Chemical potential and phase equilibrium

A melting choice at fixed temperature and pressure

A hypothetical substance can be solid or liquid near 315 K and 0.120 MPa. Hold a small assembly at 320 K and 0.120 MPa using thermal and mechanical reservoirs. Suppose both bulk phases are available and neglect their interface contribution. Which direction of phase transfer lowers the appropriate thermodynamic potential, even though melting can require an enthalpy input?

For the original local model below, the liquid’s chemical potential is 120 J/mol lower than the solid’s at this state. Transferring a small 0.004 mol from solid to liquid therefore changes the assembly’s Gibbs energy by −0.480 J. This is a thermodynamic direction test under fixed temperature and pressure, not a prediction that melting must occur immediately.

On a first reading, follow this pure-phase case, the coexistence slope, activities and the ideal-mixture transfer example. Return to the sections marked “Further” for the partial-molar proof, common-tangent stability and uncertainty reasoning. Published preparation is “Physical systems, states, boundaries, and environments”, “Algebra, functions, graphs, and logarithms”, and “Pressure, buoyancy, and fluid statics”. The needed thermodynamic definitions are supplied locally; adjacent articles need not already be available.

Name the state and the reservoirs

A phase is a macroscopically homogeneous region with its own material state, such as a solid, a liquid or a gas mixture. Several separate regions with identical intensive properties count as the same phase. A component here is one of the independent, neutral, nonreacting substances A or B. The amount n_i is measured in moles; a mole names an amount of specified entities, not a mass or a volume.

Write internal energy as U, enthalpy as H = U + pV, and Gibbs energy as G = H − TS. Entropy S is a state property in J/K; T is positive absolute temperature in K, p is pressure in Pa and V is volume in m³. These definitions make G an energy in J. A subscript m denotes a molar quantity, such as g = G/n in J/mol.

For a closed assembly at fixed T and p, with fixed total amounts of its components, only pressure–volume work and negligible endpoint bulk-motion and interface terms, equilibrium minimizes G over the states allowed by those constraints. Heat and volume work can pass to the reservoirs: the assembly’s energy need not remain fixed. An isolated assembly instead uses an entropy maximum at fixed energy and other conserved quantities.

All numerical models here are hypothetical bulk equilibrium models. Equal pressure between phases is used for planar interfaces or negligible surface effects and no external-field gradient. Surface tension, gravity, electric fields and coherency strain can change the appropriate balance; they are not hidden in the given numbers.

Chemical potential is a constrained derivative

In an equilibrium simple compressible bulk phase, dG = −S dT + V dp + Σ_i μ_i dn_i. The chemical potential of component i is μ_i = (∂G/∂n_i) at fixed T, p, every other component amount n_j and any other specified work coordinate. Its unit is J/mol. It describes the marginal change of G when that named component is added under those conditions.

Holding the other component amounts fixed generally changes the composition when n_i changes. Holding all mole fractions fixed would instead add a proportionate mixture and would define a different directional derivative. Chemical potential is not an unspecified “energy needed to add matter” regardless of constraints.

For one pure bulk phase, extensivity gives G = nμ, so μ = G/n. For a mixture, G = Σ_i n_i μ_i and g = Σ_i x_i μ_i, where x_i = n_i/Σ_j n_j. Thus a component’s partial molar Gibbs energy generally differs from the mixture’s molar Gibbs energy. A chemical-potential reference must be consistent between the states being compared.

Allowed transfer and equilibrium

Let dn_i > 0 move component i from phase α to phase β while total n_i is conserved and both phases share fixed T and p. Then dn_i^α = −dn_i and dn_i^β = +dn_i. Adding their Gibbs-energy changes gives dG_total = (μ_i^β − μ_i^α)dn_i. Transfer toward lower μ_i lowers G for this allowed small change.

At an interior equilibrium where the component is present in both phases and can move in either direction, μ_i^α = μ_i^β. Each exchangeable component has its own equality. This does not require μ_A = μ_B: the equality compares the same component between phases. A membrane that blocks A removes the A-transfer variation and its corresponding equality requirement.

Equality is a necessary stationary condition, not a complete global-stability test. A candidate pair can be beaten by another allowed phase or combination with lower G. If an amount is constrained to zero or a species is excluded, the allowed variations are one-sided or absent; do not impose an interior equality mechanically. Thermal and mechanical equilibrium must also be satisfied under the stated boundary model.

