Is it correct to say: Assuming order equals stoichiometric coefficient?

Order is empirical; determined from experiment. Molecularity is the theoretical coefficient. They are different.

Is it correct to say: Using integrated rate law with wrong units (t in min but expecting k in s⁻¹)?

Units of k depend on t and order. State consistently.

Is it correct to say: Applying first-order half-life formula to zero- or second-order reactions?

t₁/₂ = 0.693/k is ONLY for first order. Zero: [A]₀/(2k); second: 1/(k[A]₀).

Is it correct to say: Forgetting factor 2.303 when converting between ln and log?

2.303 log x = ln x. Always keep track of the base.

Is it correct to say: Using Celsius in Arrhenius equation?

T must be in kelvin. Otherwise (1/T) is meaningless.

Is it correct to say: Assuming rate constant k depends on concentration?

k depends only on temperature (via Arrhenius). It is independent of concentration.

Chemical Kinetics

Rate laws, order, molecularity, integrated rate equations, Arrhenius equation — Maharashtra HSC Chemistry Ch 6

📐 11 formulas✏️ 3 examples🎯 5 practice⚖️ 5 marks🏫 Maharashtra State Board📚 Class 12✓ 2025–26 syllabus

💡

Board Exam Tips

→Derive the integrated rate law for a first-order reaction (k = (2.303/t) log [A₀]/[A]) — asked almost every HSC year for 3 marks.
→Distinguish order and molecularity in a tabular form for 2 marks — Balbharati insists on this clarity.
→Arrhenius equation and evaluation of activation energy E_a from graph of ln k vs 1/T (Ea = −slope × R) is a Maharashtra HSC HOTS problem.
→For half-life, remember first-order t₁/₂ = 0.693/k is independent of initial concentration — HSC tests this often.
→Always show units: k has different units for different orders (zero: mol·L⁻¹·s⁻¹; first: s⁻¹; second: L·mol⁻¹·s⁻¹). Board is strict about this.

📐 Formulas(11)

Rate of Reaction

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

Rate Law★ Board fav

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

Zero-Order Integrated Rate

[A] = [A]_{0} - k t

First-Order Integrated Rate★ Board fav

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

First-Order Half-Life★ Board fav

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

Second-Order Integrated Rate

\frac{1}{[ A ]} - \frac{1}{[ A ] _{0}} = k t

Second-Order Half-Life

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

Arrhenius Equation★ Board fav

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

Log Form of Arrhenius Equation

lo g k = lo g A - \frac{E _{a}}{2.303 R T}

Ratio of Rate Constants at Two Temperatures★ 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

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

✏️ Solved Examples

1Solved Exampleeasy3 steps

A first-order reaction is 30% complete in 40 minutes. Find the rate constant.

If 30% is completed, [A]/[A]₀ = 0.70.

\frac{[ A ] _{0}}{[ A ]} = \frac{1}{0.70} = 1.4286

2Solved Exampleboard4 steps

For a first-order reaction, the rate constant at 300 K is 3.2×10⁻⁴ s⁻¹ and at 320 K is 9.5×10⁻⁴ s⁻¹. Calculate the activation energy.

Use two-point Arrhenius.

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

3Solved ExampleHOTS4 steps

For the decomposition of N₂O₅ (first order), 40% decomposes in 50 min. What percent will have decomposed in 100 min?

Find k using 40% completion at 50 min.

k = \frac{2.303}{50} lo g \frac{1}{0.60} = \frac{2.303 \times 0.2218}{50} \approx 1.021 \times 1 0^{- 2} min^{- 1}

⚠️ Traps & Common Mistakes

⚠️Common Mistakes6

1
Assuming order equals stoichiometric coefficient
✓Order is empirical; determined from experiment. Molecularity is the theoretical coefficient. They are different.
2
Using integrated rate law with wrong units (t in min but expecting k in s⁻¹)
✓Units of k depend on t and order. State consistently.
3
Applying first-order half-life formula to zero- or second-order reactions
✓t₁/₂ = 0.693/k is ONLY for first order. Zero: [A]₀/(2k); second: 1/(k[A]₀).
4
Forgetting factor 2.303 when converting between ln and log
✓2.303 log x = ln x. Always keep track of the base.
5
Using Celsius in Arrhenius equation
✓T must be in kelvin. Otherwise (1/T) is meaningless.
6
Assuming rate constant k depends on concentration
✓k depends only on temperature (via Arrhenius). It is independent of concentration.

🎯 Practice Yourself

🎯Practice Yourself5 questions

Q1
For a first-order reaction, k = 6×10⁻³ s⁻¹. Find the half-life.
Q2
A zero-order reaction has [A]₀ = 0.5 M and k = 0.01 mol·L⁻¹·s⁻¹. Time to complete?
Q3
A reaction has E_a = 60 kJ/mol. Ratio of rate at 320 K to that at 300 K?
Q4
State the difference between order and molecularity of a reaction.
Q5
In a first-order reaction, [A]₀ = 0.10 M and after 5 min [A] = 0.05 M. Find k.

📝 Notes

Chemical Kinetics — Maharashtra HSC Overview

Chapter 6 of the Balbharati Class 12 Chemistry textbook covers rate laws, order and molecularity, integrated rate expressions, half-life, and the Arrhenius equation.

Maharashtra syllabus specifics

The Balbharati book puts strong emphasis on the difference between order and molecularity — often asked as a 2-marker in tabular form.
Derivation of the first-order integrated rate law starting from −d[A]/dt = k[A] is expected in full. Every step is worth a mark.
Collision theory and effective collisions are treated qualitatively — the Maharashtra syllabus goes further than NCERT in relating the frequency factor A to collision frequency.
Activation energy determination from graph of log k vs 1/T appears in the numericals.
Pseudo-first-order reactions (e.g. hydrolysis of ester) — one specific example expected.

Numerical tips

Convert temperatures to kelvin. Match units of R with the units of E_a (J or cal).
Distinguish k units by order:
- zero-order: mol·L⁻¹·s⁻¹
- first-order: s⁻¹ (or min⁻¹)
- second-order: L·mol⁻¹·s⁻¹
For sequential-question numericals, first extract k, then use it for the second sub-part.