Chemical Potential and the Equilibrium Criterion

A puddle of water sits in humid air, unchanging in size for hours. Nothing looks like it is happening. Yet water molecules are constantly leaving the liquid and constantly returning from the vapor. The puddle looks static because the two flows balance exactly. That balance is not a coincidence of rates. It is a thermodynamic requirement written in the language of chemical potential. Chemical potential measures the tendency of a substance to escape from a phase, react, or move to another location. Equilibrium between phases means that tendency is identical everywhere the substance can go. This article builds the equilibrium criterion from the Gibbs energy of a multiphase system. You will see why unequal chemical potential forces matter to flow. You will see how equality of $\mu$ across phases falls out as a minimum-energy condition. Finally, you will use the criterion to solve real numeric problems in vapor pressure and phase composition. This sets up the Gibbs Phase Rule and the Clapeyron equation that follow.

Key takeaways

Chemical potential $\mu_i$ is the partial molar Gibbs energy of substance $i$ ; it measures escaping tendency.
At constant $T$ and $p$ , a closed multiphase system reaches equilibrium when $G$ is minimized.
Minimizing $G$ with respect to matter transfer between phases forces $\mu_i^\alpha = \mu_i^\beta$ for every phase pair.
If $\mu_i^\alpha \neq \mu_i^\beta$ , substance $i$ spontaneously flows from the high- $\mu$ phase to the low- $\mu$ phase.
This single equality criterion underlies boiling points, freezing points, solubility limits, and osmotic equilibrium.
Key idea: phases coexist in equilibrium only when every component has the same chemical potential in each phase.

The core insight

Chemical potential is the thermodynamic "pressure" that drives matter between phases, and equilibrium is reached precisely when that pressure is equalized for every species across every coexisting phase, exactly as thermal equilibrium requires equal temperature and mechanical equilibrium requires equal pressure.

The physical picture

Consider liquid water in a closed, rigid container with empty space above it, held at constant temperature. Molecules evaporate from the surface; molecules condense from the vapor. Early on, the liquid has a strong tendency to lose molecules to the vapor phase because the vapor is nearly empty and disordered — entropy favors filling it. As vapor accumulates, its molecules also feel a pull back toward the liquid. Eventually the two escaping tendencies match, and the net transfer of matter stops. This state is what we mean by liquid–vapor equilibrium at a given temperature. The abstract quantity that captures "escaping tendency" is the chemical potential,

\mu_i = \left(\frac{\partial G}{\partial n_i}\right)_{T,p,n_{j\neq i}}

the increase in Gibbs energy per mole of substance $i$ added, at fixed $T$ , $p$ , and composition otherwise. It plays the same unifying role for matter flow that temperature plays for heat flow and pressure plays for volume changes. Just as heat flows from high $T$ to low $T$ until temperatures equalize, matter flows from a phase of high $\mu_i$ to a phase of low $\mu_i$ until the chemical potentials equalize. This is a completely general statement. It applies to a pure substance distributing itself between liquid and vapor, a solute distributing itself between two immiscible solvents, or a component partitioning between solid and liquid at a melting point. In every case, the criterion for equilibrium is the same equality of chemical potential, phase by phase.

Figure

Chemical potential driving matter transfer toward equality

Notice the arrow: matter always flows from the phase of higher chemical potential to the phase of lower chemical potential, lowering the total Gibbs energy as it goes, until the two values meet. That endpoint, not any particular rate of evaporation or condensation, defines phase equilibrium. A second useful picture is a plot of $\mu$ for each phase as a function of temperature, at fixed pressure. Where two curves cross, the corresponding phases are in equilibrium; that crossing point is the melting or boiling temperature at that pressure.

Figure

Chemical potential curves crossing at the boiling point

Below the crossing temperature the liquid curve lies lower (liquid is the stable phase, lower $G$ per mole); above it the vapor curve lies lower. Exactly at the crossing, $\mu_{\text{liquid}} = \mu_{\text{vapor}}$ , and both phases coexist — this is the ordinary boiling point.

The core relations

dG = -S\,dT + V\,dp + \sum_i \mu_i\, dn_i

where: $G$ is total Gibbs energy, $S$ entropy, $V$ volume, and $\mu_i$ the chemical potential conjugate to moles $n_i$ of species $i$ .

