Defining the Chemical Potential

Introduction

What drives a substance to move between phases, dissolve or react? A useful answer is the chemical potential, which measures how the Gibbs energy changes when an infinitesimal amount of a component is added. It is a partial molar property, so it depends on the composition and surroundings. Its definition leads directly to the differential equation that connects thermal, mechanical and compositional changes in one expression.

Core explanation

For a multicomponent system with Gibbs energy G(T,p,n₁,n₂,…), define μ i = (∂G/∂n i) T,p,n j≠i. It has units of energy per mole when n i is measured in moles. This is the partial molar Gibbs energy, not simply the total G divided by total moles except in a pure, one-component phase. In a mixture, each species has its own μ i at the specified temperature, pressure and composition.

Taking the total differential of G gives dG = (∂G/∂T) p,n dT + (∂G/∂p) T,n dp + Σ i(∂G/∂n i) T,p,n j dn i. Thermodynamics identifies the first two derivatives as −S and V. Therefore dG = −S dT + V dp + Σ i μ i dn i. At fixed T and p, this simplifies to dG = Σ i μ i dn i. The equation is local: for a finite change, μ i may vary along the path and must not automatically be treated as constants.

For a pure substance, μ equals its molar Gibbs energy G m. At fixed T and p, changing its amount by dn changes G by μdn. For a binary mixture, μ A and μ B generally differ and depend on composition. Transferring a small amount of A from one phase to another changes total G by (μ A,target − μ A,source)dn. If the target has lower μ A, this transfer lowers G. Equilibrium with respect to that transfer requires equal chemical potentials in the phases.

The definition does not mean that a component's chemical potential equals its chemical energy alone. μ incorporates both energetic interactions and entropy of mixing under the chosen state conditions. Its numerical value also depends on reference conventions; differences and gradients determine spontaneous changes. A negative μ is not inherently unphysical, just as a negative Gibbs energy can depend on the reference chosen.

Step-by-step reasoning

Specify T, p, composition and the component whose amount changes. Write μ i as the held-fixed derivative of G. Expand dG in all independent variables, replacing its T and p derivatives by −S and V. At fixed T and p, use Σμ i dn i and enforce any conservation rule, such as dn source = −dn target for a transfer. Inspect the sign of dG to identify the favourable direction.

Visual explanation

Draw a surface G over axes n A and n B at fixed T and p. A tangent slope parallel to n A gives μ A; a slope parallel to n B gives μ B. Beside it show two boxes joined by an arrow for transfer of A. Label the arrow's Gibbs-energy change (μ A,right − μ A,left)dn.

Real-world analogy

The price of one extra item in a bulk order may depend on how many are already in the basket and on discounts for combinations. A partial molar quantity similarly measures the marginal effect at the current mixture, not the average price per item. Unlike a shop price, chemical potential is a thermodynamic derivative whose differences govern equilibrium.

Real-world example

Water can evaporate from a liquid into its vapour at fixed T and p when transferring a small amount to the gas lowers the total Gibbs energy. At liquid–vapour equilibrium, μ water(liquid) = μ water(vapour). Changing pressure or temperature shifts these values and can favour one phase, leading to condensation or evaporation.

Why?

At constant temperature and pressure, spontaneous changes in a closed system proceed toward lower G. A small compositional change is therefore judged by Σμ i dn i. Chemical potential supplies the coefficient that converts “how many moles moved or reacted” into the first-order change in Gibbs energy. This makes it the common language for phase equilibrium, diffusion and reaction equilibrium.

Common misconception

It is wrong to define μ i as G/n i in a mixture. That quotient includes the contributions of all components and does not isolate the response to adding i. Another mistake is to infer flow direction from the absolute sign of μ i; it is the difference between locations or phases that matters under comparable conditions.

Worked example

At fixed T and p, suppose μ A in phase α is −12 kJ mol⁻¹ and in phase β is −15 kJ mol⁻¹, using the same reference. Transfer dn = 0.010 mol of A from α to β. Then dn β = +0.010 mol and dn α = −0.010 mol. The total change is dG ≈ μ βdn + μ α(−dn) = (−15 + 12)(0.010) = −0.030 kJ, or −30 J. Transfer toward β is favourable at this state. The potentials will change if enough material moves, so this is a small-change estimate.

Quick check

1. At fixed T and p, when is a transferable component in equilibrium between two phases? Answer: Its chemical potentials are equal in the two phases. An infinitesimal transfer in either direction then has zero first-order Gibbs-energy change.

Exam focus

Write μ i = (∂G/∂n i) T,p,n j before using it. Keep the sign on −S dT and +V dp. For material transfer, include one negative and one positive amount change. Quote units such as J mol⁻¹ and compare chemical potentials only under a consistent reference convention.

Advanced insight

For a reaction with stoichiometric coefficients ν i, dn i = ν i dξ, where ξ is reaction extent. At fixed T and p, dG = (Σ iν iμ i)dξ, defining the reaction Gibbs energy Δ rG = Σ iν iμ i. Equilibrium with respect to reaction requires this derivative to vanish, connecting the partial molar definition to equilibrium constants later in the unit.

Summary

Chemical potential μ i is the partial molar Gibbs energy. The fundamental differential is dG = −S dT + V dp + Σμ i dn i. At fixed T and p, μ differences determine whether transfer or reaction lowers G, and equality of μ for a transferable species characterises phase equilibrium.

Practice questions

1. State the units of μ i when n i is in moles and G is in joules. Answer: J mol⁻¹, because μ i is the derivative of energy with respect to amount. 2. At fixed composition, temperature rises by a small dT at constant p. Which term of dG describes the first-order change? Answer: dG = −S dT because dp = 0 and all dn i = 0. For positive entropy, G falls as temperature rises locally at fixed p. 3. A species has μ left = 4 kJ mol⁻¹ and μ right = 6 kJ mol⁻¹ on a common scale. Which small transfer lowers G? Answer: Transfer from right to left lowers G, because its change per mole is μ left − μ right = −2 kJ mol⁻¹. The absolute positive values are less important than their difference.