Pure phases share one chemical potential

For a pure phase at fixed amount, divide dG = −S dT + V dp by n: dμ = −s_m dT + v_m dp. The temperature derivative of μ at fixed p is −s_m; its pressure derivative at fixed T is v_m. A coexistence state requires the solid and liquid values of μ at the same T and p to match.

Different phases generally have different molar entropy and volume, so their μ values change differently when T or p changes. Their equality therefore traces a coexistence curve in the T–p plane. Away from that curve, the lower bulk chemical potential identifies the favored phase among the supplied candidates.

A finite amount of either phase can coexist on the curve while changing the phase amounts. The pressure and temperature of a pure two-phase bulk system cannot generally be chosen independently. At a triple point, three distinct phase chemical potentials coincide under the additional material conditions; this is not three arbitrary crossings at arbitrary T and p.

Derive the Clapeyron slope

Move along a coexistence curve so both phases remain in equilibrium. Their differentials are equal: −s_m^α dT + v_m^α dp = −s_m^β dT + v_m^β dp. With Δs_m = s_m^β − s_m^α and Δv_m = v_m^β − v_m^α, this gives dp/dT = Δs_m/Δv_m when Δv_m is nonzero.

At coexistence Δμ = Δg = 0, so Δh_m = TΔs_m. The slope can therefore be written dp/dT = Δh_m/(TΔv_m). This is the Clapeyron equation for a pure bulk first-order phase boundary. Its sign depends on the volume change as well as the transition enthalpy; a positive melting enthalpy does not force a positive melting-curve slope.

The derivative uses local phase properties at the coexistence state. Treating a latent enthalpy or volume difference as constant over a finite interval is an additional model. Near a critical endpoint, the two phases lose their distinction and both numerator and denominator can vanish; the simple ratio cannot be extrapolated there without a valid limiting model.

Solve the original local phase model

Assign Δs_m = 24 J/(mol K), v_solid = 20 × 10⁻⁶ m³/mol and v_liquid = 24 × 10⁻⁶ m³/mol. The local difference law is Δμ = μ_liquid − μ_solid = −Δs_m(T − 315 K) + Δv_m(p − 120 000 Pa), with Δv_m = 4 × 10⁻⁶ m³/mol. Each term has units J/mol because Pa m³ = J.

At p = 0.120 MPa, Δμ is positive below 315 K, zero at 315 K and negative above it. At 320 K it is −120 J/mol, giving the opening −0.480 J differential change for 0.004 mol. The assembly need not lose 0.480 J of internal energy: G combines enthalpy and entropy under the reservoirs’ constraints.

The assigned slope is 24/(4 × 10⁻⁶) = 6.00 × 10⁶ Pa/K = 6.00 MPa/K. Raising pressure from 0.120 to 6.120 MPa shifts coexistence to 316 K; at 12.120 MPa it is 317 K. The coexistence enthalpy differences are respectively 7560, 7584 and 7608 J/mol because this model keeps Δs_m constant, not Δh_m.

The law compares two supplied bulk branches over 310–320 K and the stated pressure range. It does not determine their absolute chemical potentials, all heat capacities or compressibilities, a complete phase diagram, or a nucleation barrier. A common background contribution to both phase potentials leaves their difference and crossings unchanged.

A liquid–vapor approximation

For a liquid–vapor boundary, the exact bulk Clapeyron relation still uses the actual Δv_m and Δh_m. If the vapor is approximately ideal and its molar volume greatly exceeds the liquid’s, Δv_m ≈ RT/p. Then d ln p_sat/dT ≈ Δh_vap,m/(RT²). Here p_sat is the coexistence pressure, not a pressure for an arbitrary nonequilibrium vapor.

If the molar vaporization enthalpy L_m is additionally constant, integration gives ln[p_sat(T)/p_ref] = −(L_m/R)(1/T − 1/T_ref). A straight relation between ln p and 1/T is therefore a consequence of several assumptions, not a universal law. All temperatures are absolute; pressure ratios must use the same unit.

For a separate hypothetical material, take L_m = 24 000 J/mol and p_ref = 18 000 Pa at T_ref = 330 K, over 300–360 K. The model gives p_sat(345 K) ≈ 26.3292 kPa. Its local slope at 330 K is about 477.113 Pa/K. A constant pressure slope would not reproduce this exponential curve.

