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CBSE • Class XII • Chemistry • Ch 3
Estimated Time: 45 Mins
Study Progress: In Progress

Chemical Kinetics

In Class 12 Chemistry, "Chemical Kinetics" provides an authoritative, curriculum-verified master resource aligned with the 2026–27 NCERT syllabus.

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⏱️ Have You Ever Wondered?

Why do diamond gemstones take billions of years to turn into pencil graphite even though thermodynamics says they want to, while fireworks explode in a microsecond? Reaction rate, activation energy, and catalysts govern chemical speed.

Why This Chapter Matters

In Class 12 Chemistry, "Chemical Kinetics" provides an authoritative, curriculum-verified master resource aligned with the 2026–27 NCERT syllabus.

Before You Begin (Prerequisites)

  • Rates of change from Mathematics.
  • Thermodynamic equilibrium from Class 11.
  • Concentration units.

What You Will Learn (Core Objectives)

  • Define Average Rate and Instantaneous Rate of a chemical reaction: $r = -\frac{1}{a}\frac{d[A]}{dt} = \frac{1}{b}\frac{d[B]}{dt}$.
  • Distinguish between Order of a Reaction (experimentally determined) and Molecularity (theoretical collision count).
  • Derive integrated rate equations and half-life expressions for Zero Order ($k = \frac{[A]_0 - [A]}{t}, t_{1/2} = \frac{[A]_0}{2k}$) and First Order ($k = \frac{2.303}{t}\log\frac{[A]_0}{[A]}, t_{1/2} = \frac{0.693}{k}$).
  • State and apply the Arrhenius Equation: $k = A e^{-E_a/RT}$ and $\log\frac{k_2}{k_1} = \frac{E_a}{2.303R}\left(\frac{T_2 - T_1}{T_1 T_2}\right)$.
  • Explain the role of a Catalyst in lowering Activation Energy ($E_a$) without altering $\Delta H$.

Chapter Roadmap & Progression

1 1. Order vs. Molecularity & Rate La...
2 2. Zero & First Order Integrated Ra...
3 3. Arrhenius Equation & Activation...

Complete Concept Guide (100% Curriculum Coverage)

1. Order vs. Molecularity & Rate Law

The Rate Law expresses rate in terms of molar concentrations: $\text{Rate} = k [A]^x [B]^y$ where Overall Order $= x + y$.
• Order: Can be zero, fractional, or negative; strictly experimental.
• Molecularity: Number of reacting species colliding simultaneously in an elementary step; must be an integer ($1, 2, 3$), never zero or fractional!

2. Zero & First Order Integrated Rates

  • Zero Order: Rate independent of concentration ($[A] = [A]_0 - kt$). Half-life: $\mathbf{t_{1/2} = \frac{[A]_0}{2k}}$ (proportional to initial concentration).
  • First Order: $$\mathbf{k = \frac{2.303}{t}\log_{10}\left(\frac{[A]_0}{[A]}\right)} \quad \text{and} \quad \mathbf{t_{1/2} = \frac{0.693}{k}}$$ Key Property: Half-life of a first-order reaction is completely independent of initial concentration!

3. Arrhenius Equation & Activation Energy

Molecules must collide with minimum energy (Threshold Energy) to react. The energy barrier is the Activation Energy ($E_a$): $$\mathbf{k = A e^{-E_a/RT}} \implies \mathbf{\log_{10}\left(\frac{k_2}{k_1}\right) = \frac{E_a}{2.303 R}\left(\frac{T_2 - T_1}{T_1 T_2}\right)}$$ A catalyst provides an alternative pathway with a lower $E_a$, accelerating forward and backward rates equally!

Chemical Kinetics - Key Molecular Architecture & Reaction Mechanism Model

Chemical Kinetics - Molecular Architecture Electronic & Orbital Mechanisms Stereochemistry, reaction kinetics & pathways Thermodynamic & Coordination Frameworks Crystal field splitting, cell potentials & free energy High-Stakes Examination & Industrial Synthesis CBSE Class 12 Board criteria, JEE/NEET diagnostic applications & conversions

Chapter Summary & 10 Key Takeaways

Takeaway 1
Rate Constant ($k$): Characteristic reaction velocity invariant with concentration.
Takeaway 2
First-Order Invariance: Half-life $t_{1/2} = 0.693/k$ remaining constant regardless of starting concentration.
Takeaway 3
Activation Barrier ($E_a$): Energetic threshold required to form the transition state activated complex.
Takeaway 4
Arrhenius Temperature Sensitivity: Exponential rate acceleration with increasing absolute temperature.
Takeaway 5
Catalytic Acceleration: Lowering $E_a$ without shifting equilibrium constant $K_{eq}$.

Check Your Understanding (Diagnostic Practice Questions)

Diagnostic questions testing core conceptual clarity. Answers are hidden initially — solve each problem first, then click to reveal the step-by-step verified solution.

1
A first-order reaction has a rate constant of $1.15 \times 10^{-3}\text{ s}^{-1}$. How long will $5\text{ g}$ of this reactant take to reduce to $3\text{ g}$?
Reveal Answer & Explanation
Answer: $t = \frac{2.303}{k}\log\frac{[A]_0}{[A]} = \frac{2.303}{1.15 \times 10^{-3}}\log\left(\frac{5}{3}\right) = 2002.6 \times \log(1.667) = 2002.6 \times 0.2219 \approx 444.4\text{ seconds}$.
444.4 seconds.
2
Show that for a first-order reaction, the time required for 99% completion is twice the time required for 90% completion.
Reveal Answer & Explanation
Answer: $t_{90\%} = \frac{2.303}{k}\log\frac{100}{10} = \frac{2.303}{k}(1)$. $t_{99\%} = \frac{2.303}{k}\log\frac{100}{1} = \frac{2.303}{k}(2)$. Therefore, $t_{99\%} = 2 \times t_{90\%}$.
Proved t_99% = 2 × t_90%.
3
Differentiate between Order of a Reaction and Molecularity.
Reveal Answer & Explanation
Answer: Order is the sum of powers of concentrations in the experimental rate law (can be zero, fractional, or negative); Molecularity is the theoretical number of reacting species in an elementary step (always a non-zero positive integer, max 3).
Experimental sum of powers vs theoretical colliding species count.
4
The rate of a reaction quadruples when the temperature changes from $293\text{ K}$ to $313\text{ K}$. Calculate the activation energy of the reaction ($R = 8.314\text{ J/mol}\cdot\text{K}$).
Reveal Answer & Explanation
Answer: $\log\frac{k_2}{k_1} = \log 4 = 0.6021$. $\frac{E_a}{2.303 \times 8.314}\left(\frac{313 - 293}{293 \times 313}\right) = 0.6021 \implies \frac{E_a}{19.147}\left(\frac{20}{91709}\right) = 0.6021 \implies E_a = \frac{0.6021 \times 19.147 \times 91709}{20} \approx 52,863\text{ J/mol} \approx 52.86\text{ kJ/mol}$.
E_a ≈ 52.86 kJ/mol.
5
What is a Pseudo First-Order Reaction? Give an example.
Reveal Answer & Explanation
Answer: A reaction that is bimolecular in mechanism but follows first-order kinetics because one of the reactants is present in large excess such that its concentration remains practically constant. Example: Acid-catalyzed hydrolysis of ethyl acetate: $\text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O}(\text{excess}) \xrightarrow{\text{H}^+} \text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH}$.
Bimolecular reaction behaving as first order due to excess reactant.
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