Phase Stability and Chemical Potential

Introduction

At a given temperature and pressure, a pure substance may be capable of appearing as solid, liquid or vapour. Which phase is stable? For a pure phase, chemical potential equals molar Gibbs energy, so the stable phase has the lowest μ among accessible phases under the stated conditions. Equality of μ marks coexistence, while a phase with higher μ may still persist temporarily because changing phase can require nucleation.

Core explanation

For one pure substance at T and p, consider possible phases α, β and γ. Each has a chemical potential μ^α(T,p), μ^β(T,p) and μ^γ(T,p). If a small amount transfers from α to β at fixed T and p, dG = (μ^β − μ^α)dn. If μ^β < μ^α, the transfer lowers G; α is not the globally stable phase when β can form. The phase with the lowest μ is thermodynamically stable. On a phase boundary between α and β, μ^α = μ^β, so either can coexist without a first-order Gibbs-energy preference for an infinitesimal transfer.

Equality of two chemical potentials is necessary for their transfer equilibrium. For a one-component system with three coexisting phases, μ^solid = μ^liquid = μ^vapour at the triple point. Such equalities constrain the allowed T and p rather than holding everywhere. Inside a single-phase region, one phase has lower μ than the others. Phase diagrams map where the ordering changes.

The word “accessible” matters. A higher-μ phase can be metastable when an activation or nucleation barrier prevents the lower-μ phase from appearing promptly. Supercooled liquid water may remain liquid below its equilibrium freezing point until a crystal nucleus forms. Thermodynamics says solid has lower μ under those conditions; kinetics explains the delay. Metastability should not be mistaken for true global equilibrium, nor should spontaneous conversion be assumed instantaneous.

For mixtures, comparing total phase μ values as single numbers is insufficient. Every transferable component i must have equal μ i between coexisting phases at equilibrium, and compositions may differ across phases. This page's “lowest μ wins” phrase applies most directly to a pure one-component comparison. The next pages develop how μ changes with T and p and why phase boundaries have particular slopes.

Step-by-step reasoning

Fix temperature and pressure, then make sure the phases describe the same pure substance. Compare their molar Gibbs energies, which are their chemical potentials. Assign the lowest as stable and equal values as potential coexistence. If a higher-μ phase is observed, ask whether nucleation or another kinetic constraint makes it metastable. For a mixture, repeat the equality condition for each transferable component rather than comparing one total number.

Visual explanation

Plot μ against temperature at one pressure for solid, liquid and vapour. Each curve has a slope; crossings mark melting or boiling at that pressure. Below or above a crossing, shade the curve with the lowest μ to identify the stable phase. A second tiny sketch shows a local barrier separating a metastable state from the global minimum.

Real-world analogy

Two valleys can have different depths. A ball in the shallower valley is not at the lowest possible height, yet a ridge can keep it there until it receives a push. A metastable phase similarly can persist despite having higher Gibbs energy because a nucleation barrier delays transformation. The analogy is about energy landscape and barriers, not literal mechanical motion.

Real-world example

Water can remain liquid below 0 °C at near-standard pressure if it is clean and undisturbed. Ice is thermodynamically favoured under those conditions, but forming an initial crystal may be kinetically difficult. A seed crystal or disturbance can trigger rapid freezing, revealing the distinction between phase stability and persistence.

Why?

At constant T and p, the stable equilibrium state minimises Gibbs energy under allowed changes. Transferring a mole from one pure phase to another changes G by the difference in their μ values. The sign of that difference determines the preferred direction; zero difference establishes coexistence. This local derivative criterion unifies melting, boiling and sublimation boundaries.

Common misconception

Seeing a phase does not prove it is the most stable phase. Supercooled or supersaturated systems can persist. Conversely, equal μ does not mean equal phase amounts; coexistence proportions depend on total material and constraints. Also, the comparison is between the same substance in different phases at the same T and p, not between unrelated chemicals with arbitrary μ references.

Worked example

At one T and p, suppose a hypothetical pure substance has μ solid = −12.0, μ liquid = −11.5 and μ vapour = −9.0 kJ mol⁻¹ on one common reference. Solid has the lowest μ and is stable. Converting 0.010 mol of liquid to solid changes G by (−12.0 − (−11.5))(0.010) = −0.0050 kJ = −5.0 J. The liquid may still persist if no solid nucleus forms, but its persistence would be metastability rather than the global minimum.

Quick check

1. What equality holds at liquid–vapour equilibrium for a pure substance? Answer: μ liquid(T,p) = μ vapour(T,p). An infinitesimal transfer between the two then has zero first-order Gibbs-energy change.

Exam focus

State fixed T and p before comparing μ values. For a pure phase, μ = G m. Use the transfer expression to justify why lower μ is stable. Distinguish global thermodynamic stability from kinetic persistence, and for mixtures apply equality separately to each species that can transfer.

Advanced insight

At a phase boundary, equal chemical potentials do not imply equal entropies or volumes. In a first-order transition, their differences determine latent heat and the slope of the coexistence curve. The equality constraint differentiated along that curve gives the Clapeyron equation, developed shortly after this page.

Summary

At fixed T and p, a pure substance's stable phase has the lowest chemical potential, equal to its molar Gibbs energy. Two phases coexist when their μ values are equal. A higher-μ phase may persist metastably because nucleation is slow. Mixtures require equal chemical potentials for each transferable component across coexisting phases.

Practice questions

1. At one T and p, μ α = 4.0 and μ β = 3.5 kJ mol⁻¹ for the same pure substance. Which phase is stable, and what is dG for 0.020 mol α → β? Answer: β is stable because its μ is lower. dG = (3.5 − 4.0)(0.020) = −0.010 kJ = −10 J. 2. Why may a liquid remain present below its equilibrium freezing temperature? Answer: It may be supercooled and metastable. Ice has lower μ, but a nucleation barrier can delay formation of a crystal. 3. If two pure phases coexist, must their densities be equal? Answer: No. Equality of μ is the equilibrium condition; their molar volumes and densities can differ substantially, as liquid and vapour demonstrate.