1.11.3: Gibbs-Duhem Equation- Solvent and Solutes- Aqueous Solutions

A given aqueous solution at temperature \(\mathrm{T}\) and pressure \(\mathrm{p}\) (which is close to the standard pressure) is prepared using water (\(1 \mathrm{~kg}\)) and \(\mathrm{m}_{\mathrm{j}}\) moles of solute-\(\mathrm{j}\). The chemical potential of solute–j is given by equation (a) where \(\mu_{\mathrm{j}}^{0}(\mathrm{aq})\) is the chemical potential of solute-\(\mathrm{j}\) in an aqueous solutions where \(\mathrm{m}_{\mathrm{j}} = 1 \mathrm{~mol kg}^{-1}\) and the thermodynamic properties of the solution are ideal.

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

At all \(\mathrm{T}\) and \(\mathrm{p}\),

\[\operatorname{limit}\left(\mathrm{m}_{\mathrm{j}} \rightarrow 0\right) \gamma_{\mathrm{j}}=1\]

The chemical potential of solvent, water \(\mu_{1}(\mathrm{aq})\) is given by equation (c) where \(\mu_{1}^{*}(\ell)\) is the chemical potential of water(\(\ell\)) at the same \(\mathrm{T}\) and \(\mathrm{p}\); \(\phi\) is the molal osmotic coefficient.

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

At all \(\mathrm{T}\) and \(\mathrm{p}\),

\[\operatorname{limit}\left(\mathrm{m}_{\mathrm{j}} \rightarrow 0\right) \phi=1\]

For a solution at fixed \(\mathrm{T}\) and \(\mathrm{p}\), the Gibbs-Duhem equation relates \(\mu_{1}(\mathrm{aq})\) and \(\mu_{\mathrm{j}}(\mathrm{aq})\) using equation (e).

\[\left(1 / \mathrm{M}_{1}\right) \, \mathrm{d} \mu_{1}(\mathrm{aq})+\mathrm{m}_{\mathrm{j}} \, \mathrm{d} \mu_{\mathrm{j}}(\mathrm{aq})=0\]

Therefore

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

Thus,

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

\[-\phi \, d m_{j}-m_{j} \, d \phi+m_{j} \, d \ln \left(m_{j}\right)+m_{j} \, d \ln \left(\gamma_{j}\right)=0\]

From equation (h) (dividing by \(\mathrm{m}_{\mathrm{j}}\))

\[-\phi \, d \ln \left(m_{j}\right)-d \phi+d \ln \left(m_{j}\right)+d \ln \left(\gamma_{j}\right)=0\]

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

The latter equation is integrated between the limits ‘\(\mathrm{m}_{\mathrm{j}} = 0\)’ and \(\mathrm{m}_{\mathrm{j}}\); equivalent to limits ‘\(\phi =1\)’ and \(\phi\).

\[\int_{0}^{m(j)} d \ln \left(\gamma_{j}\right)=\int_{\phi=1}^{\phi} \mathrm{d} \phi-\int_{0}^{m(j)}(1-\phi) \, d \ln \left(m_{j}\right)\]

Then,

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

From equation (g)

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

or,

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

Following integration from ‘\(\mathrm{m}_{\mathrm{j}} = 0\)’ to \(\mathrm{m}_{\mathrm{j}}\),

\[\phi \, \mathrm{m}_{\mathrm{j}}=\mathrm{m}_{\mathrm{j}}+\int_{0}^{\mathrm{m}_{\mathrm{j}}} \mathrm{m}_{\mathrm{j}} \, \mathrm{d} \ln \left(\gamma_{\mathrm{j}}\right)\]

or,

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

Hence,

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

In other words, \((\phi-1)\) is related to the integral of \(\mathrm{m}_{\mathrm{j}} \, \mathrm{d} \ln \gamma_{\mathrm{j}}\) between the limits ‘\(\mathrm{m}_{\mathrm{j}} = 0\)’ and \(\mathrm{m}_{\mathrm{j}}\). Equation (q) marks the limit of the thermodynamics analysis. However we explore the significance of the equation by adopting an equation relating \(\phi\) and \(\mathrm{m}_{\mathrm{j}}\). Equation (r) signals one assumption in which \(\mathrm{r}\) is a parameter characteristic of the solution under examination. Thus

\[\phi-1=\alpha \,\left(\mathrm{m}_{\mathrm{j}}\right)^{\mathrm{r}}\]

Then [1]

\[-\ln \left(\gamma_{\mathrm{j}}\right)=(1-\phi) \,(1+\mathrm{r}) / \mathrm{r}\]

Or,

\[(1-\phi)=[r /(1+r)] \,\left\{-\ln \left(\gamma_{j}\right)\right\}\]

From equation (m),

\[\phi=1-\alpha \,\left(\mathrm{m}_{\mathrm{j}}\right)^{\mathrm{r}}\]

If for example, \(\alpha > 1\), then \(\phi < 1\) for all solutions. According to equation (c),

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

If therefore \(\phi>1, \mu_{1}(\mathrm{aq})<\mu_{1}^{*}(\ell)\); relative to the chemical potential of the pure solvent, the solvent in the solution is stabilised. For the solute according to equation (n),

\[-\ln \left(\gamma_{\mathrm{j}}\right)=[(1+\mathrm{r}) / \mathrm{r}] \,\left[\alpha \,\left(\mathrm{m}_{\mathrm{j}}\right)^{\mathrm{r}}\right]\]

Or, [Bjerrum’s Equation]

\[-\ln \left(\gamma_{\mathrm{j}}\right)=[(1+\mathrm{r}) / \mathrm{r}] \,[1-\phi]\]

Footnotes

[1] From equations (q) and (s),

\[-\ln \left(\gamma_{\mathrm{j}}\right)=\alpha \,\left(\mathrm{m}_{\mathrm{j}}\right)^{\mathrm{r}}+\alpha \, \int_{0}^{\mathrm{m}(\mathrm{j})}\left(\mathrm{m}_{\mathrm{j}}\right)^{\mathrm{r}-1} \, d \mathrm{~m}_{\mathrm{j}}\]

Then,

\[\begin{aligned}
-\ln \left(\gamma_{\mathrm{j}}\right) &=\alpha \,\left(\mathrm{m}_{\mathrm{j}}\right)^{\mathrm{r}}+\alpha \,\left(\mathrm{m}_{\mathrm{j}}\right)^{\mathrm{r}} / \mathrm{r} \\
&=\alpha \,\left(\mathrm{m}_{\mathrm{j}}\right)^{\mathrm{r}} \,[1+(1 / \mathrm{r})]
\end{aligned}\]

Hence, \(-\ln \left(\gamma_{\mathrm{j}}\right)=\alpha \,\left(\mathrm{m}_{\mathrm{j}}\right)^{\mathrm{r}} \,[1+\mathrm{r}] / \mathrm{r}\)