Chemistry · Lesson

Chemical potential and conditions for equilibrium

Use a partial molar Gibbs energy to describe a component’s response to material transfer or reaction. Match the same transferable species across phases, and retain the distinction between stationarity and stability.

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By the end, you can

Define μᵢ as a partial molar derivative rather than each component’s G/n.
Use the chemical-potential difference for a small material transfer.
Distinguish phase-transfer equilibrium from reaction equilibrium.

Ask what adding a small amount changes

The chemical potential is μᵢ = (∂G/∂nᵢ)T,p,{nⱼ≠ᵢ}. It describes the first-order change in total Gibbs energy when species i is added while temperature, pressure and other species amounts are held fixed. Its amount basis gives J mol⁻¹.

For an ordinary homogeneous extensive bulk mixture, Euler homogeneity gives G = Σᵢ nᵢ μᵢ. Extensivity means scaling all amounts and volume together scales the total energy, without changing intensive properties. Surface contributions, finite-size effects or long-range fields can require a different treatment. In a mixture, an individual μᵢ is generally not the average G/Σnᵢ.

A difference determines the first-order transfer direction

Suppose a neutral species can move from phase α to phase β at fixed T,p, with no additional electric or gravitational contribution. Moving a small amount dn from α to β gives dG = (μᵢβ − μᵢα) dn. A negative difference favors that transfer thermodynamically.

For phases in material equilibrium, the same transferable species has equal chemical potential across them. Different chemical species do not all need the same μ. Impermeable constraints can prevent exchange. Charged-species transfer between phases with different electric potentials instead requires the corresponding electrochemical potential, including the electrical contribution.

A reaction changes several amounts in a fixed ratio

For a written reaction, use signed stoichiometric numbers νᵢ, positive for products and negative for reactants. A small reaction-extent change dξ changes the amounts by dnᵢ = νᵢ dξ. At fixed T,p, dG = ΔᵣG dξ, with ΔᵣG = Σᵢ νᵢ μᵢ.

For an unconstrained interior reaction equilibrium, ΔᵣG = 0 is the stationarity condition. For A ⇌ B with unit coefficients, this means μB − μA = 0. For A + B ⇌ C, it means μC − μA − μB = 0; it does not require all three potentials to be equal. Boundary compositions and additional constraints need their own permitted-variation conditions.

Three different chemical-potential statements

Question	Relation	Specified example	Interpretation
Add species i	dG = μᵢ dnᵢ at fixed other amounts	μᵢ is in J mol⁻¹	A partial molar response
Move the same neutral species α → β	dG = (μᵢβ − μᵢα) dn	Difference −2.00 kJ mol⁻¹; dn = 1.00 mmol	First-order dG = −2.00 J
Advance A ⇌ B	ΔᵣG = μB − μA	μA = 4.00; μB = 1.00 kJ mol⁻¹	Forward ΔᵣG = −3.00 kJ mol⁻¹
Advance A + B ⇌ C	ΔᵣG = μC − μA − μB	Interior equilibrium: sum equals zero	Different species need not have equal μ

Molar chemical potentials and fixed-T,p permitted variations. Supplied numerical potentials are invented local values. Small-transfer energy changes are first-order estimates unless reservoirs keep the potentials constant.

Component potentials change together

Combining the bulk Euler relation with the Gibbs differential gives S dT − V dp + Σᵢ nᵢ dμᵢ = 0, the Gibbs–Duhem relation. At fixed T,p, Σᵢ xᵢ dμᵢ = 0, where xᵢ is mole fraction. This prevents assigning the composition dependence of every component potential independently.

Matching chemical potentials is necessary for the allowed equilibrium exchanges. Stable equilibrium also requires that allowed nearby changes do not lower the appropriate potential, and that no competing state has a lower value. A stationary but unstable composition can satisfy a first-derivative condition.

Worked examples

Evaluate an infinitesimal transfer tendency

At the current fixed T,p, a neutral transferable species has μα = 5.00 and μβ = 3.00 kJ mol⁻¹. Estimate the first-order Gibbs change for 1.00 mmol moving α → β.

The receiving-minus-donating difference is 3.00 − 5.00 = −2.00 kJ mol⁻¹.
Convert 1.00 mmol to 0.00100 mol.
Multiply the current difference by the transferred amount.
For a finite transfer, μ can change with composition; this is the local first-order estimate, not an exact finite integration.

dG ≈ −0.00200 kJ = −2.00 J. The supplied local state favors transfer toward β, without determining its speed.

Distinguish a mixture average from each partial molar value

For an invented extensive bulk state, nA = 1.00 mol, nB = 2.00 mol, μA = −10.0 kJ mol⁻¹ and μB = −15.0 kJ mol⁻¹. Evaluate total G and its amount average.

Use G = nA μA + nB μB at this state.
G = 1.00(−10.0) + 2.00(−15.0) = −40.0 kJ.
The total amount is 3.00 mol.
G/(nA + nB) ≈ −13.3 kJ mol⁻¹.

G = −40.0 kJ and the average is −13.3 kJ mol⁻¹. Neither component’s individual μ equals that mixture average.

Try it yourself

In the derivative defining μA, are the other species amounts held fixed?
Show the answer and reasoning
Yes.
Temperature and pressure are also fixed for the partial molar Gibbs derivative.
For A + B ⇌ C, supplied μA = 2, μB = 3 and μC = 5 kJ mol⁻¹ satisfy which reaction condition?
Show the answer and reasoning
ΔᵣG = 5 − 2 − 3 = 0.
This is reaction stationarity, with unequal individual chemical potentials.
Which chemical potentials match across phases at neutral-species transfer equilibrium?
Show the answer and reasoning
Those of the same species that can transfer.
The condition is species-specific and depends on the allowed exchanges.
Does a zero first derivative alone prove globally stable equilibrium?
Show the answer and reasoning
No.
Allowed nearby variations and competing states must also be considered.

What to carry forward

μᵢ is a partial molar Gibbs response.
Compare the same transferable species across phases.
Reaction equilibrium uses a stoichiometric sum.
Bulk extensivity and Gibbs–Duhem impose consistency.

References

Teaching resources and reference material supporting this lesson’s explanations and examples.