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 231 questions; sign in to attempt the remaining 226 in the exam simulator.

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$^{-1}$
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   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>
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   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)
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   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 (12 Apr, Shift 1)Rate law and rate constant
   The reaction $2NO+{Br}_{2}\rightarrow 2NOBr$ takes place through the mechanism given below $NO+{Br}_{2}\rightleftharpoons ^{}{NOBr}_{2}\left(fast\right)$ ${NOBr}_{2}+NO\rightarrow 2NOBr\left(slow\right)$ The overall order of the reaction is _____.
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   Answer: 3

   On applying Law of mass action over slowest step, $R=k[{NOBr}_{2}][NO]$ The slowest step is the rate determining step. On applying Law of mass action on equilibrium of Step I $k=\dfrac{\left[{NOBr}_{2}\right]}{\left[NO\right]\left[{Br}_{2}\right]}$ $\left[{NOBr}_{2}\right]=k\left[NO\right]\left[{Br}_{2}\right]$ Hence, overall rate equation will be $R=k{\left[NO\right]}^{2}\left[{Br}_{2}\right]$ $=k'{\left[NO\right]}^{2}\left[{Br}_{2}\right]$ The overall order is equal to sum of the powers of the concentration terms in the rate law. The overall order of the reaction is 3.
5. Q5JEE Main 2022 (25 Jun, Shift 1)Rate of reaction
   For a given chemical reaction ${\gamma }_{1}A+{\gamma }_{2}B\rightarrow {\gamma }_{3}C+{\gamma }_{4}D$. Concentration of $C$ changes from $10mmol{dm}^{-3}$ to $20mmol{dm}^{-3}$ in $10s.$ Rate of appearance of $D$ is $1.5$ times the rate of disappearance of $B$ which is twice the rate of disappearance $A$. The rate of appearance of $D$ has been experimentally determined to be $9mmol{dm}^{-3}{s}^{-1}.$ Therefore the rate of reaction is____$mmol{dm}^{-3}{s}^{-1}.$ (Nearest Integer)
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   Answer: 1

   $\dfrac{d\left[c\right]}{dt}=\left(\dfrac{20-10}{10}\right)=1milimole/{dm}^{3}\sec$ ${\gamma }_{1}A+{\gamma }_{2}B\rightarrow {\gamma }_{3}C+{\gamma }_{4}D$ $\dfrac{d\left[D\right]}{dt}=1.5\left(-\dfrac{d}{dt}\left[B\right]\right)=9milimole/{dm}^{3}\sec$ $\left(-\dfrac{d}{dt}\left[B\right]\right)=2\left(-\dfrac{d}{dt}\left[A\right]\right)$ Rate $=-\dfrac{1}{{\gamma }_{1}}\dfrac{d\left[A\right]}{dt}=-\dfrac{1}{{\gamma }_{2}}\dfrac{d\left[B\right]}{dt}=\dfrac{1}{{\gamma }_{3}}\dfrac{d\left[C\right]}{dt}=\dfrac{1}{{\gamma }_{4}}\dfrac{d\left[D\right]}{dt}$ (i) $\dfrac{d\left[D\right]}{dt}=\dfrac{{\gamma }_{4}}{{\gamma }_{2}}\left[-\dfrac{d\left[B\right]}{dt}\right]=1.5\left(-\dfrac{d}{dt}\left[B\right]\right)$ $\dfrac{{\gamma }_{4}}{{\gamma }_{2}}=1.5$ (ii) $\dfrac{{\gamma }_{2}}{{\gamma }_{1}}\left[-\dfrac{d\left[A\right]}{dt}\right]=\left(-\dfrac{d}{dt}\left[B\right]\right)=2\left(-\dfrac{d}{dt}\left[A\right]\right)$ $\dfrac{{\gamma }_{2}}{{\gamma }_{1}}=2$ (iii) $\dfrac{d\left[C\right]}{dt}=\dfrac{{\gamma }_{3}}{{\gamma }_{4}}\left[\dfrac{d\left[D\right]}{dt}\right]$ $1=\dfrac{{\gamma }_{3}}{{\gamma }_{4}}\times 9$ $\dfrac{{\gamma }_{4}}{{\gamma }_{3}}=9$ So, if ${\gamma }_{3}=1than{\gamma }_{4}=9 {\gamma }_{2}=6and{\gamma }_{1}=3 Rateofreaction=\dfrac{1}{{\gamma }_{3}}\dfrac{d[C]}{dt}=1mmol{dm}^{-3}{s}^{-1}$