How much of JEE Main comes from Chemical Kinetics?

JEEGenie estimates about 4% of JEE Main marks come from Chemical Kinetics, based on recent papers. The share varies from year to year.

Which formulas matter most in Chemical Kinetics?

The Chemical Kinetics sheet on this page has 18 formula entries, starting with Rate of Reaction, Rate Law and Order of Reaction, Integrated Rate Law: Zero Order, Integrated Rate Law: First Order, Integrated Rate Law: Second Order. Each one comes with when to use it and the trap to avoid.

What topics does Chemical Kinetics cover for JEE?

Chemical Kinetics covers Rate of Reaction, Order and Molecularity, Integrated Rate Laws, Arrhenius Equation, Collision Theory, Half-Life.

How should I revise Chemical Kinetics before the exam?

Learn each formula together with its condition of use, read the trap listed beside it, then practise questions and review every mistake. The practice link on this page opens Chemical Kinetics questions on JEEGenie.

How many previous-year questions are there on Chemical Kinetics?

JEEGenie has 329 previous-year questions on Chemical Kinetics, and this page shows a few with worked explanations.

Is practice on Chemical Kinetics free on JEEGenie?

Sign-up is free and includes a 7-day Pro trial. After the trial, free accounts keep a limited daily quota.

Chemical Kinetics for JEE Main & Advanced — formulas, what is asked, and traps

JEEGenie estimates about 4% of JEE Main marks come from Chemical Kinetics. This page covers 18 formula entries, each with when to use it, what JEE asks and the traps. A common slip: Forgetting to divide by stoichiometric coefficients when comparing rates of different species.

Key points to know

Rate of Reaction. Rate of reaction is the change in concentration of a reactant or product per unit time. For a reaction $a A \to b B$ , rate = $- \frac{1}{a} \frac{d [A]}{d t} = \frac{1}{b} \frac{d [B]}{d t}$ .
Rate Law and Order of Reaction. Rate law expresses rate as $Rate = k [A]^{m} [B]^{n}$ , where $m$ and $n$ are orders with respect to each reactant. Overall order = $m + n$ .
Molecularity of a Reaction. Molecularity is the number of molecules colliding in an elementary step. It is always a positive integer (1, 2, 3) and cannot be zero or fractional.
Pseudo-First Order Reaction. When one reactant is in large excess, its concentration remains nearly constant, making the reaction appear first order overall. Example: hydrolysis of ester in excess water.
Activation Energy and Transition State. Activation energy (E $_{a}$ ) is the minimum energy required for a reaction to occur. The transition state is the maximum-energy configuration along the reaction path (not isolable), unlike an intermediate, which sits in an energy well. A catalyst lowers E $_{a}$ by providing an alternative pathway.
Effect of Temperature on Rate. Increasing temperature increases rate constant exponentially (Arrhenius). A common rule: rate doubles for every 10°C rise (approx), but this is not exact.
Collision Theory. For a reaction to occur, molecules must collide with sufficient energy (≥ E $_{a}$ ) and proper orientation. Rate = Z × f × p, where Z is collision frequency, f is fraction with sufficient energy, p is steric factor.
Rate-Determining Step (RDS). In a multi-step reaction, the slowest step determines the overall rate. The rate law is determined by the RDS and its stoichiometry.
Catalysis and Activation Energy. A catalyst provides an alternative pathway with lower activation energy, increasing the rate constant. It does not affect the equilibrium constant or the enthalpy change.
Parallel and Consecutive Reactions. In parallel reactions, a reactant forms multiple products simultaneously. In consecutive reactions, the product of one step is the reactant of the next. Rate laws are more complex.
Radioactive Decay as First Order. Radioactive decay follows first order kinetics: $N = N_{0} e^{- λ t}$ , where $λ$ is decay constant. Half-life: $t_{1 / 2} = \frac{\ln 2}{λ}$ .
Units of Rate Constant. For overall order n, units of k are (concentration) $^{1 - n}$ time $^{- 1}$ . For zero order: M s $^{- 1}$ ; first order: s $^{- 1}$ ; second order: M $^{- 1}$ s $^{- 1}$ .

