1.18.3: Liquid Mixtures: Series Functions for Activity Coefficients
A given binary liquid mixture is prepared using liquid-1 and liquid-2 at temperature \(\mathrm{T}\) and pressure \(\mathrm{p}\), the latter being close to the standard pressure. The chemical potentials, \(\mu_{1}\left(\operatorname{mix} ; x_{1}\right)\) and \(\mu_{2}\left(\operatorname{mix} ; \mathrm{x}_{2}\right)\) are related to the mole fraction composition, \(\mathrm{x}_{1}\) and \(\mathrm{x}_{2}\left(=1-\mathrm{x}_{1}\right)\) using equations (a) and (c) where \(\mu_{1}^{*}(\ell)\) and \(\mu_{2}^{*}(\ell)\) are the chemical potentials of the two pure liquid components at the same \(\mathrm{T}\) and \(\mathrm{p}\);
\[\mu_{1}\left(\operatorname{mix} ; \mathrm{x}_{1}\right)=\mu_{1}^{*}(\ell)+R \, T \, \ln \left(\mathrm{x}_{1} \, \mathrm{f}_{1}\right) \nonumber \]
where
\[\operatorname{limit}\left(x_{1} \rightarrow 1\right) f_{1}=1 \nonumber \]
\[\mu_{2}\left(\operatorname{mix} ; \mathrm{x}_{2}\right)=\mu_{2}^{*}(\ell)+\mathrm{R} \, \mathrm{T} \, \ln \left(\mathrm{x}_{2} \, \mathrm{f}_{2}\right) \nonumber \]
where
\[\operatorname{limit}\left(\mathrm{x}_{2} \rightarrow 1\right) \mathrm{f}_{2}=1 \nonumber \]
A quite general approach to understanding the properties of binary liquid mixtures expresses, for example, \(\ln \left(f_{1}\right)\) as a series function in terms of mole fraction \(x_{2}\) at fixed \(\mathrm{T}\) and \(\mathrm{p}\). Using only three terms we obtain equation (e).
\[\ln \left(f_{1}\right)=\alpha_{2} \, x_{2}^{2}+\alpha_{3} \, x_{2}^{3}+\alpha_{4} \, x_{2}^{4} \nonumber \]
As required,
\[\operatorname{limit}\left(x_{2} \rightarrow 0\right) \ln \left(f_{1}\right)=0 ; f_{1}=1 \nonumber \]
Hence [1],
\[\begin{aligned}
\ln \left(f_{2}\right) &=\left[\alpha_{2}+(3 / 2) \, \alpha_{3}+2 \, \alpha_{4}\right] \, x_{1}^{2} \\
&-\left[\alpha_{3}+(8 / 3) \, \alpha_{4}\right] \, x_{1}^{3}+\alpha_{4} \, x_{1}^{4}
\end{aligned} \nonumber \]
As required, \(\operatorname{limit}\left(x_{1} \rightarrow 0\right) \ln \left(f_{2}\right)=0 ; f_{2}=1\)
Footnotes
[1] From, \(\ln \left(f_{1}\right)=\alpha_{2} \, x_{2}^{2}+\alpha_{3} \, x_{2}^{3}+\alpha_{4} \, x_{2}^{4}\)
\[\ln \left(f_{1}\right)=\alpha_{2} \,\left(1-x_{1}\right)^{2}+\alpha_{3} \,\left(1-x_{1}\right)^{3}+\alpha_{4} \,\left(1-x_{1}\right)^{4} \nonumber \]
Then
\[\frac{\mathrm{d} \ln \left(f_{1}\right)}{d x_{1}}=-2 \, \alpha_{2} \,\left(1-x_{1}\right)-3 \, \alpha_{3} \,\left(1-x_{1}\right)^{2}-4 \, \alpha_{4} \,\left(1-x_{1}\right)^{3} \nonumber \]
But from the Gibbs-Duhem equation (at fixed \(\mathrm{T}\) and \(\mathrm{p}\))
\[x_{1} \, \frac{d \ln \left(f_{1}\right)}{d x_{1}}+x_{2} \, \frac{d \ln \left(f_{2}\right)}{d x_{1}}=0 \nonumber \]
Or,
\[\frac{\mathrm{d} \ln \left(f_{2}\right)}{\mathrm{dx}_{1}}=-\frac{\mathrm{x}_{1}}{\mathrm{x}_{2}} \, \frac{\mathrm{d} \ln \left(\mathrm{f}_{1}\right)}{\mathrm{dx}_{1}} \nonumber \]
Or,
\[\frac{d \ln \left(f_{2}\right)}{d x_{1}}=-\frac{x_{1}}{\left(1-x_{1}\right)} \, \frac{d \ln \left(f_{1}\right)}{d x_{1}} \nonumber \]
Then,
\[\frac{\mathrm{d} \ln \left(\mathrm{f}_{2}\right)}{\mathrm{dx}_{1}}=2 \, \alpha_{2} \, \mathrm{x}_{1}+3 \, \mathrm{x}_{1} \, \alpha_{3} \,\left(1-\mathrm{x}_{1}\right)+4 \, \alpha_{4} \, \mathrm{x}_{1} \,\left(1-\mathrm{x}_{1}\right)^{2} \nonumber \]
\[\frac{\mathrm{d} \ln \left(f_{2}\right)}{d x_{1}}=2 \, \alpha_{2} \, x_{1}+3 \, x_{1} \, \alpha_{3}-3 \, \alpha_{3} \, x_{1}^{2}+4 \, \alpha_{4} \, x_{1}-8 \, \alpha_{4} \, x_{1}^{2}+4 \, \alpha_{4} \, x_{1}^{3} \nonumber \]
Or,
\[\frac{\mathrm{d} \ln \left(\mathrm{f}_{2}\right)}{\mathrm{dx}_{1}}=\left[2 \, \alpha_{2}+3 \, \alpha_{3}+4 \, \alpha_{4}\right] \, \mathrm{x}_{1}-\left[3 \, \alpha_{3}+8 \, \alpha_{4}\right] \, \mathrm{x}_{1}^{2}+4 \, \alpha_{4} \, \mathrm{x}_{1}^{3} \nonumber \]
The latter equation is integrated.
\[\ln \left(f_{2}\right)=\left[\alpha_{2}+(3 / 2) \, \alpha_{3}+2 \, \alpha_{4}\right] \, x_{1}^{2}-\left[\alpha_{3}+(8 / 3) \, \alpha_{4}\right] \, x_{1}^{3}+\alpha_{4} \, x_{1}^{4} \nonumber \]