8.5 Chemical Potential, Partial Molar Quantities, Fugacity, and Activity

Key Takeaways

  • In open systems and variable-composition mixtures, the fundamental Gibbs equation becomes dG=−SdT+VdP+∑μidnidG = -S dT + V dP + \sum \mu_i dn_i, where the chemical potential μi=(∂G∂ni)T,P,nj≠i\mu_i = (\frac{\partial G}{\partial n_i})_{T, P, n_{j \neq i}} is the partial molar Gibbs energy.

  • Partial molar quantities (Xˉi\bar{X}_i) quantify the incremental response of extensive properties to composition changes; total system properties satisfy Euler's theorem: X=∑niXˉiX = \sum n_i \bar{X}_i, as illustrated by the volumetric contraction of ethanol-water mixtures.

  • The Gibbs-Duhem equation, ∑nidμi=0\sum n_i d\mu_i = 0 at constant TT and PP, imposes a strict mathematical constraint among chemical potentials in a mixture, proving that component chemical potentials cannot vary independently.

  • Real gas deviations are described by the compressibility factor (Z=PVmRTZ = \frac{P V_m}{RT}), van der Waals parameters (aa for attraction, bb for excluded volume), and fugacity (f=ϕPf = \phi P); chemical potential follows μi=μi∘+RTln⁡fiP∘\mu_i = \mu_i^\circ + RT \ln\frac{f_i}{P^\circ}.

  • Non-ideal solutions replace mole fractions or molalities with activity (ai=γixia_i = \gamma_i x_i), referenced either to Raoult's law standard state (for solvents, pure liquid where γi→1\gamma_i \to 1 as xi→1x_i \to 1) or Henry's law standard state (for solutes, unit molality where γi→1\gamma_i \to 1 as mi→0m_i \to 0).

Last updated: October 2026

Chemical Potential, Partial Molar Quantities, Fugacity, and Activity

Pure-substance thermodynamics must be extended to multicomponent, open, and chemically reacting systems where composition varies. On the PRC Chemist Licensure Examination, mastery of this domain requires an exact physical and mathematical understanding of partial molar properties, the Gibbs-Duhem equation, real-gas fugacity (ff), and solution activity (aa).


1. Open Systems and Partial Molar Quantities

An extensive thermodynamic property XX of a multicomponent mixture depends on temperature, pressure, and the mole amounts of each constituent: X=X(T,P,n1,n2,…,nk)X = X(T, P, n_1, n_2, \dots, n_k).

The partial molar quantity (Xˉi\bar{X}_i) of component ii is defined as the rate of change of property XX with respect to the addition of component ii at constant temperature, pressure, and amounts of all other components:

Xˉi≡(∂X∂ni)T,P,nj≠i\bar{X}_i \equiv \left(\frac{\partial X}{\partial n_i}\right)_{T, P, n_{j \neq i}}

Euler's Homogeneous Function Theorem

Because extensive thermodynamic properties (V,U,H,S,A,GV, U, H, S, A, G) are homogeneous functions of degree 1 with respect to mole numbers (i.e., scaling all mole numbers by λ\lambda scales the property by λ\lambda: X(T,P,λn1,…,λnk)=λX(T,P,n1,…,nk)X(T, P, \lambda n_1, \dots, \lambda n_k) = \lambda X(T, P, n_1, \dots, n_k) Euler's theorem of homogeneous functions requires that:

X=∑i=1kniXˉiX = \sum_{i=1}^k n_i \bar{X}_i Dividing both sides by total moles n=∑nin = \sum n_i gives the mean molar property: Xm=∑i=1kxiXˉiX_m = \sum_{i=1}^k x_i \bar{X}_i

Physical Demonstration: Partial Molar Volume (Vˉi\bar{V}_i)

Total volume of a binary solution is given by: V=n1Vˉ1+n2Vˉ2V = n_1 \bar{V}_1 + n_2 \bar{V}_2 The Ethanol-Water Anomaly: Pure water has a molar volume of Vm,H2O=18.07 cm3/molV_{m, H_2O} = 18.07\text{ cm}^3/\text{mol}, while pure liquid ethanol has Vm,EtOH=58.68 cm3/molV_{m, EtOH} = 58.68\text{ cm}^3/\text{mol}. When 1.00 mol1.00\text{ mol} of ethanol (58.68 cm358.68\text{ cm}^3) is added to an infinitely large pool of pure water, the total volume increases by only 54.2 cm354.2\text{ cm}^3 (thus VˉEtOH=54.2 cm3/mol\bar{V}_{EtOH} = 54.2\text{ cm}^3/\text{mol} at infinite dilution).

