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

9.1 Rate Laws and Order: Integrated Rate Equations, Half-Life, Reversible, Parallel and Consecutive Reactions

★★★ 2021–2025: 25 parts, 211 marks (Civil Services 8 parts / 70 m · IFoS 6 / 64 m · Geo-Scientist 11 / 77 m). Every paper of every year has at least one rate-law derivation or half-life numerical; consecutive and reversible reactions are the 10–15-markers.

Syllabus lines. Civil Services and IFoS: “differential and integral rate equations for zeroth, first, second and fractional order reactions; rate equations involving reverse, parallel, consecutive and chain reactions … Study of fast reactions by stop-flow and relaxation methods”. Geo-Scientist: “Second order reactions. Determination of order of reactions. Parallel and consecutive reactions.” Arrhenius, collision and transition-state theory, steady state and chain reactions are in 9.2.

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

9.1.1 Rate, rate law, order and the units of k ★★

(9.1.1)
  • The rate law is found by experiment. The orders m, n need not equal the coefficients of the equation and may be zero or fractional; molecularity (1, 2, rarely 3) belongs only to an elementary step.SN
  • Units of k = (concentration)1−n time⁻¹: zero order mol dm⁻³ s⁻¹; first s⁻¹; second dm³ mol⁻¹ s⁻¹; order 5/2: dm9/2 mol−3/2 s⁻¹ CSE 2025 Q7(d) 10m.SN
  • Molecule units to molar units: multiply by NA and convert cm³ to dm³. O + O₃: 8.0 × 10⁻¹⁵ cm³ molecule⁻¹ s⁻¹ × 6.022 × 10²³ × 10⁻³ = 4.8 × 10⁶ dm³ mol⁻¹ s⁻¹ Geo 2024 Q1(g) 5m.

Solved PYQs IFoS 2023 Q6(a) 15m Geo 2024 Q5(a) 5m

  1. Initial rates. 2A + 2B → C + 2D: [A] 0.004 → 0.008 at fixed [B], rate × 4 ⇒ m = 2; [B] 0.0015 → 0.0030 at fixed [A], rate × 2 ⇒ n = 1. Rate = k[A]²[B]; k = 1.2 × 10⁻⁵/[(0.004)²(0.0015)] = 500 L² mol⁻² s⁻¹; at [A] = 0.012, [B] = 0.006: rate = 500(0.012)²(0.006) = 4.32 × 10⁻⁴ M s⁻¹.SN
  2. Order from two rates. Rate 3.70 at 10 % reacted, 2.60 at 25 %: 3.70/2.60 = (0.90/0.75)n, n = ln 1.423/ln 1.20 = 1.94 ≈ 2.

9.1.2 Integrated rate laws and half-lives ★★★ every year

OrderIntegrated formStraight lineHalf-life
0[A] = [A]₀ − kt[A] vs t, slope −k[A]₀/2k (∝ [A]₀)
1ln([A]₀/[A]) = ktln [A] vs t, slope −k0.693/k (independent)
2 (A + A)1/[A] − 1/[A]₀ = kt1/[A] vs t, slope +k1/k[A]₀
n ≠ 1[A]1−n − [A]₀1−n = (n − 1)kt∝ [A]₀1−n
(9.1.2)
0123400.250.51t / t½[A]/[A]0zero orderfirstsecond
Fig. 9.1.1 Fraction left against time (units of the first half-life) for orders 0, 1, 2.
Draw it in 60 s: three curves from 1 at t = 0, all through ½ at 1; straight line to zero at 2 (zero order), ¼ at 2 (first), ¼ at 3 (second).
  • Zero order: −d[A]/dt = k, so [A] = [A]₀ − kt; t½ = [A]₀/2k ∝ [A]₀; the reaction ends at t = [A]₀/k. Example: NH₃ on a saturated hot W or Pt surface CSE 2021 Q5(a) 10m.SN
  • First order: t½ = 0.693/k, [A]₀ cancels; every half-life is equal Geo 2021 Q1(h) 5m. t99 = ln 100/k = 6.64 t½; t99.9 = ln 1000/k = 9.97 ≈ 10 t½: the “10 half-lives” result holds for 99.9 %, not 99 % Geo 2024 Q11(a) 10m.SN
  • Second order: t½ = 1/k[A]₀ depends on [A]₀; each half-life is twice the previous one; t¾ = 3t½ (first order: 2t½) Geo 2022 Q1(g) 5m Geo 2025 Q12(a) 5m.SN
(9.1.3)
  • Derivation of (9.1.3): dx/dt = k(a − x)(b − x); partial fractions 1/[(a − x)(b − x)] = [1/(a − b)][1/(b − x) − 1/(a − x)]; integrate from x = 0 CSE 2025 Q7(d).SN
  • Product growth (first order N₂O₂ → 2NO): [N₂O₂] = [N₂O₂]₀e−kt and [NO] = 2[N₂O₂]₀(1 − e−kt); the factor 2 comes from the stoichiometry CSE 2022 Q5(a) 10m.

