CBSE Class 12 Chemistry Chapter 4 — Chemical Kinetics: NCERT Notes, Formulas and Important Questions 2027

Last Updated: April 2026

CBSE BOARDS 2027 | CLASS 12 CHEMISTRY

Master Chemical Kinetics — the rate laws, integrated equations, Arrhenius equation and Activation Energy concepts that appear every year in CBSE board exams

1. Rate of Reaction — Definitions and Units

What is Rate of Reaction?

The rate of a chemical reaction is defined as the change in concentration of a reactant or product per unit time.

For a general reaction: A → B

Rate = −d[A]/dt = +d[B]/dt

The negative sign for reactant shows concentration is decreasing; the positive sign for product shows concentration is increasing.

Units of Rate: mol L⁻¹ s⁻¹ (or mol L⁻¹ min⁻¹, depending on the time unit used)

Average Rate vs Instantaneous Rate

• Average rate: Calculated over a finite time interval — Δ[concentration]/Δt
• Instantaneous rate: Rate at a particular instant — d[concentration]/dt (slope of tangent to concentration-time curve)

Rate in Terms of Stoichiometry

For the reaction: 2HI(g) → H₂(g) + I₂(g)

Rate = −½ d[HI]/dt = +d[H₂]/dt = +d[I₂]/dt

Always divide by the stoichiometric coefficient of each species. This gives a unique rate for the reaction regardless of which species you measure.

2. Rate Law, Order of Reaction and Rate Constant

Rate Law Expression

The rate law (rate expression) is determined experimentally and cannot be derived from the balanced equation:

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

Where: k = rate constant, m = order with respect to A, n = order with respect to B, (m+n) = overall order of reaction

Order of Reaction

Order is the sum of the exponents (m + n) in the rate law expression. It is experimentally determined and can be:

• Zero order (rate independent of concentration)
• First order (most common in nature)
• Second order
• Fractional order (e.g., 1.5, 0.5)

Molecularity vs Order

Pseudo First-Order Reactions

When one reactant is in large excess, its concentration remains nearly constant, making a higher-order reaction appear as first-order. Example: Hydrolysis of ester in dilute acid

CH₃COOC₂H₅ + H₂O → CH₃COOH + C₂H₅OH

Rate = k[ester][H₂O] — but since [H₂O] is constant (large excess), Rate = k'[ester] — pseudo first-order.

3. Units of Rate Constant (k)

General formula: Units of k = (mol L⁻¹)^(1-n) × s⁻¹, where n = order of reaction

4. Integrated Rate Equations

Zero Order Reaction

Rate = k[A]⁰ = k (constant rate, independent of concentration)

Integrated form: [A] = [A]₀ − kt

Half-life: t½ = [A]₀ / 2k (depends on initial concentration)

Graph: [A] vs time gives a straight line with slope = −k

Example: Decomposition of ammonia on platinum surface, some enzyme-catalysed reactions

First Order Reaction (Most Important)

Rate = k[A]

Integrated form: ln[A] = ln[A]₀ − kt, OR [A] = [A]₀ e^(−kt)

In log form: log[A] = log[A]₀ − kt/2.303

k = (2.303/t) × log([A]₀/[A])

Half-life: t½ = 0.693/k = ln2/k (CONSTANT — independent of initial concentration)

Graph: ln[A] vs time → straight line, slope = −k; log[A] vs time → slope = −k/2.303

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

Second Order Reaction

Rate = k[A]²

Integrated form: 1/[A] = 1/[A]₀ + kt

Half-life: t½ = 1/(k[A]₀) — depends on initial concentration

Graph: 1/[A] vs time → straight line, slope = k

5. Arrhenius Equation — Temperature and Rate Constant

The Arrhenius Equation

k = A × e^(−Ea/RT)

Where:

• k = rate constant at temperature T
• A = frequency factor (pre-exponential factor) — related to collision frequency and orientation
• Ea = activation energy (in J/mol or kJ/mol)
• R = gas constant = 8.314 J K⁻¹ mol⁻¹
• T = absolute temperature in Kelvin

Taking the log form:

ln k = ln A − Ea/RT

log k = log A − Ea/(2.303RT)

Plot of log k vs 1/T gives a straight line with slope = −Ea/2.303R

Arrhenius Equation for Two Temperatures (Most Important for Board Numericals)

log(k₂/k₁) = (Ea/2.303R) × [(T₂ − T₁)/(T₁T₂)]

This formula is used when you need to find: k at a new temperature, or calculate Ea given k values at two different temperatures.

Activation Energy (Ea)

Activation energy is the minimum energy required for reactant molecules to collide and form products. It represents the energy barrier between reactants and products:

• Higher Ea → reaction is slower (fewer molecules have sufficient energy)
• Lower Ea → reaction is faster
• A catalyst provides an alternative pathway with lower Ea, increasing the rate
• Increasing temperature provides more molecules with energy ≥ Ea

Frequency Factor (A)

A = collision frequency × steric factor (orientation factor). It accounts for the fact that not all collisions occur with the right orientation. Molecules must collide with correct geometry for reaction to occur.

6. Effect of Various Factors on Rate

• Concentration: Increasing concentration → more collisions → higher rate (as per rate law)
• Temperature: Every 10°C rise approximately doubles the rate constant. This is explained by the Arrhenius equation — higher T means more molecules have energy ≥ Ea.
• Catalyst: Provides an alternative reaction mechanism with lower activation energy. Does not change Ea of the original pathway, just opens a new easier path. The catalyst is not consumed in the overall reaction.
• Surface area: For heterogeneous reactions (solid-liquid/solid-gas), increasing surface area increases rate.
• Light: Photochemical reactions are accelerated by light providing energy to reactant molecules.