Physical Rationale: Water's open, hydrogen-bonded tetrahedral cage structure accommodates smaller ethanol alkyl groups within interstitial cavities, accompanied by attractive water-ethanol dipole-dipole and hydrogen-bonding contractions.


2. Chemical Potential (μi\mu_i) and the Gibbs-Duhem Equation

The chemical potential of component ii is the partial molar Gibbs free energy:

μi≡Gˉi=(∂G∂ni)T,P,nj≠i\mu_i \equiv \bar{G}_i = \left(\frac{\partial G}{\partial n_i}\right)_{T, P, n_{j \neq i}} In an open system, the fundamental differential equation for Gibbs free energy becomes:

dG=−S dT+V dP+∑i=1kμi dnidG = -S\,dT + V\,dP + \sum_{i=1}^k \mu_i\,dn_i Chemical potential represents the escaping tendency of a chemical species from a phase: at phase or chemical equilibrium, the chemical potential of each component must be identical in all accessible phases: μi(α)=μi(β)=μi(γ)\mu_i^{(\alpha)} = \mu_i^{(\beta)} = \mu_i^{(\gamma)}

Derivation of the Gibbs-Duhem Equation

Applying Euler's theorem to Gibbs free energy: G=∑i=1kniμiG = \sum_{i=1}^k n_i \mu_i. Differentiating this expression completely using the product rule: dG=∑i=1kni dμi+∑i=1kμi dnidG = \sum_{i=1}^k n_i\,d\mu_i + \sum_{i=1}^k \mu_i\,dn_i Equating this to the fundamental equation dG=−S dT+V dP+∑μi dnidG = -S\,dT + V\,dP + \sum \mu_i\,dn_i: ∑i=1kni dμi+∑i=1kμi dni=−S dT+V dP+∑i=1kμi dni\sum_{i=1}^k n_i\,d\mu_i + \sum_{i=1}^k \mu_i\,dn_i = -S\,dT + V\,dP + \sum_{i=1}^k \mu_i\,dn_i Canceling ∑μi dni\sum \mu_i\,dn_i yields the general Gibbs-Duhem Equation:

−S dT+V dP−∑i=1kni dμi=0-S\,dT + V\,dP - \sum_{i=1}^k n_i\,d\mu_i = 0

At constant temperature (dT=0dT = 0) and constant pressure (dP=0dP = 0):

∑i=1kni dμi=0  ⟺  ∑i=1kxi dμi=0\sum_{i=1}^k n_i\,d\mu_i = 0 \iff \sum_{i=1}^k x_i\,d\mu_i = 0

Significance for Binary Mixtures

For a two-component mixture (x1+x2=1x_1 + x_2 = 1): x1 dμ1+x2 dμ2=0  ⟹  dμ2=−x1x2 dμ1x_1\,d\mu_1 + x_2\,d\mu_2 = 0 \implies d\mu_2 = -\frac{x_1}{x_2}\,d\mu_1 Thermodynamic Consequence: The chemical potentials of components in a solution cannot vary independently. If component 1 undergoes an increase in chemical potential (dμ1>0d\mu_1 > 0), the chemical potential of component 2 must simultaneously decrease (dμ2<0d\mu_2 < 0). This constraint underpins the Duhem-Margules equation relating vapor pressures of binary mixtures: dln⁡P1dln⁡x1=dln⁡P2dln⁡x2\frac{d \ln P_1}{d \ln x_1} = \frac{d \ln P_2}{d \ln x_2}


3. Real Gases: Van der Waals, Compressibility Factor, and Fugacity

The Van der Waals Equation of State

To account for non-ideal gas behavior, Johannes van der Waals introduced corrections for intermolecular forces (aa) and finite molecular volume (bb):

(P+an2V2)(V−nb)=nRT  ⟺  (P+aVm2)(Vm−b)=RT\left(P + \frac{a n^2}{V^2}\right)(V - nb) = nRT \iff \left(P + \frac{a}{V_m^2}\right)(V_m - b) = RT

  • Parameter aa (Cohesive / Attractive Parameter): Units: bar⋅L2⋅mol−2\text{bar}\cdot\text{L}^2\cdot\text{mol}^{-2} or Pa⋅m6⋅mol−2\text{Pa}\cdot\text{m}^6\cdot\text{mol}^{-2}. Accounts for attractive van der Waals forces between molecules, reducing the pressure exerted on the container walls.
  • Parameter bb (Excluded Volume / Co-volume): Units: L⋅mol−1\text{L}\cdot\text{mol}^{-1} or m3⋅mol−1\text{m}^3\cdot\text{mol}^{-1}. Represents the non-compressible volume occupied by 1 mole of molecules (b=4NA⋅vmoleculeb = 4 N_A \cdot v_{molecule}). Incompressible hard cores prevent the gas volume from reaching zero at high pressure.

