Chapters

Everything NCERT Class 12 Unit 3 (Chemical Kinetics) says that can be asked, plus what JEE Main adds beyond the book (marked Beyond NCERT). Nuclear chemistry is not in NCERT or the JEE Main 2027 syllabus, but 2020–2026 papers still ask radioactive decay as a first-order rate, so §9 covers that much. The short note stays for the last week; this is for a proper revision pass.

Formula sheet

Every formula in the chapter on one screen, for the last pass. Details and examples are in the sections below.

Quantity	Formula	Watch
Rate of aA + bB → cC + dD	−(1/a)Δ[A]/Δt = (1/c)Δ[C]/Δt	Divide by the coefficient
Rate law	rate = k[A]ˣ[B]ʸ; order = x + y	x, y from experiment only
Units of k	(mol L⁻¹)¹⁻ⁿ s⁻¹	0: mol L⁻¹ s⁻¹; 1: s⁻¹; 2: L mol⁻¹ s⁻¹
Zero order	[R] = [R]₀ − kt; t½ = [R]₀/2k	t½ ∝ [R]₀
First order	k = (2.303/t) log([R]₀/[R]); [R] = [R]₀e⁻ᵏᵗ	t½ = 0.693/k
First order, gas	k = (2.303/t) log(pᵢ / (2pᵢ − pₜ))	For A → B + C only
Second order · Beyond NCERT	1/[R] − 1/[R]₀ = kt; t½ = 1/(k[R]₀)	t½ ∝ 1/[R]₀
Any order · Beyond NCERT	t½ ∝ [R]₀¹⁻ⁿ
Arrhenius	k = A e^(−Ea/RT); ln k = ln A − Ea/RT	ln k vs 1/T: slope −Ea/R
Two temperatures	log(k₂/k₁) = (Ea/2.303R)(T₂ − T₁)/(T₁T₂)	R = 8.314 J; Ea in J
Catalyst · Beyond NCERT	ln(k(cat)/k) = ΔEa/RT	Same A
Collision theory	rate = P Z(AB) e^(−Ea/RT)	P = steric factor
Reverse Ea · Beyond NCERT	Ea(back) = Ea(forward) − ΔH
Radioactive decay · Beyond NCERT	N = N₀e^(−λt); t½ = 0.693/λ	First order, any T

0 · What JEE Main asks from this chapter

175 questions in JEE Main 2019–2026, 1 to 2 per paper. Grouped by type (counts are approximate):

Type	Share	Typical question
First order: k, t½, time for x%, two decaying substances, radioactive decay	~30%	75% done in 90 min → time for 60%
Arrhenius: Ea from two k's or a plot, catalyst, reverse Ea	~26%	k rises 5 times from 27 °C to 52 °C → Ea
Rate law from data, order, units of k, molecularity, mechanisms	~23%	Initial-rates table → order in each reactant
Zero order, half-life against [R]₀, graphs	~9%	t½ ∝ p₀ → order
Rate in terms of each species	~6%	2N₂O₅ → 4NO₂ + O₂: rate of NO₂ formation
Gas-phase first order from total pressure	~6%	A(g) → 2B + C, 800 → 1600 mm Hg in 10 min

Most numericals are rounded to the nearest integer, often as "x × 10⁻²". Read the multiplier before you answer.

1 · Rate of a reaction

• Thermodynamics says whether a reaction is feasible (ΔG < 0); equilibrium says how far it goes; kinetics says how fast. Diamond → graphite is feasible but immeasurably slow. (kinesis, Greek: movement.)
• Speeds: ionic reactions are very fast (AgNO₃ + NaCl precipitates at once); rusting is very slow; inversion of cane sugar and hydrolysis of starch are moderate.
• Rate = change in concentration of a reactant or product per unit time. For R → P:

• Δ[R] is negative, hence the minus sign: rate is always positive.
• Average rate is over an interval; it falls as the reaction goes on (butyl chloride hydrolysis: 1.90 × 10⁻⁴ falling to 0.4 × 10⁻⁴ mol L⁻¹ s⁻¹). Instantaneous rate is the slope of the tangent to the concentration–time curve at that moment.
• Units: concentration time⁻¹, e.g. mol L⁻¹ s⁻¹; for gases, atm s⁻¹ (partial pressure ∝ concentration at constant T).

