# 24.2: The Gibbs-Duhem Equation Relates Chemical Potential and Composition at Equilibrium

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At equilibrium, there is no change in chemical potential for the system:

$\sum_i n_i d\mu_i = 0 \label{eq1}$

This is the Gibbs-Duhem relationship and it places a compositional constraint upon any changes in the chemical potential in a mixture at constant temperature and pressure for a given composition. This result is easily derived when one considers that $$\mu_i$$ represents the partial molar Gibbs function for component $$i$$. And as with other partial molar quantities:

$G_text{tot} = \sum_i n_i \mu_i$

Taking the derivative of both sides yields:

$dG_text{tot} = \sum_i n_i d \mu_i + \sum_i \mu_i d n_i$

But $$dG$$ can also be expressed as:

$dG = Vdp - sdT + \sum_i \mu_i d n_i$

Setting these two expressions equal to one another:

$\sum_i n_i d \mu_i + \cancel{ \sum_i \mu_i d n_i } = Vdp - sdT + \cancel{ \sum_i \mu_i d n_i}$

And after canceling terms, one gets:

$\sum_i n_i d \mu_i = Vdp - sdT \label{eq41}$

For a system at constant temperature and pressure:

$Vdp - sdT = 0 \label{eq42}$

Substituting Equation \ref{eq42} into \ref{eq41} results in the Gibbs-Duhem equation (Equation \ref{eq1}). This expression relates how the chemical potential can change for a given composition while the system maintains equilibrium.

## Gibbs-Duhem for Binary Systems

For a binary system consisting of components two components, $$A$$ and $$B$$:

$n_Bd\mu_B + n_Ad\mu_A = 0$

Rearranging:

$d\mu_B = -\dfrac{n_A}{n_B} d\mu_A$

Consider a Gibbs free energy that only includes $$μ_n$$ conjugate variables as we obtained it from our scaling experiment at $$T$$ and $$P$$ constant:

$G = \mu_An_A + \mu_Bn_B \nonumber$

Consider a change in $$G$$:

$dG = d(\mu_An_A) + d(\mu_Bn_B) \nonumber$

$dG = n_Ad\mu_A+\mu_Adn_A + n_Bd\mu_B+\mu_Bdn_B \nonumber$

However, if we simply write out a change in $$G$$ due to the number of moles we have:

$dG = \mu_Adn_A +\mu_Bdn_B \nonumber$

Consequently the other terms must add up to zero:

$0 = n_Ad\mu_A+ n_Bd\mu_B \nonumber$

$d\mu_A= - \dfrac{n_B}{n_A}d\mu_B \nonumber$

$d\mu_A= - \dfrac{x_B}{x_A}d\mu_B \nonumber$

In the last step we have simply divided both denominator and numerator by the total number of moles. This expression is the Gibbs-Duhem equation for a 2-component system. It relates the change in one thermodynamic potential ($$d\mu_A$$) to the other ($$d\mu_B$$).

The Gibbs-Duhem equation relates the change in one thermodynamic potential ($$d\mu_A$$) to the other ($$d\mu_B$$).

## Gibbs-Duhem in the Ideal Case

In the ideal case we have:

$\mu_B = \mu^*_B + RT \ln x_B \nonumber$

Gibbs-Duhem gives:

$d\mu_A = - \dfrac{x_B}{x_A} d\mu_B \nonumber$

As:

$d\mu_B = 0 + \dfrac{RT}{x_B} \nonumber$

with $$x_B$$ being the only active variable at constant temperature, we get:

$d\mu_A = - \dfrac{x_B}{x_A} \dfrac{RT}{x_B} = \dfrac{RT}{x_A} \nonumber$

If we now wish to find $$\mu_A$$ we need to integrate $$d\mu_A$$, e.g. form pure 1 to $$x_A$$. This produces:

$\mu_A = \mu^*_A + RT \ln x_A \nonumber$

This demonstrates that Raoult's law can only hold over the whole range for one component if it also holds for the other over the whole range.

24.2: The Gibbs-Duhem Equation Relates Chemical Potential and Composition at Equilibrium is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts.