Chemical Kinetics

Kinetics describes how fast reactions go. Advanced tests rate laws and order determination, integrated first-order (and half-life) equations, the Arrhenius temperature dependence, and mechanism/rate-determining-step reasoning. Numerical work on half-lives and activation energy is standard.

Key results

• Rate law: rate = k[A]^m[B]^n; order = m + n (experimental, not from stoichiometry). Molecularity is mechanistic.
• Zero order: [A] = [A]0 − kt; t½ = [A]0/2k. First order: ln[A] = ln[A]0 − kt; k = (2.303/t)log([A]0/[A]); t½ = 0.693/k (independent of concentration).
• Arrhenius: k = A e^(−Ea/RT); ln(k2/k1) = (Ea/R)(1/T1 − 1/T2). Rate roughly doubles per 10°C rise.
• Mechanism: the slowest (rate-determining) step sets the rate law; catalysts lower Ea.

Order	Integrated form	t½
Zero	[A]0 − kt	[A]0/2k
First	ln([A]0/[A]) = kt	0.693/k
Second	1/[A] − 1/[A]0 = kt	1/(k[A]0)

Exam Tricks & Tips

• 🎯 First-order half-life is independent of concentration (0.693/k) — the signature test to identify first order from data.
• 🎯 Order is determined experimentally, never assumed from the balanced equation; use the method of initial rates.
• 🎯 Arrhenius: a plot of ln k vs 1/T is a straight line with slope −Ea/R — read activation energy from the slope.
• 🎯 A catalyst lowers Ea, speeding both forward and reverse rates equally, so it does not shift equilibrium, only reaches it faster.
• 🎯 The rate law follows the slowest (rate-determining) step; intermediates are eliminated using fast pre-equilibria.
• ❌ Common mistake: confusing order (empirical) with molecularity (mechanistic, always a small whole number for an elementary step) — they can differ for overall reactions.

Expected exam pattern

1–2 questions: a first-order half-life/integrated-rate numerical and an Arrhenius activation-energy or mechanism/order problem. Half-life ratio integers are common.

Quick recap

rate = k[A]^m[B]^n; order is experimental. First-order t½ = 0.693/k, concentration-independent. Arrhenius k = Ae^(−Ea/RT); slope of ln k vs 1/T gives −Ea/R. Rate law = slowest step; catalyst lowers Ea only.

Cover the answer, recall, then check. 12 cards on chemical kinetics for JEE Advanced.

Q1. How is the order of a reaction determined?
A1. Experimentally (e.g. initial-rates method); it is not read off the balanced equation.

Q2. Integrated first-order rate equation.
A2. k = (2.303/t) log([A]0/[A]); ln[A] falls linearly with time.

Q3. Half-life of a first-order reaction.
A3. t½ = 0.693/k, independent of initial concentration.

Q4. Half-life of a zero-order reaction.
A4. t½ = [A]0/2k (depends on initial concentration).

Q5. Arrhenius equation.
A5. k = A e^(−Ea/RT).

Q6. How to get Ea from rate data at two temperatures.
A6. ln(k2/k1) = (Ea/R)(1/T1 − 1/T2).

Q7. Slope of a ln k vs 1/T plot.
A7. −Ea/R.

Q8. Effect of a catalyst on activation energy and equilibrium.
A8. Lowers Ea (speeds both directions); does not change the equilibrium position.

Q9. Difference between order and molecularity.
A9. Order is experimental (can be fractional); molecularity is the number of species in an elementary step (whole number).

Q10. What sets the overall rate law of a multistep reaction?
A10. The slowest (rate-determining) step.

Q11. Integrated second-order rate law.
A11. 1/[A] − 1/[A]0 = kt.

Q12. Rule of thumb for rate vs temperature.
A12. Rate roughly doubles for every 10°C rise (temperature coefficient ≈ 2).

Kinetics answers "how fast", complementing thermodynamics' "how far". JEE Advanced tests rate laws, the integrated forms (especially first order), the Arrhenius temperature dependence, and mechanism reasoning — with a recurring emphasis on determining order from data.

Core concept: reaction rate depends on concentration through a rate law rate = k[A]^m[B]^n whose orders are found experimentally, and on temperature through the Arrhenius equation.

