The Chemical Potential

Drop a crystal of sugar into water and it dissolves — spontaneously, without being pushed. Something is driving that flow, the same way a temperature difference drives heat and a height difference drives water downhill. That something is the chemical potential, $\mu$ , the "molar Gibbs energy" of a substance in whatever mixture it happens to be in. Chemical potential matters because it is the single quantity that governs every kind of matter transfer: dissolution, evaporation, osmosis, and chemical reaction itself. Once you know how $\mu$ varies with composition, pressure, and temperature, equilibrium conditions fall out automatically — no separate rule is needed for each type of process. This article builds $\mu$ from the partial molar Gibbs energy, writes the open-system fundamental equation, and shows how equality of $\mu$ across phases defines equilibrium. It sets up the next steps: the ideal-gas formula for $\mu$ , the Gibbs–Duhem equation, and the reaction Gibbs energy.

Key takeaways

$\mu_i$ is the partial molar Gibbs energy of substance $i$ : the Gibbs energy added per mole of $i$ added, at constant $T$ , $p$ , and all other mole numbers.
The open-system fundamental equation is $dG = -SdT + Vdp + \sum_i \mu_i\, dn_i$ .
Matter flows spontaneously from high $\mu$ to low $\mu$ , exactly as heat flows from high $T$ to low $T$ .
At equilibrium between phases or across a reaction, the chemical potentials of every mobile species must be equal (or balanced according to stoichiometry).
$\mu$ depends on concentration or pressure, so a pure substance and that same substance in a dilute mixture have different $\mu$ even at the same $T$ and $p$ .
Key idea: the chemical potential is the "escaping tendency" of a substance — the thermodynamic force pushing it out of one phase or state and into another.

The core insight

The chemical potential $\mu_i$ measures how much the Gibbs energy of a system changes when one mole of substance $i$ is added at constant temperature, pressure, and composition of everything else; spontaneous processes always move each substance toward lower $\mu$ , which is why equal chemical potentials — not equal concentrations — define equilibrium.

The physical picture

Imagine a beaker of water sitting open to moist air. Water molecules leave the liquid and enter the vapor, and vapor molecules condense back into liquid. At equilibrium the rates balance, but rates are hard to use as a criterion because they depend on kinetics, surface area, and stirring. Thermodynamics needs a state function instead — something that depends only on $T$ , $p$ , and composition, and that tells us, without any kinetic information, which way matter will flow. That function is $\mu$ . Just as a ball rolls downhill because gravitational potential energy decreases, a substance moves from a region of high chemical potential to a region of low chemical potential. When $\mu_{\text{liquid}} = \mu_{\text{vapor}}$ , there is no net driving force left, and the phases are in equilibrium. The same logic applies to a solute distributing between two solvents, a proton moving between an acid and its conjugate base, and reactants converting to products. Formally, $\mu_i$ is a partial molar quantity — the same family as partial molar volume, which you have already met when studying the pressure dependence of the Gibbs energy. For any extensive property $X$ of a mixture,

\bar{X}_i = \left(\frac{\partial X}{\partial n_i}\right)_{T,p,n_{j\neq i}}

is the partial molar version of $X$ . Applying this definition to the Gibbs energy $G$ gives the chemical potential directly: $\mu_i \equiv \bar{G}_i$ . It is the slice of the total Gibbs energy that "belongs" to substance $i$ , per mole, in the specific mixture at hand. Because $G$ is extensive and depends on composition, adding a mole of substance $i$ changes $G$ by an amount that depends on how much $i$ and how much of everything else is already present. That dependence is exactly what makes $\mu$ a function of composition, not just of $T$ and $p$ — a crucial point when comparing a pure liquid to a solution.

Figure

Chemical potential as the driving force for matter flow

The core relations

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

where: $\mu_i$ is the chemical potential of species $i$ , and the derivative is taken holding $T$ , $p$ , and every other mole number fixed.

