Chemical Potential in Mixtures

Pour cream into black coffee and watch the swirl vanish into uniform tan. Nothing pushed it — no stirrer, no pressure gradient. Something else drove the mixing, and that something is chemical potential. It measures how much a substance "wants" to leave one location, phase, or chemical form for another. Chemical potential is the single most important quantity in the thermodynamics of mixtures. It governs phase equilibria, chemical equilibria, osmosis, and electrochemistry, all with one unifying rule. Material flows spontaneously toward lower chemical potential, exactly as heat flows toward lower temperature. This article defines chemical potential as a partial molar Gibbs energy. You will see why it is the correct driving force for matter transfer. Then you will derive its equation of state dependence, work two full numerical examples, and see how it sets up the Gibbs–Duhem equation and Raoult's law.

Key takeaways

Chemical potential $\mu_i$ is the partial molar Gibbs energy of component $i$ in a mixture.
It equals the rate of change of any energy function with $n_i$ , but Gibbs energy at fixed $T, p$ is the practical choice.
Matter moves spontaneously from high $\mu$ to low $\mu$ , just as heat moves from high $T$ to low $T$ .
At equilibrium between phases or in a reaction, the chemical potentials of each species must be equal (or balanced by stoichiometry).
For ideal gases, $\mu$ depends logarithmically on pressure or mole fraction, giving the template for all activity-based descriptions.
Key idea: Chemical potential is the "escaping tendency" of a substance — the thermodynamic force per mole that drives diffusion, phase change, and reaction toward equilibrium.

The core insight

Chemical potential $\mu_i = (\partial G/\partial n_i)_{T,p,n_{j\neq i}}$ is the partial molar Gibbs energy of species $i$ , and because $G$ is minimized at equilibrium, matter always flows spontaneously from regions or phases of high $\mu_i$ to regions of low $\mu_i$ until all $\mu_i$ are equal.

The physical picture

Imagine a closed system held at constant temperature and pressure, containing several chemical species. The total Gibbs energy $G$ depends on $T$ , $p$ , and the amount of each substance present, $n_1, n_2, \ldots$ . If you add a tiny amount of substance $i$ while holding everything else fixed, $G$ changes by an amount proportional to how much you added. The proportionality constant is the chemical potential of that substance in that particular mixture, at that particular composition. This is exactly the logic of partial molar quantities: just as partial molar volume tells you how volume responds to adding a component, chemical potential tells you how Gibbs energy responds. The difference is that Gibbs energy is the "master" thermodynamic potential for processes at constant $T$ and $p$ — the natural conditions of the benchtop and the living cell — so $\mu_i$ inherits a special status. It is not merely one partial molar quantity among many; it is the one that dictates which way matter moves. Consider two bulbs of gas connected by a valve, one at high pressure and one at low pressure. Gas flows from high pressure to low pressure until the pressures equalize — but from a thermodynamic standpoint, gas flows from high chemical potential to low chemical potential. Since $\mu$ for an ideal gas increases monotonically with pressure at fixed $T$ , the two descriptions agree. This mechanical intuition extends the same rule to situations where pressure is irrelevant, such as a solute partitioning between water and octanol, or a solid subliming into vapor. In every case, equilibrium means equal chemical potential of each freely-exchanging species in every phase where it appears.

Figure

Chemical potential drives transfer between two phases until equal

Figure

Chemical potential of an ideal gas versus pressure at 298 K

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 $i$ , defined as the partial molar Gibbs energy at constant temperature, pressure, and all other mole numbers.

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

where: this is the fundamental equation of open-system thermodynamics; the last term shows $\mu_i$ as the "work" done per mole of $i$ added to the system.

G = \sum_i n_i \mu_i \quad \text{(constant composition ratios, Euler's theorem)}

where: total Gibbs energy is the mole-weighted sum of chemical potentials, since $G$ is a first-order homogeneous function of the $n_i$ .

