Chemical Kinetics JEE Advanced previous year questions with solutions

5 solved JEE Advanced questions on Chemical Kinetics, free to read — no sign-in needed. The full chapter has 36 questions; sign in to attempt the remaining 31 in the exam simulator.

1. Q1JEE Advanced Adv 2025 (Paper 2)
   Consider a reaction $A+R \rightarrow$ Product. The rate of this reaction is measured to be $k[A][R]$. At the start of the reaction, the concentration of $R,[R]_0$, is 10 -times the concentration of $A,[A]_0$. The reaction can be considered to be a pseudo first order reaction with assumption that $k[R]=k^{\prime}$ is constant. Due to this assumption, the relative error (in \%) in the rate when this reaction is $40 \%$ complete, is _______ [ $k$ and $k^{\prime}$ represent corresponding rate constants]
   Show answer & solution

   Answer: 4.16

   $\begin{aligned} & \qquad \quad \mathrm{A}+\mathrm{R} \rightarrow \text { Product } \\ & \mathrm{t}=0 \quad \mathrm{~A}_0 \quad 10 \mathrm{~A}_0 \\ & \mathrm{t}=\mathrm{t} \quad 0.6 \mathrm{~A}_0 \quad 9.6 \mathrm{~A}_0 \\ & \text { Rate }=\mathrm{k}[\mathrm{A}][\mathrm{R}] \\ & \text { Rate }_1=\mathrm{k}\left(0.6 \mathrm{~A}_0\right) \times 9.6 \mathrm{~A}_0 \\ & \qquad \quad \mathrm{~A} \quad+\quad \mathrm{R} \rightarrow \text { Product } \\ & \mathrm{t}=0 \quad \mathrm{~A}_0 \qquad 10 \mathrm{~A}_0(\text { excess }) \\ & \mathrm{t}=\mathrm{t} \quad 0.6 \mathrm{~A}_0 \quad 10 \mathrm{~A}_0 \\ & \text { Rate }=\mathrm{k}^{\prime}[\mathrm{A}], \mathrm{k}^{\prime}=\mathrm{k}[\mathrm{R}] \\ & \text { Rate }_2=\left(\mathrm{k} \times 10 \mathrm{~A}_0\right) \times\left(0.6 \mathrm{~A}_0\right) \\ & 100 \times \frac{\Delta \text { Rate }}{\text { Rate }_1}=\frac{(0.6 \times 10-0.6 \times 9.6}{0.6 \times 9.6} \times 100=4.1666\end{aligned}$
2. Q2JEE Advanced Adv 2024 (Paper 2)
   A sample initially contains only U-238 isotope of uranium. With time, some of the U-238 radioactively decays into $\mathrm{Pb}-206$ while the rest of it remains undisintegrated. When the age of the sample is $\mathbf{P} \times 10^8$ years, the ratio of mass of $\mathrm{Pb}-206$ to that of $\mathrm{U}-238$ in the sample is found to be 7 . The value of $\mathbf{P}$ is ______ [Given: Half-life of U-238 is $4.5 \times 10^9$ years; $\log _c 2=0.693$ ]
   Show answer & solution

   Answer: 143

   ${ }_{92} \mathrm{U}^{238} \quad \longrightarrow \quad{ }_{82} \mathrm{~Pb}^{206}+8{ }_2 \mathrm{He}^4+6_{-1} \beta^0$ $\mathrm{t}=0 \quad \mathrm{~N}_0=\left(\frac{1}{238}+\frac{7}{206}\right)$ moles $\mathrm{t}=\mathrm{tN}_{\mathrm{t}}=\frac{1}{238}$ moles $\mathrm{x}=\frac{7}{206}$ moles Life of sample $\rightarrow t$ years $[A]_0 \propto$ Initial mole of $\mathrm{U}-238$ $[A]]_t \propto$ Final mole of U-238 $\begin{aligned} & \frac{|A|_0}{|A|_t}=\frac{\frac{1}{228}+\frac{7}{206}}{\frac{1}{238}} \\ & =\frac{0.0042+0.0340}{0.0042} \\ & =9.1 \\ & =\frac{2.303 \log 2 \times t}{4.5 \times 10^9}=2.303 \log 9.1 \\ & t=14.27 \times 10^9 \text { years } \\ & =142.7 \times 10^8 \text { years } \\ & P=142.7 \\ & P=143 \end{aligned}$
3. Q3JEE Advanced Adv 2021 (Paper 2)
   For the following reaction $2X+Y\rightarrow ^{k}P$ the rate of reaction is $\dfrac{d[P]}{dt}=k[X]$. Two moles of $X$ are mixed with one mole of $Y$ to make $1.0L$ of solution. At $50s,0.5$ mole of $Y$ is left in the reaction mixture. The correct statement(s) about the reaction is(are) $($Use: $\ln 2=0.693)$
   1. A.The rate constant, $k$, of the reaction is $13.86\times {10}^{-4}{s}^{-1}$.
   2. B.Half-life of $X$ is $50s$.
   3. C.At $50s,-\dfrac{d[x]}{dt}=13.86\times {10}^{-3}mol{L}^{-1}{s}^{-1}$.
   4. D.At $100s,-\dfrac{d\left[Y\right]}{dt}=3.46\times {10}^{-3}mol{L}^{-1}{s}^{-1}$
   Show answer & solution

