Class 12 Chemistry Chemical Kinetics Notes

Unit 3

General and Physical Chemistry

Class 12 Chemistry

Chemical Kinetics

Class 12 Chemistry – Chemical Kinetics Notes PDF

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Introduction

DefinitionChemical kinetics is the study of the rate of chemical reactions, the factors affecting reaction rate and the mathematical relationships between concentration and time.

This unit covers average and instantaneous rates, rate law, rate constant, order and molecularity, integrated equations for zero- and first-order reactions, half-life, collision theory, activation energy, Arrhenius equation, catalysts and related numerical problems.

1. Rate of Reaction

For a reaction aA + bB → cC + dD:

Rate = −(1/a)d[A]/dt = −(1/b)d[B]/dt = (1/c)d[C]/dt = (1/d)d[D]/dt

Average Rate

Average rate = change in concentration ÷ time interval

Instantaneous Rate

The instantaneous rate is the slope of the concentration–time curve at a particular instant.

Fig. 1 – Average vs Instantaneous Rate

Tangent → instantaneous rateSecant → average rateTimeReactant concentration

2. Rate Law and Rate Constant

For a reaction involving reactants A and B, an experimentally determined rate law may be:

Rate = k[A]m[B]n

Here k is the rate constant, while m and n are experimentally determined orders with respect to A and B.

Overall order = m + n

The unit of k depends on overall reaction order.

Overall OrderTypical Unit of k
Zeromol L⁻¹ s⁻¹
Firsts⁻¹
SecondL mol⁻¹ s⁻¹

Fig. 2 – Components of a Rate Law

Rate = k[A]ᵐ[B]ⁿkrate constant[A], [B]concentrationsm, nreaction orders

3. Order and Molecularity

FeatureOrderMolecularity
MeaningSum of powers of concentration terms in experimentally determined rate law.Number of reacting species involved in a single elementary step.
How obtainedExperimentally.From elementary mechanism step.
Possible valuesCan be zero, integer or sometimes fractional.Positive integer for an elementary step.
Applies toOverall reaction rate expression.Elementary reaction step.

4. Zero-Order Reaction

Rate = k[A]⁰ = k
[A]t = [A]0 − kt
t1/2 = [A]0 / (2k)

A plot of [A] versus t is linear with slope −k.

Fig. 3 – Zero-Order Integrated Plot

Time, t[A]slope = −k
ExampleIf [A]₀ = 0.80 mol L⁻¹ and k = 0.10 mol L⁻¹ min⁻¹, then t1/2 = 0.80/(2×0.10) = 4.0 min.

5. First-Order Reaction

Rate = k[A]
ln([A]₀/[A]ₜ) = kt
k = (2.303/t) log([A]₀/[A]ₜ)
t1/2 = 0.693/k

For a first-order reaction, half-life is independent of initial concentration.

Fig. 4 – First-Order Plots

t[A] tln[A]slope = −k
ExampleIf k = 0.231 min⁻¹, then t1/2 = 0.693/0.231 = 3.00 min.

6. Collision Theory, Activation Energy and Activated Complex

Collision theory states that reacting particles must collide with sufficient energy and suitable orientation for reaction to occur.

  • Effective collision: collision that leads to product formation.
  • Activation energy, Ea: minimum energy barrier that reacting particles must overcome.
  • Activated complex: unstable, high-energy arrangement near the top of the energy barrier.

Fig. 5 – Energy Profile and Activation Energy

EₐReactantsProductsActivated complexReaction coordinate

7. Temperature Effect and Arrhenius Equation

k = A e−Eₐ/(RT)
ln k = ln A − Eₐ/(RT)
log(k₂/k₁) = [Eₐ/(2.303R)](1/T₁ − 1/T₂)

As temperature increases, a larger fraction of molecules has energy equal to or greater than Ea, so k generally increases.

Fig. 6 – Maxwell–Boltzmann Concept and Activation Energy

Eₐlower Thigher TMolecular energy

8. Catalysis

DefinitionA catalyst changes the rate of a reaction by providing an alternative pathway with lower activation energy and is regenerated overall.
TypeDescriptionExample Concept
HomogeneousCatalyst and reactants are in the same phase.Acid-catalyzed reactions in solution.
HeterogeneousCatalyst and reactants are in different phases.Gas reaction on a solid metal surface.
Enzyme catalysisBiological catalysts accelerate biochemical reactions with high specificity.Enzyme + substrate → products.

Fig. 7 – Catalyst Lowers Activation Energy

UncatalyzedCatalyzedReaction coordinate

9. Worked Numerical Patterns

Zero Order[A]₀ = 0.60 M, k = 0.020 M s⁻¹, t = 10 s. Then [A]ₜ = 0.60 − (0.020×10) = 0.40 M.
First OrderIf [A] falls from 0.80 M to 0.20 M in 10 min, k = (2.303/10)log(0.80/0.20) = 0.2303×0.6021 ≈ 0.139 min⁻¹.
Half-LifeFor k = 2.0 × 10⁻³ s⁻¹, first-order t₁/₂ = 0.693/k = 346.5 s.
Order from Rate DataIf doubling [A] doubles the rate while other concentrations remain constant, the reaction is first order in A.

10. Quick Revision & Exam Points

Important Questions
  • Define chemical kinetics, average rate and instantaneous rate.
  • Explain rate law and rate constant.
  • Differentiate order and molecularity.
  • Derive integrated rate equation for zero-order reaction.
  • Derive integrated rate equation for first-order reaction.
  • Derive half-life expressions for zero and first order.
  • Explain collision theory, activation energy and activated complex.
  • State and use Arrhenius equation.
  • Explain effects of concentration, temperature and catalyst on rate.
  • Differentiate homogeneous, heterogeneous and enzyme catalysis.
  • Rate law is determined experimentally.
  • Order = sum of rate-law exponents.
  • Zero-order [A] vs t is linear.
  • First-order ln[A] vs t is linear.
  • First-order t₁/₂ = 0.693/k.
  • Effective collision needs sufficient energy and orientation.
  • Eₐ is the reaction energy barrier.
  • Higher T generally raises k.
  • Catalyst lowers activation energy.

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