Gibbs–Duhem equation

Deriving the Gibbs–Duhem equation from the fundamental thermodynamic equation is straightforward.[3] The total differential of the extensive Gibbs free energy ${\displaystyle \mathbf {G} }$ in terms of its natural variables is

${\displaystyle \mathrm {d} \mathbf {G} =\left.{\frac {\partial \mathbf {G} }{\partial p}}\right|_{T,N}\mathrm {d} p+\left.{\frac {\partial \mathbf {G} }{\partial T}}\right|_{p,N}\mathrm {d} T+\sum _{i=1}^{I}\left.{\frac {\partial \mathbf {G} }{\partial N_{i}}}\right|_{p,T,N_{j\neq i}}\mathrm {d} N_{i}.}$

Since the Gibbs free energy is the Legendre transformation of the internal energy, the derivatives can be replaced by its definitions transforming the above equation into:[4]

${\displaystyle \mathrm {d} \mathbf {G} =V\mathrm {d} p-S\mathrm {d} T+\sum _{i=1}^{I}\mu _{i}\mathrm {d} N_{i}}$

The chemical potential is simply another name for the partial molar Gibbs free energy (or the partial Gibbs free energy, depending on whether N is in units of moles or particles). Thus the Gibbs free energy of a system can be calculated by collecting moles together carefully at a specified T, P and at a constant molar ratio composition (so that the chemical potential doesn't change as the moles are added together), i.e.

${\displaystyle \mathbf {G} =\sum _{i=1}^{I}\mu _{i}N_{i}}$.

The total differential of this expression is[4]

${\displaystyle \mathrm {d} \mathbf {G} =\sum _{i=1}^{I}\mu _{i}\mathrm {d} N_{i}+\sum _{i=1}^{I}N_{i}\mathrm {d} \mu _{i}}$

Combining the two expressions for the total differential of the Gibbs free energy gives

${\displaystyle \sum _{i=1}^{I}\mu _{i}\mathrm {d} N_{i}+\sum _{i=1}^{I}N_{i}\mathrm {d} \mu _{i}=V\mathrm {d} p-S\mathrm {d} T+\sum _{i=1}^{I}\mu _{i}\mathrm {d} N_{i}}$

which simplifies to the Gibbs–Duhem relation:[4]

${\displaystyle \sum _{i=1}^{I}N_{i}\mathrm {d} \mu _{i}=-S\mathrm {d} T+V\mathrm {d} p}$

By normalizing the above equation by the extent of a system, such as the total number of moles, the Gibbs–Duhem equation provides a relationship between the intensive variables of the system. For a simple system with ${\displaystyle I}$ different components, there will be ${\displaystyle I+1}$ independent parameters or "degrees of freedom". For example, if we know a gas cylinder filled with pure nitrogen is at room temperature (298 K) and 25 MPa, we can determine the fluid density (258 kg/m³), enthalpy (272 kJ/kg), entropy (5.07 kJ/kg⋅K) or any other intensive thermodynamic variable.[5] If instead the cylinder contains a nitrogen/oxygen mixture, we require an additional piece of information, usually the ratio of oxygen-to-nitrogen.

If multiple phases of matter are present, the chemical potentials across a phase boundary are equal.[6] Combining expressions for the Gibbs–Duhem equation in each phase and assuming systematic equilibrium (i.e. that the temperature and pressure is constant throughout the system), we recover the Gibbs' phase rule.

One particularly useful expression arises when considering binary solutions.[7] At constant P (isobaric) and T (isothermal) it becomes:

${\displaystyle 0=N_{1}\mathrm {d} \mu _{1}+N_{2}\mathrm {d} \mu _{2}}$

or, normalizing by total number of moles in the system ${\displaystyle N_{1}+N_{2},}$ substituting in the definition of activity coefficient ${\displaystyle \gamma }$ and using the identity ${\displaystyle x_{1}+x_{2}=1}$:

${\displaystyle 0=x_{1}\mathrm {d} \ln(\gamma _{1})+x_{2}\mathrm {d} \ln(\gamma _{2})}$ [8]

This equation is instrumental in the calculation of thermodynamically consistent and thus more accurate expressions for the vapor pressure of a fluid mixture from limited experimental data.

Lawrence Stamper Darken has shown that the Gibbs-Duhem equation can be applied to the determination of chemical potentials of components from a multicomponent system from experimental data regarding the chemical potential ${\displaystyle {\bar {G_{2}}}}$ of only one component (here component 2) at all compositions. He has deduced the following relation[9]

${\displaystyle {\bar {G_{2}}}=G+(1-x_{2})\left({\frac {\partial G}{\partial x_{2}}}\right)_{\frac {x_{1}}{x_{3}}}}$

x_i, amount (mole) fractions of components.