Formula sheet

Name	Formula	When to use
Rate of Reaction	$Rate = - \frac{1}{a} \frac{d [A]}{d t} = \frac{1}{b} \frac{d [B]}{d t}$	Use to define and calculate the speed of a chemical reaction in terms of concentration changes.
Rate Law and Order of Reaction	$Rate = k [A]^{m} [B]^{n}$	Use to mathematically describe how concentration of reactants affects reaction rate.
Integrated Rate Law: Zero Order	$[A] = [A]_{0} - k t$ ; $t_{1 / 2} = \frac{[A]_{0}}{2 k}$	Use for reactions where rate is independent of concentration (e.g., some photochemical or surface reactions).
Integrated Rate Law: First Order	$[A] = [A]_{0} e^{- k t}$ ; $t_{1 / 2} = \frac{\ln 2}{k}$	Use for reactions like radioactive decay, decomposition of N $_{2}$ O $_{5}$ , or any reaction with rate proportional to one reactant.
Integrated Rate Law: Second Order	$\frac{1}{[A]} = \frac{1}{[A]_{0}} + k t$ ; $t_{1 / 2} = \frac{1}{k [A]_{0}}$	Use for reactions like 2A → products or A + B → products when [A] = [B] initially.
Determining Order from Experimental Data	$Rate = k [A]^{m} \Rightarrow \log (Rate) = \log k + m \log [A]$	Use when given a table of initial concentrations and initial rates to find the rate law.
Pseudo-First Order Reaction	$Rate = k^{'} [ester]$ where $k^{'} = k [H_{2} O]$	Use when one reactant is present in much higher concentration than others, simplifying kinetics.
Arrhenius Equation	$k = A e^{- E_{a} / R T}$ ; $\ln k = \ln A - \frac{E_{a}}{R T}$	Use to calculate activation energy, rate constant at different temperatures, or predict temperature dependence.
Arrhenius Equation: Two-Point Form	$\ln \frac{k_{2}}{k_{1}} = \frac{E_{a}}{R} (\frac{1}{T_{1}} - \frac{1}{T_{2}})$	Use when given k at two temperatures to find E $_{a}$ , or to find k at another temperature.
Effect of Temperature on Rate	$k \propto e^{- E_{a} / R T}$	Use to predict how rate changes with temperature.
Collision Theory	$f = e^{- E_{a} / R T}$	Use to understand the molecular basis of reaction rates and the Arrhenius pre-exponential factor.
Half-Life Method to Determine Order	$t_{1 / 2} \propto [A]_{0}^{1 - n}$	Use when given half-life data at different initial concentrations to find order.
Deriving Rate Law from Mechanism	Example: $A ⇌ B$ (fast), $B \to C$ (slow) gives rate = $k K [A]$	Use when a reaction mechanism is given and you need to find the overall rate law.
Catalysis and Activation Energy	$k_{catalyzed} = A e^{- E_{a}^{'} / R T}$ with $E_{a}^{'} < E_{a}$	Use to explain how catalysts increase reaction rates.
Temperature Coefficient (Q $_{10}$ )	$Q_{10} = \frac{k_{T + 10}}{k_{T}}$	Use as a rough estimate of temperature effect on rate.
Radioactive Decay as First Order	$N = N_{0} e^{- λ t}$ ; $t_{1 / 2} = \frac{0.693}{λ}$	Use for any radioactive decay problem.
Using Initial Rates to Find Rate Constant	$k = \frac{Rate}{[A]^{m} [B]^{n}}$	Use when given initial rates at various concentrations.
Units of Rate Constant	$Units of k = M^{1 - n} s^{- 1}$	Use to verify the order from given k units or to determine k units from order.