With an assigned liquid molar volume 22 × 10⁻⁶ m³/mol, the ratio ε = v_liquid p_sat/(RT) stays below 2.75 × 10⁻⁴ in this interval. That checks the neglected-volume term for the assigned curve. It does not independently prove vapor ideality, constant L_m or agreement with measured saturation pressures. No critical-region or outside-interval prediction is made.

Activities and fugacities need standards

For a component in a mixture, write μ_i = μ_i^ref + RT ln a_i. The activity a_i is dimensionless; its definition depends on the specified reference state and composition convention. An ideal solution using a pure-component standard at the same T and p has a_i = x_i. A nonideal solution on this basis uses a_i = γ_i x_i, where γ_i is an activity coefficient, not a universal correction independent of composition.

For a gas mixture, fugacity f_i has pressure units. With an ideal-gas standard at pressure p°, μ_i = μ_i°(T,p°) + RT ln(f_i/p°). An ideal gas mixture has f_i = y_i p; a nonideal mixture can be written f_i = φ_i y_i p with a specified fugacity coefficient φ_i. The gas mole fraction is denoted y_i here to keep it distinct from a liquid composition.

For an assigned gas state at 310 K, p = 150 kPa, y_B = 0.25 and φ_B = 0.92, f_B = 34.5 kPa. Relative to p° = 100 kPa, a_B = 0.345 and μ_B − μ_B° ≈ −2742.986 J/mol. The logarithm is of 0.345, not of a pressure carrying units.

At phase equilibrium, the chemical potentials of the same exchangeable component must match. Equal numerical activities follow only when the chosen standard chemical potentials also match. A liquid pure-component standard and an ideal-gas standard generally differ. If a single consistent fugacity convention is used for all phases, equality of that component’s fugacity is an equivalent statement; changing standards cannot create a driving force.

Follow one component between two regions

Consider two ideal A/B mixture regions at 310 K and a common fixed pressure. Initially the left contains 0.85 mol A and 0.15 mol B; the right contains 0.45 mol A and 0.55 mol B. A selective route lets B move while each region retains its A. Mechanical boundaries can adjust to maintain the imposed pressure; this is not a rigid isolated-volume problem.

Both regions use the same pure-B standard μ_B* at this T and p. Their initial B mole fractions are 0.15 and 0.55, so μ_B,left − μ_B,right = RT ln(0.15/0.55) ≈ -3348.88 J/mol. A small B transfer from right to left lowers G. For 0.002 mol the differential estimate is -6.697761 J; a finite transfer updates both compositions.

Let ξ denote moles of B allocated from right to left. Then n_B,left = 0.15 mol + ξ and n_B,right = 0.55 mol − ξ. For each region G_mix = RT[n_A ln x_A + n_B ln x_B]. The unchanged total pure-component terms cancel when taking ΔG(ξ) from the initial state. The curve versus ξ compares constrained equilibrium states; it is not a time record.

Minimizing the combined G makes μ_B,left = μ_B,right. The result is ξ_eq = 0.307692308 mol and x_B,left = x_B,right = 0.350000. The combined G falls by about 474.374 J. The second derivative is positive throughout the stated positive-amount interval, making this the unique minimum of this assigned ideal-transfer model. A was not required to cross the selective boundary.

Further: extensivity and Gibbs–Duhem

For a homogeneous bulk phase, scale every component amount by the same positive factor at fixed T and p. Extensivity scales G by that factor while preserving composition. Euler’s relation is G = Σ_i n_i μ_i. Differentiating it and comparing with dG = −S dT + V dp + Σ_i μ_i dn_i gives Σ_i n_i dμ_i = −S dT + V dp.

At fixed T and p, Σ_i x_i dμ_i = 0. This is the Gibbs–Duhem relation: chemical potentials in one phase cannot be varied independently by an arbitrary change in composition. It constrains changes, not an equality of the values μ_A and μ_B. Activity-coefficient models for the same phase must be thermodynamically consistent with it.

For a binary phase write n = n_A + n_B and x = n_B/n, so G = n g(x). Adding B at fixed n_A gives dn = dn_B and dx = (1−x)dn_B/n. Therefore μ_B = g + (1−x)g′. Adding A at fixed n_B similarly gives μ_A = g − xg′. The prime means a derivative with respect to the dimensionless mole fraction x at fixed T and p.

