In this guide
Chemical kinetics rewards students who keep a few equations straight. JEE Main questions come in three shapes: find the order from a table of initial rates, use an integrated equation or half-life, and find the activation energy from rate constants at two temperatures. Most are quick if you know which order you are dealing with and use logarithms carefully.
The official 2026 syllabus lists the rate of reaction and the factors affecting it, elementary and complex reactions, order and molecularity, rate law, rate constant and its units, the differential and integrated forms of zero- and first-order reactions with their half-lives, the effect of temperature, Arrhenius theory, activation energy and its calculation, and collision theory of bimolecular gaseous reactions (no derivation). The practical unit adds a kinetic study of iodide ions with hydrogen peroxide.
Rate of reaction
For aA + bB → cC + dD, the rate is the same whichever species you track, once you divide by coefficients:
rate = −(1/a) d[A]/dt = −(1/b) d[B]/dt = (1/c) d[C]/dt = (1/d) d[D]/dt
For 2N₂O₅ → 4NO₂ + O₂: −½ d[N₂O₅]/dt = ¼ d[NO₂]/dt = d[O₂]/dt.
Rate depends on concentration (or pressure for gases), temperature, the catalyst and, for solids, surface area.
Rate law, order and molecularity
The rate law is found by experiment: rate = k[A]ˣ[B]ʸ. The overall order is x + y. It need not match the coefficients in the balanced equation.
| Feature | Order | Molecularity |
|---|---|---|
| Defined for | Any reaction (from experiment) | An elementary step only |
| Values | 0, whole number or fraction | 1, 2 or rarely 3; never 0 or fractional |
| Found by | Experiment | The mechanism |
For a complex reaction, the slowest step controls the rate, and its molecularity sets the order. A pseudo-first-order reaction is really second order, but one reactant is in such excess that its concentration barely changes. Acid hydrolysis of ethyl acetate in water and inversion of cane sugar are the standard examples.
Units of k for order n: (mol L⁻¹)¹⁻ⁿ s⁻¹.
| Order | Units of k |
|---|---|
| 0 | mol L⁻¹ s⁻¹ |
| 1 | s⁻¹ |
| 2 | L mol⁻¹ s⁻¹ |
| 3 | L² mol⁻² s⁻¹ |
Zero- and first-order reactions
| Feature | Zero order | First order |
|---|---|---|
| Rate law | rate = k | rate = k[A] |
| Integrated form | [A] = [A]₀ − kt | k = (2.303/t) log([A]₀/[A]) |
| Straight-line plot | [A] vs t, slope −k | ln[A] vs t, slope −k |
| Half-life | t½ = [A]₀/(2k) | t½ = 0.693/k |
| t½ depends on [A]₀? | Yes, proportional | No |
| Examples | NH₃ decomposing on hot platinum at high pressure | Radioactive decay; N₂O₅ decomposition |
A zero-order reaction finishes completely at t = [A]₀/k. A first-order reaction never quite finishes, but some milestones are worth knowing:
- 50% done in one t½, 75% in two, 87.5% in three
- 99.9% done in about 10 half-lives (t = 6.909/k)
- For gas-phase A(g) → B(g) + C(g) at constant volume, with initial pressure p₀ and total pressure p_t: k = (2.303/t) log[p₀/(2p₀ − p_t)]
Temperature, Arrhenius and collision theory
k = A exp(−Ea/RT), so log k = log A − Ea/(2.303RT). A plot of log k against 1/T is a straight line with slope −Ea/(2.303R).
For two temperatures:
log(k₂/k₁) = [Ea/(2.303R)] × [(T₂ − T₁)/(T₁T₂)]
Collision theory says a reaction needs collisions with energy at least Ea and the right orientation: rate = P Z_AB exp(−Ea/RT), where P is the steric (probability) factor.
A catalyst gives an alternative path with lower Ea. It speeds up forward and backward reactions equally, so it does not change ΔH, ΔG or the equilibrium constant.