Making some rearrangements and dividing by (1 – x₂)² gives:

${\displaystyle {\frac {G}{(1-x_{2})^{2}}}+{\frac {1}{1-x_{2}}}\left({\frac {\partial G}{\partial x_{2}}}\right)_{\frac {x_{1}}{x_{3}}}={\frac {\bar {G_{2}}}{(1-x_{2})^{2}}}}$

or

${\displaystyle \left({\mathfrak {d}}{\frac {G}{\frac {1-x_{2}}{{\mathfrak {d}}x_{2}}}}\right)_{\frac {x_{1}}{x_{3}}}={\frac {\bar {G_{2}}}{(1-x_{2})^{2}}}}$

or

${\displaystyle \left({\frac {\frac {\partial G}{1-x_{2}}}{\partial x_{2}}}\right)_{\frac {x_{1}}{x_{3}}}={\frac {\bar {G_{2}}}{(1-x_{2})^{2}}}}$ as formatting variant

The derivative with respect to one mole fraction x₂ is taken at constant ratios of amounts (and therefore of mole fractions) of the other components of the solution representable in a diagram like ternary plot.

The last equality can be integrated from ${\displaystyle x_{2}=1}$ to ${\displaystyle x_{2}}$ gives:

${\displaystyle G-(1-x_{2})\lim _{x_{2}\to 1}{\frac {G}{1-x_{2}}}=(1-x_{2})\int _{1}^{x_{2}}{\frac {\bar {G_{2}}}{(1-x_{2})^{2}}}dx_{2}}$

Applying LHopital's rule gives:

${\displaystyle \lim _{x_{2}\to 1}{\frac {G}{1-x_{2}}}=\lim _{x_{2}\to 1}\left({\frac {\partial G}{\partial x_{2}}}\right)_{\frac {x_{1}}{x_{3}}}}$.

This becomes further:

${\displaystyle \lim _{x_{2}\to 1}{\frac {G}{1-x_{2}}}=-\lim _{x_{2}\to 1}{\frac {{\bar {G_{2}}}-G}{1-x_{2}}}}$.

Express the mole fractions of component 1 and 3 as functions of component 2 mole fraction and binary mole ratios:

${\displaystyle x_{1}={\frac {1-x_{2}}{1+{\frac {x_{3}}{x_{1}}}}}}$

${\displaystyle x_{3}={\frac {1-x_{2}}{1+{\frac {x_{1}}{x_{3}}}}}}$

and the sum of partial molar quantities

${\displaystyle G=\sum _{i=1}^{3}x_{i}{\bar {G_{i}}},}$

gives

${\displaystyle G=x_{1}({\bar {G_{1}}})_{x_{2}=1}+x_{3}({\bar {G_{3}}})_{x_{2}=1}+(1-x_{2})\int _{1}^{x_{2}}{\frac {\bar {G_{2}}}{(1-x_{2})^{2}}}dx_{2}}$

${\displaystyle ({\bar {G_{1}}})_{x_{2}=1}}$ and ${\displaystyle ({\bar {G_{3}}})_{x_{2}=1}}$ are constants which can be determined from the binary systems 1_2 and 2_3. These constants can be obtained from the previous equality by putting the complementary mole fraction x₃ = 0 for x₁ and vice versa.

Thus

${\displaystyle ({\bar {G_{1}}})_{x_{2}=1}=-\left(\int _{1}^{0}{\frac {\bar {G_{2}}}{(1-x_{2})^{2}}}dx_{2}\right)_{x_{3}=0}}$

and

${\displaystyle ({\bar {G_{3}}})_{x_{2}=1}=-\left(\int _{1}^{0}{\frac {\bar {G_{2}}}{(1-x_{2})^{2}}}dx_{2}\right)_{x_{1}=0}}$

The final expression is given by substitution of these constants into the previous equation:

${\displaystyle G=(1-x_{2})\left(\int _{1}^{x_{2}}{\frac {\bar {G_{2}}}{(1-x_{2})^{2}}}dx_{2}\right)_{\frac {x_{1}}{x_{3}}}-x_{1}\left(\int _{1}^{0}{\frac {\bar {G_{2}}}{(1-x_{2})^{2}}}dx_{2}\right)_{x_{3}=0}-x_{3}\left(\int _{1}^{0}{\frac {\bar {G_{2}}}{(1-x_{2})^{2}}}dx_{2}\right)_{x_{1}=0}}$

• Margules activity model
• Darken's equations
• Gibbs-Helmholtz equation

• J. Phys. Chem. Gokcen 1960
• A lecture from www.chem.neu.edu
• A lecture from www.chem.arizona.edu
• Encyclopædia Britannica entry