\mu_i = \left(\frac{\partial G}{\partial n_i}\right)_{T,p,n_{j\neq i}}

where: $\mu_i$ is the partial molar Gibbs energy, the Gibbs energy cost of adding one mole of $i$ at constant $T$ , $p$ , composition.

\mu_i^{\alpha} = \mu_i^{\beta} = \cdots = \mu_i^{\pi}

where: for every species $i$ present, its chemical potential must be identical in all $\pi$ coexisting phases at equilibrium.

dG_{T,p} = \sum_i \left(\mu_i^{\beta}-\mu_i^{\alpha}\right) dn_i^{\beta} \le 0

where: for spontaneous transfer of $dn_i^\beta$ moles of $i$ into phase $\beta$ from phase $\alpha$ at fixed $T,p$ , the total Gibbs energy must not increase.

Derivation AEquality of chemical potential from Gibbs energy minimization

Goal. Show that a closed system at fixed $T$ and $p$ , containing a substance distributed between two phases $\alpha$ and $\beta$ , reaches equilibrium only when the chemical potential of that substance is equal in both phases.

Step 1. Set up the composite system. Let phases $\alpha$ and $\beta$ coexist in a closed container held at constant $T$ and $p$ . Let $n^\alpha$ moles of substance $i$ sit in phase $\alpha$ and $n^\beta$ moles sit in phase $\beta$ . Because the container is closed to the surroundings but matter can pass freely between the two phases,

n^\alpha + n^\beta = n_{\text{total}} = \text{constant}

so that $dn^\alpha = -dn^\beta$ : whatever leaves $\alpha$ enters $\beta$ .

Step 2. Write the total Gibbs energy change. At constant $T$ and $p$ , the Gibbs energy of the whole two-phase system changes only through matter transfer,

dG = \mu^{\alpha}\,dn^{\alpha} + \mu^{\beta}\,dn^{\beta}

Substituting $dn^\alpha = -dn^\beta$ gives

dG = \left(\mu^{\beta} - \mu^{\alpha}\right) dn^{\beta}

Step 3. Apply the criterion for spontaneity. At fixed $T$ and $p$ , the second law requires that any spontaneous process lowers $G$ (or leaves it unchanged at equilibrium),

dG \le 0 \quad \text{for any allowed } dn^{\beta}

If $\mu^{\beta} < \mu^{\alpha}$ , then choosing $dn^{\beta} > 0$ (transfer into $\beta$ ) makes $dG < 0$ : transfer from $\alpha$ into $\beta$ is spontaneous. This continues, changing $n^\alpha$ and $n^\beta$ and hence $\mu^\alpha$ and $\mu^\beta$ themselves, until the driving force disappears.

Step 4. Identify the stationary point. The process stops — reaches equilibrium — only when $dG = 0$ for arbitrary infinitesimal $dn^\beta$ in either direction. That requires the coefficient itself to vanish,

\mu^{\beta} - \mu^{\alpha} = 0

If instead $\mu^\beta \neq \mu^\alpha$ at some point, matter would keep flowing (lowering $G$ further), so that point cannot be a genuine equilibrium. This is a necessary and sufficient stationarity condition for a two-phase, one-component exchange at fixed $T,p$ .

Step 5. Generalize to many phases and many components. With phases $\alpha,\beta,\ldots,\pi$ all coexisting and several species $i=1,2,\ldots$ each free to redistribute, the same argument applies independently to each species between each pair of phases, because $dn_i$ for different species and different phase pairs are independent variations. The stationarity condition becomes, for every species $i$ present in more than one phase,

\boxed{\mu_i^{\alpha} = \mu_i^{\beta} = \cdots = \mu_i^{\pi}}

Species that do not appear in a given phase impose no constraint there; the equality applies phase by phase only where the species is actually present and free to move.

