6.2 Chemical Potential, Partial Molar Quantities, Gibbs–Duhem Equation, Fugacity and Activity

6.2.1 Partial molar quantities ★★

• Y is any extensive property (V, U, H, S, G). Its partial molar value Ȳ_i is the change in Y when 1 mol of i is added to so large a mixture that the composition does not change. It is an intensive property and depends on composition.SN
• Example: 1 mol of water added to a large volume of water raises V by 18 cm³, but added to a large volume of ethanol by only about 14 cm³: each H₂O molecule packs among ethanol molecules. Partial molar volumes can even be negative (MgSO₄ in water, −1.4 cm³ mol⁻¹ at infinite dilution: ions shrink the solvent round them).SN
• Determination: (i) slope method: measure V of solutions with fixed n_A and varying n_B; the slope of V against n_B is V̄_B (Fig. 6.2.1); (ii) intercept method: plot the mean molar volume against x_B; the tangent at any composition cuts the two axes at V̄_A and V̄_B; (iii) from apparent molar volumes. Then Eq. 6.2.1 adds them back: V = n_AV̄_A + n_BV̄_B.SN

6.2.2 Chemical potential: definitions and meaning ★★★

• Meaning: the partial molar Gibbs energy; the escaping tendency of i. Matter moves from high μ to low μ (phase to phase, place to place, reactants to products), as heat moves from high T to low T. For a pure substance μ = G_m Geo 2023 Q1(e) 5m Geo 2024 Q1(e) 5m.SN
• Equilibrium: μ_i is the same in every phase in which i occurs; for a reaction Σν_iμ_i = 0.SN

6.2.3 Variation of μ with temperature and pressure ★★★

• Derivation: dG is exact, so cross-derivatives are equal: ∂/∂n_i of (∂G/∂T) = −S gives (∂μ_i/∂T) = −S̄_i; of (∂G/∂p) = V gives (∂μ_i/∂p) = V̄_i Geo 2022 Q4 10m.SN
• With T: S_m > 0, so μ falls on heating; S_m(g) ≫ S_m(l) > S_m(s), so the gas line is steepest and (∂μ_s/∂T) > (∂μ_l/∂T) ≫ (∂μ_g/∂T) (all negative). The lowest line is the stable phase; the crossings are T_f and T_b IFoS 2023 Q4(b) 10m.SN
• With p: V̄ > 0, so μ rises with pressure. Liquid water, 1 → 1000 atm: Δμ = V_mΔp = 18.07 × 10⁻⁶ × 999 × 101 325 = 1.83 kJ mol⁻¹; an ideal gas over the same range: RT ln 1000 = 17.1 kJ mol⁻¹, nine times more. So pressure lifts the gas line most and raises T_b (Fig. 6.2.2).SN

• Moving the standard state to 2 atm: each ideal gas gains RT ln 2 = 1.72 kJ mol⁻¹, so Δ_fG° changes by Δn_gasRT ln 2. HCl (½H₂ + ½Cl₂ → HCl, Δn = 0): unchanged, −95.3 kJ mol⁻¹. N₂O (N₂ + ½O₂ → N₂O, Δn = −½): 103.7 − 0.86 = +102.8 kJ mol⁻¹; the reaction loses less entropy at the higher pressure CSE 2025 Q4(d) 10m.

6.2.4 Gibbs–Duhem equation ★★★ 10–15 marks

1. At constant T, p: dG = Σμ_idn_i (1).
2. Build the mixture up at constant composition: every μ_i stays constant, so integrating gives G = Σn_iμ_i (2) (Euler’s theorem for a function homogeneous of degree one in the n_i).SN
3. Differentiate (2) in general: dG = Σn_idμ_i + Σμ_idn_i. Compare with the full dG = −SdT + Vdp + Σμ_idn_i Geo 2023 Q12(a) 15m:SN

• The same steps work for any extensive Y (V, H, S): Σn_idȲ_i = 0 at constant T, p Geo 2021 Q12(a) 10m.SN
• Meaning: the partial molar quantities of a mixture are not independent. In a binary, x_Adμ_A = −x_Bdμ_B: if μ_B rises, μ_A must fall, weighted by the mole fractions. Measuring one component fixes the other.SN
• Uses: the Duhem–Margules equation (x_A d ln p_A/dx_A = x_B d ln p_B/dx_B): if A obeys Raoult’s law, B must obey Henry’s law; testing activity data for consistency; activity of a non-volatile solute from the solvent’s vapour pressure; it leads to the phase rule (one Gibbs–Duhem relation per phase).