Unequal coefficients: divide each species' rate by its coefficient, so the rate of reaction is one number.

For 5Br⁻ + BrO₃⁻ + 6H⁺ → 3Br₂ + 3H₂O: rate = −(1/5)Δ[Br⁻]/Δt = −Δ[BrO₃⁻]/Δt = −(1/6)Δ[H⁺]/Δt = (1/3)Δ[Br₂]/Δt = (1/3)Δ[H₂O]/Δt.

Example (NCERT 3.2) · 2N₂O₅ → 4NO₂ + O₂; [N₂O₅] 2.33 → 2.08 mol L⁻¹ in 184 min. Rate = ½ × 0.25/184 = 6.79 × 10⁻⁴ mol L⁻¹ min⁻¹ = 4.07 × 10⁻² per hour = 1.13 × 10⁻⁵ per second. Rate of NO₂ formation = 4 × 6.79 × 10⁻⁴ = 2.72 × 10⁻³ mol L⁻¹ min⁻¹.

2 · Rate law, order and rate constant

Rate depends on concentration (pressure for gases), temperature and catalyst.

Rate law (rate equation, rate expression):

• k is the rate constant. x and y may or may not equal the coefficients a and b; they must be found by experiment, never read off the equation.
• Order with respect to A is x; overall order is x + y. It can be 0, 1, 2, 3 or a fraction (even negative). Zero order: rate independent of concentration.

Finding the order from initial rates: pick two runs where only one concentration changes; the rate's ratio is (concentration ratio)^order.

Example (NCERT Table 3.2) · 2NO + O₂ → 2NO₂. Doubling [NO] (0.30 → 0.60) at fixed [O₂] makes the rate 0.096 → 0.384, four times: second order in NO. Doubling [O₂] at fixed [NO] doubles it: first order in O₂. Rate = k[NO]²[O₂].

Reaction	Experimental rate law
2NO + O₂ → 2NO₂	k[NO]²[O₂] (matches coefficients)
CHCl₃ + Cl₂ → CCl₄ + HCl	k[CHCl₃][Cl₂]¹ᐟ²
CH₃COOC₂H₅ + H₂O → CH₃COOH + C₂H₅OH	k[ester]¹[H₂O]⁰

Example (NCERT 3.3) · rate = k[A]¹ᐟ²[B]³ᐟ² → order 2. rate = k[A]³ᐟ²[B]⁻¹ → order ½.

Example (2021) · Elementary 2A + B₂ → 2AB: rate = k[A]²[B₂]. Cut the volume to ⅓ and every concentration triples: rate × 3³ = 27.

Units of k for order n (concentration in mol L⁻¹, time in s):

Order	Units of k
0	mol L⁻¹ s⁻¹
1	s⁻¹
2	L mol⁻¹ s⁻¹

So k = 2.3 × 10⁻⁵ L mol⁻¹ s⁻¹ is second order; k = 3 × 10⁻⁴ s⁻¹ is first order (NCERT 3.4).

3 · Molecularity and mechanism

• Elementary reaction: happens in one step. Complex reaction: a sequence of elementary steps (the mechanism). Complex reactions may be consecutive (ethane → alcohol → aldehyde → acid → CO₂ + H₂O), reversible, or have side reactions (nitration of phenol gives o- and p-nitrophenol).
• Molecularity: the number of species (atoms, ions, molecules) that must collide simultaneously in an elementary step.
  • Unimolecular: NH₄NO₂ → N₂ + 2H₂O.
  • Bimolecular: 2HI → H₂ + I₂.
  • Termolecular: 2NO + O₂ → 2NO₂. Rare and slow; more than three colliding at once is very unlikely.
• KClO₃ + 6FeSO₄ + 3H₂SO₄ → … looks tenth order but is second order: it goes in steps.
• Rate-determining step: the slowest step controls the overall rate (a relay team is as fast as its slowest runner).