Core ideas

Beginner — rate and order

Rate = −(1/a)d[A]/dt = k[A]^m[B]^n. The order (m+n) is experimental, not from stoichiometry. Units of k reveal the order: zero order mol L⁻¹ s⁻¹, first order s⁻¹, second order L mol⁻¹ s⁻¹. Molecularity (number of species in an elementary step) is a theoretical integer, distinct from order.

Intermediate — integrated rate laws

• Zero order: [A] = [A]₀ − kt; half-life t½ = [A]₀/2k (depends on initial concentration).
• First order: ln[A] = ln[A]₀ − kt, or k = (2.303/t)log([A]₀/[A]); half-life t½ = 0.693/k, independent of concentration — the hallmark of first order. Radioactive decay is first order.
• Second order: 1/[A] = 1/[A]₀ + kt.
  A constant half-life across the reaction is the fastest way to identify first order from data.

Advanced — Arrhenius, mechanisms, and catalysis

The Arrhenius equation k = A e^(−Ea/RT): plotting ln k vs 1/T gives slope −Ea/R. A rule of thumb: rate roughly doubles per 10 °C rise. The rate-determining step (slowest elementary step) sets the observed rate law; intermediates are eliminated using the steady-state or pre-equilibrium approximation. A catalyst lowers Ea (provides an alternative path), speeding both forward and reverse rates equally without changing ΔH or K. Collision theory and the Maxwell–Boltzmann tail (fraction of molecules exceeding Ea) explain the exponential T-dependence.

Worked example

A first-order reaction is 75% complete in 60 minutes. Find its half-life. For first order, k = (2.303/t)log([A]₀/[A]). At 75% completion, [A]/[A]₀ = 0.25, so k = (2.303/60)log(1/0.25) = (2.303/60)log4 = (2.303/60)(0.602) = 0.0231 min⁻¹. Then t½ = 0.693/k = 0.693/0.0231 ≈ 30 min. Sanity check: 75% complete = two half-lives (100→50→25%), and 60/2 = 30 min. ✓ Recognising "75% done = 2 half-lives" gives the answer in seconds.

How JEE Advanced tests this

Determining order and rate constant from initial-rate or concentration–time data; first-order half-life and fraction-remaining problems; Arrhenius Ea from two temperatures or a graph; mechanism problems (deriving the rate law from a proposed mechanism via the slow step); and catalyst/energy-profile interpretation.

Exam tricks & shortcuts

• First-order half-life 0.693/k is concentration-independent — use it to spot first order.
• Units of k reveal the order at a glance.
• Fraction remaining after n half-lives = (½)ⁿ.
• Arrhenius two-point: ln(k₂/k₁) = (Ea/R)(1/T₁ − 1/T₂).

Key formulas

• Rate law: rate = k[A]ˣ[B]ʸ; overall order = x + y.
• Zero order: [R] = [R]₀ − kt; t½ = [R]₀/2k.
• First order: k = (2.303/t)log([R]₀/[R]); t½ = 0.693/k.
• Arrhenius: k = A e^(−Ea/RT); log(k₂/k₁) = (Ea/2.303R)(1/T₁ − 1/T₂).
• k units: zero mol L⁻¹s⁻¹, first s⁻¹, second L mol⁻¹s⁻¹.

Insert integrated rate laws to find k and half-life; Arrhenius describes the temperature dependence of k.

Worked Example

Problem: The rate constant of a reaction doubles when the temperature is raised from 300 K to 310 K. Using the Arrhenius equation, estimate the activation energy E_a. (R = 8.314 J/mol·K.)

Solution:
The Arrhenius equation in two-temperature form:
ln(k₂/k₁) = (E_a/R)(1/T₁ − 1/T₂).

Given k₂/k₁ = 2, T₁ = 300 K, T₂ = 310 K:
1/T₁ − 1/T₂ = 1/300 − 1/310 = (310 − 300)/(300 × 310) = 10/93000 = 1.075 × 10⁻⁴ K⁻¹.

So:
ln 2 = (E_a/8.314)(1.075 × 10⁻⁴)
0.693 = E_a × (1.293 × 10⁻⁵)
E_a = 0.693 / (1.293 × 10⁻⁵) ≈ 5.36 × 10⁴ J/mol ≈ 53.6 kJ/mol.

Answer: Activation energy E_a ≈ 53.6 kJ/mol.