Chemical Kinetics — JEE Main Chemistry Notes & Formulas

1. Chapter at a Glance

Chemical kinetics studies how fast reactions occur and what affects their speed. Thermodynamics tells us whether a reaction can happen; kinetics tells us how quickly it will happen. This chapter covers the rate of reaction, rate laws and order, integrated rate equations (especially first order), half-life and the effect of temperature through the Arrhenius equation. JEE Main almost always asks one question, often a first-order numerical or an Arrhenius calculation.

What JEE asks most

Relating the rates of disappearance and appearance of different species.
Finding order from initial-rate data; units of the rate constant.
First-order integrated law and half-life calculations.
Arrhenius equation: activation energy from rate constants at two temperatures.

2. Rate of Reaction

For aA + bB → cC + dD, the rate of reaction is the same whichever species we use, once divided by its coefficient:

Rate = −(1/a) d[A]/dt = −(1/b) d[B]/dt = (1/c) d[C]/dt = (1/d) d[D]/dt

Average rate = Δ[concentration]/Δt; instantaneous rate is the slope of the concentration–time curve at that moment.
Rate depends on concentration, temperature, catalyst, surface area and (for some reactions) light.
Unit of rate: mol L⁻¹ s⁻¹.

3. Rate Law, Order and Molecularity

The rate law is found by experiment: Rate = k[A]^x[B]^y. The order is x + y; it can be zero, fractional or even negative. The powers need not equal the stoichiometric coefficients.

Feature	Order	Molecularity
Definition	Sum of powers in the experimental rate law	Number of species colliding in an elementary step
Values	0, fractional, integer, negative	Whole numbers only (1, 2, rarely 3)
Found by	Experiment	Theory, for elementary steps
Complex reaction	Defined for the overall reaction	Defined only for each step

For a complex reaction, the slowest step (rate-determining step) decides the rate law. A pseudo first-order reaction looks first order because one reactant is in large excess (e.g. acid hydrolysis of ester with excess water, inversion of cane sugar).

Units of k

For an n^th-order reaction: unit of k = (mol L⁻¹)¹⁻ⁿ s⁻¹. Zero order: mol L⁻¹ s⁻¹; first: s⁻¹; second: L mol⁻¹ s⁻¹.

4. Integrated Rate Equations

Order	Integrated law	Half-life	Straight-line plot
Zero	[A] = [A]₀ − kt	[A]₀/2k	[A] vs t (slope −k)
First	k = (2.303/t) log([A]₀/[A])	0.693/k	ln[A] vs t (slope −k)
Second (one reactant)	1/[A] − 1/[A]₀ = kt	1/(k[A]₀)	1/[A] vs t (slope k)

First-order half-life does not depend on the initial concentration. After n half-lives, the fraction remaining is (1/2)ⁿ.
Time for 75% completion = 2 half-lives; 87.5% = 3; 99.9% ≈ 10 half-lives.
Radioactive decay is always first order.
Zero-order examples: decomposition of NH₃ on a hot platinum surface; photochemical H₂ + Cl₂.
General: t½ ∝ [A]₀¹⁻ⁿ.

5. Effect of Temperature: Arrhenius Equation

For most reactions, the rate roughly doubles for every 10 °C rise in temperature. Arrhenius explained this:

k = A e^−Ea/RT log(k₂/k₁) = (Ea/2.303R)(1/T₁ − 1/T₂)

E_a = activation energy (minimum extra energy needed to form the activated complex); A = frequency factor.
A plot of ln k against 1/T is a straight line with slope −E_a/R.
e^−Ea/RT is the fraction of molecules with energy ≥ E_a.
Collision theory: rate = PZ_ABe^−Ea/RT, where P is the steric (orientation) factor.

A catalyst provides a new path with lower activation energy. It does not change ΔH or the equilibrium constant.

Role of a catalyst

A catalyst lowers E_a for both the forward and backward reactions equally, so it speeds up both and helps reach equilibrium faster. It is not consumed and does not change ΔH, ΔG or K.

6. Formula Master Table

Quantity	Formula
Rate relation	−(1/a)d[A]/dt = (1/c)d[C]/dt
Units of k (order n)	(mol L⁻¹)¹⁻ⁿ s⁻¹
Zero order	[A] = [A]₀ − kt; t½ = [A]₀/2k
First order	k = (2.303/t) log([A]₀/[A]); t½ = 0.693/k
Fraction left after n half-lives	(1/2)ⁿ
Arrhenius	k = Ae^−Ea/RT
Two temperatures	log(k₂/k₁) = (Ea/2.303R)(T₂ − T₁)/(T₁T₂)
Temperature coefficient	k_T+10/k_T ≈ 2 to 3

7. Mind Map

Chemical Kinetics

Rate

Average vs instantaneous
Stoichiometric relation

Rate law

Order vs molecularity
Units of k
Rate-determining step

Integrated laws

Zero, first, second
Half-life
Linear plots

Temperature

Arrhenius
Collision theory
Catalysis

8. Tips and Tricks

Shortcuts

From initial-rate data: if doubling [A] multiplies rate by 2^x, the order in A is x (×2 → 1, ×4 → 2, ×1 → 0, ×√2 → ½).
For first-order "percentage" questions, try half-lives first: 50% → 1, 75% → 2, 87.5% → 3, 93.75% → 4.
t_99% = 2 × t_90% for a first-order reaction.
Check units of k given in the question — they reveal the order immediately.

