Thermodynamic Potentials and Equilibrium

Thermodynamic potentials (enthalpy, Helmholtz free energy, Gibbs free energy, grand potential) answer the question: which state is stable under a given set of constraints? This page is relevant for predicting phase coexistence, chemical reaction direction, battery voltage, and material stability. It unifies the first and second laws into a single minimization principle.

Authoritative sources

Thermodynamic potential — Wikipedia

Gibbs free energy — Wikipedia

Callen, Thermodynamics and an Introduction to Thermostatistics, Ch. 5

The fundamental potentials

Potential	Definition	Natural variables	Minimized at equilibrium when
U(S,V,N)	Internal energy	S, V, N	(S, V, N fixed)
H(S,P,N) = U + PV	Enthalpy	S, P, N	Constant P, S, N (e.g., throttling)
F(T,V,N) = U − TS	Helmholtz free energy	T, V, N	Constant T, V, N (e.g., system in a heat bath at fixed volume)
G(T,P,N) = U + PV − TS	Gibbs free energy	T, P, N	Constant T, P, N (most common: open to atmosphere, in a heat bath)
Ω(T,V,μ) = F − μN	Grand potential	T, V, μ	Constant T, V, μ (system in contact with both a heat bath and a particle reservoir)

Details

Why these combinations? Each potential is a Legendre transform of $U$ that replaces an extensive variable with its conjugate intensive variable. For example, $F = U - TS$ replaces $S$ (extensive) with $T$ (intensive, its conjugate via $T = (\partial U/\partial S)_{V,N}$ ). At fixed $T, V$ , minimizing $F$ is equivalent to maximizing $S_{\text{total}}$ (system + reservoir), which is the second law.

Differential forms:

dU = T\,dS - P\,dV + \mu\,dN dH = T\,dS + V\,dP + \mu\,dN dF = -S\,dT - P\,dV + \mu\,dN dG = -S\,dT + V\,dP + \mu\,dN d\Omega = -S\,dT - P\,dV - N\,d\mu

These are the fundamental relations. All thermodynamic properties can be derived from any one potential as a function of its natural variables (it is a generating function).

Microscopic derivation. The connection to statistical mechanics is direct: $F = -k_B T \ln Z$ (canonical partition function) and $G = -k_B T \ln \Xi$ (isothermal–isobaric partition function). See Statistical Foundations of Thermodynamics.

Stability criteria

At equilibrium the relevant potential is at a minimum with respect to internal variables (composition, phase fractions). For the minimum to be stable (not a saddle point), the potential must satisfy convexity/concavity conditions:

For $U(S,V,N)$ : $(\partial^2 U/\partial S^2)_{V,N} > 0$ (equivalently $C_V > 0$ ) and $(\partial^2 U/\partial V^2)_{S,N} > 0$ (equivalently $\kappa_S > 0$ ).
For $G(T,P,N)$ : $(\partial^2 G/\partial T^2)_{P,N} < 0$ (equivalently $C_P > 0$ ) and $(\partial^2 G/\partial P^2)_{T,N} < 0$ (equivalently $\kappa_T > 0$ ).

These conditions ensure that the equilibrium is physically realizable: a positive heat capacity means adding heat raises temperature (not lowers it), and a positive compressibility means compressing raises pressure (not lowers it).

Chemical potential and reaction equilibrium

The chemical potential $\mu_i = (\partial G / \partial N_i)_{T,P,N_{j\neq i}}$ is the molar Gibbs free energy of species $i$ in a mixture.

For a reaction $\sum_i \nu_i A_i = 0$ at constant $T, P$ :

\Delta_r G = \sum_i \nu_i \mu_i

$\Delta_r G < 0$ : reaction proceeds forward.
$\Delta_r G = 0$ : equilibrium.
$\Delta_r G > 0$ : reaction proceeds in reverse.

Details

Standard state and equilibrium constant. Writing $\mu_i = \mu_i^\circ + RT\ln a_i$ (where $a_i$ is activity), at equilibrium:

\Delta_r G^\circ = -RT\ln K

where $K$ is the equilibrium constant. This connects thermodynamics to chemical kinetics and to electrochemistry (Nernst equation: $\mathcal{E} = \mathcal{E}^\circ - (RT/nF)\ln Q$ ).

Batteries. The maximum electrical work a battery can deliver at constant $T, P$ is $\Delta G$ of the cell reaction. The open-circuit voltage is $\mathcal{E} = -\Delta G / (nF)$ . See also Electrochemistry (future page) for the full treatment of electrode potentials and the Nernst equation.

