Partial molar properties and Gibbs-Duhem equation

Why it matters

Mix 50 mL of ethanol with 50 mL of water and you get about 96 mL, not 100 mL; dissolving acid in water releases heat. Mixtures do not simply add up their pure-component properties. Partial molar properties tell you how much each component really contributes, and the chemical potential (the partial molar Gibbs energy) is the quantity that must be equal across phases at equilibrium, so this topic is the foundation of all solution thermodynamics, VLE and reaction equilibrium.

Key ideas

Definition. For any extensive property nM of a mixture (V, U, H, S, G), the partial molar property of species i is the rate at which nM changes when a small amount of i is added at constant T, P and amounts of all other species: M̄ᵢ = [∂(nM)/∂nᵢ]_{T,P,nⱼ}. It is an intensive property and depends on composition. For an ideal-gas mixture or an ideal solution, the partial molar volume equals the pure-component molar volume; in real solutions it differs and can even be negative (for example some salts at high dilution in water).

Summability. Because nM is homogeneous of degree one in the mole numbers at fixed T and P, the whole equals the sum of its partial parts: nM = Σnᵢ·M̄ᵢ, or M = Σxᵢ·M̄ᵢ.

Chemical potential. μᵢ = Ḡᵢ = [∂(nG)/∂nᵢ]_{T,P,nⱼ}. The fundamental relation for an open phase becomes d(nG) = −(nS)dT + (nV)dP + Σμᵢ dnᵢ. Equality of μᵢ in all phases is the criterion for phase equilibrium.

Gibbs-Duhem equation. Differentiating summability and comparing with the general change in nM gives a constraint among the partial properties. At constant T and P: Σxᵢ dM̄ᵢ = 0. The partial properties of a mixture are therefore not independent: in a binary, if you know M̄₁ as a function of composition you can find M̄₂ by integration. Applied to G it gives Σxᵢ dμᵢ = 0 and, in excess form, Σxᵢ d ln γᵢ = 0, which is used to test experimental activity-coefficient data for thermodynamic consistency and to build models such as Margules and van Laar.

Binary mixtures and the tangent-intercept method. For a binary, if M is known as a function of x₁ at fixed T and P:

• M̄₁ = M + x₂·dM/dx₁
• M̄₂ = M − x₁·dM/dx₁ Graphically, draw the tangent to the M versus x₁ curve at the composition of interest. Its intercept at x₁ = 1 is M̄₁ and its intercept at x₁ = 0 is M̄₂.

Limits. As xᵢ → 1, M̄ᵢ → Mᵢ (the pure-component value). As xᵢ → 0, M̄ᵢ → M̄ᵢ^∞, the value at infinite dilution, which can be very different from Mᵢ.

Property change of mixing. ΔM_mix = M − ΣxᵢMᵢ = Σxᵢ(M̄ᵢ − Mᵢ). ΔV_mix and ΔH_mix are zero for an ideal solution, but ΔS_mix and ΔG_mix are not (they are covered under ideal solutions).

Formulas

M̄ᵢ = [∂(nM)/∂nᵢ]_{T,P,nⱼ} nM = Σ nᵢ·M̄ᵢ, M = Σ xᵢ·M̄ᵢ (summability) μᵢ = Ḡᵢ d(nG) = −(nS) dT + (nV) dP + Σ μᵢ dnᵢ Σ xᵢ dM̄ᵢ = 0 (Gibbs-Duhem, constant T and P) x₁ dM̄₁/dx₁ + x₂ dM̄₂/dx₁ = 0 (binary, constant T and P) M̄₁ = M + x₂·dM/dx₁, M̄₂ = M − x₁·dM/dx₁ (binary) ΔM_mix = M − Σ xᵢ·Mᵢ

Symbols: n total moles (mol); nᵢ moles of i; xᵢ mole fraction; M molar property of the mixture (for example V in m³/mol or cm³/mol, H in J/mol); M̄ᵢ partial molar property of i (same units as M); Mᵢ pure-component molar property; μᵢ chemical potential (J/mol); T (K); P (Pa). Gibbs-Duhem as written holds at constant T and P for any mixture, ideal or not.

Worked examples

Example 1 (standard): partial molar volumes from a fitted curve. Given: for a binary liquid at 25 °C and 1 bar, V = 109.4 − 16.8x₁ − 2.64x₁² cm³/mol. Find V̄₁ and V̄₂ at x₁ = 0.4, and check summability.