Worked examples

Example 1 — Foundational: verifying equilibrium from tabulated chemical potentials. At $T = 298.15$ K and $p = 1$ bar, suppose the molar Gibbs energy of formation data give an effective chemical potential for liquid water of $\mu(\mathrm{l}) = -237.13$ kJ/mol (this is $\Delta_f G^\circ$ for liquid water). At the same temperature and pressure, water vapor at 1 bar has $\mu(\mathrm{g}) = -228.58$ kJ/mol (its $\Delta_f G^\circ$ ). Compare the two:

\mu(\mathrm{g}) - \mu(\mathrm{l}) = -228.58 - (-237.13) = +8.55\ \text{kJ/mol}

Since $\mu(\mathrm{g}) > \mu(\mathrm{l})$ at 1 bar vapor pressure and 298.15 K, the vapor phase has higher chemical potential. By the equilibrium criterion, matter should spontaneously flow from vapor into liquid — condensation is favored, not evaporation, at these conditions. This is exactly correct: 298.15 K is below the normal boiling point of 373.15 K, so water at 1 bar of vapor pressure is thermodynamically unstable relative to liquid, and indeed the actual equilibrium vapor pressure of water at 25°C is only about 0.032 bar (3.2 kPa), far below 1 bar. The two phases are not in equilibrium under the stated (hypothetical, 1 bar vapor) conditions; the calculation correctly diagnoses that condensation must occur, consistent with everyday observation that liquid water at room temperature does not spontaneously convert to 1 bar of vapor.

Sanity check: the sign makes physical sense — raising the temperature would raise $\mu(\mathrm{g})$ less steeply-negative relative to $\mu(\mathrm{l})$ is not quite the comparison; rather, increasing $T$ decreases both, but $\mu(\mathrm{g})$ decreases faster (gas has larger molar entropy, and $(\partial\mu/\partial T)_p = -S_m$ ), so eventually the curves cross at 373.15 K, matching Figure 2 above.

Example 2 — Applied: two-component liquid–vapor equilibrium and mole fraction from Raoult's law. Consider an ideal liquid mixture of benzene (component A) and toluene (component B) at $T = 363.15$ K (90 °C). At this temperature the pure-component vapor pressures are

p_A^{*} = 136.1\ \text{kPa}, \qquad p_B^{*} = 54.0\ \text{kPa}

Suppose the liquid mixture has mole fraction $x_A = 0.40$ (so $x_B = 0.60$ ). We want the equilibrium vapor composition, using the chemical-potential equality condition applied through Raoult's law, which itself is a direct consequence of $\mu_A^{\text{liquid}} = \mu_A^{\text{vapor}}$ for each component.

Step 1: apply Raoult's law for each component. For an ideal solution, equality of chemical potential across phases for component $i$ reduces to

p_i = x_i\, p_i^{*}

p_A = x_A p_A^{*} = 0.40 \times 136.1\ \text{kPa} = 54.44\ \text{kPa}

p_B = x_B p_B^{*} = 0.60 \times 54.0\ \text{kPa} = 32.40\ \text{kPa}

Step 2: total vapor pressure.

p_{\text{total}} = p_A + p_B = 54.44 + 32.40 = 86.84\ \text{kPa}

Step 3: vapor-phase mole fractions from Dalton's law. Since $\mu_A$ equal in both phases fixes $p_A$ , and the vapor is treated as ideal gas mixture, $y_A = p_A/p_{\text{total}}$ :

y_A = \frac{54.44}{86.84} = 0.627

y_B = \frac{32.40}{86.84} = 0.373

Sanity check. $y_A + y_B = 0.627+0.373 = 1.000$ ✓. Also $y_A = 0.627 > x_A = 0.40$ : the vapor is richer in benzene, the more volatile component (higher $p_A^*$ ), exactly as expected — the phase in equilibrium with the liquid is enriched in whichever component more readily achieves equal $\mu$ by escaping into the vapor. This illustrates that the single criterion $\mu_i^{\text{liquid}}=\mu_i^{\text{vapor}}$ for every component simultaneously, not just for the mixture as a whole, generates the entire quantitative machinery of distillation.

