1.10.3: Gibbs Energies- Solutions- Solvent and Solute
- Page ID
- 379744
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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 \(\mathrm{m}_{\mathrm{j}}\).
\[\mathrm{G}_{\mathrm{m}}^{\mathrm{E}}=\mathrm{R} \, \mathrm{T} \, \mathrm{m}_{\mathrm{j}} \,\left[1-\phi+\ln \left(\gamma_{\mathrm{j}}\right)\right]\]
Hence, at fixed \(\mathrm{T}\) and \(\mathrm{p}\),
\[\begin{aligned}
(1 / \mathrm{R} \, \mathrm{T}) \,\left[\mathrm{dG}^{\mathrm{E}} / \mathrm{dm}_{\mathrm{j}}\right]=\left[1-\phi+\ln \left(\gamma_{\mathrm{j}}\right)\right]-\mathrm{m}_{\mathrm{j}} \,\left[\mathrm{d} \phi / \mathrm{dm}_{\mathrm{j}}\right] \\
&+\mathrm{m}_{\mathrm{j}} \,\left[\mathrm{d} \ln \left(\gamma_{\mathrm{j}}\right) / \mathrm{dm}_{\mathrm{j}}\right]
\end{aligned}\]
But according to the Gibbs-Duhem equation,
\[-\phi-\mathrm{m}_{\mathrm{j}} \,\left[\mathrm{d} \phi / \mathrm{dm}_{\mathrm{j}}\right]+1+\mathrm{m}_{\mathrm{j}} \,\left[\mathrm{d} \ln \left(\gamma_{\mathrm{j}}\right) / \mathrm{dm}_{\mathrm{j}}\right]=0\]
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]
\[\ln \left(\gamma_{\mathrm{j}}\right)=(1 / \mathrm{R} \, \mathrm{T}) \,\left[\mathrm{dG}^{\mathrm{E}} / \mathrm{dm}_{\mathrm{j}}\right]\]
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.
\[1-\phi=(1 / \mathrm{R} \, \mathrm{T}) \,\left[\mathrm{G}^{\mathrm{E}} / \mathrm{m}_{\mathrm{j}}-\mathrm{dG}^{\mathrm{E}} / \mathrm{dm}_{\mathrm{j}}\right]\]
A more elegant derivation of equation (e) starts out with the equation (a) for the excess Gibbs energy written in the following form.
\[\left[\mathrm{G}^{\mathrm{E}} / \mathrm{m}_{\mathrm{j}}\right] / \mathrm{R} \, \mathrm{T}=1-\phi+\ln \left(\gamma_{\mathrm{j}}\right)\]
Then at fixed \(\mathrm{T}\) and \(\mathrm{p}\),
\[(1 / \mathrm{R} \, \mathrm{T}) \,\left\{\mathrm{d}\left[\mathrm{G}^{\mathrm{E}} / \mathrm{m}_{\mathrm{j}}\right] / \mathrm{dm}_{\mathrm{j}}\right\}=-\left(\mathrm{d} \phi / \mathrm{dm}_{\mathrm{j}}\right)+\mathrm{d} \ln \left(\gamma_{\mathrm{j}}\right) / \mathrm{dm}_{\mathrm{j}}\]
But according to the Gibbs-Duhem equation,
\[-\left(\mathrm{d} \phi / d m_{\mathrm{j}}\right)+\left(\mathrm{d} \ln \left(\gamma_{\mathrm{j}}\right) / \mathrm{dm} \mathrm{m}_{\mathrm{j}}\right)=(\phi-1) / \mathrm{m}_{\mathrm{j}}\]
Then,
\[1-\phi=-(1 / \mathrm{R} \, \mathrm{T}) \,\left\{\mathrm{d}\left[\mathrm{G}^{\mathrm{E}} / \mathrm{m}_{\mathrm{j}}\right] / \mathrm{dm}_{\mathrm{j}}\right\} \, \mathrm{m}_{\mathrm{j}}\]
Or,
\[1-\phi=-(1 / \mathrm{R} \, \mathrm{T}) \,\left[\mathrm{dG}^{\mathrm{E}} / \mathrm{dm}_{\mathrm{j}}\right] \, \mathrm{m}_{\mathrm{j}}\]
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.
Footnotes
[1] \(\ln \left(\gamma_{\mathrm{j}}\right)=\left[\mathrm{J} \mathrm{K}^{-1} \mathrm{~mol}^{-1}\right]^{-1} \,\left[\mathrm{K}^{-1} \,\left[\mathrm{J} \mathrm{kg}^{-1}\right] \,\left[\mathrm{mol} \mathrm{kg}^{-1}\right]^{-1}=[1]\right.\)
[2] \((1-\phi)=\left[\frac{1}{\left[\mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right] \,[\mathrm{K}]}\right] \,\left[\frac{\left[\mathrm{J} \mathrm{kg}^{-1}\right]}{\left[\mathrm{mol} \mathrm{kg}^{-1}\right.}+\frac{\left[\mathrm{J} \mathrm{kg}^{-1}\right]}{\left[\mathrm{mol} \mathrm{kg}^{-1}\right]}\right]=[1]\)