Further: a common supporting tangent

The tangent to g(x) at a composition x has intercept g − xg′ at x = 0 and value g + (1−x)g′ at x = 1. These are μ_A and μ_B. A line tangent to two candidate phase curves at x_α and x_β therefore gives μ_A^α = μ_A^β and μ_B^α = μ_B^β. The tangent slope is μ_B − μ_A, not either chemical potential by itself.

For a given overall composition, a two-phase combination has molar G on the straight chord between its two phase compositions, weighted by their mole amounts. Stable equilibrium selects the lowest allowed combination. Thus the equilibrium construction is a supporting common tangent: no other supplied allowed homogeneous state or phase combination lies below it.

A merely common tangent that is beaten by a third phase or another part of the free-energy surface is not a global-equilibrium proof. A stable bulk equilibrium free-energy curve is the lower convex envelope of the homogeneous candidates under the stated constraints. Interfaces and finite-size effects add terms and may change which combinations are accessible or favorable.

Further: a regular solution with a miscibility gap

Use a separate symmetric regular-solution model at 310 K and fixed p. With x the B mole fraction, define f(x) = g_mix/(RT) = (1−x)ln(1−x) + x ln x + χx(1−x), with χ = 2.8. The linear pure-component reference has been subtracted; it does not change phase-composition or convex-envelope tests. This assigned mean-field model is not a material fit.

For 0 < x < 1, f′ = ln[x/(1−x)] + χ(1−2x) and f″ = 1/[x(1−x)] − 2χ. At x = 0 or 1, only f has its continuous 0 ln 0 endpoint limit. The absent-species logarithmic chemical potential is not assigned a finite endpoint value.

Symmetry makes the coexistence tangent horizontal. Solving f′(x_α) = 0 below 1/2 gives x_α ≈ 0.092735734 and x_β = 1−x_α ≈ 0.907264266. The line f = -0.073241743 supports the whole supplied curve. These are binodal compositions: the two bulk phases that coexist within this phase family.

The same G model gives μ_A − μ_A* = RT[ln(1−x) + χx²] and μ_B − μ_B* = RT[ln x + χ(1−x)²]. Hence ln γ_A = χx² and ln γ_B = χ(1−x)² on the common pure-component standard convention. Differentiating verifies (1−x)dμ_A + x dμ_B = 0. These activities are model outputs, not separately adjustable inputs.

Further: local curvature is not global stability

For this model, f″ = 0 at x ≈ 0.232738758 and 0.767261242. These spinodal compositions delimit the negative-curvature interval. A small composition redistribution with conserved mean can lower bulk G when f″ < 0. This is a local composition-stability test, not a transport-rate calculation.

Between each binodal and its neighboring spinodal, f″ > 0 yet the supporting two-phase combination has lower G. Such a homogeneous state is metastable against phase separation in this bulk model: sufficiently small composition perturbations cost G, while a finite change can reach a lower-G state. Interface energy can impose a nucleation barrier that is absent from the bulk curve.

At overall x = 0.200000 the homogeneous curvature f″ = 0.65 is positive. Nevertheless, for 1.6 mol the equilibrium two-phase mixing G is about −302.047 J, compared with −216.106 J for the homogeneous state. The two-phase combination is lower by 85.941 J. At x = 0.5 the homogeneous curvature is −1.6, so even local composition stability is lost.

The binodal, spinodal and equilibrium phase amounts answer different questions. None supplies a nucleation probability, domain size, mixing time or growth speed. Also χ is specified at one temperature only: although a symmetric model changes its curvature threshold at χ = 2, no critical temperature follows without a justified χ(T).

Use a lever rule for phase amounts

Let F_α and F_β be the fractions of the total mole amount in phases α and β. Conservation gives F_α + F_β = 1 and x̄ = F_α x_α + F_β x_β. Solving yields F_β = (x̄−x_α)/(x_β−x_α). This amount rule is valid when x̄ lies between the coexisting phase compositions; outside that interval the formula would request a negative phase amount and is not a two-phase solution.

For x̄ = 0.20 and n_total = 1.6 mol, F_β ≈ 0.131688776. The phase amounts are n_α ≈ 1.389297958 mol and n_β ≈ 0.210702042 mol. Their B amounts add to 0.320000 mol. The compositions of the two phases are still the binodal values; changing the overall composition within the gap primarily changes their amounts at the same T and p.

The fractions here are mole-amount fractions. They are not automatically mass fractions, volume fractions or fractions of the diagram’s drawn area. Converting to mass or volume needs the corresponding molar masses or phase molar volumes. A schematic depiction of the two phase regions must not suggest a numerical volume ratio that has not been provided.