The Compressibility Factor (ZZ)

The compressibility factor quantifies deviation from ideal gas behavior:

Z≡PVmRTZ \equiv \frac{P V_m}{RT}

Value of ZZDominant Physical RegimeIntermolecular ForcesPressure Range
Z=1Z = 1Ideal Gas behaviorAttractive and repulsive forces exactly balance.Very low pressures (P→0P \rightarrow 0) or at the Boyle temperature (TBT_B).
Z<1Z < 1More compressible than ideal gasLong-range attractive forces dominate (aa dominates), drawing molecules together.Moderate pressures (1−200 bar1 - 200\text{ bar}) below TBT_B.
Z>1Z > 1Less compressible than ideal gasShort-range repulsive forces dominate (bb dominates), resisting volume reduction.Extremely high pressures (>300 bar>300\text{ bar}).

The Boyle Temperature (TBT_B)

The Boyle temperature is the temperature at which the initial slope of ZZ versus PP vanishes as P→0P \rightarrow 0: lim⁡P→0(∂Z∂P)T=0\lim_{P \to 0} \left(\frac{\partial Z}{\partial P}\right)_T = 0 For a van der Waals gas, expanding ZZ in terms of virial coefficients yields:

TB=aRbT_B = \frac{a}{Rb} At T=TBT = T_B, attractive and repulsive forces cancel over a broad pressure range, causing the real gas to obey the ideal gas law up to substantial pressures.

Fugacity (ff) and Fugacity Coefficient (ϕ\phi)

For an ideal gas, chemical potential is μi=μi∘+RTln⁡(PiP∘)\mu_i = \mu_i^\circ + RT \ln(\frac{P_i}{P^\circ}). Gilbert N. Lewis introduced fugacity (ff) as an "effective thermodynamic pressure" to retain this exact mathematical form for real gases:

μi=μi∘+RTln⁡(fiP∘)with boundary condition: lim⁡P→0fP=1\mu_i = \mu_i^\circ + RT \ln\left(\frac{f_i}{P^\circ}\right) \quad \text{with boundary condition: } \lim_{P \to 0} \frac{f}{P} = 1

The dimensionless fugacity coefficient (ϕ\phi) is: ϕ≡fP\phi \equiv \frac{f}{P} μ=μ∘+RTln⁡(ϕPP∘)=μideal+RTln⁡ϕ\mu = \mu^\circ + RT \ln\left(\frac{\phi P}{P^\circ}\right) = \mu_{ideal} + RT \ln \phi

Exact Thermodynamic Evaluation of ϕ\phi

From the fundamental relation dμ=Vm dPd\mu = V_m\,dP at constant TT: ln⁡ϕ=∫0PZ−1P dP\ln \phi = \int_{0}^{P} \frac{Z - 1}{P}\,dP

Worked Step-by-Step Fugacity Calculation

Scenario: A gas at 300.0 K300.0\text{ K} and P=50.0 barP = 50.0\text{ bar} has its compressibility factor modeled by the truncated virial equation: Z=1+B′Pwhere B′=−2.00×10−3 bar−1Z = 1 + B'P \quad \text{where } B' = -2.00 \times 10^{-3}\text{ bar}^{-1} Calculate the fugacity coefficient ϕ\phi and the fugacity ff.

Step 1: Evaluate the integral for ln⁡ϕ\ln \phi: Z−1P=(1+B′P)−1P=B′=−2.00×10−3 bar−1\frac{Z - 1}{P} = \frac{(1 + B'P) - 1}{P} = B' = -2.00 \times 10^{-3}\text{ bar}^{-1} ln⁡ϕ=∫050.0B′ dP=B′⋅P=(−2.00×10−3 bar−1)(50.0 bar)=−0.1000\ln \phi = \int_{0}^{50.0} B'\,dP = B' \cdot P = (-2.00 \times 10^{-3}\text{ bar}^{-1})(50.0\text{ bar}) = -0.1000

Step 2: Solve for ϕ\phi and ff: ϕ=e−0.1000=0.90484≈0.905\phi = e^{-0.1000} = 0.90484 \approx 0.905 f=ϕ⋅P=(0.90484)(50.0 bar)=45.24 bar≈45.2 barf = \phi \cdot P = (0.90484)(50.0\text{ bar}) = 45.24\text{ bar} \approx 45.2\text{ bar} Thermodynamic Deduction: Because attractive forces dominate (Z<1Z < 1), the fugacity coefficient is less than unity (ϕ=0.905\phi = 0.905). The real gas exerts a chemical escaping tendency equivalent to an ideal gas at only 45.2 bar45.2\text{ bar} rather than 50.0 bar50.0\text{ bar}.


