Class 12 Chemical Kinetics Complete Guide 2027 — Rate Laws, Order of Reaction, Arrhenius Equation

In short: Chemical Kinetics (Chapter 4) carries 5 marks in CBSE Class 12 Chemistry 2027 and is a formula-driven Physical Chemistry chapter. It covers rate of reaction, rate law expressions, order of reaction (zeroth, first, second), integrated rate equations, half-life formulas, the Arrhenius equation (k = Ae⁻ᴱᵃ/ᴿᵀ), and collision theory. Expect 1 numerical on integrated rate equations or Arrhenius equation and 1 conceptual question on order vs molecularity. Learn the five core formulas, practise the graph-based questions, and this chapter becomes an easy 5-mark pickup.

Chemical Kinetics is Chapter 4 in CBSE Class 12 Chemistry and one of the most predictable chapters on the board paper. It carries 5 marks, typically split between one numerical (3 marks) and one or two conceptual questions (2 marks). Unlike equilibrium or electrochemistry, kinetics questions follow a tight pattern — once you know the formulas and graphs, you can answer almost anything the paper throws at you.

For the full marks breakdown across all chapters, see the Class 12 Chemistry chapter-wise weightage. For all formulas in one place, check the Class 12 Chemistry formulas and reactions list.

Marks Distribution

Rate of a Chemical Reaction

The rate of reaction measures how quickly reactants are consumed or products are formed.

For a general reaction: aA + bB → cC + dD

Average rate = -1/a × Δ[A]/Δt = -1/b × Δ[B]/Δt = +1/c × Δ[C]/Δt = +1/d × Δ[D]/Δt

Instantaneous rate = -1/a × d[A]/dt (slope of concentration-time curve at a given instant)

Units of rate: mol L⁻¹ s⁻¹ (or mol L⁻¹ min⁻¹)

Key point: The negative sign for reactants ensures the rate is always positive. The stoichiometric coefficients make the rate unique regardless of which species you measure.

Rate Law and Rate Constant

The rate law is an experimentally determined expression that relates the rate to concentrations of reactants.

Rate = k[A]ˣ[B]ʸ

Where:

• k = rate constant (specific rate constant)
• x = order with respect to A
• y = order with respect to B
• (x + y) = overall order of reaction

Important distinctions:

Memory trick — "OEM": Order is Experimental, Molecularity is Mechanistic. You can never look at a balanced equation and write the order — you must have experimental data.

Integrated Rate Equations

These are the most important formulas in the chapter. They connect concentration with time.

Zeroth Order Reaction

Rate = k (rate is constant, independent of concentration)

Integrated equation: [A]₀ - [A] = kt → [A] = [A]₀ - kt

Half-life: t₁/₂ = [A]₀ / 2k

Graph patterns:

• [A] vs t → straight line with slope = -k
• Rate vs [A] → horizontal line (rate is constant)

Example: Decomposition of NH₃ on a hot platinum surface, photochemical reactions.

First Order Reaction

Rate = k[A]

Integrated equation: k = (2.303/t) log([A]₀/[A])

Or equivalently: ln[A] = ln[A]₀ - kt → [A] = [A]₀ e⁻ᵏᵗ

Half-life: t₁/₂ = 0.693/k (independent of initial concentration)

Graph patterns:

• ln[A] vs t → straight line with slope = -k
• log[A] vs t → straight line with slope = -k/2.303
• [A] vs t → exponential decay curve

Examples: Radioactive decay, decomposition of N₂O₅, decomposition of H₂O₂.

Second Order Reaction

Rate = k[A]²

Integrated equation: 1/[A] - 1/[A]₀ = kt

Half-life: t₁/₂ = 1/(k[A]₀) (depends on initial concentration)

Graph pattern: 1/[A] vs t → straight line with slope = k

Summary Table — The Big Comparison

Memory trick — "012 trick" for units of k: The unit of k is always (mol L⁻¹)^(1-n) s⁻¹ where n is the order. For 0th order: mol L⁻¹ s⁻¹. For 1st order: s⁻¹. For 2nd order: L mol⁻¹ s⁻¹.

Arrhenius Equation

This equation connects the rate constant to temperature and activation energy.

k = Ae⁻ᴱᵃ/ᴿᵀ

Where:

• k = rate constant
• A = pre-exponential factor (frequency factor)
• Eₐ = activation energy (J/mol)
• R = gas constant = 8.314 J mol⁻¹ K⁻¹
• T = temperature in Kelvin

Logarithmic form: ln k = ln A - Eₐ/RT

log k = log A - Eₐ/(2.303RT)

Graph: Plot of log k vs 1/T gives a straight line with slope = -Eₐ/(2.303R) and intercept = log A.

Two-Temperature Form

When given k values at two different temperatures:

log(k₂/k₁) = (Eₐ/2.303R) × (1/T₁ - 1/T₂)

This is the most common numerical format in board exams. You are given k₁, k₂, T₁, T₂ and asked to find Eₐ (or vice versa).