Open the Chemical Kinetics formula page

What JEE asks

Rate of Reaction. JEE often asks to calculate rate from given concentration-time data, or to relate rates of different species using stoichiometric coefficients.
Rate Law and Order of Reaction. JEE frequently asks to determine order from experimental data (initial rates method) or to write rate law for elementary reactions.
Molecularity of a Reaction. JEE tests the distinction between order and molecularity, often in multiple-choice questions.
Integrated Rate Law: Zero Order. JEE often asks to calculate half-life or concentration after time t for zero order reactions.
Integrated Rate Law: First Order. JEE frequently asks to calculate k, t $_{1 / 2}$ , or time for a given fraction to react using first order equations.
Integrated Rate Law: Second Order. JEE tests the inverse relationship of half-life with initial concentration and calculation of k from concentration data.
Determining Order from Experimental Data. JEE often provides a data table and asks to determine order and rate constant.
Pseudo-First Order Reaction. JEE frequently asks to identify pseudo-first order conditions or calculate k from such data.
Arrhenius Equation. JEE often asks to calculate E $_{a}$ from a plot of ln k vs 1/T or from two data points using the two-point form.
Arrhenius Equation: Two-Point Form. JEE frequently tests this formula in numerical problems.
Activation Energy and Transition State. JEE tests conceptual understanding of E $_{a}$ and catalysts in multiple-choice questions.
Effect of Temperature on Rate. JEE may ask to calculate the factor by which rate increases for a given temperature rise using Arrhenius.
Collision Theory. JEE tests conceptual understanding of collision theory and its limitations.
Half-Life Method to Determine Order. JEE may ask to determine order from half-life vs concentration data.
Rate-Determining Step (RDS). JEE often gives a mechanism and asks to identify the RDS or derive the rate law.
Deriving Rate Law from Mechanism. JEE frequently tests this in problems with pre-equilibrium steps.
Catalysis and Activation Energy. JEE tests the effect of catalysts on E $_{a}$ and rate, often in conceptual questions.
Temperature Coefficient (Q $_{10}$ ). JEE may ask to calculate Q $_{10}$ from given rate constants.
Graphical Method for Order Determination. JEE may ask to identify order from a given graph or to plot data.
Parallel and Consecutive Reactions. JEE occasionally tests the concept of parallel reactions (e.g., percentage yield from rate constants).
Radioactive Decay as First Order. JEE often asks to calculate age or remaining amount using decay equations.
Using Initial Rates to Find Rate Constant. JEE frequently tests this in numerical problems.
Units of Rate Constant. JEE often asks to identify order from units of k.

Common traps

Forgetting to divide by stoichiometric coefficients when comparing rates of different species. Rate of Reaction
Assuming order equals stoichiometric coefficients; order is experimental, not from balanced equation. Rate Law and Order of Reaction
Confusing molecularity with order; molecularity applies only to elementary steps, not overall reactions. Molecularity of a Reaction
Forgetting that half-life depends on initial concentration for zero order (unlike first order). Integrated Rate Law: Zero Order
Using natural log (ln) instead of log base 10; the formula $\log [A] = \log [A]_{0} - \frac{k t}{2.303}$ is common. Integrated Rate Law: First Order
Confusing with first order; second order half-life depends on initial concentration. Integrated Rate Law: Second Order
Assuming order from stoichiometry; always use experimental data. Determining Order from Experimental Data
Forgetting that the excess reactant's concentration is absorbed into the rate constant. Pseudo-First Order Reaction
Using log base 10 instead of natural log; the formula $\log k = \log A - \frac{E_{a}}{2.303 R T}$ is common. Arrhenius Equation
Using Celsius instead of Kelvin for temperature. Arrhenius Equation: Two-Point Form
Thinking E $_{a}$ is the energy difference between reactants and products; it's the barrier height. Activation Energy and Transition State
Assuming the 10°C rule is universal; it's an approximation and depends on E $_{a}$ . Effect of Temperature on Rate
Thinking all collisions lead to reaction; only effective collisions do. Collision Theory
Forgetting that this method works only for reactions with a single reactant or when other reactants are in excess. Half-Life Method to Determine Order

Topics in Chemical Kinetics

Rate of Reaction
Order and Molecularity
Integrated Rate Laws
Arrhenius Equation
Collision Theory
Half-Life

Previous-year questions on Chemical Kinetics

JEEGenie has 329 previous-year questions on Chemical Kinetics. A few worked examples follow.

