1.10.12: Gibbs Energies- Solutions- Two Neutral Solutes

A given aqueous solution containing two neutral solutes, \(\mathrm{i}\) and \(\mathrm{j}\), (e.g. urea and sucrose) was prepared using \(1 \mathrm{~kg}\) of water at temperature \(\mathrm{T}\) and pressure \(\mathrm{p}\) ( which was close to ambient). The molalities of the two solutes were \(\mathrm{m}_{\mathrm{i}}\) and \(\mathrm{m}_{\mathrm{j}}\) The chemical potential of the solvent in the mixed aqueous solution is given by equation (a) where for an ideal solution the practical osmotic coefficient \(\phi\) is unity. Then,

\[\mu_{1}(\mathrm{aq})=\mu_{1}^{*}(\ell)-\phi \, \mathrm{R} \, \mathrm{T} \, \mathrm{M}_{1} \,\left(\mathrm{m}_{\mathrm{i}}+\mathrm{m}_{\mathrm{j}}\right)\]

The chemical potentials of the two solutes are related to their molalities using equations (b) and (d).

\[\mu_{\mathrm{i}}(\mathrm{aq})=\mu_{\mathrm{i}}^{0}(\mathrm{aq})+\mathrm{R} \, \mathrm{T} \, \ln \left(\mathrm{m}_{\mathrm{i}} \, \gamma_{\mathrm{i}} / \mathrm{m}^{0}\right)\]

where

\[\left.\operatorname{limit}\left(\mathrm{m}_{\mathrm{i}} \rightarrow 0 ; \mathrm{m}_{\mathrm{j}} \rightarrow 0\right) \gamma_{\mathrm{i}}=1.0 \quad \text { (at all } \mathrm{T} \text { and } \mathrm{p}\right)\]

Similarly,

\[\mu_{\mathrm{j}}(\mathrm{aq})=\mu_{\mathrm{j}}^{0}(\mathrm{aq})+\mathrm{R} \, \mathrm{T} \, \ln \left(\mathrm{m}_{\mathrm{j}} \, \gamma_{\mathrm{j}} / \mathrm{m}^{0}\right)\]

where

\[\left.\operatorname{limit}\left(\mathrm{m}_{\mathrm{i}} \rightarrow 0 ; \mathrm{m}_{\mathrm{j}} \rightarrow 0\right) \gamma_{\mathrm{j}}=1.0 \quad \text { (at all T and } \mathrm{p}\right)\]

Therefore the Gibbs energy of a solution prepared using \(1 \mathrm{~kg}\) of water is given by equation (f).

\[\begin{aligned}
\mathrm{G}\left(\mathrm{aq} ; \mathrm{w}_{1} / \mathrm{kg}=1.0\right)=\left(1 / \mathrm{M}_{1}\right) \, & {\left[\mu_{1}^{*}(\ell)-\phi \, \mathrm{R} \, \mathrm{T} \, \mathrm{M}_{1} \,\left(\mathrm{m}_{\mathrm{i}}+\mathrm{m}_{\mathrm{j}}\right)\right] } \\
+\mathrm{m}_{\mathrm{i}} \,\left[\mu_{\mathrm{i}}^{0}(\mathrm{aq})+\right.&\left.\mathrm{R} \, \mathrm{T} \, \ln \left(\mathrm{m}_{\mathrm{i}} \, \gamma_{\mathrm{i}} / \mathrm{m}^{0}\right)\right] \\
&+\mathrm{m}_{\mathrm{j}} \,\left[\mu_{\mathrm{j}}^{0}(\mathrm{aq})+\mathrm{R} \, \mathrm{T} \, \ln \left(\mathrm{m}_{\mathrm{j}} \, \gamma_{\mathrm{j}} / \mathrm{m}^{0}\right)\right]
\end{aligned}\]

For the corresponding solution having ideal thermodynamic properties

\[\begin{aligned}
\mathrm{G}\left(\mathrm{aq} ; \mathrm{w}_{1} / \mathrm{kg}=1.0\right)=\left(1 / \mathrm{M}_{1}\right) \,\left[\mu_{1}^{*}(\ell)-\mathrm{R} \, \mathrm{T} \, \mathrm{M}_{1} \,\left(\mathrm{m}_{\mathrm{i}}+\mathrm{m}_{\mathrm{j}}\right)\right] \\
+\mathrm{m}_{\mathrm{i}} \,\left[\mu_{\mathrm{i}}^{0}(\mathrm{aq})+\mathrm{R} \, \mathrm{T} \, \ln \left(\mathrm{m}_{\mathrm{i}} / \mathrm{m}^{0}\right)\right] \\
&+\mathrm{m}_{\mathrm{j}} \,\left[\mu_{\mathrm{j}}^{0}(\mathrm{aq})+\mathrm{R} \, \mathrm{T} \, \ln \left(\mathrm{m}_{\mathrm{j}} / \mathrm{m}^{0}\right)\right]
\end{aligned}\]

