The Gibbs-Duhem Equation

Pour a splash of ethanol into water and the total volume does not simply add up. The two partial molar volumes shift together, and neither is free to change on its own. The Gibbs-Duhem equation is the precise mathematical statement of that coupling. It matters because every mixture property, chemical potential, volume, enthalpy, entropy, obeys the same constraint. Once you know how one component's partial molar quantity varies with composition, the equation fixes how the other's must vary. This is not a modeling assumption; it follows directly from the definition of extensive state functions. This article derives the Gibbs-Duhem equation from the total differential of Gibbs energy. It shows how the equation constrains chemical potentials and partial molar volumes in a binary mixture. It ends with two worked numerical examples, common pitfalls, and a preview of how the equation lets you build one activity coefficient from another.

Key takeaways

The Gibbs-Duhem equation states that at constant $T$ and $P$ , mole-weighted changes in chemical potential across all components must sum to zero.
It is a direct consequence of Euler's theorem for extensive, first-order homogeneous functions like $G$ .
It applies to any partial molar quantity, not just chemical potential: volume, enthalpy, entropy all obey the same form.
In a binary mixture it lets you compute one component's activity coefficient curve from the other's, a technique used constantly in vapor-liquid equilibrium work.
It also explains why partial molar volumes of the two components in a binary mixture always move in opposite directions relative to mole fraction.
Key idea: Because total $G$ (or $V$ , $H$ , ...) is fully determined by the mole-weighted sum of partial molar quantities, those partial molar quantities cannot vary independently of composition.

The core insight

Because the Gibbs energy of a mixture is exactly $G = \sum_i n_i \mu_i$ with no leftover term, differentiating this identity forces the mole-weighted sum of chemical potential changes to vanish at fixed $T$ and $P$ , so the components' properties are mathematically chained together rather than free to vary independently.

The physical picture

Start with the chemical potential of a mixture component, defined as the partial molar Gibbs energy,

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

This is a prerequisite idea: chemical potential tells you how much the total Gibbs energy changes per mole of $i$ added, holding everything else fixed. Partial molar volume is defined the same way for $V$ . The subtlety is that $\mu_i$ is not a fixed number for a substance. It depends on composition. In an ideal solution, $\mu_i = \mu_i^{\star} + RT\ln x_i$ , so as $x_i$ changes, $\mu_i$ changes too. The Gibbs-Duhem equation asks: if $\mu_1$ changes by some amount as composition shifts, how much must $\mu_2$ change in response? The answer comes from a bookkeeping fact. Total Gibbs energy of a mixture is exactly the mole-weighted sum of the chemical potentials,

G = n_1\mu_1 + n_2\mu_2 + \dots

with no extra term. This is because $G$ is an extensive, first-order homogeneous function of the $n_i$ at fixed $T,P$ — doubling every mole number exactly doubles $G$ . Euler's theorem then guarantees the sum above is exact, not approximate. Differentiating that exact identity and comparing it against the general total differential of $G$ forces a constraint. That constraint is the Gibbs-Duhem equation. Physically: in a binary liquid mixture, if adding component 2 lowers $\mu_1$ (say, by dilution), the equation guarantees $\mu_2$ must simultaneously rise in exactly the compensating amount, weighted by mole fractions. Neither chemical potential curve is independent; one determines the other up to an integration constant.

Figure

Coupled variation of two partial molar quantities across composition

The core relations

\sum_i n_i\, d\mu_i = 0 \quad (\text{at constant } T, P)

where: $n_i$ is moles of component $i$ , $d\mu_i$ the differential change in its chemical potential.

\sum_i x_i\, d\mu_i = 0 \quad (\text{at constant } T, P)

where: $x_i = n_i/n_{\text{tot}}$ is mole fraction; dividing through by total moles gives the intensive, per-mole form.

x_1\,d\mu_1 + x_2\,d\mu_2 = 0 \quad \text{(binary system)}

where: this is the two-component version used in essentially all worked problems.

x_1\left(\frac{\partial \mu_1}{\partial x_1}\right)_{T,P} + x_2\left(\frac{\partial \mu_2}{\partial x_1}\right)_{T,P} = 0

where: rewriting $d\mu_i$ as a derivative with respect to composition at fixed $T,P$ , the practical form used to relate activity coefficient slopes.

