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Chemistry · Class 12 · Chapter 3

Chemical Kinetics

Thermodynamics tells you whether a reaction can happen; kinetics tells you how fast it happens and what controls the speed. The chapter is short, but its first-order and Arrhenius numericals are among the most dependable marks in physical chemistry.

In this chapter: average and instantaneous rate, factors affecting rate, rate law and rate constant, order and molecularity, integrated rate equations for zero and first order reactions, half-life, pseudo first order reactions, effect of temperature and the Arrhenius equation, effect of a catalyst, and collision theory.

Rate of a reaction

The rate of a reaction is the change in concentration of a reactant or product per unit time. Its unit is mol L−1 s−1 (for gases, atm s−1 can also be used). The average rate over an interval is Δ[R]/Δt; the instantaneous rate is the slope of the tangent to the concentration-time curve at that instant, d[R]/dt.

To get a single rate for the whole reaction, each rate term is divided by its stoichiometric coefficient. For aA + bB → cC + dD:

Rate = −(1/a) d[A]/dt = −(1/b) d[B]/dt = +(1/c) d[C]/dt = +(1/d) d[D]/dtThe minus sign makes the rate positive, since reactant concentration falls.

For 2HI → H2 + I2, rate = −½ d[HI]/dt = d[H2]/dt. HI disappears twice as fast as H2 appears. Rate depends on the concentration of reactants (and pressure for gases), temperature and the presence of a catalyst.

Rate law, order and molecularity

The rate law expresses rate in terms of reactant concentrations, each raised to a power that must be found by experiment:

Rate = k[A]x[B]y; order = x + yk is the rate constant (the rate when all concentrations are 1 mol L−1). x and y need not equal the coefficients in the balanced equation.

For example, 2NO(g) + O2(g) → 2NO2(g) has rate = k[NO]2[O2], which happens to match the coefficients, but CHCl3 + Cl2 → CCl4 + HCl has rate = k[CHCl3][Cl2]½, order 1.5. Order can be 0, a whole number or a fraction.

OrderUnit of k
Zeromol L−1 s−1
Firsts−1
SecondL mol−1 s−1
n (general)(mol L−1)1−n s−1

Molecularity is the number of reacting species (atoms, ions or molecules) that must collide at the same time in an elementary reaction. It can be 1 (unimolecular, e.g. NH4NO2 → N2 + 2H2O), 2 (bimolecular, e.g. 2HI → H2 + I2) or, rarely, 3. It is always a whole number and never zero; reactions with molecularity above three are very unlikely because a simultaneous collision of that many particles is rare.

A complex reaction goes through several elementary steps, and the slowest step decides the overall rate (the rate determining step). The decomposition of H2O2 catalysed by iodide ion in alkaline medium has rate = k[H2O2][I−]. It is explained by two steps:

H2O2 + I− → H2O + IO− (slow)H2O2 + IO− → H2O + I− + O2 (fast)

IO− is an intermediate. Order applies to both elementary and complex reactions; molecularity has meaning only for an elementary step. For a complex reaction, the molecularity of the slowest step is generally the same as the overall order.

Integrated rate equations

Zero order reactions

[R] = [R]0 − kt so k = ([R]0 − [R]) ÷ tt½ = [R]0 ÷ 2k

A plot of [R] against t is a straight line with slope −k. Zero order reactions are uncommon and usually happen on metal surfaces, where the surface is saturated with reactant. Examples: decomposition of NH3 on a hot platinum surface at high pressure (2NH3 → N2 + 3H2), and thermal decomposition of HI on a gold surface.

First order reactions

ln [R] = ln [R]0 − kt or [R] = [R]0 e−ktk = (2.303 ÷ t) log ([R]0 ÷ [R])t½ = 0.693 ÷ k

The half-life of a first order reaction does not depend on the initial concentration. A plot of ln [R] against t is a straight line with slope −k; a plot of log ([R]0/[R]) against t has slope k/2.303. Examples: hydrogenation of ethene, decomposition of N2O5 and N2O, and all natural radioactive decay.

For a gas phase reaction A(g) → B(g) + C(g) followed by total pressure, with initial pressure pi and total pressure pt at time t: pA = 2pi − pt, so k = (2.303/t) log [pi ÷ (2pi − pt)].

First order reaction: concentration and ln concentration against timewww.iitmedicoguide.com[R] against tln [R] against t[R]₀[R]₀/2[R]₀/4t½2t½time →[R] = [R]₀e⁻ᵏᵗintercept = ln [R]₀slope = −ktime →www.iitmedicoguide.com
In a first order reaction the concentration halves in every half-life, whatever the starting value. Taking the natural log turns the exponential decay into a straight line of slope −k.
Worked example: A first order reaction is 75% complete in 60 minutes. Find its half-life and rate constant.
Solution: 75% complete means [R] = [R]0/4, which is two half-lives. So t½ = 60/2 = 30 min, and k = 0.693/30 = 0.0231 min−1. Check with the formula: k = (2.303/60) log 4 = (2.303 × 0.602)/60 = 0.0231 min−1.

Pseudo first order reactions

When one reactant is present in large excess, its concentration hardly changes and the reaction behaves as first order. Acid hydrolysis of ethyl acetate in a large excess of water (CH3COOC2H5 + H2O → CH3COOH + C2H5OH) has rate = k′[CH3COOC2H5]. Inversion of cane sugar (C12H22O11 + H2O → C6H12O6 (glucose) + C6H12O6 (fructose), in H+) is another example.