But the excess Gibbs energy is defined by equation (h).

\[\mathrm{G}^{\mathrm{E}}(\mathrm{aq})=\mathrm{G}(\mathrm{aq})-\mathrm{G}(\mathrm{aq} ; \mathrm{id})\]

Hence

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

Actually \(\phi\), \(\gamma_{\mathrm{i}}\) and \(\gamma_{\mathrm{j}}\) are linked. They cannot change independently. According to the Gibbs-Duhem equation (at fixed \(\mathrm{T}\) and \(\mathrm{p}\)),

\[\left(1 / M_{1}\right) \, d \mu_{1}(a q)+m_{j} \, d \mu_{j}+m_{i} \, d \mu_{i}=0\]

We consider the case where the solution is perturbed by a change in molality, \(\mathrm{dm}_{\mathrm{j}}\), recognising that the chemical potentials of solvent and solutes change. Thus,

\[\left(1 / M_{1}\right) \,\left(d \mu_{1}(a q) / d m_{j}\right)+m_{j} \,\left(d \mu_{j} / d m_{j}\right)+m_{i} \,\left(d \mu_{i} / d m_{j}\right)=0\]

Therefore from equation (j),

\[\begin{aligned}
&\left(1 / \mathrm{M}_{1}\right) \, \frac{\mathrm{d}}{\mathrm{dm}_{\mathrm{j}}}\left[\mu_{1}^{*}(\ell)-\phi \, \mathrm{R} \, \mathrm{T} \, \mathrm{M}_{1} \,\left(\mathrm{m}_{\mathrm{i}}+\mathrm{m}_{\mathrm{j}}\right)\right] \\
&+\mathrm{m}_{\mathrm{i}} \, \frac{\mathrm{d}}{\mathrm{dm}_{\mathrm{j}}}\left[\mu_{\mathrm{i}}^{0}(\mathrm{aq})+\mathrm{R} \, \mathrm{T} \, \ln \left(\mathrm{m}_{\mathrm{i}} \, \gamma_{\mathrm{i}} / \mathrm{m}^{0}\right)\right] \\
&\quad+\mathrm{m}_{\mathrm{j}} \, \frac{\mathrm{d}}{\mathrm{dm}_{\mathrm{j}}}\left[\mu_{\mathrm{j}}^{0}(\mathrm{aq})+\mathrm{R} \, \mathrm{T} \, \ln \left(\mathrm{m}_{\mathrm{j}} \, \gamma_{\mathrm{j}} / \mathrm{m}^{0}\right)\right]=0
\end{aligned}\]

In the case considered here molality \(\mathrm{m}_{\mathrm{i}}\) does not change. Then (at constant, \(\mathrm{m}_{\mathrm{i}}\), \(\mathrm{T}\), \(\mathrm{p}\) and mass of solvent),

\[-\left(\mathrm{m}_{\mathrm{i}}+\mathrm{m}_{\mathrm{j}}\right) \,\left[\frac{\partial \phi}{\partial \mathrm{m}_{\mathrm{j}}}\right]_{\mathrm{m}(\mathrm{i})}+(1-\phi)+\mathrm{m}_{\mathrm{j}} \,\left[\frac{\partial \ln \left(\gamma_{\mathrm{j}}\right)}{\partial \mathrm{m}_{\mathrm{j}}}\right]_{\mathrm{m}(\mathrm{i})}+\mathrm{m}_{\mathrm{i}} \,\left[\frac{\partial \ln \left(\gamma_{\mathrm{i}}\right)}{\partial \mathrm{m}_{\mathrm{j}}}\right]_{\mathrm{m}(\mathrm{i})}=0\]

With the help of the latter equation we explore how the Gibbs energy depends on the molality of solute \(j\) at constant \(\mathrm{m}_{\mathrm{i}}\). Then from equation (i),

\[\begin{aligned}
\frac{1}{R \, T} \,\left[\frac{\partial G^{\mathrm{E}}(\mathrm{aq})}{\partial \mathrm{m}_{\mathrm{j}}}\right] &=-\left(\mathrm{m}_{\mathrm{i}}+\mathrm{m}_{\mathrm{j}}\right) \,\left[\frac{\partial \phi}{\partial \mathrm{m}_{\mathrm{j}}}\right]_{\mathrm{m}(\mathrm{i})}+(1-\phi) \\
&+\mathrm{m}_{\mathrm{j}} \,\left[\frac{\partial \ln \left(\gamma_{\mathrm{j}}\right)}{\partial \mathrm{m}_{\mathrm{j}}}\right]_{\mathrm{m}(\mathrm{i})}+\ln \left(\gamma_{\mathrm{j}}\right)+\mathrm{m}_{\mathrm{i}} \,\left[\frac{\partial \ln \left(\gamma_{\mathrm{i}}\right)}{\partial \mathrm{m}_{\mathrm{j}}}\right]_{\mathrm{m}(\mathrm{i})}
\end{aligned}\]