\sum_i n_i\, d\bar{Y}_i = 0 \quad (\text{general partial molar quantity } Y)

where: the same relation holds for any partial molar property $\bar{Y}_i$ (volume, enthalpy, entropy), not only chemical potential.

Derivation AFrom Euler's theorem to the Gibbs-Duhem constraint

Goal. Show that the mole-weighted differentials of chemical potential must sum to zero at constant $T$ and $P$ .

Step 1. Write the general differential of $G$ . For a mixture, $G$ depends on $T$ , $P$ , and all mole numbers $n_i$ . The total differential is

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

This is the fundamental equation of mixture thermodynamics, following from the definitions of $S=-(\partial G/\partial T)_{P,n}$ , $V=(\partial G/\partial P)_{T,n}$ , and $\mu_i$ .

Step 2. Write the exact Euler sum for $G$ . Because $G$ is extensive and first-order homogeneous in the $n_i$ at fixed $T,P$ — scaling all $n_i$ by $\lambda$ scales $G$ by $\lambda$ — Euler's theorem gives an exact identity, not an approximation:

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

Step 3. Differentiate the Euler sum completely. Take the total differential of $G = \sum_i n_i\mu_i$ , treating both $n_i$ and $\mu_i$ as variables (product rule on each term):

dG = \sum_i \mu_i\, dn_i + \sum_i n_i\, d\mu_i

Step 4. Compare the two expressions for $dG$ . At constant $T$ and $P$ , Step 1 reduces to $dG = \sum_i \mu_i\,dn_i$ . Setting this equal to the Step 3 result:

\sum_i \mu_i\, dn_i = \sum_i \mu_i\, dn_i + \sum_i n_i\, d\mu_i

Step 5. Cancel and isolate the constraint. The $\sum_i \mu_i\,dn_i$ terms cancel from both sides, leaving

0 = \sum_i n_i\, d\mu_i

Dividing by total moles $n_{\text{tot}}$ converts this to intensive mole fractions, since $n_i/n_{\text{tot}} = x_i$ :

\boxed{\sum_i x_i\, d\mu_i = 0 \quad \text{at constant } T, P}

The identical argument applied to $V = \sum_i n_i \bar V_i$ or $H = \sum_i n_i \bar H_i$ gives the same structure for any partial molar quantity, since none of the steps used any property specific to $G$ or $\mu_i$ .

Worked examples

Example 1 — Partial molar volumes in a binary liquid mixture (foundational). An experimentalist measures the partial molar volume of water in an ethanol-water mixture at $x_{\text{water}} = 0.60$ and finds $\bar V_{\text{water}} = 17.0\ \text{cm}^3/\text{mol}$ . Nearby, at $x_{\text{water}} = 0.65$ , $\bar V_{\text{water}} = 16.7\ \text{cm}^3/\text{mol}$ (a decrease of $0.3\ \text{cm}^3/\text{mol}$ as water mole fraction rises, typical of the contraction seen in this system). Estimate the corresponding change in $\bar V_{\text{ethanol}}$ over the same interval. Step 1 — set up the Gibbs-Duhem relation for volume.

x_1\, d\bar V_1 + x_2\, d\bar V_2 = 0

Let component 1 = water, component 2 = ethanol. At the midpoint of the interval, $x_1 \approx 0.625$ , so $x_2 \approx 0.375$ . Step 2 — insert the measured change.

d\bar V_1 = 16.7 - 17.0 = -0.3\ \text{cm}^3/\text{mol}

Step 3 — solve for $d\bar V_2$ .

d\bar V_2 = -\frac{x_1}{x_2}\, d\bar V_1 = -\frac{0.625}{0.375}\times(-0.3\ \text{cm}^3/\text{mol})

d\bar V_2 = +0.50\ \text{cm}^3/\text{mol}

Sanity check. Water's partial molar volume fell while ethanol's rose — exactly the compensating behavior Gibbs-Duhem requires. The magnitudes differ because they are weighted by the other component's mole fraction ( $x_1/x_2 = 1.67$ ), not by symmetry. This matches the qualitative shape in the figure above: as $x_2$ (ethanol) increases from this point, $\bar V_1$ should keep gently falling while $\bar V_2$ rises.

