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Chemical Kinetics

Study the speed of reactions and what affects it: concentration, temperature, and catalysts. Covers rate law, order of reaction, the Arrhenius equation, and reaction mechanisms step by step.

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Reading time~8 min
Revision time~3 min
Last updated2026-07-19
1 Read the chapter ~8 min

🎯 Key Points

  • Rate Law: Rate = k[A]^m[B]^n — m, n found experimentally, NOT from balanced equation coefficients
  • Order can be zero/fractional/whole; Molecularity is always a whole number (≥1) and only applies to elementary steps
  • First-order half-life t½ = 0.693/k is CONSTANT, independent of starting concentration — a key identifying feature
  • Arrhenius equation: k = Ae^(−Ea/RT); higher T or lower Ea means faster reaction
  • Pseudo-first-order: a higher-order reaction that behaves as first-order because one reactant is in large excess
  • For multi-step reactions, the overall rate = rate of the slowest (rate-determining) step
First-Order Decay: Constant Half-Lifetime[A][A]₀[A]₀/22 × t½[A]₀/4Each successive half-life takes the SAME amount of time, regardless of starting concentration

First-order decay curve: concentration halves every t½, and that half-life is constant — a defining test for first-order kinetics.

Rate of Reaction

Rate measures how fast reactants are consumed or products are formed per unit time.

Rate = -Δ[reactant]/Δt = +Δ[product]/Δt

Factors Affecting Rate

  • Concentration: More concentration = more collisions = faster rate
  • Temperature: Higher temperature gives more energy to molecules
  • Catalyst: Provides an alternative lower-energy pathway
  • Surface area: More surface = more reaction sites

Rate Law

Rate = k[A]^m[B]^n

  • k = rate constant; m and n are determined experimentally (not from stoichiometry)
  • Overall order = m + n

Half-Life

  • First-order: t½ = 0.693/k (constant; does not depend on concentration)
  • Zero-order: t½ = [A]₀/2k

Arrhenius Equation

k = A × e^(-Ea/RT)

  • Ea = activation energy (minimum energy needed for reaction)
  • log(k₂/k₁) = (Ea/2.303R) × (1/T₁ - 1/T₂)
Maxwell-Boltzmann distribution curves of molecular speeds at 100 K, 1200 K and 3000 K; higher temperature broadens and flattens the curve and shifts the peak to higher speed

The Maxwell–Boltzmann distribution of molecular speeds at three temperatures. Raising the temperature broadens and flattens the curve and shifts it to higher speeds, so a much larger fraction of molecules has energy exceeding the activation energy — the reason reaction rate rises steeply with temperature. Image: MikeRun, CC BY-SA 4.0, via Wikimedia Commons.

Quick Tips

  • A catalyst lowers activation energy without being consumed
  • Every 10°C rise roughly doubles the reaction rate
  • Half-life of first-order reactions is independent of initial concentration

Order vs Molecularity

  • Order of reaction: Sum of powers of concentration terms in the experimental rate law; can be zero, fractional, or a whole number
  • Molecularity: Number of reacting species colliding simultaneously in an elementary step; always a whole number, never zero or fractional
  • For elementary reactions, order = molecularity; for complex multi-step reactions, order is decided by the slowest (rate-determining) step

Integrated Rate Equations

  • Zero order: [A] = [A]₀ - kt; a plot of [A] vs t is a straight line with negative slope k
  • First order: k = (2.303/t) log([A]₀/[A]t); a plot of log[A] vs t gives a straight line
  • Units of k: zero order = mol L⁻¹ s⁻¹; first order = s⁻¹ (independent of concentration units)

Pseudo-First-Order Reactions

A reaction that is actually higher order but behaves as first order because one reactant is present in large excess, so its concentration stays effectively constant.