1. V at x₁ = 0.4: 109.4 − 6.72 − 0.4224 = 102.258 cm³/mol.
2. dV/dx₁ = −16.8 − 5.28x₁ = −16.8 − 2.112 = −18.912 cm³/mol.
3. V̄₁ = V + x₂·dV/dx₁ = 102.258 + 0.6 × (−18.912) = 90.910 cm³/mol.
4. V̄₂ = V − x₁·dV/dx₁ = 102.258 − 0.4 × (−18.912) = 109.822 cm³/mol.
5. Check: x₁V̄₁ + x₂V̄₂ = 0.4 × 90.910 + 0.6 × 109.822 = 36.364 + 65.893 = 102.258 cm³/mol. Matches V. Answer: V̄₁ = 90.91 cm³/mol, V̄₂ = 109.82 cm³/mol

Example 2 (GATE level): partial molar enthalpies and infinite dilution. Given: for a binary at fixed T and P, H = 400x₁ + 600x₂ + x₁x₂(40x₁ + 20x₂) J/mol. Find H̄₁ and H̄₂ as functions of x₁, the pure-component and infinite-dilution values, and ΔH_mix at x₁ = 0.3.

1. Substitute x₂ = 1 − x₁ and expand: H = 600 − 180x₁ − 20x₁³ J/mol, so dH/dx₁ = −180 − 60x₁².
2. H̄₁ = H + (1 − x₁)·dH/dx₁ = 600 − 180x₁ − 20x₁³ + (1 − x₁)(−180 − 60x₁²) = 420 − 60x₁² + 40x₁³.
3. H̄₂ = H − x₁·dH/dx₁ = 600 − 180x₁ − 20x₁³ + 180x₁ + 60x₁³ = 600 + 40x₁³.
4. Pure values: H₁ = H̄₁(x₁ = 1) = 400 J/mol; H₂ = H̄₂(x₁ = 0) = 600 J/mol.
5. Infinite dilution: H̄₁^∞ = H̄₁(x₁ = 0) = 420 J/mol; H̄₂^∞ = H̄₂(x₁ = 1) = 640 J/mol.
6. Gibbs-Duhem check: x₁(−120x₁ + 120x₁²) + (1 − x₁)(120x₁²) = −120x₁² + 120x₁³ + 120x₁² − 120x₁³ = 0. Satisfied.
7. At x₁ = 0.3: H = 600 − 54 − 0.54 = 545.46 J/mol; ΔH_mix = 545.46 − (0.3 × 400 + 0.7 × 600) = 545.46 − 540 = 5.46 J/mol. Answer: H̄₁ = 420 − 60x₁² + 40x₁³, H̄₂ = 600 + 40x₁³ (J/mol); H̄₁^∞ = 420 J/mol, H̄₂^∞ = 640 J/mol; ΔH_mix(0.3) = 5.46 J/mol

Common mistakes

• Writing M̄₁ = M + x₁·dM/dx₁; the multiplier is the other component's mole fraction, x₂.
• Differentiating with respect to x₁ while x₂ is still written separately; substitute x₂ = 1 − x₁ first.
• Treating the partial molar volume as the pure molar volume in a non-ideal solution.
• Applying Gibbs-Duhem in the form Σxᵢ dM̄ᵢ = 0 when T or P is changing.
• Mixing up the intercepts: the tangent's intercept at x₁ = 1 gives M̄₁, not M̄₂.

For GATE CH

• NAT problems: given M(x₁) as a polynomial, find M̄₁ or M̄₂ at a composition, at infinite dilution, or the property change of mixing.
• Gibbs-Duhem consistency: given an expression for M̄₁ (or ln γ₁), find M̄₂ (or ln γ₂), or decide whether a proposed pair is thermodynamically consistent.
• Conceptual MCQs on summability, chemical potential as partial molar Gibbs energy and the tangent-intercept construction.
• Practise the algebra quickly: substitute x₂ = 1 − x₁, expand, differentiate, and check with summability.

Quick check

1. Which variables are held constant in the definition of a partial molar property?
2. State the summability relation for a binary.
3. The chemical potential is the partial molar value of which property?
4. If H̄₁ = a + b·x₂², what form must H̄₂ take (up to a constant) to satisfy Gibbs-Duhem?
5. As x₁ → 1, what does M̄₁ approach?

Answers: 1. T, P and the moles of all other species. 2. M = x₁M̄₁ + x₂M̄₂. 3. Gibbs energy. 4. H̄₂ = constant + b·x₁². 5. The pure-component molar property M₁.