Gibbs-Duhem Equation: Application to Binary Solutions
Chemical thermodynamics is central in examinations including CSIR NET, GATE Chemistry, IIT JAM, and CUET-PG. In non-ideal mixtures, chemical potentials and partial molar quantities do not vary independently. When one component changes, the other responds in an exact manner governed by the Gibbs-Duhem equation. Mastering its binary forms enables students to solve partial volume problems, find activity coefficients, and verify data consistency.
Derivation from Euler's Theorem
At constant temperature and pressure, total Gibbs energy G is extensive. By Euler's theorem, G = ∑ ni μi, where ni is mole number and μi is chemical potential (μi = (∂G / ∂ni)T, P, nj ≠ i). Differentiating gives dG = ∑ μi dni + ∑ ni dμi. Because dG = ∑ μi dni at constant T and P, equating both gives the Gibbs-Duhem equation:
∑ ni dμi = 0 or in mole fractions ∑ xi dμi = 0
The Gibbs-Duhem Equation for Binary Solutions
For a binary solution of components 1 and 2, the equation simplifies into a working relation:
Gibbs-Duhem Equation (Binary Solution at Constant T, P):
x1 dμ1 + x2 dμ2 = 0 and x1 dȲ1 + x2 dȲ2 = 0
Variables and Standard Units:
x1, x2 = Mole fractions of components 1 and 2 (dimensionless, x1 + x2 = 1)
μ1, μ2 = Chemical potentials (J/mol)
Ȳ1, Ȳ2 = Partial molar properties, such as partial volume V̄ (cm3/mol)
Application to Activities and Activity Coefficients
Because μi = μi° + R T ln ai, differentiating at constant temperature gives dμi = R T d(ln ai). Substituting into the binary equation gives x1 d(ln a1) + x2 d(ln a2) = 0. Because ai = xi γi and x1 d(ln x1) + x2 d(ln x2) = dx1 + dx2 = 0, activity coefficients satisfy:
x1 d(ln γ1) + x2 d(ln γ2) = 0 ⇒ d(ln γ2) = - (x1 / x2) d(ln γ1)
Measuring solvent vapour pressures yields solute activity coefficients by integration.
The Duhem-Margules Equation and Raoult's Law
For an ideal vapour phase, the activity relation gives the Duhem-Margules equation: (x1 / P1)(dP1 / dx1) = (x2 / P2)(dP2 / dx2). This proves two facts:
- If component 1 obeys Raoult's law (γ1 = 1), component 2 must also obey Raoult's law (γ2 = 1).
- If the solvent obeys Raoult's law near infinite dilution, the solute obeys Henry's law.
Worked Numerical Examples
Exam details
- Source: practice problem written for this article (not a past-paper question)
- Relevant exams: CSIR NET, GATE Chemistry, IIT JAM, CUET-PG
- Topic: Gibbs-Duhem Equation and Binary Solutions
- Difficulty: Advanced
Question: At 298 K, the molar volume of a binary mixture of A and B is Vm = 50.0 + 15.0 xB - 4.0 xB2 (in cm3/mol). Calculate partial molar volumes V̄A and V̄B at xB = 0.40.
Solution:
Step 1: Formulas: V̄B = Vm + (1 - xB)(dVm / dxB) and V̄A = Vm - xB(dVm / dxB).
Step 2: Differentiate: dVm / dxB = 15.0 - 8.0 xB.
Step 3: At xB = 0.40: dVm / dxB = 15.0 - 8.0(0.40) = 11.80 cm3/mol. With xA = 0.60, Vm = 50.0 + 15.0(0.40) - 4.0(0.40)2 = 55.36 cm3/mol.
Step 4: Compute: V̄B = 55.36 + (0.60)(11.80) = 62.44 cm3/mol; V̄A = 55.36 - (0.40)(11.80) = 50.64 cm3/mol.
Step 5: Double-check: Vm = xA V̄A + xB V̄B = (0.60)(50.64) + (0.40)(62.44) = 55.36 cm3/mol. Both methods agree.
Answer: V̄A = 50.64 cm3/mol, V̄B = 62.44 cm3/mol
Question: In a binary solution at 298 K, ln γ1 = β x22 with β = 0.85. Using the Gibbs-Duhem equation, deduce ln γ2 as a function of x1 and calculate γ2 at x2 = 0.30 (x1 = 0.70).
Solution:
Step 1: Write relation: d(ln γ2) = - (x1 / x2) d(ln γ1).
Step 2: Differentiate: d(ln γ1) = 2 β x2 dx2.
Step 3: Substitute: d(ln γ2) = - (x1 / x2)(2 β x2 dx2) = - 2 β x1 dx2 = 2 β x1 dx1, since dx2 = - dx1.
Step 4: Integrate from pure 2 (x1 = 0, ln γ2 = 0) to x1: ln γ2 = 2 β ∫0x1 x1 dx1 = β x12.
Step 5: Evaluate at x1 = 0.70: ln γ2 = 0.85 × (0.70)2 = 0.4165, giving γ2 = e0.4165 ≈ 1.52.
Step 6: Double-check: GE / (n R T) = β x1 x2 gives ∂[n GE/(RT)] / ∂n2 = β x12 = 0.4165, confirming γ2 = 1.52.
Answer: ln γ2 = β x12, γ2 = 1.52
Common Student Mistakes
- Missing Minus Sign: Writing dȲ2 / dx1 = (x1 / x2)(dȲ1 / dx1) without negative sign. A positive change in one partial quantity requires an opposing change in the other.
- Applying at Variable T or P: Using ∑ xi dμi = 0 when T or P varies. At variable T and P, the full relation -S dT + V dP - ∑ ni dμi = 0 applies.
- Incorrect Integration Limits: Forgetting that at x1 = 0, component 2 is pure liquid (γ2 = 1, ln γ2 = 0). Non-zero limits yield erroneous formulas.
- Confusing Total with Partial Volume: Assuming partial molar volume equals total volume divided by moles. Partial molar volumes depend on mixture composition.
| Competitive Examination | Question Types and Focus Areas | Preparation Strategy |
|---|---|---|
| CSIR NET Chemical Sciences | Thermodynamic consistency proofs and activity coefficient integration | Master Euler reciprocity and review excess free energy relations. |
| GATE Chemistry (CY) | Numerical Answer Type questions on partial molar volumes | Focus on the tangent-intercept method and check signs in derivatives. |
| IIT JAM Chemistry | Concept of partial molar quantities and link to Raoult's law | Learn the derivation from G = ∑ ni μi and practise binary algebra. |
| CUET-PG Chemistry | Objective questions on Gibbs-Duhem conditions and Duhem-Margules results | Review chemical potential definitions and ideal solution conditions. |
Turn this concept into marks: solve previous-year CSIR-NET questions and take a timed quiz.
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