   Answer: B,C,D

   $\begin{array}{lllll} & 2 \mathrm{X}+ & \mathrm{y} & \longrightarrow & \mathrm{P} \\ \mathrm{t}=0 & 2 & 1 & & 0 \\ \mathrm{t}=50 & 2-2 \times 0.5 & 1-0.5 & & \\ & \text { 1mole } & 0.5 & & \end{array}$ rate $=-\frac{1}{2} \frac{d x}{d t}=-\frac{d y}{d t}=\frac{d P}{d t}=K[X]$ $-\frac{1}{2} \frac{d x}{d t}=K[X]$ $-\frac{d x}{d t}=2 K[X]=K^1[X]$ Half life is $t=50 \mathrm{sec}$ $2 K=\frac{0.653 L}{50}$ $K=\frac{0.6932}{100}=6.332 \times 10^{-3}$ $t=50 \mathrm{sec}$ $-\frac{d x}{d t}=2 K[X]$ $-\frac{d x}{d t}=2 \times 6.332 \times 10^{-3} \times 1$ $=13.864 \times 10^{-3} \mathrm{~mole} / \mathrm{L} / \mathrm{Sec}$ $-\frac{d y}{d t}=K[X]=6.332 \times 10^{-3}\left(\frac{1}{2}\right)$ $=3.46 \times 10^{-3} \mathrm{~mole} / \mathrm{L} / \mathrm{Sec}^{-1}$
4. Q4JEE Advanced Adv 2020 (Paper 1)
   $U92238$ is known to undergo radioactive decay to form $Pb82206$ by emitting alpha and bet a particles. A rock initially contained $68\times {10}^{-6}g$ of $U92238$. If the number of alpha particles that it would emit during its radioactive decay of $U92238$ to $Pb82206$ in three half-lives is $Z\times {10}^{18}$, then what is the value of $Z$?
   Show answer & solution

   Answer: 1.2

   ${}_{92}^{238}U{\rightarrow }_{82}^{206}Pb{+}_{2}^{4}He+6{ }_{-1}^{{}^{0}}\beta$ $0.286\times {10}^{-6}mole$ $\left[0.286\times {10}^{-6}\times {N}_{a}\right]$ $=0.286\times 6.02\times {10}^{17}$ $=1.72\times {10}^{17}$ After 3 half lives $N=\dfrac{No}{(2{)}^{3}}=\dfrac{1.72\times {10}^{17}}{8}$ $=0.215\times {10}^{17}$ So, no. of molecule of uranium decayed $=[1.72-0.215]\times {10}^{17}$ $=1.5\times {10}^{17-}$ So No. of $\alpha$ particle produced $=8\times 1.5\times {10}^{17}$ $=12\times {10}^{17}$ $=1.2\times {10}^{18}$
5. Q5JEE Advanced Adv 2019 (Paper 1)
   Consider the kinetic data given in the following table for the reaction $A+B+C\rightarrow$ Product. Experiment No. $\left[A\right]$ $\left(mol{dm}^{-3}\right)$ $\left[B\right]$ $\left(mold{m}^{-3}\right)$ $\left[C\right]$ $\left(mold{m}^{-3}\right)$ Rate of reaction $\left(mold{m}^{-3}{s}^{-1}\right)$ $1$ $0.2$ $0.1$ $0.1$ $6.0\times {10}^{-5}$ $2$ $0.2$ $0.2$ $0.1$ $6.0\times {10}^{-5}$ $3$ $0.2$ $0.1$ $0.2$ $1.2\times {10}^{-4}$ $4$ $0.3$ $0.1$ $0.1$ $9.0\times {10}^{-5}$ The rate of the reaction for $\left[A\right]=0.15mold{m}^{-3},\left[B\right]=0.25mold{m}^{-3}$ and $\left[C\right]=0.15mold{m}^{-3}$ is found to be $Y\times {10}^{-5}mold{m}^{-3}{s}^{-1}.$ The value of $Y$ is __________
   Show answer & solution

   Answer: 6.75

   $\text{r}=\text{k}[\text{A}]{ }^{\text{a}} {[\text{B}]}^{\text{b}} {[\text{C}]}^{\text{c}}$ $6\times {10}^{-5}=\text{k}{(0.2)}^{\text{a}} {(0.1)}^{\text{b}} {(0.1)}^{\text{c}}$ ..... (1) $6\times {10}^{-5}=k{(0.2)}^{\text{a}}{(0.2)}^{\text{b}}{(0.1)}^{\text{c}}$ ...... (2) $1.2\times {10}^{-4}=k{(0.2)}^{\text{a}}{(0.1)}^{\text{b}} {(0.2)}^{\text{c}}$ ...... (3) $9\times {10}^{-5}=k{(0.3)}^{\text{a}}{(0.1)}^{\text{b}} {(0.1)}^{\text{c}}$ ...... (4) From equation (1) and (ii) : $b=0$ From equation (1) and (iii) : $c=1$ From equation (1) and (iv) : $a=1$ Put $x,y$ and $z$ in equation (i) we get; $6\times {10}^{-5}=k{(0.2)}^{1} {(0.1)}^{0}{(0.1)}^{1}$ $\text{r}=3\times {10}^{-3}[\text{A}][\text{C}]$ For $[\text{A}]=0.15\text{mol}{\text{dm}}^{-3}$ $[\text{B}]=0.25\text{mol}{\text{dm}}^{-3}$ $[\text{C}]=0.15\text{mol}{\text{dm}}^{-3}$ $\text{r}=3\times {10}^{-3}\times 0.15\times 0.15$ $\text{r}=6.75\times {10}^{-5}$ $∴\text{r}\text{=}\text{6.75}$