\mu_i(T,p) = \mu_i^{\circ}(T) + RT\ln\left(\frac{p_i}{p^{\circ}}\right) \quad \text{(ideal gas, component } i\text{)}

where: $\mu_i^{\circ}$ is the standard chemical potential at $p^\circ = 1\ \text{bar}$ , $p_i$ is the partial pressure, $R = 8.314\ \text{J K}^{-1}\text{mol}^{-1}$ .

d\mu_i = -\bar{S}_i\, dT + \bar{V}_i\, dp

where: $\bar{S}_i$ and $\bar{V}_i$ are the partial molar entropy and volume of $i$ ; this shows how $\mu_i$ itself responds to temperature and pressure changes.

Derivation AFrom the fundamental equation to the ideal-gas chemical potential

Goal. Show that chemical potential is the partial molar Gibbs energy, then derive its explicit pressure dependence for an ideal gas at fixed temperature.

Step 1. Start from the combined first and second laws for an open system. For a closed system undergoing reversible change, $dU = T\,dS - p\,dV$ . Allowing the system to exchange matter, each species $i$ contributes an extra term proportional to $dn_i$ :

dU = T\,dS - p\,dV + \sum_i \mu_i\, dn_i

This equation defines $\mu_i$ as the coefficient linking energy change to matter change, at constant $S$ and $V$ .

Step 2. Convert to the Gibbs energy representation. Since $G = U + pV - TS$ , differentiate:

dG = dU + p\,dV + V\,dp - T\,dS - S\,dT

Substituting $dU$ from Step 1, the $T\,dS$ and $p\,dV$ terms cancel pairwise, leaving

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

Reading off the coefficient of $dn_i$ at fixed $T$ and $p$ confirms

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

which is precisely the definition of a partial molar Gibbs energy — the same operational recipe used for partial molar volume, just applied to $G$ .

Step 3. Fix composition ratios and integrate using Euler's theorem. $G$ is a first-order homogeneous function of the $n_i$ at fixed $T, p$ : doubling all mole numbers doubles $G$ . Euler's theorem for homogeneous functions then gives directly

G = \sum_i n_i \mu_i

This is not a new assumption — it follows automatically once $\mu_i$ is defined as a partial derivative of an extensive, first-order homogeneous $G$ .

Step 4. Specialize to one mole of pure ideal gas at temperature $T$ . For one mole, $\mu = G_m$ , the molar Gibbs energy. From $dG_m = -S_m\,dT + V_m\,dp$ at constant $T$ ,

d\mu = V_m\,dp

For an ideal gas $V_m = RT/p$ , so integrating from the standard pressure $p^{\circ}$ to $p$ :

\mu(p) - \mu^{\circ} = \int_{p^{\circ}}^{p} \frac{RT}{p'}\, dp' = RT \ln\left(\frac{p}{p^{\circ}}\right)

Step 5. Extend to a component in a gas mixture. For an ideal gas mixture, each component behaves as if it alone occupied the container at its own partial pressure $p_i = x_i p$ (Dalton's law), so the same integration applies component-by-component:

\boxed{\mu_i = \mu_i^{\circ} + RT \ln\left(\frac{p_i}{p^{\circ}}\right) = \mu_i^{\circ} + RT\ln\left(\frac{p}{p^{\circ}}\right) + RT\ln x_i}

This boxed result is the seed from which Raoult's law, Henry's law, and all activity descriptions of real mixtures grow — replace $p_i/p^\circ$ with an activity $a_i$ and the entire machinery of non-ideal solutions follows the same logarithmic template.

