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VKC Notes · Chemical kinetics · Civil Services P-I · IFoS P-I · Geo-Scientist P-II

9.2 Theories of Reaction Rate: Arrhenius, Collision and Transition-State Theory

★★★ 2021–2025: 21 parts, 190 marks (Civil Services 5 parts / 65 m · IFoS 7 / 63 m · Geo-Scientist 9 / 62 m). An Arrhenius numerical is set almost every year; transition-state theory is a 15–20-marker.

Syllabus lines. Civil Services and IFoS: “rate equations involving … chain reactions; branching chain and explosions; effect of temperature and pressure on rate constant … Collisions and transition state theories”. Geo-Scientist: “Temperature dependence of reaction rate, energy of activation. Collision theory and transition state theory of reaction rates. Enthalpy of activation, entropy of activation, effect of dielectric constant and ionic strength on reaction rate, kinetic isotope effect”. Rate laws and orders have their own notes.

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CSE 2024 Q7(a) = asked there; tap to open the solved answerSN = line kept in the short notes

9.2.1 The Arrhenius equation ★★★ numerical every year

(9.2.1)
EₐΔHactivated complex ‡catalysedRPreaction coordinateEslope = −Eₐ/Rintercept ln A1/Tln k
Fig. 9.2.1 Energy profile and Arrhenius plot.
Draw it in 60 s: left: a hump from R down to P with a lower dashed hump (catalyst), arrows Eₐ and ΔH; right: a straight line falling against 1/T, label slope −Eₐ/R.
  • Eₐ = −R d ln k/d(1/T): slope of the Arrhenius plot × (−R). Slope −1.2 × 10⁴ K gives Eₐ = 1.2 × 10⁴ × 8.314 = 99.8 kJ mol⁻¹ Geo 2024 Q11(b) 5m.SN
  • With base-10 logs, divide by 2.303: log k = 14.34 − 1.25 × 10⁴/T gives Eₐ = 2.303 × 8.314 × 1.25 × 10⁴ = 239 kJ mol⁻¹, A = 1014.34 = 2.19 × 10¹⁴ s⁻¹, k(700 K) = 3.04 × 10⁻⁴ s⁻¹ IFoS 2024 Q2(c) 10m.SN
  • Units of A = units of k; a rule of thumb: Eₐ ≈ 50 kJ mol⁻¹ doubles k for a 10 K rise near room temperature.

Solved PYQs CSE 2025 Q5(a) 5m Geo 2023 Q1(g) 5m Geo 2021 Q12(b) 10m IFoS 2025 Q6(b)(i) 10m CSE 2024 Q1(e) 10m

  1. CH₄ + 2S₂: k = 1.1 × 10⁻³ (823 K) and 6.4 × 10⁻³ m³ mol⁻¹ s⁻¹ (898 K): ln 5.82 = 1.761; 1/823 − 1/898 = 1.015 × 10⁻⁴ K⁻¹; Eₐ = 8.3145 × 1.761/1.015 × 10⁻⁴ = 144 kJ mol⁻¹.
  2. 2HI: 1.33 × 10⁻⁶ (575 K), 2.50 × 10⁻³ (716 K): ln 1880 = 7.539; Eₐ = 183 kJ mol⁻¹; A = 5.6 × 10¹⁰ dm³ mol⁻¹ s⁻¹.
  3. N₂O₅: A = 4.3 × 10¹³ s⁻¹, k = 4.329 × 10⁻⁵ s⁻¹ at 300 K: Eₐ = RT ln(A/k) = 2494 × 41.44 = 103 kJ mol⁻¹.
  4. t½ = 60 s, Eₐ = 104.5 kJ mol⁻¹, A = 5 × 10¹³ s⁻¹: k = 0.693/60 = 0.01155 s⁻¹; T = Eₐ/[R ln(A/k)] = 349 K.
  5. Catalase lowers Eₐ from 42 to 7.0 kJ mol⁻¹; same A, so equal k needs Eₐ/T = Eₐ′/T₀: T = 293.15 × 42/7 = 1759 K; at 20 °C the enzyme speeds the reaction e14.4 ≈ 1.7 × 10⁶ times.SN

9.2.2 Energy profiles, catalysts and odd activation energies ★★

  • Exothermic: products below reactants, Eₐ(forward) < Eₐ(back); endothermic: the reverse, and Eₐ ≥ ΔH. Always: Eₐ(f) − Eₐ(b) = ΔH IFoS 2024 Q3(a) 15m.SN
  • A catalyst gives a new path of lower Eₐ; it is consumed in one step and regenerated in a later one; it does not appear in the overall equation and does not change K. Intermediates are made and then used up. Mechanism A + B → C + D; 2C → F; F + B → 2A + G: overall 3B → 2D + G; A catalyst; C, F intermediates Geo 2025 Q1(g) 5m.SN
  • Specific acid catalysis: kobs = kH⁺[H⁺], so log kobs = log kH⁺ − pH: a line of slope −1, intercept log kH⁺ (at pH 0); it flattens at high pH if an uncatalysed path exists Geo 2025 Q1(h) 5m.
  • Heterogeneous catalysis (Haber: N₂ + 3H₂ on Fe): adsorb, dissociate, react on the surface, desorb; see the adsorption notes (11.2) IFoS 2025 Q6(b)(ii) 5m.
  • Can Eₐ be zero or negative? Zero: radical recombination (CH₃• + CH₃•) has no barrier. Negative (apparent): when a fast exothermic pre-equilibrium comes first, kobs = K₁k₂ and Eₐ = E₂ + ΔH₁ < 0 if ΔH₁ is large and negative: 2NO + O₂ → 2NO₂ goes slower as T rises IFoS 2022 Q8(c)(ii) 5m.SN

9.2.3 Collision theory of bimolecular gas reactions ★★★

  1. Molecules are hard spheres; reaction needs a collision. Collision cross-section σ = πdAB², dAB = (dA + dB)/2; mean relative speed (8kBT/πμ)½.
  2. Collision density ZAB = σ(8kBT/πμ)½ NANB (number densities).
  3. Only collisions with energy along the line of centres ≥ E₀ react: fraction e−E₀/RT.
  4. Only correctly oriented ones react: steric factor P Geo 2021 Q5 10m.SN
(9.2.2)

Limitations CSE 2021 Q6(b) 20m:SN

  • Molecules have no internal structure; P (10⁻⁶ to 1) cannot be calculated, only fitted; E₀ must come from experiment.
  • No entropy term, so no link with K = eΔS°/Re−ΔH°/RT; fails for complex molecules and for reactions in solution; predicts A ∝ T½ only.

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