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

where: this is the fundamental equation for an open system — the closed-system relation $dG = -SdT + Vdp$ extended with a term for each species that can enter or leave.

G = \sum_i n_i \mu_i \qquad (T, p \text{ fixed})

where: because $G$ is a first-order homogeneous function of the $n_i$ , Euler's theorem lets the total Gibbs energy be written as a mole-weighted sum of chemical potentials.

\mu_i^{\alpha} = \mu_i^{\beta} \quad \text{(equilibrium between phases } \alpha, \beta\text{)}

where: at equilibrium the chemical potential of any species free to move between phases is the same in every phase it occupies.

Derivation AFrom the fundamental equation to the equilibrium condition

Goal. Starting from the closed-system relation for $G$ , build the open-system fundamental equation, then show why equal $\mu$ across phases is required for equilibrium.

Step 1: the closed-system starting point. For a closed system of fixed composition, combining the first and second laws gives the familiar

dG = -S\,dT + V\,dp .

This was established when studying the pressure dependence of $G$ ; it holds only if no matter crosses the system boundary and no reaction changes the amounts present.

Step 2: allow composition to change. Now let $G$ depend on the mole numbers $n_1, n_2, \dots$ as well as $T$ and $p$ :

G = G(T, p, n_1, n_2, \dots).

The total differential of any such function is

dG = \left(\frac{\partial G}{\partial T}\right)_{p,n_i} dT + \left(\frac{\partial G}{\partial p}\right)_{T,n_i} dp + \sum_i \left(\frac{\partial G}{\partial n_i}\right)_{T,p,n_{j\neq i}} dn_i .

The first two partial derivatives are unchanged from the closed-system case, so they equal $-S$ and $V$ respectively. The third set of derivatives is new — it captures the energy cost of adding material — and each one is defined as the chemical potential $\mu_i$ .

Step 3: write the open-system fundamental equation. Substituting these identifications gives

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

This equation is completely general: it applies to any single phase, however many components it contains, as long as $T$ , $p$ , and the $n_i$ are the independent variables.

Step 4: integrate at constant intensive state. Hold $T$ and $p$ fixed and imagine building the system by adding all its components simultaneously, always in the same proportions, so that every $\mu_i$ stays constant along the way (this is legitimate because $\mu_i$ depends only on the intensive state — mole fractions, $T$ , $p$ — not on the total size of the system). Integrating $dG = \sum_i \mu_i\, dn_i$ under these conditions from zero material to the final amounts gives

G = \sum_i n_i \mu_i .

Step 5: apply to a two-phase system. Consider a substance $i$ present in two phases $\alpha$ and $\beta$ (say liquid and vapor) in thermal and mechanical contact, so $T$ and $p$ are equal throughout. Let $dn_i$ moles move from $\alpha$ to $\beta$ , with the total amount of $i$ conserved: $dn_i^\alpha = -dn_i^\beta$ . The change in total Gibbs energy is

dG = \mu_i^\alpha\, dn_i^\alpha + \mu_i^\beta\, dn_i^\beta = (\mu_i^\beta - \mu_i^\alpha)\, dn_i^\beta .

Step 6: impose the equilibrium condition. At constant $T$ and $p$ , spontaneous change requires $dG \le 0$ , and equilibrium requires $dG = 0$ for any small displacement $dn_i^\beta$ . If $\mu_i^\alpha \neq \mu_i^\beta$ , the system can always lower $G$ by moving matter from the higher- $\mu$ phase to the lower- $\mu$ phase, so $dG < 0$ and the system is not at equilibrium. Only when

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

is $dG = 0$ for arbitrary $dn_i^\beta$ , which is the condition for phase equilibrium.

