JEE (Main) · Chemistry, Class 12
Chemical Kinetics: JEE Main PYQ, 2020–2025
Chemical Kinetics (Chemistry, Class 12) has been asked 116 times in the 111 JEE (Main) shift papers from 2020–2025: about 1 questions a paper, 4% of all Chemistry questions. It is the 14th most-asked of 21 Chemistry chapters. The 2025 shifts had 22 from it in all. Last checked 28 October 2026.
Questions per year, all shifts
Year by year.
| Year | Questions |
|---|---|
| 2020 | 13in 14 papers |
| 2021 | 26in 26 papers |
| 2022 | 24in 24 papers |
| 2023 | 13in 12 papers |
| 2024 | 18in 18 papers |
| 2025 | 22in 17 papers |
JEE (Main) is sat in many shifts a year, each with its own paper; the counts add up every shift. 2021–2024 papers had 30 questions a subject, the others 25.
Inside the chapter
Topics asked most.
- Integrated rate laws and half-life48
- Arrhenius equation33
- Rate laws18
- Order of reaction17
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10 Chemical Kinetics questions to practise.
Our own questions, written in the JEE Main pattern and checked, each with an explanation. Pick an option to see the answer.
Q1 · Chemistry · Chemical Kinetics
Which of the following statements about the pre-exponential factor A in the Arrhenius equation k = A e^(−Ea/RT) is correct?
Answer: (A)
In the Arrhenius equation, the exponential term e^(−Ea/RT) gives the fraction of effective collisions, and A gives the collision frequency and orientation factor. Neither depends on the concentrations of the reactants, which affect only the rate. So changing concentrations leaves A unchanged, and A is treated as a constant for a given reaction at a fixed temperature.
Q2 · Chemistry · Chemical Kinetics
For a reaction A → products, a plot of log₁₀(rate) against log₁₀[A] is a straight line with slope 2. It cuts the log(rate) axis at −3.0, that is, at log₁₀[A] = 0. Concentration is in mol L⁻¹ and rate is in mol L⁻¹ s⁻¹. What is the rate of the reaction when [A] = 0.10 mol L⁻¹?
Answer: (D)
Taking logarithms of rate = k[A]ⁿ gives log(rate) = log k + n log[A]. The slope of the line is the order, n = 2. The intercept at log[A] = 0 is log k = −3, so k = 10⁻³ L mol⁻¹ s⁻¹. At [A] = 0.10 M, rate = 10⁻³ × (0.10)² = 1.0 × 10⁻⁵ mol L⁻¹ s⁻¹.
Q3 · Chemistry · Chemical Kinetics
The gas-phase reaction A(g) → 2B(g) is carried out at constant temperature in a rigid vessel, starting with pure A at a pressure of 0.40 atm. The total pressure in the vessel is 0.60 atm after 10 min and 0.70 atm after 20 min. Assuming the reaction follows simple zero-, first- or second-order kinetics in A, what is its order?
Answer: (B)
The total pressure rises because each mole of A gives two moles of B. If x is the pressure of A consumed, pA = 0.40 − x, pB = 2x and P_total = 0.40 + x. So pA = 0.80 − P_total. At 10 min, pA = 0.80 − 0.60 = 0.20 atm. At 20 min, pA = 0.80 − 0.70 = 0.10 atm. Starting from 0.40 atm, the pressure halves to 0.20 atm in 10 min and halves again to 0.10 atm in the next 10 min. A constant half-life independent of concentration is the signature of a first-order reaction. The zero-order and second-order predictions (0.80 atm and 0.667 atm) do not fit.
Q4 · Chemistry · Chemical Kinetics
A reaction has a rate constant k whose units are L mol⁻¹ s⁻¹. What is the order of this reaction?
Answer: (C)
The rate law for an nth-order reaction is rate = k[A]^n, so k = rate / [A]^n. Rate has units mol L⁻¹ s⁻¹ and concentration has units mol L⁻¹. So the units of k are (mol L⁻¹ s⁻¹)/(mol L⁻¹)^n = (mol L⁻¹)^(1−n) s⁻¹. Setting this equal to L mol⁻¹ s⁻¹ = (mol L⁻¹)⁻¹ s⁻¹ requires 1 − n = −1, so n = 2. The reaction is second order.
Q5 · Chemistry · Chemical Kinetics
The reaction NO₂(g) + CO(g) → NO(g) + CO₂(g) is believed to occur by the mechanism: Step 1 (slow): 2NO₂ → NO₃ + NO; Step 2 (fast): NO₃ + CO → NO₂ + CO₂. At constant temperature, [NO₂] is doubled and [CO] is tripled. By what factor does the rate change?
Answer: (A)
The overall rate is controlled by the slowest (rate-determining) step. Step 1 is an elementary bimolecular step involving two NO₂ molecules, so rate = k[NO₂]². CO enters only in the fast second step and is zero order. Doubling [NO₂] gives 2² = 4, and tripling [CO] has no effect, so the rate rises by a factor of 4.
