Chemical Kinetics Previous Year Questions (JEE Main 2026)
Solved JEE Main 2026 previous year questions on Chemical Kinetics with concise, step-by-step solutions covering order of reaction, first-order kinetics, half-life and the Arrhenius equation.
Solved JEE Main 2026 previous year questions on Chemical Kinetics, worked out step by step so you can benchmark your problem-solving and revise the high-yield formulas fast.
Solutions are AI-generated and pending review.
Solution
For a first order reaction, the integrated rate law is:
$$[R] = [R]_0\, e^{-k_1 t}$$Fraction remaining:
$$\frac{[R]}{[R]_0} = e^{-k_1 t}$$Fraction decomposed = total minus fraction remaining:
$$f = 1 - \frac{[R]}{[R]_0} = 1 - e^{-k_1 t}$$Answer: D
Solution
The reaction is first order, so the fraction remaining after $n$ half-lives is $(1/2)^n$.
Number of half-lives in 6 hours:
$$n = \frac{6}{t_{1/2}} = \frac{6}{3} = 2$$Fraction remaining:
$$\left(\frac{1}{2}\right)^{2} = \frac{1}{4} = 0.25$$Percentage remaining:
$$0.25 \times 100 = 25\%$$Answer: 25
Solution
Zero order: $[A] = [A]_0 - kt$. The reaction finishes when $[A] = 0$:
$$t_{100\%} = \frac{[A]_0}{k}, \qquad t_{1/2} = \frac{[A]_0}{2k}$$$$\Rightarrow t_{100\%} = 2\,t_{1/2}$$First order: $[A] = [A]_0 e^{-kt}$. The concentration approaches zero only as $t \to \infty$, so the reaction is never truly 100% complete in finite time:
$$t_{100\%} \to \infty = (t_{1/2})^{\infty}$$Hence zero order gives $t_{100\%} = 2t_{1/2}$ and first order gives $t_{100\%} = (t_{1/2})^{\infty}$.
Answer: B
Solution
Using the two-temperature Arrhenius form:
$$\ln\frac{k_2}{k_1} = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$$Here $\dfrac{k_2}{k_1} = \dfrac{4.5\times10^3}{1.5\times10^3} = 3$, so:
$$\ln 3 = \ln 10 \times \log 3 = 2.3 \times 0.48 = 1.104$$With $T_1 = 300\,\text{K}$, $E_a = 60000\,\text{J mol}^{-1}$, $R = 8.3$:
$$\frac{1}{T_2} = \frac{1}{T_1} - \frac{R\,\ln 3}{E_a} = \frac{1}{300} - \frac{8.3 \times 1.104}{60000}$$$$\frac{1}{T_2} = 3.333\times10^{-3} - 1.527\times10^{-4} = 3.181\times10^{-3}$$$$T_2 \approx 314.4\,\text{K} = 41.4\,°\text{C}$$Answer: 41
Solution
For a first order reaction:
$$k = \frac{1}{t}\ln\frac{[A]_0}{[A]}$$From $t = 0$ to $t = 20$ min, the concentration drops from $0.6500$ to $0.00065$, a factor of $1000$:
$$k = \frac{1}{20}\ln(1000) = \frac{\ln 10^3}{20} = \frac{3 \times 2.303}{20} = 0.3454\ \text{min}^{-1}$$For the drop from $0.6500$ to $0.0650$ (a factor of $10$):
$$x = \frac{1}{k}\ln(10) = \frac{2.303}{0.3454} \approx 6.67\ \text{min}$$Equivalently, since $0.00065$ is $10^3$ times smaller reached in 20 min, a $10^1$ drop takes one-third of that time: $x = 20/3 \approx 6.67$.
Answer: 7
Solution
The units of $k$ are $\text{mol}^{-1}\,\text{L}\,\text{s}^{-1} = \text{L mol}^{-1}\text{s}^{-1}$, which correspond to a second order reaction. For an $n$-th order reaction, units of $k$ are $(\text{mol L}^{-1})^{1-n}\text{s}^{-1}$; matching gives $1-n = -1 \Rightarrow n = 2$.
- A: Rate $= k[X]^2$. If $[X] \to 4[X]$, rate becomes $4^2 = 16$ times. True.
- B: Second order, as shown above. True.
- C: For second order, $t_{1/2} = \dfrac{1}{k[X]_0}$, which depends on initial concentration. False.