Further: infer a slope with uncertainty

A separate hypothetical reading pair gives p₁ = 14 000 Pa at T₁ = 325 K and p₂ = 24 000 Pa at T₂ = 345 K. Under ideal-vapor, negligible-liquid-volume and constant-enthalpy assumptions, L̂ = R ln(p₂/p₁)/(1/T₁−1/T₂) ≈ 25.1242 kJ/mol. This pair was supplied independently of the earlier 24.000 kJ/mol model; it is a separate inference example.

Let D = 1/T₁−1/T₂. The sensitivities are ∂L̂/∂p₁ = −R/(Dp₁), ∂L̂/∂p₂ = R/(Dp₂), ∂L̂/∂T₁ = L̂/(DT₁²), and ∂L̂/∂T₂ = −L̂/(DT₂²). For uncorrelated input standard uncertainties, u_L² = Σ_j[(∂L̂/∂z_j)u_j]². Correlated inputs require the covariance form u_L² = Σ_jΣ_k(∂L̂/∂z_j)(∂L̂/∂z_k)cov(z_j,z_k).

Assign independent pressure standard uncertainties 70 and 120 Pa and independent temperature standard uncertainties 0.05 K each. Their propagated u_L is about 0.341446 kJ/mol. A separate shared additive thermometer offset with standard uncertainty 0.08 K contributes through (∂L̂/∂T₁ + ∂L̂/∂T₂)u_b, because a common offset does not cancel from reciprocal temperatures.

Including that independent shared-offset term gives u_L ≈ 0.341657 kJ/mol, a first-order standard uncertainty of about 1.36%. One can instead encode the same shared offset in the temperature covariance matrix, but must not count it twice. A narrow statistical uncertainty does not cover model bias from changing latent enthalpy, vapor nonideality, wrong pressure calibration or failure to establish coexistence.

Further: count freedom and check the boundary

For an ordinary nonreacting bulk system in complete thermal, mechanical and component-transfer equilibrium, with C independent components and P distinct phases, the Gibbs phase rule is F = C−P+2. F counts independent intensive variables that can vary while those phases remain, not their amounts. Membranes, reactions, external fields and additional constraints require the appropriate revised counting.

A pure two-phase system has F = 1, consistent with a coexistence curve. A pure triple point has F = 0. A binary two-phase system has F = 2; at a chosen common T and p the coexisting compositions are fixed by its phase relations, while the overall composition determines amounts within the compatible two-phase region.

Do not apply the complete-equilibrium phase-rule count unchanged to the earlier selective-B apparatus: its retained-A constraint removes some transfer variations. Likewise a curved droplet can have a pressure difference across its interface, and charged or gravitationally separated matter needs the appropriate generalized or electrochemical potential. The simple equal-p, equal-μ formulas here have explicit bulk, neutral, negligible-field conditions.

Use equilibrium with its limits

To use chemical potential, identify the components, amount basis, allowed exchanges and held variables first. Compare the same component with a consistent reference convention. Check all equilibrium conditions, then test the competing allowed states and enforce total-component balances. Finally ask whether neglected interfaces, fields, composition changes or model uncertainty alter the conclusion.

The neighboring “Enthalpy, Helmholtz energy, and Gibbs energy” and “Maxwell relations and thermodynamic stability” articles develop thermodynamic potentials and response stability more broadly. “Phase transformations and microstructure” treats the material mechanisms and geometry; “Phase transitions and order parameters” develops statistical descriptions; “Irreversible thermodynamics and heat and mass transfer” addresses transport.

A phase boundary tells which bulk states can coexist. A convex envelope tells which supplied combinations minimize the appropriate potential. Neither replaces a kinetic law or establishes that a finite experiment has equilibrated. Those are separate physical questions.

Worked examples

Try it yourself

1. In μ_B = (∂G/∂n_B), which ordinary variables and amounts are held fixed?

   Show the answer and reasoning

   T, p and every other component amount, plus any other specified work coordinate.

   Adding B alone generally changes composition. Adding a proportionate mixture at fixed composition is a different variation.

2. Must μ_B equal G/(n_A+n_B) in a binary mixture?

   Show the answer and reasoning

   No.

   The molar mixture value is g = (1−x)μ_A+xμ_B. The partial molar values are μ_A=g−xg′ and μ_B=g+(1−x)g′.