Q6 · Chemistry · Chemical Kinetics
The initial rates of the reaction A + B → products were measured at constant temperature in three experiments. Experiment 1: [A] = 0.04 M, [B] = 0.10 M, rate = 1.5 × 10⁻³ mol L⁻¹ s⁻¹. Experiment 2: [A] = 0.16 M, [B] = 0.10 M, rate = 1.2 × 10⁻² mol L⁻¹ s⁻¹. Experiment 3: [A] = 0.04 M, [B] = 0.30 M, rate = 4.5 × 10⁻³ mol L⁻¹ s⁻¹. What is the initial rate (in mol L⁻¹ s⁻¹) when [A] = 0.01 M and [B] = 0.20 M?
Answer: (A)
Compare experiments 1 and 2, where [B] is constant: [A] rises 4 times and rate rises 12 / 1.5 = 8 times, so 4^m = 8 and m = 3/2. Compare experiments 1 and 3, where [A] is constant: [B] rises 3 times and rate rises 3 times, so n = 1. The rate law is rate = k[A]^1.5[B]. For the new conditions, [A] falls to 1/4 of its Experiment 1 value, so the rate factor is (1/4)^1.5 = 1/8. [B] doubles, so the factor is 2. Rate = 1.5 × 10⁻³ × (1/8) × 2 = 3.75 × 10⁻⁴ mol L⁻¹ s⁻¹.
Q7 · Chemistry · Chemical Kinetics
A first-order reaction has a rate constant of 1.5 × 10⁻² s⁻¹. What is its rate constant expressed in min⁻¹?
Answer: (A)
The unit of a first-order rate constant is time⁻¹. Converting from per second to per minute means one unit of time⁻¹ in s⁻¹ corresponds to 60 units in min⁻¹. So k = 1.5 × 10⁻² s⁻¹ × 60 s/min = 0.9 min⁻¹. The numerical value of k rises when the time unit is made larger.
Q8 · Chemistry · Chemical Kinetics
For a reaction X + Y → Z at constant temperature, three initial-rate experiments gave: (1) [X] = 0.10 M, [Y] = 0.10 M, rate = 4.0 × 10⁻⁴ mol L⁻¹ s⁻¹; (2) [X] = 0.20 M, [Y] = 0.10 M, rate = 1.6 × 10⁻³ mol L⁻¹ s⁻¹; (3) [X] = 0.20 M, [Y] = 0.30 M, rate = 4.8 × 10⁻³ mol L⁻¹ s⁻¹. What is the initial rate, in mol L⁻¹ s⁻¹, when [X] = 0.30 M and [Y] = 0.20 M?
Answer: (A)
Compare experiments 1 and 2, where only [X] changes. [X] doubles and the rate rises 4-fold, so the order in X is 2. Compare experiments 2 and 3, where only [Y] changes. [Y] triples and the rate triples, so the order in Y is 1. The rate law is rate = k[X]²[Y]. From experiment 1, k = 4.0 × 10⁻⁴ / (0.10² × 0.10) = 0.40 L² mol⁻² s⁻¹. At [X] = 0.30 M and [Y] = 0.20 M, rate = 0.40 × 0.09 × 0.20 = 7.2 × 10⁻³ mol L⁻¹ s⁻¹.
Q9 · Chemistry · Chemical Kinetics
A first-order reaction has a half-life of 6.93 min. What is the rate of the reaction at the instant when the concentration of the reactant is 0.5 mol L⁻¹? (ln 2 = 0.693)
Answer: (A)
For a first-order reaction, k = 0.693 / t½ = 0.693 / 6.93 = 0.1 min⁻¹. Rate = k[R] = 0.1 × 0.5 = 0.05 mol L⁻¹ min⁻¹.
Q10 · Chemistry · Chemical Kinetics
For a reaction whose activation energy is zero, how does the rate constant k change when the temperature is increased?
Answer: (A)
Write k = A e^(−Ea/RT). If Ea = 0, the exponent is zero at all temperatures, so e^(−Ea/RT) = 1 and k = A. The rate constant then does not change with temperature. Real reactions have positive Ea, which is why rate constants usually rise with temperature, but the limiting case Ea = 0 shows that the temperature dependence comes entirely from the exponential term.
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All 116 real JEE Main questions from Chemical Kinetics.
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About Chemical Kinetics in JEE Main.
How many questions come from Chemical Kinetics in JEE Main?
Chemical Kinetics (Chemistry, Class 12) has been asked 116 times in the 111 JEE (Main) shift papers from 2020–2025: about 1 questions a paper, 4% of all Chemistry questions. It is the 14th most-asked of 21 Chemistry chapters. The 2025 shifts had 22 from it in all.
Which topics of Chemical Kinetics are asked most in JEE Main?
Integrated rate laws and half-life (48), Arrhenius equation (33), Rate laws (18), Order of reaction (17), counted over the papers from 2020–2025.
Is Chemical Kinetics important for JEE Main 2027?
It is in the syllabus in force, so it can be asked. By past papers it ranks 14 of 21 in Chemistry; use that to decide how much time it gets, not whether to study it.
Where can I solve Chemical Kinetics JEE Main previous year questions?
On Nikaalle, free: sign up and practise every past JEE Main question from Chemical Kinetics with an explanation for each, or sit the full papers timed. The ten questions on this page are our own, written in the same pattern.
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