Example (NCERT) · 2H₂O₂ → 2H₂O + O₂, catalysed by I⁻ in alkaline medium. Rate = k[H₂O₂][I⁻]. Step 1 (slow): H₂O₂ + I⁻ → H₂O + IO⁻. Step 2: H₂O₂ + IO⁻ → H₂O + I⁻ + O₂. IO⁻ is an intermediate: made and used up, absent from the overall equation.

	Order	Molecularity
Found by	experiment	the mechanism
Values	0, fraction, whole numbers	whole numbers 1, 2, 3 only
Applies to	elementary and complex reactions	elementary steps only
For a complex reaction	given by the slowest step	meaningless overall; that of the slow step equals the order

Fast pre-equilibrium · Beyond NCERT. If an intermediate forms in a fast reversible step before the slow one, replace it using the equilibrium constant.

Example (2023) · 2NO + Br₂ → 2NOBr: NO + Br₂ ⇌ NOBr₂ (fast), NOBr₂ + NO → 2NOBr (slow). Rate = k[NOBr₂][NO] = kK[NO]²[Br₂] → order 3.

Steady state and composite k · Beyond NCERT

• A → B → C with d[B]/dt = 0: [B] = (k₁/k₂)[A] (2019, 2024).
• If k = k₁k₂/k₃, then Ea = Ea₁ + Ea₂ − Ea₃ (2024: 40 + 50 − 60 = 30 kJ). If k = (k₁k₃/k₂)¹ᐟ², Ea = ½(Ea₁ + Ea₃ − Ea₂) (2025).
• Parallel first-order reactions of one reactant: k = k₁ + k₂, so 1/t½ = 1/t½₁ + 1/t½₂ (2023: half-lives 12 and 3 min → 2.4 min).

4 · Integrated rate equations

Zero order (R → P, rate = k):

• [R] against t: straight line, slope −k, intercept [R]₀.
• Uncommon; happens under special conditions: some enzyme-catalysed reactions and reactions on metal surfaces.
• 2NH₃ → N₂ + 3H₂ on hot Pt (1130 K) at high pressure: the surface is saturated, so rate = k[NH₃]⁰ = k. Also HI decomposing on gold.
• t½ ∝ [R]₀: successive half-lives get shorter (each starts from less). The reaction is complete at t = [R]₀/k = 2t½.

Example (2023) · Zero order, t½ = 50 min. [R]₀/k = 100 min; to fall to ¼ means losing ¾[R]₀: t = 0.75 × 100 = 75 min.

First order (rate = k[R]):

• ln[R] against t: straight line, slope −k, intercept ln[R]₀. log([R]₀/[R]) against t: straight line through the origin, slope k/2.303.
• Between two times: k = [2.303/(t₂ − t₁)] log([R]₁/[R]₂).
• t½ is independent of [R]₀. Fraction decomposed by time t = 1 − e⁻ᵏᵗ (2026).
• Examples: hydrogenation of ethene, rate = k[C₂H₄]; decomposition of N₂O₅ and N₂O; all radioactive decay, e.g. ²²⁶Ra → ²²²Rn + ⁴He, rate = k[Ra].

Example (NCERT 3.5) · N₂O₅ from 1.24 × 10⁻² to 0.20 × 10⁻² mol L⁻¹ in 60 min. k = (2.303/60) log 6.2 = 0.0304 min⁻¹.