  • Classic example: acidic hydrolysis of ethyl acetate (water is in large excess): CH₃COOC₂H₅ + H₂O → CH₃COOH + C₂H₅OH
  • Inversion of cane sugar in dilute acid is another textbook example

Methods to Determine Order of Reaction

  • Initial rate method: Measure rate at different initial concentrations, keeping others constant
  • Integrated rate law method: Test which integrated equation gives a constant k for the data
  • Half-life method: t½ independent of concentration indicates first order; t½ ∝ 1/[A]₀ⁿ⁻¹ for other orders
  • Van't Hoff differential method: Uses log(rate) vs log(concentration) plots; slope gives the order

Collision Theory

  • Reaction occurs only when molecules collide with sufficient energy (activation energy) and proper orientation
  • Rate = pZAB e^(-Ea/RT), where Z is collision frequency and p is the steric (probability) factor
  • Threshold energy = activation energy + average kinetic energy of reactants

Average vs Instantaneous Rate

  • Average rate = change in concentration over a measurable time interval (Δ[X]/Δt); it is only an approximation because the rate changes continuously
  • Instantaneous rate = rate at a particular instant, found as the slope of the tangent to the concentration–time curve (the limit of the average rate as Δt → 0)
  • For a reaction aA + bB → cC + dD, the unique rate = −(1/a)d[A]/dt = −(1/b)d[B]/dt = +(1/c)d[C]/dt = +(1/d)d[D]/dt

Reaction Mechanism and Rate-Determining Step

  • Most reactions occur through a sequence of elementary steps (the mechanism); the sum of these steps gives the overall balanced equation
  • The rate-determining step is the slowest step; the overall rate law is governed by it and by any steps before it
  • This is why the experimental order need not match the stoichiometric coefficients — those come from the overall equation, not the slow step
  • Reaction intermediates are produced and consumed within the mechanism and do not appear in the overall equation

Effect of Temperature: Temperature Coefficient

  • The temperature coefficient is the ratio of rate constants over a 10°C rise: k(T+10)/k(T), usually between 2 and 3 (rate roughly doubles or triples per 10°C)
  • Raising temperature increases the fraction of molecules with energy ≥ Ea (the tail of the Maxwell–Boltzmann distribution), so many more effective collisions occur
  • The Arrhenius plot of ln k vs 1/T is a straight line of slope −Ea/R, giving a graphical route to the activation energy

Activation Energy and the Role of a Catalyst

Energy Profile: With and Without CatalystEnergyReaction progressreactantsproductsEa (uncatalysed)Ea (catalysed, lower)

A catalyst provides an alternative pathway with a lower activation energy, so more collisions succeed and the rate rises. It lowers Ea for BOTH forward and backward reactions equally, so it does not shift the position of equilibrium — only speeds up its attainment.

  • A catalyst is not consumed and does not change ΔH or the equilibrium constant; it only lowers Ea
  • Because Ea drops for forward and reverse reactions alike, both are accelerated equally

Units of the Rate Constant

OrderRate lawUnits of k
ZeroRate = kmol L⁻¹ s⁻¹
FirstRate = k[A]s⁻¹
SecondRate = k[A]²L mol⁻¹ s⁻¹

General rule: units of k = (mol L⁻¹)^(1−order) s⁻¹, so the units themselves reveal the overall order.

🚀 JEE Advanced Edge

Steric factor and orientation: The steric factor p in collision theory accounts for the fact that molecules must collide with the CORRECT orientation, not just sufficient energy — this is why collision theory alone often overestimates rates for complex molecules, and why p < 1 for most real reactions (only a fraction of correctly-energised collisions are also correctly oriented).

Parallel and consecutive reactions: When a reactant can form two different products via two competing pathways (parallel/side reactions), the ratio of products formed equals the ratio of their individual rate constants, NOT the ratio of activation energies directly — useful for selectivity problems in synthesis.

Graphical determination tricks: A straight line on a log[A] vs t plot confirms first order; a straight line on [A] vs t confirms zero order; a straight line on 1/[A] vs t confirms second order. Recognising which linearized plot is given in a JEE graph question is the fastest way to identify the order without doing any calculation.

