Is it correct to say: Confusing order and molecularity — treating stoichiometric coefficients as orders.?

Order is experimental, molecularity is theoretical (for elementary steps only). E.g. 2NO + O₂ has order 2 (not 3) as determined experimentally.

Is it correct to say: Using log base e in first-order equation without the 2.303 factor.?

k = (2.303/t) log₁₀([A]₀/[A]_t) OR k = (1/t) ln([A]₀/[A]_t). Do not mix ln and log.

Is it correct to say: Assuming half-life is always independent of concentration.?

Only first-order has t₁/₂ = 0.693/k (concentration-independent). Zero-order: t₁/₂ ∝ [A]₀; second-order: t₁/₂ ∝ 1/[A]₀.

Is it correct to say: Forgetting to convert Celsius to kelvin in Arrhenius numericals.?

Always add 273 to Celsius temperatures before using them in Arrhenius equation.

Is it correct to say: Substituting Ea in kJ/mol without converting to J/mol when using R = 8.314 J/(mol·K).?

Keep Ea in J/mol OR use R = 8.314 × 10⁻³ kJ/(mol·K). Consistent units are mandatory.

Is it correct to say: Writing units of k the same for every order.?

Zero: mol L⁻¹ s⁻¹; First: s⁻¹; Second: L mol⁻¹ s⁻¹. Deducible from rate = k·[A]^order.

Chemical Kinetics

Rate laws, order and molecularity, integrated rate equations, half-life, Arrhenius equation and activation energy — NCERT Class 12 Chemistry Ch 3

📐 9 formulas✏️ 3 examples🎯 6 practice⚖️ 5-7 marks🏫 CBSE📚 Class 12✓ 2025–26 syllabus

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Board Exam Tips

→Distinguish ORDER (experimental, may be fractional or zero) from MOLECULARITY (theoretical integer, elementary steps only). Board loves this distinction.
→First-order integrated rate law k = (2.303/t) log([A]₀/[A]_t) — used in almost every 3-mark numerical.
→Arrhenius two-temperature form appears frequently. Convert T to kelvin, use R = 8.314 J/(mol·K).
→Half-life of a first-order reaction is INDEPENDENT of initial concentration — a very common MCQ.
→Units of k give away the order: zero (mol L⁻¹ s⁻¹), first (s⁻¹), second (L mol⁻¹ s⁻¹).

📐 Formulas(9)

Rate of Reaction

Rate = - \frac{1}{a} \frac{d [ A ]}{d t} = + \frac{1}{b} \frac{d [ B ]}{d t}

Symbol	Meaning
$[A], [B]$	Molar concentrations (mol/L)
$a, b$	Stoichiometric coefficients
$t$	Time (s)

Rate Law and Order★ Board fav

Rate = k [A]^{m} [B]^{n}, Order = m + n

Zero-Order Integrated Rate Law

[A]_{t} = [A]_{0} - k t, t_{1/2} = \frac{[ A ] _{0}}{2 k}

Symbol	Meaning
$[A]_{t}$	Concentration at time t
$[A]_{0}$	Initial concentration
$k$	Zero-order rate constant

First-Order Integrated Rate Law★ Board fav

k = \frac{2.303}{t} lo g \frac{[ A ] _{0}}{[ A ] _{t}}

Half-Life of First-Order Reaction★ Board fav

t_{1/2} = \frac{0.693}{k}

Half-Life of Second-Order Reaction

t_{1/2} = \frac{1}{k [ A ] _{0}}

Arrhenius Equation★ Board fav

k = A e^{- E_{a} / R T}

Symbol	Meaning
$A$	Frequency factor (same units as k)
$E_{a}$	Activation energy (J/mol)
$R$	8.314 J·mol⁻¹·K⁻¹
$T$	Absolute temperature (K)

Arrhenius Two-Temperature Form★ Board fav

lo g \frac{k _{2}}{k _{1}} = \frac{E _{a}}{2.303 R} (\frac{1}{T _{1}} - \frac{1}{T _{2}})

Temperature Coefficient (rule of thumb)

μ = \frac{k _{T + 10}}{k _{T}} \approx 2 to 3

✏️ Solved Examples

1Solved Exampleeasy3 steps

For the reaction 2N₂O₅ → 4NO₂ + O₂, the rate of formation of NO₂ is 0.008 mol L⁻¹ s⁻¹. What is the rate of disappearance of N₂O₅ and the rate of reaction?

General rate expression

Rate = - \frac{1}{2} \frac{d [ N _{2} O _{5} ]}{d t} = \frac{1}{4} \frac{d [ NO _{2} ]}{d t}

2Solved Exampleboard4 steps

A first-order reaction has rate constant k = 1.386 × 10⁻³ s⁻¹. Find (i) the half-life and (ii) the time for 80% completion.