Example (NCERT 3.7, 3.8) · k = 5.5 × 10⁻¹⁴ s⁻¹ → t½ = 0.693/k = 1.26 × 10¹³ s. 99.9% done: t = (2.303/k) log 1000 = 6.909/k = 10 t½.

First-order shortcuts (t = (2.303/k) log(1/fraction left)):

Done	Time	In half-lives
50%	0.693/k	1
75%	1.386/k	2
87.5%	2.079/k	3
90%	2.303/k	3.32
99%	4.606/k	6.64
99.9%	6.909/k	10

After n half-lives, (½)ⁿ is left.

Example (2020) · 75% in 90 min → t½ = 45 min. 60% done leaves 0.4: t = 45 × log 2.5/log 2 = 45 × 0.40/0.30 = 60 min.

Example (2021) · A (t½ 54 min) and B (t½ 18 min), equal at start. [A]/[B] = 2^(t/18 − t/54) = 2^(t/27) = 16 → t = 108 min.

Gas-phase first order from total pressure. For A(g) → B(g) + C(g) with initial pressure pᵢ and total pₜ: pₜ = pᵢ + x, so p(A) = 2pᵢ − pₜ.

For other equations, redo the bookkeeping. A → 2B + C: pₜ = pᵢ + 2x. 2N₂O₅ → 2N₂O₄ + O₂: pₜ = pᵢ + x with p(N₂O₅) = pᵢ − 2x.

Example (NCERT 3.6) · 2N₂O₅(g) → 2N₂O₄(g) + O₂(g); 0.5 atm, 0.512 atm after 100 s. p(N₂O₅) = 1.5 − 2pₜ = 0.476 atm. k = (2.303/100) log(0.5/0.476) ≈ 4.9 × 10⁻⁴ s⁻¹.

Example (2023) · A(g) → 2B(g) + C(g), 800 → 1600 mm Hg in 10 min. pₜ = 800 + 2x → x = 400 = half of A, so t½ = 10 min. At 30 min, A = 800/8 = 100, x = 700, pₜ = 2200 mm Hg.

Second order · Beyond NCERT (one reactant): 1/[R] − 1/[R]₀ = kt; 1/[R] against t is a straight line of slope k; t½ = 1/(k[R]₀).

Finding the order from half-lives · Beyond NCERT: t½ ∝ [R]₀¹⁻ⁿ. Doubling [R]₀ doubles t½ → zero order; leaves it alone → first; halves it → second.

Example (2022) · t½ = 340 s at 55.5 kPa and 170 s at 27.8 kPa: t½ ∝ p₀ → zero order. (Another 2022 paper: 240 s at 500 Torr and 4.0 min at 250 Torr → unchanged → first order.)

Order	Straight-line plot	t½
0	[R] vs t (slope −k)	[R]₀/2k
1	ln[R] vs t (slope −k)	0.693/k
2 · Beyond NCERT	1/[R] vs t (slope +k)	1/(k[R]₀)

Pseudo first-order reactions: truly higher order, but one reactant is in such excess that its concentration hardly changes.

• Hydrolysis of ethyl acetate with excess water (0.01 mol ester, 10 mol water: water only falls to 9.99 mol): rate = k[CH₃COOC₂H₅].
• Inversion of cane sugar: C₁₂H₂₂O₁₁ + H₂O → glucose + fructose (H⁺), rate = k[C₁₂H₂₂O₁₁].

5 · Temperature and the Arrhenius equation

• Most reactions speed up on heating. N₂O₅'s half-life: 12 min at 50 °C, 5 h at 25 °C, 10 days at 0 °C. KMnO₄ + oxalic acid decolourises faster when warm.
• Rule of thumb: a 10° rise nearly doubles the rate constant.
• First proposed by van 't Hoff; Arrhenius gave the physical meaning.

