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Class 12 Chemistry Notes

Chemical Kinetics Class 12 Notes

Complete, exam-ready notes on chemical kinetics: rate of reaction, rate laws, order and molecularity, integrated rate equations, half-life, the Arrhenius equation and catalysis — written for CBSE boards, JEE and NEET revision.

Class12SubjectChemistryCoversCBSE · JEE · NEET

Written byDeep Narayan· Science & Mathematics EducatorReviewed byPushpanjali

What is chemical kinetics in one line?

Chemical kinetics is the branch of chemistry that studies the rate of a reaction, the factors that control it, and the mechanism by which reactants are converted into products.

Rate of a Reaction

Rate of reaction

The change in concentration of a reactant or product per unit time. It is always positive: Rate=−Δ[R]Δt=Δ[P]Δt\text{Rate} = -\frac{\Delta[R]}{\Delta t} = \frac{\Delta[P]}{\Delta t}. Average rate over a time interval becomes the instantaneous rate as the interval tends to zero.

rate=−1ad[A]dt=1bd[B]dt\text{rate} = -\frac{1}{a}\frac{d[A]}{dt} = \frac{1}{b}\frac{d[B]}{dt}
Rate for aA + bB → products
  • Instantaneous rate is found from the slope of concentration vs time graph.
  • Units of rate are mol L⁻¹ s⁻¹ (or pressure per time for gases).

Rate Law, Order and Molecularity

Rate law

rate=k[A]x[B]y\text{rate} = k[A]^x[B]^y

The dependence of rate on the concentrations of reactants. The exponents x, yx,\ y are determined experimentally and combined give the order of the reaction.

  • Order:\text{Order}:
  • the sum of the exponents in the rate law (x + y); it can be zero, fractional or negative and must be found by experiment.
  • Molecularity:\text{Molecularity}:
  • the number of reactant molecules in the slowest step; always a whole number and can never be zero or fractional.

Rate constant k

The rate constant k is the rate when all concentrations are unity. Its unit varies with order: s−1\text{s}^{-1} for first order, L mol−1s−1\text{L mol}^{-1}\text{s}^{-1} for second order.

Integrated Rate Equations and Half-Life

[A]=[A]0e−kt,ln⁡[A]0[A]=kt,t1/2=0.693k[A] = [A]_0 e^{-kt},\quad \ln\frac{[A]_0}{[A]} = kt,\quad t_{1/2} = \frac{0.693}{k}
First-order reaction
  • Zero order:\text{Zero order}:
  • [A]=[A]0−kt,t1/2=[A]02k[A] = [A]_0 - kt,\quad t_{1/2} = \frac{[A]_0}{2k}
  • .
  • Second order:\text{Second order}:
  • 1[A]=1[A]0+kt,t1/2=1k[A]0\frac{1}{[A]} = \frac{1}{[A]_0} + kt,\quad t_{1/2} = \frac{1}{k[A]_0}
  • .

Half-life is constant for first order

For a first-order reaction the half-life is independent of initial concentration — a distinguishing feature. For zero order it depends on [A]₀, for second order on 1/[A]₀.

Radioactive decay and most unimolecular gas decompositions follow first-order kinetics, so the same equations apply.

The Arrhenius Equation

k=Ae−Ea/RT,ln⁡k=ln⁡A−EaRTk = A e^{-E_a/RT},\quad \ln k = \ln A - \frac{E_a}{RT}
Arrhenius equation

Activation energy

EaE_a

The minimum energy that colliding molecules must have for an effective collision. A larger EaE_a gives a smaller rate constant and a rate that rises more steeply with temperature.

Comparing two temperatures

For the same reaction at two temperatures: log⁡k2k1=Ea2.303R(T2−T1T1T2)\log\frac{k_2}{k_1} = \frac{E_a}{2.303 R}\left(\frac{T_2 - T_1}{T_1 T_2}\right). A catalyst lowers EaE_a, speeding the reaction without being consumed.

Solved Examples

Example: The half-life of a first-order reaction is 346.5 s. Find the rate constant and the time to complete 90% of the reaction.

Solution: k=0.693346.5=2×10−3 s−1k = \frac{0.693}{346.5} = 2\times 10^{-3}\,\text{s}^{-1}. For 90% completion ln⁡10010=kt⇒t=ln⁡102×10−3≈1151 s\ln\frac{100}{10} = kt \Rightarrow t = \frac{\ln 10}{2\times 10^{-3}} \approx 1151\,\text{s}.

Revision

Key formulas at a glance

Memorise these before attempting numericals — most exam questions hinge on one of them.

Rate law

rate=k[A]x[B]y\text{rate} = k[A]^x[B]^y

First order

ln⁡[A]0[A]=kt\ln\frac{[A]_0}{[A]} = kt

Half-life (1st order)

t1/2=0.693kt_{1/2} = \frac{0.693}{k}

Zero order

[A]=[A]0−kt[A] = [A]_0 - kt

Arrhenius

ln⁡k=ln⁡A−EaRT\ln k = \ln A - \frac{E_a}{RT}

Two temperatures

log⁡k2k1=Ea2.303R(T2−T1T1T2)\log\frac{k_2}{k_1} = \frac{E_a}{2.303R}\left(\frac{T_2-T_1}{T_1T_2}\right)

Exam tips

How this chapter is asked

Where this topic appears in CBSE, JEE Main and NEET papers.

  • Order is determined experimentally; molecularity is a property of the slowest step.
  • First-order half-life is independent of initial concentration.
  • Units of k: s⁻¹ (1st order), L mol⁻¹ s⁻¹ (2nd order).
  • A catalyst lowers activation energy, not the equilibrium constant.
  • ln[A] vs t gives a straight line for a first-order reaction.

FAQ

Common questions

What is the difference between order and molecularity?

Order is the sum of the concentration exponents in the rate law and is found experimentally; molecularity is the count of molecules in the slowest elementary step and is always a small whole number.

How can I tell a first-order reaction?

Its half-life is constant regardless of initial concentration, and a graph of ln[A] against time is a straight line.

What does the Arrhenius equation tell us?

It relates the rate constant to temperature and activation energy: higher activation energy or lower temperature gives a smaller rate constant.

How does a catalyst speed up a reaction?

A catalyst provides an alternative path with a lower activation energy, raising the rate constant without altering the position of equilibrium or being consumed.

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