The Gibbs–Duhem Equation

Introduction

It might seem that every component in a mixture can have its chemical potential changed independently. Thermodynamics says otherwise. At a fixed temperature and pressure, raising the chemical potential of one component through a composition change must be accompanied by changes in the others. The Gibbs–Duhem equation expresses this connection and prevents impossible models of solution behaviour.

Core explanation

Gibbs energy G is extensive: at fixed T, p and composition, doubling every component amount doubles G. Euler's theorem for such a function gives G = Σ i n iμ i, where μ i = (∂G/∂n i) T,p,n j. Differentiate the Euler expression: dG = Σ i μ i dn i + Σ i n i dμ i. Separately, the fundamental Gibbs differential gives dG = −S dT + V dp + Σ i μ i dn i. Subtracting the common amount-change terms yields Σ i n i dμ i = −S dT + V dp. Equivalently, S dT − V dp + Σ i n i dμ i = 0.

At fixed T and p, the result simplifies to Σ i n i dμ i = 0. For a binary mixture, n A dμ A + n B dμ B = 0, or x A dμ A + x B dμ B = 0 after dividing by total amount. If a composition change makes μ A rise slightly, μ B must fall in a proportion set by their amounts, assuming both are present and the system follows an equilibrium mixture path. This relationship does not state that μ A = μ B; different chemical species generally have different chemical potentials.

The equation is a differential restriction, not a claim that each μ i remains constant whenever T and p are fixed. Composition may vary, causing individual μ values to change. It also does not mean one can choose arbitrary changes satisfying the sum and thereby create a physically valid mixture; the actual μ i depend on a consistent Gibbs-energy function. The restriction is especially useful when deriving activity relations or checking whether measured partial molar properties are thermodynamically compatible.

For a pure one-component phase, fixed T and p imply n dμ = 0 and thus dμ = 0 along a path that changes only the amount. This is consistent with chemical potential being intensive. In a mixture, changing a single n while the other n values remain fixed changes composition, so μ values may shift. Extensivity and intensivity therefore fit together through the Gibbs–Duhem relation.

Step-by-step reasoning

Start with G = Σn iμ i, valid at a specified equilibrium state for an extensive system. Differentiate every product using the product rule. Write the independent fundamental differential of G, then cancel Σμ i dn i. Only after obtaining the general equation set dT = dp = 0. For a binary mixture, divide by total moles if mole fractions are more convenient.

Visual explanation

Plot μ A and μ B against x A at one T and p. As composition moves, both curves vary. At a selected composition, draw tangent arrows with opposite weighted contributions so that x A dμ A + x B dμ B = 0. A second diagram shows G as an extensive surface whose tangent plane has slopes μ A and μ B.

Real-world analogy

Imagine a shared budget with two variable expenditures constrained to keep the total change zero. If one expenditure rises, the other must change to offset it, with weights depending on how many units of each are purchased. The mixture constraint is more precise: its weights are mole numbers and its variables are chemical potentials, not arbitrary prices.

Real-world example

Activity coefficients of two components in a liquid mixture cannot be assigned completely independent composition curves. Experimental vapour–liquid equilibrium data can be tested for consistency with the Gibbs–Duhem relation. A proposed model that fits one component but violates the weighted differential relation for the other cannot arise from one coherent mixture Gibbs energy.

Why?

The relation follows from combining two ways to describe the same dG. The fundamental differential accounts for changes in amounts, T and p; Euler's expression accounts for how total G is built from partial molar Gibbs energies. Their agreement requires a restriction on how the intensive μ i can vary. Without it, scaling a system could change a supposedly intensive chemical potential.

Common misconception

The equation at fixed T and p does not say Σn iμ i = 0; that sum equals G. It says Σn i dμ i = 0 for infinitesimal changes. Nor does it imply μ A = μ B in a mixture; equality is required for the same species in phases between which it can transfer, not for two different species simply because they mix.

Worked example

At fixed T and p, a binary mixture has x A = 0.25 and x B = 0.75. Suppose a small composition change gives dμ A = +12 J mol⁻¹. The Gibbs–Duhem relation requires 0.25(12) + 0.75dμ B = 0. Thus dμ B = −4 J mol⁻¹. The numbers are local changes along a composition path, not the absolute chemical potentials of the two components.

Quick check

1. At fixed T and p in a binary mixture, can both chemical potentials increase for the same infinitesimal composition change? Answer: No, if both amounts are positive. The weighted sum n A dμ A + n B dμ B must be zero, so two strictly positive changes would violate the Gibbs–Duhem equation.

Exam focus

Differentiate G = Σn iμ i with a product rule before comparing it with dG. Keep the full −S dT + V dp form until the conditions are specified. Use dμ i, not μ i, in the fixed-T,p restriction. Explain the physical consequence as linked composition dependence of partial molar Gibbs energies.

Advanced insight

For ideal mixing, μ i = μ i + RT ln x i at fixed T and p. Then dμ i = RT dln x i, and Σx i dln x i = Σdx i = 0 because mole fractions sum to one. Thus the ideal-solution formulas automatically satisfy Gibbs–Duhem. Real-solution activity coefficients must collectively preserve the same thermodynamic consistency.

Summary

Extensivity gives G = Σn iμ i. Comparing its differential with dG = −S dT + V dp + Σμ i dn i yields Σn i dμ i = −S dT + V dp. At fixed temperature and pressure, chemical potentials in a mixture are linked through Σn i dμ i = 0 and cannot vary independently.

Practice questions

1. State the fixed-T,p Gibbs–Duhem equation for a three-component mixture. Answer: n A dμ A + n B dμ B + n C dμ C = 0. It may also be divided by total amount to give x A dμ A + x B dμ B + x C dμ C = 0. 2. In a binary mixture with x A = x B = 0.5, dμ A = −6 J mol⁻¹. Find dμ B. Answer: 0.5(−6) + 0.5dμ B = 0, so dμ B = +6 J mol⁻¹. 3. Why may a pure phase have its amount doubled at fixed T and p without changing μ? Answer: μ is intensive. Doubling amount doubles G but leaves G per mole and the derivative μ unchanged; fixed-T,p Gibbs–Duhem for one component gives n dμ = 0.