- D: Decomposition of $N_2O_5$ is a well-known first order reaction. False.
- E: $\ln\dfrac{[R_0]}{[R]}$ vs $t$ being linear is characteristic of first order, not this second order reaction. False.
Only A and B are true.
Answer: A
Solution
Let the initial pressure of $A$ be $p_0$ (only $A$ present at $t = 0$).
Stoichiometry: 2 mol $A$ produce $4 + 1 = 5$ mol of gas. When all $A$ is consumed, total pressure becomes:
$$P_\infty = \frac{5}{2}p_0 = 600 \Rightarrow p_0 = 240\ \text{mm Hg}$$Let $2\alpha$ be the pressure of $A$ reacted at 30 min. Then:
$$A: p_0 - 2\alpha,\quad B: 4\alpha,\quad C: \alpha$$$$P_{30} = (p_0 - 2\alpha) + 4\alpha + \alpha = p_0 + 3\alpha = 300$$$$240 + 3\alpha = 300 \Rightarrow \alpha = 20\ \text{mm Hg}$$The pressure of $C$ at 30 min is $\alpha = 20$ mm Hg.
Answer: 20
Solution
The Arrhenius equation is:
$$k = A\, e^{-E_a/RT}$$Comparing the exponent with the given expression:
$$\frac{E_a}{RT} = \frac{28000\,\text{K}}{T} \Rightarrow \frac{E_a}{R} = 28000\,\text{K}$$$$E_a = 28000 \times 8.3 = 232400\ \text{J mol}^{-1} = 232.4\ \text{kJ mol}^{-1}$$Answer: 232
Solution
Let $x$ be the pressure of $A$ that has decomposed at time $t$.
| Species | $A$ | $B$ | $C$ |
|---|---|---|---|
| Pressure at $t$ | $p_i - x$ | $x$ | $x$ |
Total pressure:
$$p_t = (p_i - x) + x + x = p_i + x \Rightarrow x = p_t - p_i$$Pressure of $A$ remaining:
$$p_A = p_i - x = p_i - (p_t - p_i) = 2p_i - p_t$$For a first order reaction:
$$k = \frac{1}{t}\ln\frac{p_i}{p_A} = \frac{1}{t}\ln\frac{p_i}{2p_i - p_t}$$Answer: A
Solution
Rate constant from half-life:
$$k = \frac{0.693}{t_{1/2}} = \frac{0.693}{6.93} = 0.1\ \text{min}^{-1}$$For 99% completion, $[A]/[A]_0 = 0.01$, i.e. $[A]_0/[A] = 100$:
$$t = \frac{2.303}{k}\log\frac{[A]_0}{[A]} = \frac{2.303}{0.1}\log 100 = \frac{2.303}{0.1}\times 2$$$$t = 23.03 \times 2 = 46.06 \approx 46\ \text{min}$$Answer: 46
Solution
The plot of $t_{1/2}$ versus $[A]_0$ is a straight line through the origin, so:
$$t_{1/2} \propto [A]_0 \Rightarrow \frac{t_{1/2}}{[A]_0} = \text{constant}$$(This is the signature of a second order reaction, where $t_{1/2} = \dfrac{1}{k[A]_0}$… but here the direct proportionality is read straight from the linear plot.)
Using the given data point:
$$\frac{240}{4\times10^{-3}} = \frac{x}{1.5\times10^{-3}}$$$$x = 240 \times \frac{1.5\times10^{-3}}{4\times10^{-3}} = 240 \times 0.375 = 90\ \text{s}$$Answer: 90
Solution
Statement I: Convert $E_a = 12.6\ \text{kcal mol}^{-1} = 12.6 \times 4.2 \times 10^3 = 52920\ \text{J mol}^{-1}$.
$$\ln\frac{k_2}{k_1} = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right) = \frac{52920}{8.314}\left(\frac{1}{298} - \frac{1}{308}\right)$$$$= 6365.9 \times (1.0895\times10^{-4}) \approx 0.693$$$$\frac{k_2}{k_1} = e^{0.693} \approx 2$$The rate constant doubles. Statement I is true.
Statement II: For a first order reaction, $t_{1/2} = \dfrac{0.693}{k}$, which is independent of $[A]_0$. Its plot against $[A]_0$ is a horizontal line, not a line through the origin. Statement II is false.
Answer: C