Solved PYQs CSE 2023 Q5(a) 10m IFoS 2021 Q5(a) 8m

  1. AB₂ → AB + B, k = 2.8 × 10⁻⁷ s⁻¹: t½ = 0.693/k = 2.48 × 10⁶ s = 28.7 days; 1 g → 0.6 g: t = ln(1/0.6)/k = 21.1 days; after 35 days m = e−0.847 = 0.43 g. Molar mass cancels: use masses directly.SN
  2. Second order, 30 % in 30 min: k[A]₀ = (1/0.70 − 1)/30 = 1/70 min⁻¹; at 60 min [A]/[A]₀ = 1/(1 + 60/70) = 7/13 = 0.538. A number for k itself needs [A]₀.

9.1.3 Finding the order ★★★

t[A]zero orderslope = −ktln [A]first orderslope = −kt1/[A]second orderslope = +k
Fig. 9.1.2 The three straight-line (graphical) tests.
Draw it in 60 s: three small axes: [A] falling line (0), ln [A] falling line (1), 1/[A] rising line (2); the one that is straight gives the order.
  1. Integration (substitution) method: put the data into each integrated law; the one that gives a constant k (or a straight line) is the order.SN
  2. Half-life method: t½ ∝ [A]₀1−n, so n = 1 + log(t1/t2)/log(a2/a1). Two runs suffice and no k is needed.SN
  3. Initial rates (differential method): log v₀ = log k + n log [A]₀; slope of log v₀ vs log [A]₀ is n.
  4. Isolation (Ostwald): all reactants but one in large excess, so their concentrations stay constant; the reaction becomes pseudo-nth order in the isolated one. A catalyst ([H⁺]) is constant in a run in the same way.SN
  • Is the half-life method “the best”? Justify: quick, needs only two experiments, works for any n. Criticise: valid only for one reactant (or all at equal concentration); fails for reversible, parallel and consecutive reactions where t½ is not defined by one n; t½ read from a graph has error. So it is good for simple reactions, not the best in general Geo 2023 Q9(a) 2m.
  • Gas pressure data: for A(g) → 2B(g) at constant volume, p = p₀ + x, so pA = 2p₀ − p; then test ln(pA0/pA) against t IFoS 2025 Q5(a) 8m.SN

Solved PYQs CSE 2025 Q5(d) 5m IFoS 2025 Q5(a) Geo 2025 Q12(b) 10m

  1. NH₄NO₂ → N₂ + 2H₂O, V∞ = 35.05 cc: V∞ − Vt = 28.80, 26.05, 23.65, 21.40 at 10, 15, 20, 25 min; k = (1/t) ln[V∞/(V∞ − Vt)] = 1.96, 1.98, 1.97, 1.97 × 10⁻² min⁻¹: constant, so first order, t½ = 35 min.SN
  2. ClCO₂CCl₃(g) → 2COCl₂ (p/mmHg 15.76, 18.88, 22.79, 27.08 at 0, 181, 513, 1164 s): pA = 15.76, 12.64, 8.73, 4.44; ln(pA0/pA) = 0, 0.221, 0.591, 1.267, nearly linear in t: first order, k ≈ 1.1 × 10⁻³ s⁻¹.
  3. Sucrose, −d[S]/dt = k[S]a[H⁺]b: t½ independent of [S]₀ ⇒ a = 1; t½ 500 min (pH 5) and 50 min (pH 4): 10 = 10b ⇒ a = 1, b = 1.

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