Limiting cases & common traps

Situation	What the criterion says	Common error
Pure substance, two phases	$\mu^\alpha=\mu^\beta$ defines a single $T$ at each $p$ (melting/boiling point)	Forgetting $\mu$ depends on both $T$ and $p$ ; forgetting the line, not a region, is where phases coexist
Component absent from a phase	No constraint on $\mu_i$ in that phase	Wrongly requiring $\mu_i$ equal in a phase where species $i$ does not exist
Mixture, several components	Each species independently satisfies $\mu_i^\alpha=\mu_i^\beta$	Assuming only the total Gibbs energy, not each $\mu_i$ , must balance
Far from equilibrium	$\mu_i^\alpha \neq \mu_i^\beta$ ; net flow persists	Confusing "phases present together" with "phases at equilibrium" — coexistence in time is not automatic equilibrium
Metastable states	Supercooled liquid, superheated liquid: $\mu$ not minimized globally	Assuming any observed stable-looking state satisfies the equality; metastability can persist without true equilibrium

The chemist behind it

The chemical potential and the equilibrium criterion trace directly to Josiah Willard Gibbs, whose 1876–1878 memoir "On the Equilibrium of Heterogeneous Substances" founded chemical thermodynamics essentially single-handedly. Gibbs introduced the function we now call the Gibbs energy, defined the chemical potential as its partial molar derivative, and proved — using variational arguments almost identical to the derivation above — that phase and reaction equilibrium require equality of chemical potential. His work was so far ahead of experimental chemistry of the time that it was barely read for two decades, until continental European scientists like Wilhelm Ostwald and Henri Le Chatelier translated and popularized it, whereupon it became the backbone of physical chemistry.

Origins · 1 minJosiah Willard Gibbs and how this was discovered

Josiah Willard Gibbs

1839–1903 · American mathematical physicist who founded chemical thermodynamics and defined chemical potential as the criterion for phase and reaction equilibrium

Ask the tutor: How does this equality of chemical potential turn into the Clapeyron equation for the slope of a phase boundary?

Why is chemical potential the right variable, rather than concentration or pressure, for the equilibrium criterion?

Chemical potential is built directly from Gibbs energy, the quantity that is minimized at fixed $T$ and $p$ for any spontaneous process. Concentration and partial pressure influence $\mu_i$ but are not themselves conserved driving forces; two phases can have very different concentrations of a species yet identical $\mu_i$ , as happens in any partition equilibrium. Only $\mu_i$ directly measures the Gibbs energy cost of moving one mole of species $i$ from one phase to another, so only its equality guarantees $dG = 0$ and therefore true equilibrium.

Does the equality of chemical potential apply to reacting systems too, not just phase equilibrium?

Yes. The same Gibbs-energy-minimization argument, applied to a reaction extent $\xi$ rather than an interphase transfer $dn^\beta$ , gives $\sum_i \nu_i \mu_i = 0$ at chemical equilibrium, where $\nu_i$ are stoichiometric coefficients. Phase equilibrium and reaction equilibrium are the same underlying idea: $G$ is stationary with respect to every internal variable the system is free to adjust, whether that variable is moles transferred between phases or moles converted by reaction.

How does this relate to the Gibbs Phase Rule that comes next?

The Gibbs Phase Rule counts how many intensive variables (like $T$ , $p$ , composition) can be independently varied while keeping a given number of phases in equilibrium. It is derived by counting the equality constraints $\mu_i^\alpha = \mu_i^\beta = \cdots$ established here and comparing them to the total number of independent composition and state variables. The equilibrium criterion in this article supplies the constraint equations; the phase rule counts degrees of freedom left over.

What happens physically if a system is prepared with unequal chemical potentials across phases?

Matter flows spontaneously from the phase of higher $\mu_i$ to the phase of lower $\mu_i$ , exactly as heat flows from hot to cold. This flow continues, changing local composition and hence $\mu_i$ in both phases, until the values converge. The rate of that approach to equilibrium is a kinetic question outside classical thermodynamics, but the direction and endpoint are dictated entirely by the sign and eventual vanishing of $\mu_i^\alpha - \mu_i^\beta$ .

Can chemical potential be measured directly like temperature with a thermometer?

Not with a single universal instrument, but it can be inferred from measurable quantities: vapor pressure, solubility, electrode potential, or osmotic pressure all provide routes to differences in $\mu_i$ . For an ideal gas, $\mu = \mu^\circ + RT\ln(p/p^\circ)$ links $\mu$ directly to measured pressure, which is how the earlier worked examples convert tabulated $\Delta_f G^\circ$ values and vapor pressures into equilibrium statements.