9.4: Composition Dependence of ΔG

\[ \left(\dfrac{\partial G}{\partial n_i} \right)_{T,P}=\mu_i \nonumber \]

The third and final coefficient gives the chemical potential as the dependence of \(G\) on the chemical composition at constant \(T\) and \(P\). Similarly to the previous cases, we can take the definition of the coefficient and integrate it directly between the initial and final stages of a reaction. If we consider a reaction product, pure substance \(i\), at the beginning of the reaction there will be no moles of it \(n_i=0\), and consequently \(G=0\).¹ We can then integrate the left-hand side between zero and the number of moles of product at the end of the reaction, \(n\), and the right-hand side between zero and the Gibbs free energy of the product, \(G\). The integral will become:

\[ \int_0^G d G = \int_0^n \mu^* dn, \label{9.4.1} \]

where \(\mu^*\) indicates the chemical potential of a pure substance, which is independent on the number of moles by definition. As such, Equation \ref{9.4.1} becomes:

\[ \int_0^G d G = \mu^* \int_0^n dn \quad \rightarrow \quad G = \mu^* n \quad \rightarrow \quad \mu^* = \dfrac{G}{n}, \label{9.4.2} \]

which gives a straightforward interpretation of the chemical potential of a pure substance as the molar Gibbs free energy.

We can start from Equation 9.3.3 and write for a pure substance that is brought from \(P_i=P^{-\kern-6pt{\ominus}\kern-6pt-}\) to \(P_f=P\) at constant \(T\):

\[ G - G^{-\kern-6pt{\ominus}\kern-6pt-}= nRT \ln \dfrac{P}{P^{-\kern-6pt{\ominus}\kern-6pt-}}, \label{9.4.3} \]

dividing both sides by \(n\), we obtain:

\[ \dfrac{G}{n} - \dfrac{G^{-\kern-6pt{\ominus}\kern-6pt-}}{n} = RT \ln \dfrac{P}{P^{-\kern-6pt{\ominus}\kern-6pt-}}, \label{9.4.4} \]

which, for a pure substance at \(P^{-\kern-6pt{\ominus}\kern-6pt-}= 1 \;\text{bar}\), becomes:

\[ \mu^* - \mu^{-\kern-6pt{\ominus}\kern-6pt-}= RT \ln P. \label{9.4.5} \]

Notice that, while we use the pressure of the gas inside the logarithm in Equation \ref{9.4.5}, the quantity is formally divided by the standard pressure \(P^{-\kern-6pt{\ominus}\kern-6pt-}= 1 \;\text{bar}\), and therefore it is a dimensionless quantity, as it should be. For simplicity of notation, however, we will omit the division by \(P^{-\kern-6pt{\ominus}\kern-6pt-}\) in the remaining of this textbook, especially wherever it does not create confusion. Let’s now consider a mixture of ideal gases, and let’s try to find out whether the chemical potential of a pure gas inside the mixture, \(\mu_i^{\text{mixture}}\), is the same as its chemical potential outside the mixture, \(\mu^*\). To do so, we can use Equation \ref{9.4.5} and replace the pressure \(P\) with the partial pressure \(P_i\):

\[ \mu_i^{\text{mixture}} = \mu_i^{-\kern-6pt{\ominus}\kern-6pt-}+ RT \ln P_i, \label{9.4.6} \]

where the partial pressure \(P_i\) can be obtained from the simple relation that is known as Dalton’s Law:

\[ P_i = y_i P, \label{9.4.7} \]

with \(y_i\) being the concentration of gas \(i\) measured as a mole fraction in the gas phase \(y_i=\dfrac{n_i}{n_{\text{TOT}}} < 1\). Replacing Equation \ref{9.4.7} into Equation \ref{9.4.6}, we obtain:

\[\begin{equation} \begin{aligned} \mu_i^{\text{mixture}} &= \mu_i^{-\kern-6pt{\ominus}\kern-6pt-}+ RT \ln (y_i P) \\ &= \underbrace{\mu_i^{-\kern-6pt{\ominus}\kern-6pt-}+ RT \ln P}_{\mu_i^*} + RT \ln y_i, \end{aligned} \end{equation} \label{9.4.8} \]

which then reduces to the following equation:

\[ \mu_i^{\text{mixture}} = \mu_i^* + RT \ln y_i. \label{9.4.9} \]

Analyzing Equation \ref{9.4.9}, we can immediately see that, since \(y_i < 1\):

\[ \mu_i^{\text{mixture}} < \mu_i^*,\label{9.4.10} \]

or, in other words, the chemical potential of a substance in the mixture is always lower than the chemical potential of the pure substance. If we consider a process where we start from two separate pure ideal gases and finish with a mixture of the two, we can calculate the change in Gibbs free energy due to the mixing process with:

\[ \Delta_{\text{mixing}} G = \sum n_i \left( \mu_i^{\text{mixture}} - \mu_i^* \right) < 0, \label{9.4.11} \]

or, in other words, the process is spontaneous under all circumstances, and pure ideal gases will always mix.

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1. For reactants, the same situation usually applies but in reverse. More complicated cases where the reaction does not consume all reactants are possible, but insignificant for the following treatment.