Comparison of equations (m) and (n) shows that the differential dependence of \(\mathrm{G}^{\mathrm{E}}(\mathrm{aq})\) on \(\mathrm{m}_{\mathrm{j}}\) at constant \(\mathrm{m}_{\mathrm{i}}\) is related to \(\ln \left(\gamma_{\mathrm{s}}\right)\);

\[\ln \left(\gamma_{\mathrm{j}}\right)=\frac{1}{\mathrm{R} \, \mathrm{T}} \,\left[\frac{\partial \mathrm{G}^{\mathrm{E}}(\mathrm{aq})}{\partial \mathrm{m}_{\mathrm{j}}}\right]\]

If the solution is dilute then \(\mathrm{G}^{\mathrm{E}}(\mathrm{aq})\) can be described using pairwise Gibbs energy interaction parameters. Thus

\[\mathrm{G}^{\mathrm{E}}(\mathrm{aq})=\mathrm{g}_{\mathrm{ji}} \,\left(\mathrm{m}_{\mathrm{j}} / \mathrm{m}^{0}\right)^{2}+2 \, \mathrm{g}_{\mathrm{ij}} \, \mathrm{m}_{\mathrm{i}} \, \mathrm{m}_{\mathrm{j}} \,\left(\mathrm{m}^{0}\right)^{-2}+\mathrm{g}_{\mathrm{ii}} \,\left(\mathrm{m}_{\mathrm{i}} / \mathrm{m}^{0}\right)^{2}\]

Here \(\mathrm{g}_{i i}\) and \(\mathrm{g}_{j j}\) are homotactic pairwise Gibbs energy interaction parameters whereas \(\mathrm{g}_{i j}\) is the corresponding heterotactic parameter. According to equation (p) the differential dependence of \(\mathrm{G}^{\mathrm{E}}(\mathrm{aq})\) on molality \(\mathrm{m}_{\mathrm{j}}\) at constant \(\mathrm{m}_{\mathrm{i}}\) is given by equation (q). Thus,

\[\left[\frac{\partial \mathrm{G}^{\mathrm{E}}(\mathrm{aq})}{\partial \mathrm{m}_{\mathrm{j}}}\right]_{\mathrm{m}(\mathrm{i})}=2 \, \mathrm{g}_{\mathrm{j}} \,\left(\mathrm{m}^{0}\right)^{-2} \, \mathrm{m}_{\mathrm{j}}+2 \, \mathrm{g}_{\mathrm{ij}} \,\left(\mathrm{m}^{0}\right)^{-2} \, \mathrm{m}_{\mathrm{i}}\]

Combination of equations (o) and (q) yields an equation for \(\ln \left(\gamma_{j}\right)\) as a function of two pairwise interaction parameters. Then [1],

\[\ln \left(\gamma_{\mathrm{j}}\right)=\left[\frac{2}{R \, T}\right] \,\left[\frac{1}{m^{0}}\right]^{2} \,\left[g_{i j} \, m_{j}+g_{i j} \, m_{i}\right]\]

In other words \(\ln \left(\gamma_{\mathrm{j}}\right)\) is simply related to the molality of the two solutes. In many applications we are concerned with a solution in which \(\mathrm{m}_{\mathrm{i}} >> \mathrm{m}_{\mathrm{j}}\) such that the solution contains only a trace of solute \(j\). The activity coefficient for solute \(j\) is written \(\gamma_{\mathrm{j}}^{\mathrm{T}}\). The latter describes the effect of solute-solute interactions on solute \(j\). If we set \(\mathrm{m}_{\mathrm{j}} \cong 0\), then equation (r) yields an equation for \(\gamma_{\mathrm{j}}^{\mathrm{T}}\) in terms of \(\mathrm{m}_{\mathrm{i}}\).

\[\ln \left(\gamma_{\mathrm{j}}^{\mathrm{T}}\right)=\left[\frac{2}{\mathrm{R} \, \mathrm{T}}\right] \,\left[\frac{1}{\mathrm{~m}^{0}}\right]^{2} \, \mathrm{g}_{\mathrm{ij}} \, \mathrm{m}_{\mathrm{i}}\]

Footnote

[1] \(\ln \left(\gamma_{\mathrm{j}}\right)=\left[\frac{2}{\left[\mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right] \,[\mathrm{K}]}\right] \,\left[\frac{1}{\left[\mathrm{~mol} \mathrm{~kg}{ }^{-1}\right]}\right]^{2} \,\left[\mathrm{J} \mathrm{kg}^{-1}\right] \,\left[\mathrm{mol} \mathrm{kg}{ }^{-1}\right]\)