Example 2 — Predicting an activity coefficient curve from Gibbs-Duhem (applied, exam-level). For a binary mixture at fixed $T$ and $P$ , the activity coefficient of component 1 is found experimentally to follow the two-suffix Margules form

\ln\gamma_1 = A x_2^2

with $A = 0.85$ over the whole composition range. Use the Gibbs-Duhem equation to find the corresponding expression for $\ln\gamma_2$ , and evaluate both activity coefficients at $x_1 = 0.30$ . Step 1 — write $\mu_i$ in terms of activity coefficients.

\mu_i = \mu_i^{\star} + RT\ln x_i + RT\ln\gamma_i

d\mu_i = RT\, d\ln x_i + RT\, d\ln\gamma_i

Step 2 — insert into the Gibbs-Duhem constraint at constant $T,P$ .

x_1\, d\ln x_1 + x_1\, d\ln\gamma_1 + x_2\, d\ln x_2 + x_2\, d\ln\gamma_2 = 0

Since $x_1\,d\ln x_1 + x_2\,d\ln x_2 = dx_1 + dx_2 = 0$ (using $d\ln x_i = dx_i/x_i$ and $x_1+x_2=1$ ), those terms cancel identically, leaving the activity-coefficient-only form:

x_1\, d\ln\gamma_1 + x_2\, d\ln\gamma_2 = 0

Step 3 — differentiate the given $\ln\gamma_1$ . With $x_2 = 1-x_1$ , treat $\ln\gamma_1 = Ax_2^2$ as a function of $x_1$ :

\frac{d\ln\gamma_1}{dx_1} = 2Ax_2\left(\frac{dx_2}{dx_1}\right) = -2Ax_2

Step 4 — solve for $d\ln\gamma_2/dx_1$ .

x_2\frac{d\ln\gamma_2}{dx_1} = -x_1\frac{d\ln\gamma_1}{dx_1} = -x_1(-2Ax_2) = 2Ax_1x_2

\frac{d\ln\gamma_2}{dx_1} = 2Ax_1

Step 5 — integrate.

\ln\gamma_2 = Ax_1^2 + C

The physical boundary condition $\gamma_2 \to 1$ as $x_2\to 1$ (i.e., $x_1\to 0$ ) forces $C=0$ :

\ln\gamma_2 = Ax_1^2

Step 6 — evaluate numerically at $x_1 = 0.30$ , so $x_2 = 0.70$ .

\ln\gamma_1 = (0.85)(0.70)^2 = (0.85)(0.49) = 0.4165 \;\Rightarrow\; \gamma_1 = e^{0.4165} = 1.517

\ln\gamma_2 = (0.85)(0.30)^2 = (0.85)(0.09) = 0.0765 \;\Rightarrow\; \gamma_2 = e^{0.0765} = 1.079

Sanity check. The dilute component (component 1, $x_1=0.30$ , more "diluted" relative to its own reference than component 2 is) shows the larger deviation from ideality ( $\gamma_1 = 1.517 \gg 1$ ), while the more concentrated component 2 sits closer to $\gamma_2=1$ . Both curves derived from a single Margules parameter $A$ automatically satisfy Gibbs-Duhem — this is exactly the symmetric Margules model, and the consistency is not a coincidence but the whole point of the derivation.