• A: Arrhenius, frequency or pre-exponential factor; constant for a reaction; same units as k. Ea: activation energy, J mol⁻¹.
• e^(−Ea/RT) is the fraction of molecules with kinetic energy ≥ Ea (2026 asked it as "less than": false).
• Plot ln k against 1/T: straight line, slope −Ea/R, intercept ln A. With log k, slope −Ea/2.303R. The steeper line has the larger Ea.
• Raising T, or lowering Ea, raises k exponentially. A larger Ea means k depends more strongly on T.

The activated complex. H₂ + I₂ → 2HI goes through an unstable intermediate, the activated complex. Ea is the energy needed to form it; energy is released as it breaks into products, so ΔH depends only on reactants and products.

The catalysed path (dashed) has a lower barrier Ea′; reactants, products and ΔH are unchanged.

Energy distribution (Maxwell and Boltzmann). Molecules don't all have the same kinetic energy. The curve of fraction (N(E)/N(total)) against energy peaks at the most probable kinetic energy. On heating, the peak moves to higher energy and the curve broadens and flattens; the area stays constant (total probability 1). The fraction beyond Ea about doubles for a 10° rise.

Example (NCERT 3.9) · k = 0.02 s⁻¹ at 500 K, 0.07 s⁻¹ at 700 K. log 3.5 = 0.544 = Ea × (200/350 000)/(2.303 × 8.314) → Ea = 18.2 kJ mol⁻¹. Then A = 0.02/e^(−18 230/(8.314 × 500)) = 0.02/0.0125 = 1.61 s⁻¹.

Example (NCERT 3.10) · C₂H₅I → C₂H₄ + HI: k = 1.60 × 10⁻⁵ s⁻¹ at 600 K, Ea = 209 kJ mol⁻¹. log(k₂/k₁) = 209 000 × 100/(2.303 × 8.314 × 600 × 700) = 2.599 → k₂ = 6.36 × 10⁻³ s⁻¹ at 700 K.

Example (2021) · k rises 5 times from 27 °C to 52 °C: ln 5 = (Ea/8.314)(25/(300 × 325)) → Ea = 52 kJ mol⁻¹.

Example (2021) · log k = 20.35 − 2470/T: slope = −Ea/2.303R → Ea = 2.303 × 8.314 × 2470 = 47 kJ mol⁻¹.

Exothermic or endothermic · Beyond NCERT: Ea(backward) = Ea(forward) − ΔH. 2021: Ea = 30 kJ, ΔH = −20 kJ → reverse 50 kJ. Every reaction's k rises with T, endothermic or not (2021).

6 · Catalysts

• A catalyst raises the rate without itself being permanently changed: 2KClO₃ → 2KCl + 3O₂ with MnO₂. A substance that slows a reaction is an inhibitor, not a catalyst.
• Intermediate complex theory: the catalyst forms temporary bonds with the reactants; the short-lived intermediate complex breaks into products and the catalyst.
• It gives an alternative path with lower Ea.
• A small amount catalyses a large amount of reactant.
• It does not change ΔG or ΔH. It speeds only spontaneous reactions, never non-spontaneous ones.
• It does not change K: it speeds the forward and backward reactions equally, so equilibrium is reached sooner but sits at the same place.

Catalyst arithmetic · Beyond NCERT (same A):

Example (2022) · Ea lowered by 10 kJ mol⁻¹ at 300 K: ln(k(cat)/k) = 10 000/(8.31 × 300) = 4.0. (2020: an enzyme 10⁶ times faster lowers Ea by 6 × 2.303 RT.)

7 · Collision theory

Developed by Max Trautz and William Lewis (1916–18), from the kinetic theory of gases. Molecules are hard spheres; they react when they collide.

• Collision frequency Z: collisions per second per unit volume.
• Only collisions with enough energy (at least the threshold energy = Ea + the energy the molecules already have) and the right orientation are effective.
• Orientation counts: CH₃Br + OH⁻ → CH₃OH + Br⁻ works only when OH⁻ approaches correctly.