Worked problem: A first-order reaction is 50% complete in 20 minutes. How long will it take to be 87.5% complete? Approach: 87.5% complete means 12.5% remaining = (1/2)³ of the original, i.e., exactly 3 half-lives have passed. Since each half-life = 20 min, total time = 3 × 20 = 60 minutes.

2 Revise ~3 min before the exam

📐 Formula Sheet

  • Rate: −(1/a)d[A]/dt = (1/p)d[P]/dt for aA → pP
  • Rate law: rate = k[A]ˣ[B]ʸ; overall order = x + y
  • Zero order: [A] = [A]₀ − kt  |  t½ = [A]₀/2k
  • First order: k = (2.303/t)·log([A]₀/[A])  |  t½ = 0.693/k (independent of concentration)
  • Units of k: zero order mol L⁻¹ s⁻¹; first order s⁻¹; second order L mol⁻¹ s⁻¹
  • Arrhenius: k = A·e−Ea/RT  |  log(k₂/k₁) = (Ea/2.303R)(1/T₁ − 1/T₂)
  • Rule of thumb: rate roughly doubles for every 10°C rise
  • Catalyst: lowers Ea, speeding both directions equally; it does not shift equilibrium
3 Practice apply it

✍️ Worked Examples

Example 1 — First-order half-life
Q: A first-order reaction has k = 0.0231 min⁻¹. Find its half-life.
Step 1 — For first order, t½ = 0.693/k.
Step 2 — Substitute: t½ = 0.693/0.0231.
Step 3 — Compute: 30 min.
Answer: 30 min. Key idea: for first order the half-life is constant — it never depends on the starting concentration.

Example 2 — Determining reaction order
Q: When [A] doubles, the rate quadruples. What is the order with respect to A?
Step 1 — Rate ∝ [A]ⁿ, so doubling [A] multiplies rate by 2ⁿ.
Step 2 — We are told the factor is 4: 2ⁿ = 4.
Step 3 — Solve: n = 2.
Answer: second order in A. Note: if the rate had doubled, n = 1; if unchanged, n = 0.

Example 3 — Time for a first-order reaction
Q: How long for a first-order reaction (k = 0.0693 s⁻¹) to be 75% complete?
Step 1 — 75% done means [A] = 25% of [A]₀, so [A]₀/[A] = 4.
Step 2 — Use k = (2.303/t)·log([A]₀/[A]): t = (2.303/k)·log 4.
Step 3 — Compute: t = (2.303/0.0693)(0.602) ≈ 33.24 × 0.602 ≈ 20 s.
Answer: ≈ 20 s. Shortcut: 75% complete is exactly two half-lives, and t½ = 0.693/0.0693 = 10 s, so 2 × 10 = 20 s ✓.

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Frequently Asked Questions — Chemical Kinetics

What are the key concepts in Chemical Kinetics?
Study the speed of reactions and what affects it: concentration, temperature, and catalysts. Covers rate law, order of reaction, the Arrhenius equation, and reaction mechanisms step by step.
Is Chemical Kinetics important for NEET & JEE?
Yes. Chemical Kinetics is part of the Chemistry Class 12 NCERT syllabus and is directly tested in NEET and JEE examinations. StudyHub provides structured notes, diagrams, and practice questions covering all exam-level subtopics.
How can I practice Chemical Kinetics questions on StudyHub?
Open StudyHub and select Chemistry → Chemical Kinetics. Choose Easy, Medium, or Hard difficulty. Hard-tier questions are at NEET & JEE level with full step-by-step explanations.

References

  1. NCERT Class 12 Chemistry Textbook — Chapter: Chemical Kinetics
  2. CBSE Curriculum — Chemistry (Class 12)
  3. NTA NEET UG Official Syllabus — subject-wise topic list
  4. NTA JEE Main Official Syllabus — subject-wise topic list