Worked examples

Example 1 — Using $\mu$ to find the direction of transfer. At 298 K and 1 bar, suppose the chemical potential of water in a particular sugar solution is $\mu(\text{soln}) = -237.13\ \text{kJ mol}^{-1}$ , while the chemical potential of pure liquid water at the same $T$ and $p$ is $\mu^*(\text{water}) = -237.13\ \text{kJ mol}^{-1} + RT\ln x_{\text{H}_2\text{O}}$ with mole fraction $x_{\text{H}_2\text{O}} = 0.960$ (this logarithmic form is the ideal-solution result, previewed here and derived properly for gases in the next article). Compute $\mu(\text{soln})$ :

RT\ln x_{\text{H}_2\text{O}} = (8.314\ \text{J K}^{-1}\text{mol}^{-1})(298\ \text{K})\ln(0.960)

= 2477.6\ \text{J mol}^{-1} \times (-0.04082) = -101.1\ \text{J mol}^{-1} = -0.1011\ \text{kJ mol}^{-1}

\mu(\text{soln}) = -237.13\ \text{kJ mol}^{-1} + (-0.1011\ \text{kJ mol}^{-1}) = -237.23\ \text{kJ mol}^{-1}.

Compare this to $\mu^*(\text{pure water}) = -237.13\ \text{kJ mol}^{-1}$ . Since

\mu(\text{soln}) = -237.23\ \text{kJ mol}^{-1} < \mu^*(\text{pure water}) = -237.13\ \text{kJ mol}^{-1},

water in the solution has a lower chemical potential than pure water. If the sugar solution is placed in contact with pure water across a semipermeable membrane, water will flow spontaneously from the pure-water side (high $\mu$ ) into the solution (low $\mu$ ) — this is exactly osmosis, and the calculation shows why solute lowers water's escaping tendency. Sanity check: the shift is small (about 100 J out of 237,000 J, or 0.04%), consistent with a dilute solution producing a modest but nonzero osmotic driving force.

Example 2 — Vapor–liquid equilibrium and finding the equilibrium vapor pressure. A liquid has chemical potential $\mu^{\text{liq}} = \mu^{\ominus} + RT\ln(1) = \mu^{\ominus}$ (taking the pure liquid as the reference state, so its activity is 1). Its vapor, treated as an ideal gas, has chemical potential

\mu^{\text{vap}}(p) = \mu^{\ominus}_{\text{gas}} + RT\ln\left(\frac{p}{p^{\ominus}}\right),

a relation you will derive explicitly from the pressure-dependence of $G$ in the next article on the chemical potential of an ideal gas. Suppose at 350 K, $\mu^{\ominus}(\text{liq}) - \mu^{\ominus}_{\text{gas}} = -5.20\ \text{kJ mol}^{-1}$ , and $p^{\ominus} = 1\ \text{bar}$ . Find the equilibrium vapor pressure $p_{\text{eq}}$ . At equilibrium, $\mu^{\text{liq}} = \mu^{\text{vap}}$ :

\mu^{\ominus}(\text{liq}) = \mu^{\ominus}_{\text{gas}} + RT\ln\left(\frac{p_{\text{eq}}}{p^{\ominus}}\right)

Rearranging,

\ln\left(\frac{p_{\text{eq}}}{p^{\ominus}}\right) = \frac{\mu^{\ominus}(\text{liq}) - \mu^{\ominus}_{\text{gas}}}{RT} = \frac{-5200\ \text{J mol}^{-1}}{(8.314\ \text{J K}^{-1}\text{mol}^{-1})(350\ \text{K})}

= \frac{-5200}{2909.9} = -1.787

\frac{p_{\text{eq}}}{p^{\ominus}} = e^{-1.787} = 0.1675 \quad \Rightarrow \quad p_{\text{eq}} = 0.1675\ \text{bar} = 16.75\ \text{kPa}.