3. A membrane blocks A and permits B. Does equilibrium require an A-transfer equality across it?

   Show the answer and reasoning

   Not from a blocked A-transfer variation.

   Equality follows from allowed two-way variations. B needs its equality; the retained-A constraint changes the allowed state set.

4. At binary two-phase equilibrium, must μ_A equal μ_B?

   Show the answer and reasoning

   No. Each component matches itself between phases.

   The conditions are μ_A^α=μ_A^β and μ_B^α=μ_B^β. Different species generally have different reference values and potentials.

5. A separate local transition has Δs_m = 18 J/(mol K) and Δv_m = −1.5 × 10⁻⁶ m³/mol. Find the Clapeyron slope.

   Show the answer and reasoning

   −1.2 × 10⁷ Pa/K = −12 MPa/K.

   Use Δs_m/Δv_m. The negative volume change reverses the slope despite positive transition entropy.

6. Where does the assigned local model coexist at 12.120 MPa?

   Show the answer and reasoning

   317 K.

   The 12 MPa increase gives Δv_mΔp = 48 J/mol; divide by 24 J/(mol K) to obtain 2 K above 315 K.

7. Can Celsius readings replace 325 K and 345 K inside the reciprocal-temperature inference?

   Show the answer and reasoning

   No.

   The derivation uses absolute thermodynamic temperature. An additive Celsius offset does not cancel inside reciprocal temperatures.

8. Does ε ≪ 1 by itself validate the entire integrated vapor-pressure model?

   Show the answer and reasoning

   No.

   It checks neglect of liquid volume relative to the assigned ideal vapor volume. Vapor ideality, constant latent enthalpy and actual coexistence require separate evidence.

9. Two phases have equal μ_B but use different standard chemical potentials. Must their numerical a_B values be equal?

   Show the answer and reasoning

   No.

   Equality applies to μ_B^ref+RT ln a_B. Activities alone can be equated only with compatible identical standards; reference changes do not alter the physical equilibrium.

10. Is the initial μ difference times a finite B transfer an exact finite ΔG for the ideal apparatus?

    Show the answer and reasoning

    Generally no.

    Both region compositions and their μ values change as B moves. Integrate the derivative or evaluate the full combined G; the small-transfer product is a differential estimate.

11. Why is the regular-solution homogeneous x_B=0.20 state not globally stable even though f″=0.65?

    Show the answer and reasoning

    The compatible two-phase supporting-tangent combination has lower G.

    Positive curvature resists sufficiently small composition variations; the 85.940800 J lowering for the stated 1.6 mol is a global comparison within the supplied phase family.

12. For the same 1.6 mol regular solution at overall x_B=0.5, find both phase mole amounts.

    Show the answer and reasoning

    0.8 mol in each phase.

    The binodal compositions are symmetric about 0.5, giving equal amount fractions. Their compositions remain x_α and x_β, not 0.5 each.

13. Can χ=2.8 specified only at 310 K determine a critical temperature?

    Show the answer and reasoning

    No.

    The symmetric curvature threshold is χ=2, but converting that condition into a temperature requires a justified temperature dependence χ(T).

14. Does a common additive thermometer offset cancel from D=1/T₁−1/T₂?

    Show the answer and reasoning

    No.

    Its differential sensitivity is −1/T₁²+1/T₂², generally nonzero. Include the shared uncertainty through its summed L̂ sensitivity or through covariance, without double counting.

15. What are F for a pure triple point and an ordinary binary two-phase equilibrium, and do they specify a transition rate?

    Show the answer and reasoning

    F=0 and F=2 respectively; neither supplies a rate.

    F=C−P+2 counts independent intensive equilibrium variables under the complete bulk-equilibrium conditions. Phase amounts and kinetic mechanisms are separate information.

What to carry forward

• Chemical potential is a named component’s constrained partial molar property; it generally differs from the mixture’s molar Gibbs energy.
• Allowed transfer equilibrates the same component’s chemical potential between phases; global stability and component balances are additional requirements.
• Clapeyron uses local coexistence properties. Ideal-vapor and constant-enthalpy integrations add distinct approximations.
• Activity and fugacity equations require dimensionless logarithms and consistent standard states.
• A supporting common tangent and lever rule separate equilibrium phase compositions from their amounts.
• Positive local composition curvature can coexist with global metastability. Equilibrium data do not specify transition rates.