JEE (Advanced) · Chemistry, Class 12
Chemical Kinetics: JEE Advanced PYQ, 2007–2026
Chemical Kinetics (Chemistry, Class 12) has been asked 29 times in the 40 JEE (Advanced) papers from 2007–2026: about 0.7 questions a paper, 4% of all Chemistry questions. It is the 10th most-asked of 31 Chemistry chapters. The two 2026 papers had 2 from it. Last checked 28 October 2026.
Questions per year, both papers
Year by year.
| Year | Questions |
|---|---|
| 2007 | 1in 2 papers |
| 2008 | 2in 2 papers |
| 2009 | 2in 2 papers |
| 2010 | 2in 2 papers |
| 2011 | 1in 2 papers |
| 2012 | 1in 2 papers |
| 2013 | 1in 2 papers |
| 2014 | 1in 2 papers |
| 2015 | 0in 2 papers |
| 2016 | 2in 2 papers |
| 2017 | 1in 2 papers |
| 2018 | 2in 2 papers |
| 2019 | 3in 2 papers |
| 2020 | 2in 2 papers |
| 2021 | 1in 2 papers |
| 2022 | 1in 2 papers |
| 2023 | 1in 2 papers |
| 2024 | 2in 2 papers |
| 2025 | 1in 2 papers |
| 2026 | 2in 2 papers |
Each year has two papers, and the number of questions has fallen over time, so compare per paper and expect some drift.
Inside the chapter
Topics asked most.
- Integrated rate laws and half-life16
- Arrhenius equation6
- Order of reaction5
- Rate laws2
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10 Chemical Kinetics questions to practise.
Our own questions, written in the JEE Advanced pattern and checked, each with an explanation. Pick an option to see the answer.
Q1 · Chemistry · Chemical Kinetics
A gaseous reaction has the rate law: rate = k[A]^(3/2)[B]^(1/2). If the volume of the vessel is reduced to one quarter at constant temperature, by what factor does the rate change at that instant?
Answer: (C)
A quarter of the volume means four times each concentration. Rate ∝ [A]^(3/2)[B]^(1/2), so it changes by 4^(3/2) × 4^(1/2) = 8 × 2 = 16, which is 4 raised to the overall order, 2.
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
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.
Q5 · 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.
Q6 · 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.
Q7 · Chemistry · Chemical Kinetics
The reaction A + B → products is first order in A and first order in B, so rate = k[A][B]. It is run at constant temperature with [A]₀ = 0.010 mol L⁻¹ and [B]₀ = 2.0 mol L⁻¹. The concentration of A is found to halve every 10 min throughout the run. (ln 2 = 0.693) What is the value of k?
Answer: (C)
The constant half-life of A, independent of how much A has been used, shows pseudo-first-order behaviour. B is 200 times more concentrated than A, so [B] stays essentially at 2.0 mol L⁻¹. The rate law reduces to rate = k′[A] with k′ = k[B]₀. From the half-life, k′ = 0.693/10 min = 0.0693 min⁻¹. Then k = k′/[B]₀ = 0.0693/2.0 = 0.0347 L mol⁻¹ min⁻¹.
Q8 · Chemistry · Chemical Kinetics
A substance A decomposes by two parallel first-order paths: A → B with rate constant 2 × 10⁻³ s⁻¹, and A → C with rate constant 3 × 10⁻³ s⁻¹. What percentage of the product is B?
Answer: (A)
At every instant d[B]/dt : d[C]/dt = k₁ : k₂ = 2 : 3. So B is 2/(2 + 3) = 40% of the product, at all times.
Q9 · Chemistry · Chemical Kinetics
For a reaction A + B → products, the following initial-rate data were obtained: (i) [A] = 0.10 M, [B] = 0.10 M gives initial rate r; (ii) [A] = 0.20 M, [B] = 0.10 M gives initial rate 4r; (iii) [A] = 0.20 M, [B] = 0.40 M gives initial rate 8r. Under identical conditions, 100 mL of 0.80 M A is rapidly mixed with 100 mL of 0.80 M B (volumes additive). What is the initial rate immediately after mixing?
Answer: (A)
First find the orders. Between (i) and (ii) only [A] doubles and the rate becomes 4 times, so 2^x = 4 and x = 2. Between (ii) and (iii) only [B] quadruples and the rate doubles, so 4^y = 2 and y = 1/2. The rate law is therefore rate = k[A]²[B]^(1/2), with rate r at [A] = [B] = 0.10 M. Next, mixing equal volumes halves each concentration, so [A] = [B] = 0.40 M, which is 4 times the reference value of 0.10 M. The rate is then r × 4² × 4^(1/2) = r × 16 × 2 = 32r. A student who skips the dilution gets about 181r. A student who takes the order in B as 1 gets 64r, and one who takes it as 0 gets 16r.
Q10 · Chemistry · Chemical Kinetics
The gas-phase reaction 2NO(g) + O₂(g) → 2NO₂(g) is elementary in its rate behaviour, with rate = k[NO]²[O₂]. The volume of the reaction vessel is suddenly reduced to one-third of its original value at constant temperature. The rate of the reaction then becomes:
Answer: (C)
For a fixed amount of gas at constant temperature, concentration is inversely proportional to volume, so reducing the volume to one-third triples the concentration of each gas. Then rate' = k(3[NO])²(3[O₂]) = 3³ × k[NO]²[O₂] = 27 × rate. The factor is 3 raised to the overall order, 3.
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All 29 real JEE Advanced questions from Chemical Kinetics.
The questions from the actual papers, with the official answer and an explanation for each — chapter by chapter, or as full timed papers.
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About Chemical Kinetics in JEE Advanced.
How many questions come from Chemical Kinetics in JEE Advanced?
Chemical Kinetics (Chemistry, Class 12) has been asked 29 times in the 40 JEE (Advanced) papers from 2007–2026: about 0.7 questions a paper, 4% of all Chemistry questions. It is the 10th most-asked of 31 Chemistry chapters. The two 2026 papers had 2 from it.
Which topics of Chemical Kinetics are asked most in JEE Advanced?
Integrated rate laws and half-life (16), Arrhenius equation (6), Order of reaction (5), Rate laws (2), counted over the papers from 2007–2026.
Is Chemical Kinetics important for JEE Advanced 2027?
It is in the syllabus in force, so it can be asked. By past papers it ranks 10 of 31 in Chemistry; use that to decide how much time it gets, not whether to study it.
Where can I solve Chemical Kinetics JEE Advanced previous year questions?
On Nikaalle, free: sign up and practise every past JEE Advanced 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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