Sanity check: this is a physically reasonable sub-atmospheric vapor pressure for a moderately volatile liquid at 350 K (compare water's vapor pressure at 350 K, about 41.8 kPa — same order of magnitude, and this liquid is evidently somewhat less volatile). The negative sign of $\mu^{\ominus}(\text{liq}) - \mu^{\ominus}_{\text{gas}}$ correctly predicts $p_{\text{eq}} < p^{\ominus}$ , since the liquid is thermodynamically favored relative to the standard gas state at this temperature.

Limiting cases & common traps

Situation	What to remember
Pure substance, one phase	$\mu = \bar{G}_m$ , the ordinary molar Gibbs energy; no composition dependence to worry about.
Ideal gas	$\mu$ rises logarithmically with $p$ ; derived explicitly in the next article.
Dilute solute	$\mu$ rises logarithmically with concentration (Henry's-law region); large changes in $\mu$ for small concentration changes at very low concentration.
Multi-phase equilibrium	Equal $T$ , equal $p$ , and equal $\mu_i$ for every mobile species — all three, not just one.
Reaction equilibrium	$\sum_i \nu_i \mu_i = 0$ over products minus reactants, weighted by stoichiometric numbers $\nu_i$ — the basis of the reaction Gibbs energy, coming next.

The chemist behind it

The chemical potential was introduced by the American mathematical physicist Josiah Willard Gibbs in his monumental 1876–1878 memoir "On the Equilibrium of Heterogeneous Substances." Working largely in isolation at Yale, with almost no contemporary readership, Gibbs built the entire framework of chemical thermodynamics — Gibbs energy, phase rule, and chemical potential — from a handful of postulates about entropy and energy. It took decades, and translation efforts by Wilhelm Ostwald into German and Henri Le Chatelier into French, before chemists outside a small circle of physicists appreciated that Gibbs had already solved, in full generality, the problem of predicting equilibrium in any multi-phase, multi-component system.

Origins · 1 minJosiah Willard Gibbs and how this was discovered

Josiah Willard Gibbs

1839–1903 · American scientist who defined chemical potential and founded chemical thermodynamics

Ask the tutor: How does the chemical potential of an ideal gas get its logarithmic form from the pressure dependence of Gibbs energy?

Is chemical potential the same thing as molar Gibbs energy?

Only for a pure substance. For a pure substance in a single phase, $\mu = \bar{G}_m = G/n$ , the ordinary molar Gibbs energy. In a mixture, however, $\mu_i$ is the partial molar Gibbs energy — the rate of change of total $G$ with respect to $n_i$ holding everything else fixed — and it generally differs from $G/n_{\text{total}}$ and from the molar Gibbs energy of pure $i$ . This distinction becomes essential once you study solutions and Raoult's law.

Can chemical potential be negative?

Yes, and it usually is, because $\mu$ is measured relative to an arbitrary standard state (often $\mu^{\ominus}$ at 1 bar or 1 mol/L) plus a logarithmic correction term. The value of $\mu$ has no absolute meaning by itself — only differences in $\mu$ , or comparisons between phases at the same reference state, carry physical significance for predicting spontaneous change.

How is chemical potential related to concentration or partial pressure?

For an ideal gas, $\mu = \mu^{\ominus} + RT\ln(p/p^{\ominus})$ ; for an ideal dilute solute, $\mu = \mu^{\ominus} + RT\ln(c/c^{\ominus})$ . Both forms say the same thing: $\mu$ increases logarithmically as the substance becomes more concentrated, which is why doubling concentration has a much bigger effect near very low concentrations than near high ones. The next article derives the gas-phase formula from the pressure dependence of $G$ .

Why is equality of chemical potential, rather than equality of concentration, the correct equilibrium condition?

Concentration equality would only apply to identical phases in identical environments. Chemical potential accounts for how favorable each environment is: a solute can sit at very different concentrations in two immiscible phases and still be at equilibrium, because differences in solvation energy shift $\mu^{\ominus}$ between the phases. Equilibrium requires the total escaping tendency, $\mu$ , to match, not the raw amount present.