Worked problems
Problem 1: order from initial rates.
| Run | [A] (M) | [B] (M) | Rate (M s⁻¹) |
|---|---|---|---|
| 1 | 0.1 | 0.1 | 2 × 10⁻³ |
| 2 | 0.2 | 0.1 | 8 × 10⁻³ |
| 3 | 0.1 | 0.2 | 4 × 10⁻³ |
Runs 1 and 2: doubling [A] quadruples the rate, so the order in A is 2. Runs 1 and 3: doubling [B] doubles the rate, so the order in B is 1.
Rate = k[A]²[B], overall order 3. k = 2 × 10⁻³/(0.01 × 0.1) = 2 L² mol⁻² s⁻¹.
Problem 2: first-order milestones. A first-order reaction is 75% complete in 60 min. Find t½, k and the time for 90% completion.
75% is two half-lives, so t½ = 30 min and k = 0.693/30 = 0.0231 min⁻¹.
For 90%: t = (2.303/0.0231) log 10 = 99.7 min, about 100 min.
Problem 3: activation energy. A rate constant doubles when the temperature rises from 300 K to 310 K. Find Ea (R = 8.314 J K⁻¹ mol⁻¹).
log 2 = 0.301 = [Ea/(2.303 × 8.314)] × (10/93,000).
Ea = 0.301 × 19.147 × 9,300 = 53,600 J = 53.6 kJ mol⁻¹.
Problem 4 (numerical answer): zero order. In a zero-order reaction, [A] falls from 0.5 M to 0.3 M in 20 min. Find the half-life (for [A]₀ = 0.5 M) in minutes.
k = (0.5 − 0.3)/20 = 0.01 M min⁻¹. t½ = [A]₀/(2k) = 0.5/0.02 = 25.
Traps to watch
- Order from the balanced equation. Only for elementary steps. Otherwise use data.
- Natural and common logs. 2.303 converts ln to log; do not use it twice.
- Kelvin. Arrhenius temperatures must be in K.
- Rate of disappearance and rate of reaction. Divide by the coefficient.
- Zero-order half-life depends on [A]₀; do not use 0.693/k.
- A catalyst and equilibrium. It shortens the time to reach equilibrium but does not shift it.
Practice set
- Give the units of k for a second-order reaction.
- Find the half-life of a first-order reaction with k = 0.0231 min⁻¹.
- For 2N₂O₅ → 4NO₂ + O₂, N₂O₅ disappears at 0.02 M s⁻¹. Find the rates of formation of NO₂ and O₂.
- What fraction of a first-order reactant is left after three half-lives?
- A plot of log k against 1/T has slope −5,000 K. Find Ea.
- A zero-order reaction has [A]₀ = 0.4 M and k = 0.01 M s⁻¹. When is it complete?
- Doubling [A]₀ doubles the half-life. What is the order?
- Why can molecularity not be zero or fractional?
Answers
- L mol⁻¹ s⁻¹.
- 0.693/0.0231 = 30 min.
- NO₂: 0.04 M s⁻¹; O₂: 0.01 M s⁻¹.
- (½)³ = 1/8 (12.5%).
- Ea = 2.303 × 8.314 × 5,000 = 95,700 J ≈ 95.7 kJ mol⁻¹.
- t = [A]₀/k = 40 s.
- Zero order (t½ ∝ [A]₀).
- It counts the molecules colliding in one elementary step, so it must be a whole number of at least 1.
What to do next
- Make one table of zero- and first-order formulas and plots and revise it until you can write it from memory.
- Solve five initial-rate tables and five Arrhenius problems against the clock.
- Revise chemical equilibrium for how a catalyst affects K, and try mixed sets in the physical chemistry numericals guide.
A note on dates and numbers. Exam patterns, vacancies and schedules change from year to year. Always confirm the current details in the latest notification on the National Testing Agency website .
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