1.10.3: Gibbs Energies- Solutions- Solvent and Solute

A given solution (at temperature \(\mathrm{T}\) and pressure \(\mathrm{p}\), where the latter is close to the standard pressure) is prepared using \(1 \mathrm{~kg}\) of water and \(\mathrm{m}_{\mathrm{j}}\) moles of a simple solute. We consider the differential dependence of the excess Gibbs energy for the solution \(\mathrm{G}^{\mathrm{E}}\) on molality \(\mathrm{m}_{\mathrm{j}}\).

\[\mathrm{G}_{\mathrm{m}}^{\mathrm{E}}=\mathrm{R} \, \mathrm{T} \, \mathrm{m}_{\mathrm{j}} \,\left[1-\phi+\ln \left(\gamma_{\mathrm{j}}\right)\right] \nonumber \]

Hence, at fixed \(\mathrm{T}\) and \(\mathrm{p}\),

\[\begin{aligned}
(1 / \mathrm{R} \, \mathrm{T}) \,\left[\mathrm{dG}^{\mathrm{E}} / \mathrm{dm}_{\mathrm{j}}\right]=\left[1-\phi+\ln \left(\gamma_{\mathrm{j}}\right)\right]-\mathrm{m}_{\mathrm{j}} \,\left[\mathrm{d} \phi / \mathrm{dm}_{\mathrm{j}}\right] \\
&+\mathrm{m}_{\mathrm{j}} \,\left[\mathrm{d} \ln \left(\gamma_{\mathrm{j}}\right) / \mathrm{dm}_{\mathrm{j}}\right]
\end{aligned} \nonumber \]

But according to the Gibbs-Duhem equation,

\[-\phi-\mathrm{m}_{\mathrm{j}} \,\left[\mathrm{d} \phi / \mathrm{dm}_{\mathrm{j}}\right]+1+\mathrm{m}_{\mathrm{j}} \,\left[\mathrm{d} \ln \left(\gamma_{\mathrm{j}}\right) / \mathrm{dm}_{\mathrm{j}}\right]=0 \nonumber \]

Hence, we obtain an equation for \(\ln \left(\gamma_{j}\right)\) as a function of the differential dependence of \(\mathrm{G}^{\mathrm{E}}\) on \(\mathrm{m}_{\mathrm{j}}\).[1]

\[\ln \left(\gamma_{\mathrm{j}}\right)=(1 / \mathrm{R} \, \mathrm{T}) \,\left[\mathrm{dG}^{\mathrm{E}} / \mathrm{dm}_{\mathrm{j}}\right] \nonumber \]

If we substitute for \(\ln \left(\gamma_{j}\right)\) in the equation for \(\mathrm{G}^{\mathrm{E}}\), an equation for \(\phi\) in terms of \(\mathrm{G}^{\mathrm{E}}\) is obtained.

\[1-\phi=(1 / \mathrm{R} \, \mathrm{T}) \,\left[\mathrm{G}^{\mathrm{E}} / \mathrm{m}_{\mathrm{j}}-\mathrm{dG}^{\mathrm{E}} / \mathrm{dm}_{\mathrm{j}}\right] \nonumber \]

A more elegant derivation of equation (e) starts out with the equation (a) for the excess Gibbs energy written in the following form.

\[\left[\mathrm{G}^{\mathrm{E}} / \mathrm{m}_{\mathrm{j}}\right] / \mathrm{R} \, \mathrm{T}=1-\phi+\ln \left(\gamma_{\mathrm{j}}\right) \nonumber \]

Then at fixed \(\mathrm{T}\) and \(\mathrm{p}\),

\[(1 / \mathrm{R} \, \mathrm{T}) \,\left\{\mathrm{d}\left[\mathrm{G}^{\mathrm{E}} / \mathrm{m}_{\mathrm{j}}\right] / \mathrm{dm}_{\mathrm{j}}\right\}=-\left(\mathrm{d} \phi / \mathrm{dm}_{\mathrm{j}}\right)+\mathrm{d} \ln \left(\gamma_{\mathrm{j}}\right) / \mathrm{dm}_{\mathrm{j}} \nonumber \]

But according to the Gibbs-Duhem equation,

\[-\left(\mathrm{d} \phi / d m_{\mathrm{j}}\right)+\left(\mathrm{d} \ln \left(\gamma_{\mathrm{j}}\right) / \mathrm{dm} \mathrm{m}_{\mathrm{j}}\right)=(\phi-1) / \mathrm{m}_{\mathrm{j}} \nonumber \]

Then,

\[1-\phi=-(1 / \mathrm{R} \, \mathrm{T}) \,\left\{\mathrm{d}\left[\mathrm{G}^{\mathrm{E}} / \mathrm{m}_{\mathrm{j}}\right] / \mathrm{dm}_{\mathrm{j}}\right\} \, \mathrm{m}_{\mathrm{j}} \nonumber \]

Or,

\[1-\phi=-(1 / \mathrm{R} \, \mathrm{T}) \,\left[\mathrm{dG}^{\mathrm{E}} / \mathrm{dm}_{\mathrm{j}}\right] \, \mathrm{m}_{\mathrm{j}} \nonumber \]

The latter equation does not however require that \((1-\phi)\) is a linear function of \(\mathrm{m}_{\mathrm{j}}\). The actual form of this dependence has to be obtained by experiment.

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