10-degree rule (approximate): For many reactions, the rate roughly doubles or triples when temperature increases by 10°C. Temperature coefficient = k(T+10)/k(T) ≈ 2-3.

Memory trick — "A-E-R-T" for Arrhenius: Think of it as "A rate Equation Relating Temperature." A = frequency, E = energy barrier, R = gas constant, T = temperature. Higher T or lower Eₐ → larger k → faster reaction.

Collision Theory

Collision theory explains why not every collision leads to a reaction.

Rate = Z_AB × e⁻ᴱᵃ/ᴿᵀ × P

Where:

• Z_AB = collision frequency (number of collisions per unit volume per unit time)
• e⁻ᴱᵃ/ᴿᵀ = fraction of molecules with energy ≥ Eₐ (Boltzmann factor)
• P = steric factor (probability factor) — accounts for the correct orientation

Key concepts:

• Threshold energy: Minimum energy molecules must have for a reaction to occur
• Activation energy (Eₐ): Threshold energy minus the average kinetic energy of molecules
• Effective collision: A collision where molecules have (a) sufficient energy (≥ Eₐ) AND (b) proper orientation
• Why P < 1: Not all high-energy collisions lead to products — molecules must collide in the correct orientation

Energy profile diagram:

Reactants → [Activated complex at peak] → Products

• The height of the peak above reactants = Eₐ (forward)
• The height of the peak above products = Eₐ (backward)
• ΔH = Eₐ(forward) - Eₐ(backward)

Effect of catalyst: A catalyst provides an alternative pathway with lower Eₐ. It does not change ΔH or equilibrium — it speeds up both forward and backward reactions equally.

Solved Numericals

Numerical 1: First Order Rate Constant

Q: A first-order reaction has a rate constant of 1.15 × 10⁻³ s⁻¹. How long will 5 g of this reactant take to reduce to 3 g?

Solution: k = (2.303/t) log([A]₀/[A]) 1.15 × 10⁻³ = (2.303/t) log(5/3) 1.15 × 10⁻³ = (2.303/t) × 0.2218 t = (2.303 × 0.2218) / (1.15 × 10⁻³) t = 0.5109 / (1.15 × 10⁻³) t = 444.3 seconds ≈ 444 s

Numerical 2: Half-Life of First Order Reaction

Q: The half-life of a first-order reaction is 60 minutes. What percentage of the reactant will be left after 200 minutes?

Solution: k = 0.693/t₁/₂ = 0.693/60 = 0.01155 min⁻¹

k = (2.303/t) log([A]₀/[A]) 0.01155 = (2.303/200) log([A]₀/[A]) log([A]₀/[A]) = (0.01155 × 200)/2.303 = 1.003 [A]₀/[A] = 10^1.003 ≈ 10.07 [A] = [A]₀/10.07 ≈ 9.93% remaining

Numerical 3: Arrhenius Equation

Q: The rate constant of a reaction at 500 K and 700 K are 0.02 s⁻¹ and 0.07 s⁻¹ respectively. Calculate the activation energy. (R = 8.314 J mol⁻¹ K⁻¹)

Solution: log(k₂/k₁) = (Eₐ/2.303R)(1/T₁ - 1/T₂) log(0.07/0.02) = (Eₐ/(2.303 × 8.314))(1/500 - 1/700) log(3.5) = (Eₐ/19.147)(0.002 - 0.001428) 0.5441 = (Eₐ/19.147)(5.714 × 10⁻⁴) Eₐ = (0.5441 × 19.147)/(5.714 × 10⁻⁴) Eₐ = 10.424/5.714 × 10⁻⁴ Eₐ = 18235.9 J/mol ≈ 18.24 kJ/mol

Numerical 4: Determining Order from Data

Q: For the reaction 2A → Products, the following data was obtained:

Find the order and rate constant.

Solution: Rate = k[A]ⁿ 2 × 10⁻³ = k(0.1)ⁿ ... (1) 8 × 10⁻³ = k(0.2)ⁿ ... (2)

Dividing (2) by (1): 4 = (0.2/0.1)ⁿ = 2ⁿ → n = 2

From (1): 2 × 10⁻³ = k × (0.1)² → k = 2 × 10⁻³/0.01 = 0.2 L mol⁻¹ s⁻¹

Order = 2, k = 0.2 L mol⁻¹ s⁻¹

Board Exam Question Types

1. Derive the integrated rate equation for a first-order reaction. (3-5 marks) — High frequency, appears almost every alternate year
2. Numerical on Arrhenius equation — given k₁, k₂, T₁, T₂, find Eₐ. (3 marks) — Very common
3. Differentiate between order and molecularity of a reaction. (2 marks) — Frequent short answer
4. The rate constant for a first-order reaction is given. Find t₁/₂ or time for a given percentage completion. (2-3 marks)
5. Graph-based questions — identify order from [A] vs t or log[A] vs t plot. (1-2 marks)
6. Define rate constant. What are its units for nth order reaction? (1-2 marks)
7. Explain the effect of temperature and catalyst on rate of reaction using Arrhenius equation. (2-3 marks)

Common Mistakes Students Make

1. Confusing order with molecularity. Order is experimental and can be zero or fractional. Molecularity is theoretical and must be a positive integer (1, 2, or 3). You cannot determine order from a balanced equation.