Limiting cases & common traps

Situation	What Gibbs-Duhem tells you
Ideal mixture, all $x_i$	$d\mu_i = RT\,d\ln x_i$ automatically satisfies $\sum x_i d\mu_i=0$ ; check it as a consistency test.
Dilute limit, $x_2\to 0$	Component 2's chemical potential can vary steeply (Henry's law regime) while component 1 barely changes, since $x_1\approx1$ demands only a tiny compensating shift.
Single-component system	The equation is trivially satisfied ( $x_1=1$ , no coupling partner) — Gibbs-Duhem has no content for pure substances.
Non-isothermal, non-isobaric change	The simple form $\sum x_i d\mu_i=0$ does not apply; you must retain the $-\bar S_i\,dT + \bar V_i\,dP$ terms.
Ternary and higher mixtures	The same relation holds with an extra term per component: $\sum_{i=1}^{k} x_i\,d\mu_i = 0$ , still just one constraint equation.

The chemist behind it

The equation bears two names because it was derived independently by two physical chemists working on different sides of the Atlantic in the 1870s-1880s. Josiah Willard Gibbs established the general thermodynamic framework of chemical potential and phase equilibrium in his monumental 1876-1878 papers "On the Equilibrium of Heterogeneous Substances," effectively inventing the mathematics of multicomponent thermodynamics from scratch. Pierre Duhem, a French physicist and philosopher of science, developed closely related relations for mixtures and extended thermodynamic potential theory in his own systematic treatises around 1886. Their combined names now label this constraint precisely because both approached the same extensivity argument from independent directions, arriving at the identical mole-weighted sum rule.

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

Ask the tutor: How do I use the Gibbs-Duhem equation to derive Henry's law constant from a solvent's Raoult's law behavior?

Does the Gibbs-Duhem equation apply only to chemical potential?

No. It applies to any partial molar quantity because the derivation never used a property specific to $\mu_i$ or $G$ . The same Euler-theorem argument works for partial molar volume, enthalpy, entropy, or heat capacity. For volume the relation reads $\sum_i x_i\,d\bar V_i = 0$ at constant $T,P$ , which is exactly what Example 1 used to link water's and ethanol's partial molar volumes.

Why does the equation require constant temperature and pressure?

The full differential of $G$ includes $-S\,dT + V\,dP$ terms alongside the composition terms. Only when $dT=0$ and $dP=0$ do those extra terms vanish, leaving the clean mole-weighted sum over chemical potentials. If temperature or pressure varies during the process, the complete relation must retain $-\bar S_i\,dT$ and $\bar V_i\,dP$ contributions for each component, and the simple form no longer holds.

How is Gibbs-Duhem used to test experimental activity coefficient data for consistency?

If you measure $\gamma_1$ and $\gamma_2$ independently across a composition range, say from vapor pressure data, the Gibbs-Duhem equation predicts they cannot be arbitrary. Plotting $x_1\,d\ln\gamma_1 + x_2\,d\ln\gamma_2$ against composition should give zero everywhere. Large deviations flag experimental error or an inconsistent data set, making this one of the standard consistency checks in vapor-liquid equilibrium research.

Can Gibbs-Duhem be used to derive Henry's law?

Yes, in the dilute limit. If the solvent obeys Raoult's law behavior ( $\gamma_1\to1$ ) as $x_2\to0$ , the Gibbs-Duhem equation forces the solute's activity coefficient to approach a constant, non-unity limiting value rather than 1. That constant limiting behavior, chemical potential linear in $\ln x_2$ with a solute-specific reference state, is precisely the thermodynamic content of Henry's law, and it emerges directly from integrating the Gibbs-Duhem constraint in the dilute regime.

What happens to the equation in a pure single-component system?

It becomes trivial. With only one component, $x_1=1$ always, so there is no other mole fraction to weight against, and the "constraint" reduces to $1\cdot d\mu_1 = 0$ , meaning $\mu_1$ is fixed by $T$ and $P$ alone, which is simply the ordinary behavior of a pure substance. Gibbs-Duhem only carries genuine content once two or more components are present and can vary relative to each other.