Year	Questions
2026	27
2025	29
2024	22
2023	24
2022	24
2021	26
2020	16
2019	33
2018	8
2017	6
2016	6
2015	6
2014	12
2013	4
2012	5
2011	2
2010	4
2009	2
2008	2
2007	5
2006	7
2005	7
2004	6
2003	6
2002	8
2001	2
2000	2
1999	2
1998	1
1997	2
1995	2
1994	2
1993	2
1992	2
1991	1
1990	1
1989	1
1987	1
1986	3
1985	2
1984	1
1983	3
1982	1
1981	1

Question 1 easy

t100% is the time required for the 100% completion of the reaction while t1/2 is the time required for 50% of the reaction to be completed. Which of the following option correctly represents the relation between t100% and t1/2 for zero and first order reactions respectively?

$t_{100 %} = (t_{1 / 2})^{2} a n d t_{100 %} = (t_{1 / 2})^{- \infty}$
$t_{100 %} = 2 t_{1 / 2} a n d t_{100 %} = (t_{1 / 2})^{\infty}$
$t_{100 %} = 2 t_{1 / 2} a n d t_{100 %} = (2 t_{1 / 2})^{2}$
$t_{100 %} = (t_{1 / 2})^{\infty} a n d t_{100 %} = 2 t_{1 / 2}$

Answer: B

Explanation. For a **zero-order reaction**:

[A]_{t} = [A]_{0} - k t

t_{1 / 2}

[A]_{t} = \frac{[A]_{0}}{2}

\frac{[A]_{0}}{2} = [A]_{0} - k t_{1 / 2} \Rightarrow t_{1 / 2} = \frac{[A]_{0}}{2 k}

t_{100 %}

[A]_{t} = 0

0 = [A]_{0} - k t_{100 %} \Rightarrow t_{100 %} = \frac{[A]_{0}}{k} = 2 t_{1 / 2}

For a **first-order reaction**:

[A]_{t} = [A]_{0} e^{- k t}

t_{1 / 2}

\frac{[A]_{0}}{2} = [A]_{0} e^{- k t_{1 / 2}} \Rightarrow t_{1 / 2} = \frac{\ln 2}{k}

t_{100 %}

[A]_{t} = 0

requires

e^{- k t} = 0

, which happens only as

t \to \infty

. Hence:

t_{100 %} = \infty = (t_{1 / 2})^{\infty}

Thus the correct relation is:

B

From NTA/IIT papers: JEE Main 2026 · 2 Apr Shift 1

Question 2 hard

Consider the following reaction,

2 H_{2} (g) + 2 NO (g) \to N_{2} (g) + 2 H_{2} O (g)

which follows the mechanism given below:

2 NO (g) \underset{k_{- 1}}{\overset{k_{1}}{⇌}} N_{2} O_{2} (g)

(fast equilibrium);

N_{2} O_{2} (g) + H_{2} (g) \overset{k_{2}}{\to} N_{2} O (g) + H_{2} O (g)

(slow reaction);

N_{2} O (g) + H_{2} (g) \overset{k_{3}}{\to} N_{2} (g) + H_{2} O (g)

(fast reaction). The order of the reaction is ______.

Answer: 3

Explanation. The slow step is rate-determining:

Rate = k_{2} [N_{2} O_{2}] [H_{2}]

From the fast equilibrium:

K = \frac{k_{1}}{k_{- 1}} = \frac{[N_{2} O_{2}]}{[NO]^{2}} \Rightarrow [N_{2} O_{2}] = K [NO]^{2}

Substitute into the rate law:

Rate = k_{2} K [NO]^{2} [H_{2}]

Thus the overall order is:

2 + 1 = 3

3

From NTA/IIT papers: JEE Advanced 2024

Question 3 medium

A molecule undergoes two independent first order reactions whose respective half lives are 12 min and 3 min. If both the reactions are occurring then the time taken for the 50% consumption of the reactant is ___________ min. (Nearest integer)