2. Using the wrong integrated rate equation. Students apply the first-order formula to all problems. Always check the order first. If the question says "second order," use 1/[A] - 1/[A]₀ = kt, not the log formula.

3. Forgetting to convert temperature to Kelvin in the Arrhenius equation. The equation uses T in Kelvin, not Celsius. 25°C = 298 K. Plugging in 25 instead of 298 gives a wildly wrong answer.

4. Mixing up log and ln in the Arrhenius equation. The natural log form uses ln k = ln A - Eₐ/RT. The log₁₀ form uses log k = log A - Eₐ/(2.303RT). If your equation uses 2.303, use log₁₀. If it does not, use ln. Most board numericals use the log₁₀ version.

5. Getting the sign wrong in the two-temperature Arrhenius form. The formula is log(k₂/k₁) = (Eₐ/2.303R)(1/T₁ - 1/T₂). Note it is 1/T₁ minus 1/T₂ (lower temperature first). Swapping gives a negative Eₐ, which is physically meaningless.

6. Ignoring units of k when verifying order. If the rate constant has units of s⁻¹, the reaction is first order. If it is L mol⁻¹ s⁻¹, the reaction is second order. Checking the units of k is a quick way to verify your answer.

4-Week Practice Plan

Week 1: Foundations (Rate, Rate Law, Order)

• Read NCERT Section 4.1-4.2 thoroughly
• Make a table of rate law vs order vs molecularity differences
• Solve all NCERT in-text questions from these sections
• Practice 5 problems on determining order from experimental data

Week 2: Integrated Rate Equations & Half-Life

• Memorise all three integrated equations (0th, 1st, 2nd order)
• Derive the first-order integrated rate equation from scratch (board derivation)
• Solve 10 numericals on first-order reactions (time, percentage completion, half-life)
• Draw all six graphs (concentration vs time, rate vs concentration for each order)

Week 3: Arrhenius Equation & Collision Theory

• Memorise the Arrhenius equation in all three forms (exponential, ln, log₁₀)
• Solve 8-10 numericals using the two-temperature Arrhenius form
• Understand the energy profile diagram — draw it with and without catalyst
• Read NCERT collision theory section and note down the three conditions for effective collision

Week 4: Revision & Board-Pattern Practice

• Solve 3 previous year board papers (kinetics questions only)
• Take a 30-minute mini-test covering the full chapter
• Review the quick revision card below before your test
• Redo any problems you got wrong during Weeks 1-3

Quick Revision Card

One-liner mnemonics:

• "OEM" — Order is Experimental, Molecularity is Mechanistic
• "012 trick" — plug n = 0, 1, 2 into (mol L⁻¹)^(1-n) s⁻¹ for units
• "0.693 for first" — first-order half-life = 0.693/k (independent of concentration)
• "A-E-R-T" — Arrhenius: A = frequency factor, E = activation energy, R = gas constant, T = temperature

Frequently Asked Questions

How many marks does Chemical Kinetics carry in CBSE 2027?

Chemical Kinetics carries 5 marks out of 70 in the theory paper. It is clubbed with Surface Chemistry and Electrochemistry in the Physical Chemistry section. While not the heaviest chapter, the marks are highly predictable — almost always one numerical and one conceptual question.

Which is the most important formula in this chapter?

The first-order integrated rate equation: k = (2.303/t) log([A]₀/[A]). It appears in about 70% of board papers either as a derivation, a direct numerical, or as part of a half-life problem. The Arrhenius two-temperature form is the second most important.

How do I quickly identify the order of a reaction from a graph?

Plot concentration vs time. If it is a straight line → zeroth order. If log[A] vs t is a straight line → first order. If 1/[A] vs t is a straight line → second order. This graphical method is a common 1-2 mark question.

What is the difference between rate of reaction and rate constant?

Rate of reaction changes as concentrations change during the reaction — it is not constant. The rate constant (k) is a proportionality constant that stays fixed at a given temperature. Rate depends on concentration; k depends only on temperature and activation energy.

Is collision theory asked in board exams?

Collision theory carries 1-2 marks at most, usually as a short-answer or MCQ question. Know three things: effective collisions need sufficient energy AND proper orientation, the steric factor P accounts for orientation, and a catalyst lowers Eₐ but does not change ΔH.

Can the order of a reaction be zero or fractional?

Yes. Order is determined experimentally and can be zero, fractional (like 1.5), or even negative. Molecularity, on the other hand, must be a positive integer (1, 2, or 3). This distinction is a favourite board question.

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