Mistakes students make

Using stoichiometric coefficients as orders for a non-elementary reaction.
Forgetting to convert E_a into J mol⁻¹ when R = 8.314 J K⁻¹ mol⁻¹.
Thinking a catalyst changes the equilibrium position.

9. Important One-liners

Rate constant k changes only with temperature (and catalyst), not with concentration.
Molecularity is never zero or fractional.
Half-life of a zero-order reaction is proportional to the initial concentration.
Half-life of a first-order reaction is independent of the initial concentration.
At very high temperature (T → ∞), k approaches A.
Effective collisions need both enough energy and the correct orientation.

10. PYQ-pattern Questions with Solutions

These questions are original and modelled on concepts frequently tested in JEE Main. (log 2 = 0.301, R = 8.314 J K⁻¹ mol⁻¹.)

Q1Easy

The unit of the rate constant of a second-order reaction is:

s⁻¹
mol L⁻¹ s⁻¹
L mol⁻¹ s⁻¹
L² mol⁻² s⁻¹

Show solution

Unit = (mol L⁻¹)¹⁻² s⁻¹ = L mol⁻¹ s⁻¹.

Answer: (C)

Q2Numerical

A first-order reaction has a half-life of 10 min. How many minutes are needed for 75% of the reactant to react?

Show solution

75% reacted means 25% = (1/2)² remains, i.e. 2 half-lives = 20 min.

Answer: 20

Q3Medium

For 2N₂O₅ → 4NO₂ + O₂, the rate of disappearance of N₂O₅ is 2 × 10⁻⁴ mol L⁻¹ s⁻¹. The rate of formation of NO₂ is:

1 × 10⁻⁴
2 × 10⁻⁴
4 × 10⁻⁴
8 × 10⁻⁴

Show solution

−(1/2)d[N₂O₅]/dt = (1/4)d[NO₂]/dt ⇒ d[NO₂]/dt = 2 × (2 × 10⁻⁴) = 4 × 10⁻⁴ mol L⁻¹ s⁻¹.

Answer: (C)

Q4Medium

For A + B → products, doubling [A] (with [B] fixed) doubles the rate, and doubling [B] (with [A] fixed) makes the rate four times. The overall order is:

Show solution

Order in A = 1 (2¹ = 2); order in B = 2 (2² = 4). Rate = k[A][B]², overall order 3.

Answer: (C)

Q5Hard

The rate constant of a reaction doubles when temperature rises from 300 K to 310 K. The activation energy is about:

26.8 kJ mol⁻¹
53.6 kJ mol⁻¹
107 kJ mol⁻¹
5.36 kJ mol⁻¹

Show solution

log 2 = (Ea/2.303R) × (10/(300 × 310)).

Ea = 0.301 × 2.303 × 8.314 × 9300 ≈ 53600 J mol⁻¹ ≈ 53.6 kJ mol⁻¹.

Answer: (B)

Q6Easy

A catalyst increases the rate of a reaction by:

Increasing ΔH
Lowering the activation energy
Increasing the equilibrium constant
Increasing the energy of products

Show solution

A catalyst provides an alternative path with lower E_a. It does not change ΔH or K.

Answer: (B)

Q7Hard

For a first-order reaction, the time required for 99.9% completion is approximately how many times its half-life?

Show solution

t_99.9% = (2.303/k) log(1000) = 6.909/k; t½ = 0.693/k. Ratio = 6.909/0.693 ≈ 10.

Answer: (C)

Q8Numerical

A first-order reaction is 50% complete in 20 min. What percentage of the reactant remains after 60 min?

Show solution

60 min = 3 half-lives. Fraction remaining = (1/2)³ = 1/8 = 12.5%.

Answer: 12.5

11. Quick Revision Summary

Remember before the exam

Rate = −(1/a)d[A]/dt; order is experimental, molecularity is theoretical.
Unit of k = (mol L⁻¹)¹⁻ⁿ s⁻¹.
First order: k = (2.303/t) log([A]₀/[A]), t½ = 0.693/k, independent of [A]₀.
Zero order: t½ = [A]₀/2k.
Arrhenius: k = Ae^−Ea/RT; slope of ln k vs 1/T = −Ea/R.
Catalyst lowers E_a; ΔH and K unchanged.

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