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

6.3 Chemical Equilibrium: ΔG° and K, Kp, Kc and Kx, van ’t Hoff Isochore and Isobar

★★★ 2021–2025: 12 Mains parts, 98 marks (Civil Services 2 parts / 15 m · IFoS 2 / 23 m · Geo-Scientist 8 / 60 m) and 2 Geo-Scientist Prelims questions. Short numericals: one formula, one log, the sign read correctly.

Syllabus lines. Civil Services: “criteria for equilibrium, relation between equilibrium constant and thermodynamic quantities”. IFoS: the same, plus “chemical equilibrium in ideal gas reactions” (statistical thermodynamics). Geo-Scientist: “Thermodynamic conditions for equilibrium, degree of advancement. van ’t Hoff’s reaction isotherm. Equilibrium constant and standard Gibbs energy change. Definitions of KP, KC and Kx; van ’t Hoff’s reaction isobar and isochore”.

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

6.3.1 Reaction Gibbs energy and the equilibrium constant ★★★

(6.3.1)
  1. Extent of reaction ξ (degree of advancement): dni = νidξ (ν negative for reactants). At constant T, p: dG = Σμidni = (Σνiμi)dξ, so ΔrG is the slope of G against ξ Geo 2024 Q4 10m.SN
  2. Put μi = μi° + RT ln ai: ΔrG = ΔrG° + RT ln Q, Q = Πaiνi (the van ’t Hoff reaction isotherm).SN
  3. At equilibrium G is a minimum: ΔrG = 0 and Q = K, so ΔrG° = −RT ln K. μ° depends only on T, so K depends only on T.SN
  4. Activities: ideal gas pi/p°; ideal solute ci/c°; pure solid or liquid 1. K is dimensionless.
0.00.20.40.60.81.00-1-2-3-4extent of reaction ξ (mol of E formed)G − G(Z) / kJminimum at ξ = 0.80:Δr G = 0, Q = Kslope = Δr G < 0:forward runsdashed: no mixing term,falls straight to Δr G°
Fig. 6.3.1 G against ξ for Z-2-butene → E-2-butene at 300 K (ΔrG° = −3.44 kJ mol−1, computed): the mixing term makes a minimum short of complete conversion.
Draw it in 60 s: a curve falling from the left, a minimum, a short rise to the right; a dashed straight line from the same start to a lower end; mark the minimum ΔrG = 0.
  • Why a minimum and not complete reaction: even when ΔrG° < 0, mixing products with reactants lowers G further (RTΣx ln x). Z → E-2-butene: K = e3440/(8.314 × 300) = 3.97, xE = K/(1 + K) = 0.80 Geo Pre 2026 Q62.SN

6.3.2 Using ΔG° = −RT ln K ★★★

  • A₂ → 2A at 1200 K, K = 3.3 × 10⁻³, ΔrH° = 216.65 kJ: ΔrG° = −(8.314)(1200) ln(3.3 × 10⁻³) = +57.0 kJ mol⁻¹; ΔrS° = (216 650 − 57 006)/1200 = +133.0 J K⁻¹ mol⁻¹ (one gas molecule → two). K = 1 at ΔrH°/ΔrS° = 1628 K Geo 2023 Q4 10m.SN
  • Solids in the equilibrium: activity 1, so K contains only the gas. U(s) + 3/2 H₂(g) → UH₃(s), p(H₂) = 139 Pa at 500 K: K = (pH₂/p°)−3/2; ΔfG° = (3/2)RT ln(139/101 325) = −41.1 kJ mol⁻¹ IFoS 2025 Q1(c) 8m.SN
  • NiO(s) ⇌ Ni(s) + ½O₂(g), ΔrG° = 212 kJ: ln K = −85.57, K = 6.9 × 10⁻³⁸ = p(O₂)½, so p(O₂) = 4.8 × 10⁻⁷⁵ atm: NiO does not decompose by itself; nickel needs a reducing agent CSE 2024 Q3(c)(iii) 5m.
  • From cell potentials: ΔrG° = −nFE° = −RT ln K. Ce + Cr³⁺ → Ce³⁺ + Cr: E° = −0.41 − (−2.33) = 1.92 V; log K = 3 × 1.92/0.05916 = 97.4, K ≈ 2.3 × 10⁹⁷ Geo 2025 Q1(j) 5m.SN

6.3.3 Kp, Kc and Kx ★★★

(6.3.2)
  • Derivation: put pi = ciRT (ideal gas) and pi = xip (Dalton) into Kp = Πpiνi. All three are equal when Δn = 0 (H₂ + I₂ ⇌ 2HI). Kp and Kc depend only on T; Kx also on p Geo 2024 Q1(f) 5m.SN
  • NH₃ ⇌ ½N₂ + 3/2 H₂, Kp = 1.36 × 10⁻³ at 298 K, Δn = +1: Kc = Kp/RT = 1.36 × 10⁻³/(0.08204 × 298) = 5.56 × 10⁻⁵ mol L⁻¹ (with p in atm; 5.49 × 10⁻⁵ if Kp is on the bar scale) Geo 2025 Q11(b) 5m.SN

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