Temperature and the Arrhenius equation

For most chemical reactions, the rate constant nearly doubles for a 10 °C rise in temperature. The dependence is given by the Arrhenius equation:

k = A e−Ea/RTln k = ln A − Ea/RT; log k = log A − Ea/(2.303 RT)log (k2/k1) = [Ea ÷ (2.303 R)] × [(T2 − T1) ÷ (T1T2)]A is the Arrhenius (frequency) factor, Ea the activation energy (J mol−1), R = 8.314 J K−1 mol−1.

A plot of ln k against 1/T is a straight line with slope −Ea/R (for log k, the slope is −Ea/2.303R). The factor e−Ea/RT is the fraction of molecules with kinetic energy equal to or greater than Ea. According to the Maxwell-Boltzmann distribution, raising the temperature spreads the curve to higher energies, and the fraction of molecules above Ea roughly doubles for a 10 °C rise.

Reacting molecules first form an unstable activated complex at the top of the energy barrier. The extra energy needed to reach it from the reactants is the activation energy.

Energy profile with and without a catalystwww.iitmedicoguide.comEaEa′ΔHReactantsProductsActivated complexwithout catalystwith catalystReaction coordinate →Potential energy →www.iitmedicoguide.com
A catalyst gives a path with a lower activation energy (Ea′ instead of Ea). The reactant and product levels do not change, so ΔH, ΔG and the equilibrium constant stay the same.
Worked example: The rate constant of a reaction doubles when the temperature rises from 298 K to 308 K. Find Ea.
Solution: log 2 = [Ea ÷ (2.303 × 8.314)] × [10 ÷ (298 × 308)]. So Ea = (0.3010 × 2.303 × 8.314 × 298 × 308) ÷ 10 = 52900 J mol−1 ≈ 52.9 kJ mol−1.

Effect of a catalyst

A catalyst increases the rate without being used up. By the intermediate complex theory, it forms a temporary bond with the reactants, providing an alternative path of lower activation energy, and is regenerated at the end.

  • A small amount of catalyst can change the rate of a large amount of reactants.
  • It does not change ΔG of the reaction; it can only speed up reactions that are already spontaneous.
  • It speeds up forward and backward reactions equally, so equilibrium is reached sooner but the equilibrium constant is unchanged.

Collision theory

Reactant molecules are treated as hard spheres; a reaction happens only when they collide. The number of collisions per second per unit volume is the collision frequency, Z. A collision is effective only if the molecules have at least the threshold energy and collide with proper orientation.

Rate = P ZAB e−Ea/RTP is the probability (steric) factor, which accounts for orientation. In CH3Br + OH− → CH3OH + Br−, the OH− must approach carbon from the side opposite to Br.

The theory works well for simple atoms and molecules, but it treats molecules as hard spheres and ignores their structure, which is its main weakness.

Common mistakes: (1) Writing the order from the balanced equation. Order comes only from experiment. (2) Calling molecularity zero or fractional. (3) Using log where the formula needs ln, or forgetting the factor 2.303. (4) Thinking t½ of every reaction is independent of concentration. That is true only for first order; for zero order t½ ∝ [R]0. (5) Taking temperature in °C in the Arrhenius equation, or Ea in kJ with R in J.

JEE and NEET focus

  • Relating rates of disappearance and appearance through stoichiometric coefficients.
  • Finding order from initial rate data, and the units of k for any order.
  • Zero and first order integrated equations, graphs and half-lives; results such as 75% completion = 2 half-lives and 99.9% completion ≈ 10 half-lives for first order.
  • Arrhenius calculations at two temperatures and the ln k vs 1/T graph.
  • Energy profile diagrams: activation energy, activated complex, ΔH, and what a catalyst changes and what it does not.

Practice questions

The unit of the rate constant of a second order reaction is:

  1. s−1
  2. mol L−1 s−1
  3. L mol−1 s−1
  4. L2 mol−2 s−1
Show answer
C. (mol L−1)1−2 s−1 = L mol−1 s−1.

The half-life of a first order reaction is 20 min. The time taken for 87.5% completion is:

  1. 40 min
  2. 60 min
  3. 80 min
  4. 35 min
Show answer
B. 12.5% left = 1/8 = three half-lives = 3 × 20 min.

For a zero order reaction, the half-life is:

  1. Independent of initial concentration
  2. Directly proportional to initial concentration
  3. Inversely proportional to initial concentration
  4. Proportional to the square of initial concentration
Show answer
B. t½ = [R]0/2k.

For the rate law rate = k[A]2[B], if [A] is doubled and [B] is halved, the rate becomes:

  1. Unchanged
  2. Two times
  3. Four times
  4. Half
Show answer
B. 22 × ½ = 2.

Which of the following statements about molecularity is correct?

  1. It can be fractional
  2. It can be zero
  3. It is defined only for elementary reactions
  4. It is found only by experiment
Show answer
C. Molecularity is a theoretical whole number for a single elementary step.

The slope of a graph of log k against 1/T is:

  1. −Ea/R
  2. −Ea/2.303R
  3. +Ea/2.303R
  4. log A
Show answer
B. From log k = log A − Ea/(2.303RT).

A catalyst increases the rate of a reaction by:

  1. Increasing ΔH
  2. Decreasing ΔG
  3. Providing a path of lower activation energy
  4. Shifting the equilibrium towards products
Show answer
C. ΔH, ΔG and K are not affected.

Acid hydrolysis of ethyl acetate in a large excess of water is an example of a:

  1. Zero order reaction
  2. Pseudo first order reaction
  3. Third order reaction
  4. Reaction of molecularity zero
Show answer
B. Water is in such excess that its concentration stays practically constant.
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