Worked examples

Example 1 — Chemical potential difference across a pressure gradient (foundational). Two bulbs of nitrogen gas at $T = 298\ \text{K}$ are connected by a closed valve. Bulb A is at $p_A = 5.00\ \text{bar}$ ; bulb B is at $p_B = 1.00\ \text{bar}$ . Find $\mu_A - \mu_B$ and predict what happens when the valve opens. Step 1 — write the expression for each bulb. Treating N₂ as ideal and using the same reference state $\mu^\circ$ for both (same substance, same $T$ ):

\mu_A = \mu^{\circ} + RT\ln\left(\frac{p_A}{p^{\circ}}\right), \qquad \mu_B = \mu^{\circ} + RT\ln\left(\frac{p_B}{p^{\circ}}\right)

Step 2 — subtract; $\mu^\circ$ cancels.

\mu_A - \mu_B = RT\ln\left(\frac{p_A}{p_B}\right)

Step 3 — plug in numbers.

\mu_A - \mu_B = (8.314\ \text{J K}^{-1}\text{mol}^{-1})(298\ \text{K})\ln\left(\frac{5.00}{1.00}\right)

= (2477.8\ \text{J mol}^{-1})(1.6094) = 3987\ \text{J mol}^{-1} \approx +3.99\ \text{kJ mol}^{-1}

Step 4 — interpret. Since $\mu_A > \mu_B$ , gas flows spontaneously from A to B when the valve opens, exactly matching the everyday expectation that gas flows from high to low pressure. The chemical potential framework reproduces the mechanical answer and generalizes it to cases with no pressure at all. Sanity check: the sign is positive because $p_A > p_B$ , consistent with flow from A (high $\mu$ ) to B (low $\mu$ ).

Example 2 — Chemical potential of a solute and its concentration dependence (applied). A dilute aqueous solution of glucose has mole fraction $x_{\text{glc}} = 2.50 \times 10^{-3}$ at $T = 310\ \text{K}$ (body temperature), behaving ideally-dilute so that $\mu_{\text{glc}} = \mu^{\circ}_{\text{glc}} + RT\ln x_{\text{glc}}$ . A second compartment, separated by a semipermeable membrane, has $x'_{\text{glc}} = 8.00\times 10^{-4}$ at the same temperature. Compute $\mu_{\text{glc}} - \mu'_{\text{glc}}$ and determine which way glucose diffuses. Step 1 — set up the difference. Since both compartments share the same solvent, temperature, and standard state,

\mu_{\text{glc}} - \mu'_{\text{glc}} = RT\ln\left(\frac{x_{\text{glc}}}{x'_{\text{glc}}}\right)

Step 2 — evaluate the ratio.

\frac{x_{\text{glc}}}{x'_{\text{glc}}} = \frac{2.50\times 10^{-3}}{8.00\times 10^{-4}} = 3.125

Step 3 — compute $RT$ at 310 K.

RT = (8.314\ \text{J K}^{-1}\text{mol}^{-1})(310\ \text{K}) = 2577.3\ \text{J mol}^{-1}

Step 4 — take the logarithm and multiply.

\ln(3.125) = 1.1394

\mu_{\text{glc}} - \mu'_{\text{glc}} = (2577.3\ \text{J mol}^{-1})(1.1394) = 2937\ \text{J mol}^{-1} \approx +2.94\ \text{kJ mol}^{-1}

Step 5 — interpret. The first compartment has higher chemical potential, so glucose diffuses spontaneously from the concentrated side to the dilute side across the membrane, even though no external pressure difference drives it. This is the same logic used later for osmotic pressure and for the direction of facilitated transport across cell membranes. Sanity check: the answer is positive and of similar order to Example 1 (a few kJ/mol), reasonable for a modest ten-fold-ish concentration ratio at RT ≈ 2.5–2.6 kJ/mol.