Phase equilibrium and the Gibbs phase rule

Two phases $\alpha$ and $\beta$ of a pure substance coexist when their chemical potentials are equal:

\mu^\alpha(T,P) = \mu^\beta(T,P)

For a system with $C$ components and $P$ phases, the Gibbs phase rule gives the number of intensive degrees of freedom:

F = C - P + 2

Details

Derivation. Each phase has $C + 2$ intensive variables ( $T, P$ , and $C$ composition variables). At equilibrium, $T$ and $P$ are the same in all phases, and $\mu_i^\alpha = \mu_i^\beta$ for each component and each pair of phases. Counting constraints gives $F = C - P + 2$ .

Examples:

Pure substance ( $C=1$ ), one phase ( $P=1$ ): $F = 2$ (choose any $T, P$ ).
Pure substance, two phases (e.g., liquid + vapor): $F = 1$ (vapor pressure curve; choose $T$ , then $P$ is fixed).
Pure substance, three phases (triple point): $F = 0$ (unique $T, P$ ).
Binary alloy ( $C=2$ ), two phases: $F = 2$ (choose $T$ and one composition).

Clapeyron and Clausius–Clapeyron equations. The slope of a phase boundary:

\frac{dP}{dT} = \frac{\Delta S}{\Delta V} = \frac{\Delta H}{T\,\Delta V}

For a vapor–liquid transition ( $V_{\text{gas}} \gg V_{\text{liq}}$ , ideal gas): $dP/dT = \Delta H_{\text{mol}}/(RT^2)$ (Clausius–Clapeyron), where $\Delta H_{\text{mol}}$ is the molar latent heat (the enthalpy change per mole transferred between phases). This integrates to the familiar vapor-pressure curve.

Maxwell relations

From the equality of mixed partials of a potential, e.g. from $dG = -S\,dT + V\,dP + \mu\,dN$ :

\left(\frac{\partial S}{\partial P}\right)_{T,N} = -\left(\frac{\partial V}{\partial T}\right)_{P,N}

The four Maxwell relations (one per potential) allow you to compute entropy changes from measurable $P, V, T$ data.

Details

All four (pure substance, $N$ fixed):

\left(\frac{\partial T}{\partial V}\right)_S = -\left(\frac{\partial P}{\partial S}\right)_V \quad (U) \left(\frac{\partial T}{\partial P}\right)_S = \left(\frac{\partial V}{\partial S}\right)_P \quad (H) \left(\frac{\partial S}{\partial V}\right)_T = \left(\frac{\partial P}{\partial T}\right)_V \quad (F) \left(\frac{\partial S}{\partial P}\right)_T = -\left(\frac{\partial V}{\partial T}\right)_P \quad (G)

Use. To find $(\partial S/\partial T)_P = C_P/T$ , or to compute $(\partial U/\partial V)_T = T(\partial P/\partial T)_V - P$ (which vanishes for an ideal gas, confirming $U = U(T)$ ).

Example problem

At 298 K and 1 atm, the standard Gibbs free energies of formation are: $\Delta_f G^\circ(\text{H}_2\text{O, l}) = -237.1\,\text{kJ/mol}$ , $\Delta_f G^\circ(\text{H}_2\text{, g}) = 0$ , $\Delta_f G^\circ(\text{O}_2\text{, g}) = 0$ . For the reaction $\text{H}_2 + \tfrac{1}{2}\text{O}_2 \to \text{H}_2\text{O(l)}$ , find $\Delta_r G^\circ$ and the maximum electrical work per mole of water produced. What is the corresponding cell voltage ( $n = 2$ electrons)?

Solution

\Delta_r G^\circ = \Delta_f G^\circ(\text{H}_2\text{O}) - \Delta_f G^\circ(\text{H}_2) - \tfrac{1}{2}\Delta_f G^\circ(\text{O}_2) = -237.1 \text{ kJ/mol}

Maximum electrical work (at constant $T, P$ ): $w_{\text{elec,max}} = \Delta_r G^\circ = -237.1\,\text{kJ/mol}$ (negative sign means work is done by the system).

Cell voltage:

\mathcal{E}^\circ = \frac{-\Delta_r G^\circ}{nF} = \frac{237100}{2 \times 96485} \approx 1.23 \text{ V}

This is the standard hydrogen fuel-cell voltage.

Thermodynamic Potentials and Equilibrium §

The fundamental potentials §

Stability criteria §

Chemical potential and reaction equilibrium §

Phase equilibrium and the Gibbs phase rule §

Maxwell relations §

Example problem §

Thermodynamic Potentials and Equilibrium

The fundamental potentials

Stability criteria

Chemical potential and reaction equilibrium

Phase equilibrium and the Gibbs phase rule

Maxwell relations

Example problem