7. Methods of Determining Order of Reaction

Initial Rate Method

Keep one reactant concentration constant, vary the other, and measure initial rates. If doubling [A] doubles the rate → first order in A. If doubling [A] quadruples the rate → second order in A.

Integrated Rate Method (Graphical)

• If [A] vs time is linear → zero order
• If ln[A] vs time is linear → first order
• If 1/[A] vs time is linear → second order

Half-Life Method

• If t½ is constant (independent of initial concentration) → first order
• If t½ ∝ 1/[A]₀ → second order
• If t½ ∝ [A]₀ → zero order

8. Solved Numerical Problems

Problem 1: Finding Rate Constant for First Order Reaction

The initial concentration of a reactant is 0.8 mol/L. After 30 minutes, it decreases to 0.2 mol/L. Calculate the rate constant for this first-order reaction.

Solution:

k = (2.303/t) × log([A]₀/[A])

k = (2.303/30 min) × log(0.8/0.2)

k = (2.303/30) × log(4)

k = (0.0768) × (0.6021)

k = 0.0462 min⁻¹

Problem 2: Arrhenius Equation — Finding k at New Temperature

The rate constant for a reaction is 4.5 × 10⁻³ s⁻¹ at 25°C. If the activation energy is 60 kJ/mol, calculate the rate constant at 45°C.

Solution:

T₁ = 298 K, T₂ = 318 K, k₁ = 4.5 × 10⁻³ s⁻¹, Ea = 60,000 J/mol

log(k₂/k₁) = (Ea/2.303R) × [(T₂ − T₁)/(T₁T₂)]

log(k₂/k₁) = (60000/(2.303 × 8.314)) × [(318−298)/(298 × 318)]

log(k₂/k₁) = (60000/19.15) × [20/94764]

log(k₂/k₁) = 3133.7 × 0.000211 = 0.661

k₂/k₁ = 10^0.661 = 4.58

k₂ = 4.58 × 4.5 × 10⁻³ = 2.06 × 10⁻² s⁻¹

Problem 3: Half-Life Calculation

A first-order reaction has a half-life of 693 seconds. Calculate the rate constant and time for 90% completion.

Solution:

t½ = 0.693/k → k = 0.693/693 = 1.0 × 10⁻³ s⁻¹

For 90% completion: [A] = 10% of [A]₀ = 0.1[A]₀

t = (2.303/k) × log([A]₀/[A]) = (2.303/1.0 × 10⁻³) × log(1/0.1)

t = 2303 × log(10) = 2303 × 1

t = 2303 seconds ≈ 3.32 × t½

Problem 4: Determining Order from Initial Rate Data

For the reaction A + B → C, the following data was obtained:

Solution: Compare Exp 1 and 2 (B constant): [A] doubles, rate doubles → first order in A. Compare Exp 2 and 3 (A constant): [B] doubles, rate doubles → first order in B.

Rate = k[A][B], Overall order = 2. k = rate/([A][B]) = 2.0 × 10⁻³/(0.1 × 0.1) = 0.2 L mol⁻¹ s⁻¹

Problem 5: Finding Activation Energy from Two Rate Constants

k₁ = 1.5 × 10⁻³ s⁻¹ at 300 K; k₂ = 6.0 × 10⁻³ s⁻¹ at 320 K. Find Ea.

Solution:

log(k₂/k₁) = (Ea/2.303R) × [(T₂−T₁)/(T₁T₂)]

log(6.0/1.5) = (Ea/2.303 × 8.314) × [(320−300)/(300 × 320)]

log(4) = (Ea/19.15) × [20/96000]

0.602 = Ea × 1.089 × 10⁻⁵ / 19.15

0.602 = Ea × 5.69 × 10⁻⁷

Ea = 0.602 / (5.69 × 10⁻⁷) = 1.058 × 10⁶ J/mol ≈ 44.7 kJ/mol

9. Important Board Exam Questions (3–5 Marks)

Q1. What is the difference between order and molecularity of a reaction? Give one example of a pseudo first-order reaction.
Answer: Order is the sum of powers in experimental rate law (can be fractional/zero); molecularity is the number of molecules colliding in rate-determining step (always whole number). Example: Hydrolysis of ester in dilute HCl — appears first-order despite being second-order overall because water is in large excess.

Q2. For a first-order reaction, show that time for 99% completion is twice the time for 90% completion.
Answer: t₉₀% = (2.303/k) × log(100/10) = 2.303/k; t₉₉% = (2.303/k) × log(100/1) = 2 × 2.303/k. Hence t₉₉% = 2 × t₉₀%.

Q3. Explain the effect of temperature on rate constant using the Arrhenius equation.
Answer: k = Ae^(-Ea/RT). As T increases, the exponential factor e^(-Ea/RT) increases (becomes less negative exponent), so k increases. This means more molecules have energy ≥ Ea at higher temperatures, resulting in more effective collisions.

Q4. The half-life of a first-order reaction is 30 minutes. Calculate (a) rate constant (b) time for 75% completion.
Answer: (a) k = 0.693/30 = 0.0231 min⁻¹. (b) 75% completion means [A] = 25% of [A]₀. t = 2 × t½ = 60 minutes (because 75% is completed after exactly 2 half-lives: 50% at t½, 75% at 2t½).

Q5. What is meant by activation energy? How does a catalyst affect it?
Answer: Activation energy (Ea) is the minimum energy that reactant molecules must possess for a successful collision to form products — it is the energy barrier between reactants and transition state. A catalyst provides an alternative reaction pathway with a lower activation energy, allowing more molecules to react at the same temperature. The catalyst does not change the Ea of the uncatalysed pathway — it simply makes a new, easier pathway available.

10. Important Formulas Summary