Limiting cases & common traps

Situation	What happens to $\mu_i$	Common trap
Pure substance	$\mu = G_m$ , the molar Gibbs energy	Forgetting $\mu$ is per mole, not total $G$
Ideal gas, $p \to 0$	$\mu_i \to -\infty$ (logarithm diverges)	Treating this as unphysical rather than a reference-state artifact
$x_i \to 1$ (nearly pure)	$\mu_i \to \mu_i^{*}$ (pure-substance value)	Confusing $\mu_i^{\circ}$ (fixed standard state) with $\mu_i^{*}$ (pure liquid/solid reference, composition-dependent)
Equilibrium between two phases	$\mu_i(\alpha) = \mu_i(\beta)$ for every mobile species	Assuming equal pressure or concentration is required — only $\mu$ equality is required
Chemical reaction at equilibrium	$\sum_i \nu_i \mu_i = 0$	Forgetting stoichiometric coefficients $\nu_i$ multiply each $\mu_i$

The chemist behind it

The concept of 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 experimental apparatus, Gibbs built the entire theory of multicomponent phase and chemical equilibrium from pure thermodynamic reasoning. His definition of $\mu_i$ as a partial derivative of internal energy (and, equivalently, of Gibbs energy) with respect to the amount of a component unified problems that previously seemed unrelated: evaporation, solubility, chemical reaction, and osmosis all became instances of one governing rule, equality of chemical potential. The work was so far ahead of its time and so densely mathematical that it took decades and translation efforts by scientists like Wilhelm Ostwald before its full impact on European chemistry was felt.

Origins · 1 minJosiah Willard Gibbs and how this was discovered

Josiah Willard Gibbs

1839–1903 · American theoretical physicist and chemist who founded chemical thermodynamics and defined chemical potential

Ask the tutor: How does the Gibbs–Duhem equation constrain how the chemical potentials of two components in a binary mixture can change together?

Is chemical potential the same as Gibbs energy?

No. Gibbs energy $G$ is an extensive total-system property with units of joules, while chemical potential $\mu_i$ is an intensive, per-mole quantity with units of joules per mole. They are related by $G = \sum_i n_i \mu_i$ at fixed composition ratios, and $\mu_i$ itself is the partial derivative of $G$ with respect to the amount of component $i$ . Think of $\mu_i$ as the "molar price" of species $i$ , and $G$ as the total bill.

Why use Gibbs energy rather than internal energy to define chemical potential?

Chemical potential can be defined from any energy function, but $(\partial G/\partial n_i)_{T,p}$ is the most useful version because laboratory and biological processes typically occur at constant temperature and pressure. Since $G$ is minimized at equilibrium under those conditions, using the Gibbs-energy-based $\mu_i$ directly connects to spontaneity criteria you already use, avoiding awkward constant-volume, constant-entropy conditions that rarely apply in practice.

What happens to chemical potential as concentration approaches zero?

For an ideal or ideally-dilute species, $\mu_i = \mu_i^{\circ} + RT\ln x_i$ , and as $x_i \to 0$ the logarithm diverges toward negative infinity. This is not a physical catastrophe; it reflects the enormous entropic reward of dispersing a few particles into a huge volume. It also explains why complete removal of a solute or trace gas is thermodynamically impossible in a finite process — the chemical potential drop needed becomes infinite.

How does chemical potential relate to Raoult's law and Henry's law?

Both laws are direct consequences of the equality of chemical potential between the liquid and vapor phases. At equilibrium $\mu_i(\text{liquid}) = \mu_i(\text{vapor})$ , and substituting the ideal-gas expression for the vapor and an ideal-solution expression for the liquid produces Raoult's law $p_i = x_i p_i^{*}$ for the solvent, or Henry's law $p_i = K_i x_i$ for a dilute solute. Chemical potential is the shared currency that makes both relationships thermodynamically necessary rather than empirical curve fits.

Why must chemical potential be equal in every phase at equilibrium, but not concentration?

Equilibrium is defined by the minimum of total Gibbs energy at fixed $T$ and $p$ . If $\mu_i$ differed between two phases, moving a small amount of $i$ from the high- $\mu$ phase to the low- $\mu$ phase would lower $G$ , so the system has not yet reached equilibrium. Concentrations need not match because different phases offer different molecular environments, entropy, and interaction energies, all folded into $\mu_i^{\circ}$ and the activity term, so equal $\mu_i$ can correspond to very unequal concentrations.