• Comparing with Arrhenius, A is related to collision frequency.
• The first form fits atoms and simple molecules; complex molecules deviate. P, the probability or steric factor, accounts for orientation.
• Drawback: atoms and molecules treated as hard spheres; their structure ignored.

8 · Lab: iodide ion and H₂O₂ · Beyond NCERT

The practical asked in 2022 and 2023: H₂O₂ oxidises I⁻ to I₂; a fixed amount of sodium thiosulphate turns I₂ back to I⁻; once the thiosulphate is used up, I₂ meets starch and turns blue. Time to the blue colour measures the rate. Use fresh starch solution, keep thiosulphate below the KI concentration, and note the time as soon as the blue appears.

9 · Radioactive decay · Beyond NCERT

NCERT says only that all radioactive decay is first order. JEE asks it as a first-order rate question.

• Activity is proportional to the amount, so the activity ratio is the amount ratio.
• The decay constant does not depend on temperature, pressure or chemical form (2025: "decay constant rises with temperature" was the false statement).
• After 3 half-lives, ⅛ is left.

Example (2020) · ⁹⁰Sr, t½ = 6.93 years, to fall by 90%: t = (2.303 × 6.93/0.693) × log 10 = 23.03 years.

Example (2024) · Wood whose ¹⁴C/¹²C is ⅛ of the atmosphere's: 3 half-lives × 5730 = 17 190 years.

Growth laws: bacterial growth N = N₀eᵏᵗ is the same law with a positive sign (2025).

10 · Traps

Trap	What goes wrong	Do this
Rate from one species	Quoting −d[HI]/dt as the rate	Divide by the coefficient
Order from the equation	Using the coefficients as orders	Only experiment gives order (elementary steps excepted)
Molecularity	Saying it can be 0 or ½	Whole numbers 1–3, elementary steps only
t½ for zero order	Treating it as constant	t½ = [R]₀/2k; it shrinks as [R] falls
log vs ln	Using 0.693 with log, or 2.303 twice	ln 2 = 0.693; with log, multiply by 2.303
Arrhenius units	Ea in kJ with R = 8.314 J	Keep J throughout; T in K
ln k vs 1/T slope	Calling the slope +Ea/R	Slope = −Ea/R (−Ea/2.303R with log)
Gas pressures	Using pₜ as the pressure of A	p(A) = 2pᵢ − pₜ for A → B + C; redo for other equations
Catalyst	Saying it changes ΔH, ΔG or K	It changes Ea only; equilibrium is reached sooner
Two decaying substances	Subtracting half-lives	Ratio = 2^(t/t½B − t/t½A)
Volume change, elementary step	Multiplying by the volume factor once	Rate × (factor)^order
Radioactive decay	Letting λ change with T	λ is independent of T

11 · Numbers worth knowing

• ln 2 = 0.693; ln 10 = 2.303; log 2 = 0.301; log 3 = 0.477; log 5 = 0.699; log 7 = 0.845.
• R = 8.314 J K⁻¹ mol⁻¹ (≈ 2 cal K⁻¹ mol⁻¹); 2.303R = 19.15 J K⁻¹ mol⁻¹.
• First order: t₉₉.₉% = 10 t½; t₉₉% = 6.64 t½; t₉₀% = 3.32 t½; t₇₅% = 2 t½; t₉₉.₉% = 3 × t₉₀%.
• Rate roughly doubles per 10 K rise near room temperature.
• ¹⁴C t½ = 5730 years.

12 · For the boards only

JEE Main doesn't ask this, but the CBSE paper can.

• Kinetics answers practical questions: how fast food spoils, how to make a fast-setting dental filling, what controls the burning rate of fuel in an engine.
• "Diamond is forever" only because its conversion to graphite is imperceptibly slow.
• The butyl chloride table (NCERT Table 3.1) and its tangent method for instantaneous rates (5.12 × 10⁻⁵ mol L⁻¹ s⁻¹ at 600 s).