Answer: 2

Explanation. For two independent first-order reactions occurring simultaneously, the overall rate constant is the sum of the individual rate constants. Given:

t_{1 / 2}^{(1)} = 12

min

\Rightarrow k_{1} = \frac{\ln 2}{12}

min

^{- 1} t_{1 / 2}^{(2)} = 3

min

\Rightarrow k_{2} = \frac{\ln 2}{3}

min

^{- 1}

The overall (effective) rate constant when both reactions occur simultaneously:

k_{total} = k_{1} + k_{2} = \frac{\ln 2}{12} + \frac{\ln 2}{3} = \ln 2 (\frac{1}{12} + \frac{4}{12}) = \frac{5 \ln 2}{12} {min}^{- 1}

The effective half-life of the reactant is given by the harmonic-mean formula:

t_{1 / 2}^{eff} = \frac{\ln 2}{k_{total}} = \frac{t_{1 / 2}^{(1)} \times t_{1 / 2}^{(2)}}{t_{1 / 2}^{(1)} + t_{1 / 2}^{(2)}} = \frac{12 \times 3}{12 + 3} = \frac{36}{15} = 2.4 min

The concentration of the reactant as a function of time is:

[A] = [A]_{0} e^{- (k_{1} + k_{2}) t}

For 50% consumption,

[A] = 0.5 [A]_{0}

0.5 = e^{- (k_{1} + k_{2}) t} ⟹ t = \frac{\ln 2}{k_{1} + k_{2}} = \frac{12}{5} = 2.4 min

Therefore, the time taken for 50% consumption of the reactant is

2.4

min.

From NTA/IIT papers: JEE Main 2023 · 10 Apr Shift 1

Question 4 easy

The half life for the decomposition of gaseous compound

A

240 s

when the gaseous pressure was 500 Torr initially. When the pressure was 250 Torr, the half life was found to be

4.0

min. The order of the reaction is ______________. (Nearest integer)

Answer: 1

Explanation. **Principle:** For a reaction of order

n

, the half-life depends on initial concentration (or pressure) as

t_{1 / 2} \propto [A_{0}]^{1 - n}

**Working:** Given: -

t_{1 / 2, 1} = 240 s

P_{1} = 500

Torr -

t_{1 / 2, 2} = 4.0 min = 240 s

P_{2} = 250

Torr Since the half-life is the same (

240 s

) even though the initial pressure is halved, the half-life is independent of initial concentration. For a first-order reaction,

t_{1 / 2} = \frac{0.693}{k}

which depends only on rate constant

k

, not on initial pressure. Check with the relation:

\frac{t_{1 / 2, 1}}{t_{1 / 2, 2}} = {(\frac{P_{1}}{P_{2}})}^{1 - n}

\frac{240}{240} = {(\frac{500}{250})}^{1 - n}

1 = 2^{1 - n}

1 - n = 0 \Rightarrow n = 1

Thus the reaction is first order. Zero-order would give half-life proportional to initial pressure, and second-order would give inverse dependence, so neither fits the unchanged half-life. Answer: 1

From NTA/IIT papers: JEE Main 2022 · 25 Jul Shift 1

Frequently asked questions

How much of JEE Main comes from Chemical Kinetics?: JEEGenie estimates about 4% of JEE Main marks come from Chemical Kinetics, based on recent papers. The share varies from year to year.
Which formulas matter most in Chemical Kinetics?: The Chemical Kinetics sheet on this page has 18 formula entries, starting with Rate of Reaction, Rate Law and Order of Reaction, Integrated Rate Law: Zero Order, Integrated Rate Law: First Order, Integrated Rate Law: Second Order. Each one comes with when to use it and the trap to avoid.
What topics does Chemical Kinetics cover for JEE?: Chemical Kinetics covers Rate of Reaction, Order and Molecularity, Integrated Rate Laws, Arrhenius Equation, Collision Theory, Half-Life.
How should I revise Chemical Kinetics before the exam?: Learn each formula together with its condition of use, read the trap listed beside it, then practise questions and review every mistake. The practice link on this page opens Chemical Kinetics questions on JEEGenie.
How many previous-year questions are there on Chemical Kinetics?: JEEGenie has 329 previous-year questions on Chemical Kinetics, and this page shows a few with worked explanations.
Is practice on Chemical Kinetics free on JEEGenie?: Sign-up is free and includes a 7-day Pro trial. After the trial, free accounts keep a limited daily quota.