4. Non-Ideal Solutions: Activity, Activity Coefficients, and Standard States

Activity (aia_i) and Activity Coefficient (γi\gamma_i)

To preserve the thermodynamic equations of ideal solutions (where μi=μi∗+RTln⁡xi\mu_i = \mu_i^* + RT \ln x_i), Lewis introduced activity (aia_i) as "effective concentration":

μi=μi∘+RTln⁡ai\mu_i = \mu_i^\circ + RT \ln a_i

The activity coefficient quantifies departure from ideality:

  • Mole Fraction Scale: ai=γixia_i = \gamma_i x_i
  • Molality Scale: ai=γm(mim∘)[m∘=1 mol/kg]a_i = \gamma_m \left(\frac{m_i}{m^\circ}\right) \quad [m^\circ = 1\text{ mol/kg}]

Standard State Conventions

Standard State TypeTypical ApplicationReference BehaviorMathematical Definition of Standard StateBehavior as Limit
Convention I (Symmetric / Solvent Scale)Solvents, liquid mixtures with miscible liquids (0≤xi≤10 \le x_i \le 1).Raoult's Law: Pi=xiPi∗P_i = x_i P_i^*Pure liquid substance ii at system TT and P∘=1 barP^\circ = 1\text{ bar} (μi∘=μi∗\mu_i^\circ = \mu_i^*).As xi→1x_i \rightarrow 1: γi→1\gamma_i \rightarrow 1 and ai→xia_i \rightarrow x_i.
Convention II (Unsymmetric / Solute Scale)Dissolved solutes (salts, organic compounds in water).Henry's Law: Pi=kH,ixiP_i = k_{H, i} x_iHypothetical ideal solution at m∘=1 mol/kgm^\circ = 1\text{ mol/kg} behaving as if at infinite dilution.As mi→0m_i \rightarrow 0: γm→1\gamma_m \rightarrow 1 and ai→mi/m∘a_i \rightarrow m_i / m^\circ.

The Lewis-Randall Rule

The Lewis-Randall rule estimates the fugacity of a component in a real gas or liquid mixture assuming solution ideality without requiring empirical mixture parameters: fi=xifi∗f_i = x_i f_i^* where fi∗f_i^* is the fugacity of pure component ii at the mixture's temperature and total pressure.

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Chemical Potential, Real Gases, and Non-Ideal Solutions
Test Your Knowledge

In a binary liquid solution of components A and B at constant temperature and pressure, the Gibbs-Duhem equation requires that x_A dmu_A + x_B dmu_B = 0. If an increase in solute concentration causes the chemical potential of solvent A to decrease (dmu_A < 0), how must the chemical potential of component B change?

A

The chemical potential of component B must remain exactly constant because chemical potential is an intensive property.

B

The chemical potential of component B must increase according to the relation dmu_B = -(x_A / x_B) dmu_A.

C

The chemical potential of component B must decrease by the exact same amount: dmu_B = dmu_A.

D

The chemical potential of component B changes randomly because Gibbs-Duhem applies only to ideal ternary systems.

Test Your Knowledge

A real gas is modeled by the van der Waals equation of state with parameters a = 4.225 barL^2/mol^2 and b = 0.0371 L/mol. What is the Boyle temperature (T_B) of this gas, and how does the compressibility factor Z behave as pressure increases at temperatures below T_B? (Use R = 0.08314 barL/(mol*K))

A

T_B = 1024 K; Z remains identically 1.0 at all pressures.

B

T_B = 542 K; below T_B, Z is always greater than 1.0 because molecular repulsions dominate at all non-zero pressures.

C

T_B = 1370 K; below T_B, Z first dips below 1, then rises above 1 at high pressure.

D

T_B = 273.2 K; Z increases monotonically above 1.0 from zero pressure.

Test Your Knowledge

For a real gas at 300.0 K and 50.0 bar, the compressibility factor is accurately modeled by the truncated virial equation Z = 1 + B'P, where B' = -2.00 * 10^-3 bar^-1. What is the fugacity coefficient (phi) and the fugacity (f) of the gas under these conditions?

A

phi = 0.905 and f = 45.2 bar

B

phi = 1.000 and f = 50.0 bar

C

phi = 1.105 and f = 55.3 bar

D

phi = 0.818 and f = 40.9 bar

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