s}^{-1}) e^{-\dfrac{28000}{T}}$. Comparing the exponent of $e$ in both equations: $\dfrac{E_a}{RT} = \dfrac{28000}{T}$ $E_a = 28000 \times R$ Substituting the value of $R = 8.3 \text{ J K}^{-1} \text{ mol}^{-1}$: $E_a = 28000 \times 8.3 \text{ J mol}^{-1}$ $E_a = 232400 \text{ J mol}^{-1}$ $E_a = 232.4 \text{ kJ mol}^{-1}$ Rounding to the nearest integer, the activation energy is $232$. Answer: $232$
2. Q2JEE Main 2025 (08 Apr, Shift 2)Integrated rate laws
   <p>In a first order decomposition reaction, the time taken for the decomposition of reactant to one fourth and one eighth of its initial concentration are $t_1$ and $t_2(s)$, respectively. The ratio $t_1 / t_2$ will :</p>
   1. A.<p>$\frac{4}{3}$</p>
   2. B.<p>$\frac{3}{2}$</p>
   3. C.<p>$\frac{3}{4}$</p>
   4. D.<p>$\frac{2}{3}$</p>
   Show answer & solution

   Answer: (D)

   <p>For $\mathrm{I}^{\text {st }}$ order reaction When $\mathrm{C}_{\mathrm{t}}=\mathrm{Co} / 4$ $\begin{aligned} & \mathrm{t}_1=2 \mathrm{t}_{50 \%}. \\ & \text { when } \mathrm{C}_{\mathrm{t}}=\mathrm{Co} / 8 \\ & \mathrm{t}_2=3 \mathrm{t}_{50 \%} \\ & \text { so } \frac{\mathrm{t}_1}{\mathrm{t}_2}=\frac{2}{3} \end{aligned}$</p>
3. Q3JEE Main 2024 (06 Apr, Shift 1)Methods to determine order of reaction
   Time required for $99.9 \%$ completion of a first order reaction is _______ times the time required for completion of $90 \%$ reaction.(nearest integer)
   Show answer & solution

   Answer: 3

   $\begin{aligned} & \mathrm{K}=\frac{1}{\mathrm{t}_{99.9 \%}} \ln \left(\frac{100}{0.1}\right)=\frac{1}{\mathrm{t}_{90 \%}} \ln \left(\frac{100}{10}\right) \\ & \mathrm{t}_{99.9 \%}=\mathrm{t}_{90 \%} \frac{\ell \mathrm{n}\left(10^3\right)}{\ell \mathrm{n} 10} \\ & \mathrm{t}_{99.9 \%}=\mathrm{t}_{90 \%} \times 3\end{aligned}$
4. Q4JEE Main 2023 (29 Jan, Shift 2)
   An indicator ‘$X$’ is used for studying the effect of variation in concentration of iodide on the rate of reaction of iodide ion with ${H}_{2}{O}_{2}$ at room temp. The indicator ‘$X$’ forms blue colored complex with compound ‘$A$’ present in the solution. The indicator ‘$X$’ and compound ‘$A$’ respectively are
   1. A.Starch and iodine
   2. B.Methyl orange and ${H}_{2}{O}_{2}$
   3. C.Starch and ${H}_{2}{O}_{2}$
   4. D.Methyl orange and iodine
   Show answer & solution

   Answer: (A)

   The reaction between iodide ions and hydrogen peroxide (${H}_{2}{O}_{2}$) occurs in the acidic medium and can be represented in the following manner: $2{I}^{-}(aq)+{H}_{2}{O}_{2}(l)+2{H}^{+}(aq)\rightarrow {I}_{2}(g)+2{H}_{2}O(l)$ In this reaction, hydrogen peroxide oxidises iodide ions (I-) to molecular iodine. If a calculated amount of sodium thiosulphate is added in the presence of starch solution as an indicator to the above reaction mixture, the liberated iodine reacts with thiosulphate ions as fast as it is formed and is reduced back to iodide ions till all the thiosulphate ions are oxidised to tetrathionate ions. ${I}_{2}(g)+2{S}_{2}{O}_{3}^{2-}(aq)\rightarrow {S}_{4}{O}_{6}^{2-}(aq)+2{I}^{-}(aq)$ After the complete consumption of thiosulphate ions, the concentration of iodine liberated in the reaction of hydrogen peroxide with iodide ions increases rapidly to a point where iodine forms intense blue complex with starch.
5. Q5JEE Main 2022 (29 Jul, Shift 1)Rate law and rate constant
   The reaction between $X$ and $Y$ is first order with respect to $X$ and zero order with respect to $Y$. Experiment $\dfrac{\left[X\right]}{{molL}^{-1}}$ $\dfrac{\left[Y\right]}{{molL}^{-1}}$ $\dfrac{Initialrate}{mol{L}^{-1}{\min }^{-1}}$ I $0.1$ $0.1$ $2\times {10}^{-3}$ II $L$ $0.2$ $4\times {10}^{-3}$ III $0.4$ $0.4$ $M\times {10}^{-3}$ IV $0.1$ $0.2$ $2\times {10}^{-3}$ Examine the data of table and calculate ratio of numerical values of $M$ and $L$.
   Show answer & solution

   Answer: 40

   Given that reaction is first order with respect to X and of zero order with respect to Y. Means $r=k\left[X\right]{\left[Y\right]}^{0}$ Where k = rate constant. Using data $I$ & $II$ $\dfrac{4\times {10}^{-3}}{2\times {10}^{-3}}=\left(\dfrac{L}{0.1}\right){\left(\dfrac{0.2}{0.1}\right)}^{0} \Rightarrow L=0.2$ Using data from $I$ & $III$ $\dfrac{M\times {10}^{-3}}{2\times {10}^{-3}}=\dfrac{0.4}{0.1}\Rightarrow M=8$ Not the ratio which is asked in question $\dfrac{M}{L}=\dfrac{8}{0.2}=40$