Half-life of first-order reaction

t_{1/2} = \frac{0.693}{k} = \frac{0.693}{1.386 \times 1 0 ^{- 3}} = 500 s

3Solved ExampleHOTS5 steps

The rate constant of a reaction doubles when temperature is raised from 300 K to 310 K. Calculate the activation energy of the reaction.

Given: k₂/k₁ = 2, T₁ = 300 K, T₂ = 310 K

lo g \frac{k _{2}}{k _{1}} = \frac{E _{a}}{2.303 R} (\frac{1}{T _{1}} - \frac{1}{T _{2}})

⚠️ Traps & Common Mistakes

⚠️Common Mistakes6

1
Confusing order and molecularity — treating stoichiometric coefficients as orders.
✓Order is experimental, molecularity is theoretical (for elementary steps only). E.g. 2NO + O₂ has order 2 (not 3) as determined experimentally.
2
Using log base e in first-order equation without the 2.303 factor.
✓k = (2.303/t) log₁₀([A]₀/[A]_t) OR k = (1/t) ln([A]₀/[A]_t). Do not mix ln and log.
3
Assuming half-life is always independent of concentration.
✓Only first-order has t₁/₂ = 0.693/k (concentration-independent). Zero-order: t₁/₂ ∝ [A]₀; second-order: t₁/₂ ∝ 1/[A]₀.
4
Forgetting to convert Celsius to kelvin in Arrhenius numericals.
✓Always add 273 to Celsius temperatures before using them in Arrhenius equation.
5
Substituting Ea in kJ/mol without converting to J/mol when using R = 8.314 J/(mol·K).
✓Keep Ea in J/mol OR use R = 8.314 × 10⁻³ kJ/(mol·K). Consistent units are mandatory.
6
Writing units of k the same for every order.
✓Zero: mol L⁻¹ s⁻¹; First: s⁻¹; Second: L mol⁻¹ s⁻¹. Deducible from rate = k·[A]^order.

🎯 Practice Yourself

🎯Practice Yourself6 questions

Q1
A first-order reaction is 30% complete in 10 minutes. Calculate the rate constant.
$k = \frac{2.303}{t} lo g \frac{[ A ] _{0}}{[ A ] _{t}}$
Q2
The half-life of a first-order reaction is 200 s. What fraction of the reactant remains after 600 s?
Q3
The rate constant of a first-order reaction at 27 °C is 5 × 10⁻³ s⁻¹. At 47 °C it becomes 2 × 10⁻² s⁻¹. Find the activation energy.
Q4
The rate of a certain reaction depends on [A] as: rate = k[A]^(3/2). If [A] is halved, by what factor does the rate change?
Q5
For a zero-order reaction, k = 0.02 mol L⁻¹ s⁻¹, initial [A] = 0.5 M. Find half-life.
Q6
State the difference between order and molecularity.

📝 Notes

Chemical Kinetics

Branch of chemistry that studies how fast a reaction proceeds and which factors control its rate — concentration, temperature, catalyst, surface area and light. Thermodynamics tells us if a reaction happens; kinetics tells us how fast.

Rate — average vs instantaneous

Average rate: Δ[A]/Δt over a time interval.
Instantaneous rate: d[A]/dt at a specific instant — the slope of the concentration-vs-time curve.

Order vs molecularity

| Feature | Order | Molecularity | | --- | --- | --- | | Determined by | Experiment | Balanced elementary equation | | Values | 0, fractional, negative | Positive integer (1, 2, 3) | | Applies to | Any reaction (overall) | Elementary steps only | | Example | Order of H₂ + Cl₂ (photochemical) ≈ 0 | Molecularity of a step must equal number of reactant molecules |

Integrated rate laws — quick summary

Zero order: [A]_t = [A]₀ − kt; t₁/₂ = [A]₀/(2k); linear plot of [A] vs t.
First order: ln[A]_t = ln[A]₀ − kt; t₁/₂ = 0.693/k; linear plot of ln[A] vs t.
Second order: 1/[A]_t = 1/[A]₀ + kt; t₁/₂ = 1/(k[A]₀); linear plot of 1/[A] vs t.

Temperature — collision theory

The Arrhenius equation captures two ideas:

Molecules must collide with enough energy (Ea) to reach the transition state.
They must have the correct orientation (A, steric factor).

Raising T raises the fraction with sufficient energy, sharply increasing rate. A rule of thumb: rate roughly doubles every 10 K rise for many reactions near room temperature.

Catalysts

A catalyst provides an alternative pathway with lower Ea. It does NOT change ΔG or the equilibrium constant — only accelerates the approach to equilibrium.