1.18.2: Liquid Mixtures: General Equations

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, \(\mu_{1}\left(\operatorname{mix} ; \mathrm{x}_{1}\right)\) and \(\mu_{2}\left(\operatorname{mix} ; \mathrm{x}_{2}\right)\) are related to the mole fraction composition, \(x_{1}\) and \(x_{2} (= 1 - x_{1})\) 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 \(\mathrm{p}\);

\[\mu_{1}\left(\operatorname{mix} ; \mathrm{x}_{1}\right)=\mu_{1}^{*}(\ell)+\mathrm{R} \, \mathrm{T} \, \ln \left(\mathrm{x}_{1} \, \mathrm{f}_{1}\right)\]

where

\[\operatorname{limit}\left(x_{1} \rightarrow 1\right) f_{1}=1\]

\[\mu_{2}\left(\operatorname{mix} ; \mathrm{x}_{2}\right)=\mu_{2}^{*}(\ell)+\mathrm{R} \, \mathrm{T} \, \ln \left(\mathrm{x}_{2} \, \mathrm{f}_{2}\right)\]

where

\[\operatorname{limit}\left(\mathrm{x}_{2} \rightarrow 1\right) \mathrm{f}_{2}=1\]

A general equation for activity coefficient \(\mathrm{f}_{1}\) takes the following form [1].

\[\ln \left(f_{1}\right)=\sum_{k=1}^{k=\infty} \alpha_{k} \, x_{s}^{\lambda(k)}\]

Equation (e) satisfies the condition,

\[\operatorname{limit}\left(x_{2} \rightarrow 0\right) \ln \left(f_{1}\right)=0 ; f_{1}=1\]

The parameter \(\alpha_{\mathrm{k}}\) is characteristic of the mixture, temperature and pressure. The property \(\lambda_{\mathrm{k}}\) is a real number. In the limit that the liquid mixture is dilute in chemical substance liquid-2, equation (e) simplifies to equation (g).

\[\ln \left(f_{1}\right)=\alpha \, x_{2}^{\lambda}\]

In general terms [2],

\[x_{1} \, d \ln \left(f_{1}\right)+x_{2} \, d \ln \left(f_{2}\right)=0\]

We combine equations (e) and (h) with \(\lambda_{k} \geq 2\) [3].

\[\frac{\mathrm{d} \ln \left(\mathrm{f}_{1} / \mathrm{f}_{2}\right)}{\mathrm{dx}_{2}}=\frac{1}{\mathrm{x}_{2}} \, \frac{\mathrm{d} \ln \left(\mathrm{f}_{1}\right)}{\mathrm{dx} \mathrm{x}_{2}}=\sum_{\mathrm{k}=1}^{\mathrm{k}=\infty} \alpha_{\mathrm{k}} \, \lambda_{\mathrm{k}} \, \mathrm{x}_{2}^{\lambda(\mathrm{k})-2}\]

Equation (i) is integrated to yield equation (j) where \(\mathrm{I}\) is the constant of integration.

\[\ln \left(f_{2}\right)=\ln \left(f_{1}\right)-\sum_{k=1}^{k=\infty} \frac{\alpha_{k} \, \lambda_{k} \, x_{2}^{\lambda(k)-1}}{\lambda_{k}-1}-I\]

Hence [4,5]

\[\begin{aligned}
\ln \left(f_{2}\right) &=\ln \left(f_{1}\right)-\sum_{k=1}^{k=\infty} \frac{\alpha_{k} \, \lambda_{k}}{\lambda_{k}-1} \,\left(x_{2}^{\lambda(k)-1}-1\right)-\sum_{k=1}^{k=\infty} \alpha_{k} \\
&=\ln \left(f_{1}\right)-\sum_{k=1}^{k=\infty} \alpha_{k} \,\left[\frac{\lambda_{k}}{\lambda_{k}-1} \,\left(x_{2}^{\lambda-(k-1)}-1\right)-1\right]
\end{aligned}\]

In other words, granted that \(\ln \left(f_{1}\right)\) is known as a function of \(x_{2}\), then \(\ln \left(f_{2}\right)\) can be calculated.

Footnotes

[1] I. Prigogine and R. Defay, Chemical Thermodynamics, transl. D. H. Everett, Longmans Green, London, 1954.

[2] For a binary liquid mixture , the Gibbs-Duhem equation relates activity coefficients \(\mathrm{f}_{1}\) and \(\mathrm{f}_{2}\). Thus,

\[-S \, d T+V \, d p+n_{1} \, d \mu_{1}+n_{2} \, d \mu_{2}=0\]

At fixed \(\mathrm{T}\) and \(\mathrm{p}\), \(\mathrm{n}_{1} \, \mathrm{d} \mu_{1}+\mathrm{n}_{2} \, \mathrm{d} \mu_{2}=0\)

Divide by \(\left(n_{1}+n_{2}\right)\); \(x_{1} \, d \mu_{1}+x_{2} \, d \mu_{2}=0\)

\[\begin{gathered}
\mathrm{x}_{1} \, \mathrm{d}\left[\mu_{1}^{*}(\ell)+\mathrm{R} \, \mathrm{T} \, \ln \left(\mathrm{x}_{1} \, \mathrm{f}_{1}\right)\right]+\mathrm{x}_{2} \, \mathrm{d}\left[\mu_{2}^{*}(\ell)+\mathrm{R} \, \mathrm{T} \, \ln \left(\mathrm{x}_{2} \, \mathrm{f}_{2}\right)\right]=0 \\
\mathrm{x}_{1} \, \mathrm{d} \ln \left(\mathrm{x}_{1} \, \mathrm{f}_{1}\right)+\mathrm{x}_{2} \, \mathrm{d} \ln \left(\mathrm{x}_{2} \, \mathrm{f}_{2}\right)=0 \\
\mathrm{x}_{1} \, \mathrm{d} \ln \left(\mathrm{x}_{1}\right)+\mathrm{x}_{1} \, \mathrm{d} \ln \left(\mathrm{f}_{1}\right)+\mathrm{x}_{2} \, \mathrm{d} \ln \left(\mathrm{x}_{2}\right)+\mathrm{x}_{2} \, \mathrm{d} \ln \left(\mathrm{f}_{2}\right)=0
\end{gathered}\]

But

\[\mathrm{x}_{1} \, \mathrm{d} \ln \left(\mathrm{x}_{1}\right)+\mathrm{x}_{2} \, \mathrm{d} \ln \left(\mathrm{x}_{2}\right)=\left(\mathrm{x}_{1} / \mathrm{x}_{1}\right) \, \mathrm{dx} \mathrm{x}_{1}+\left(\mathrm{x}_{2} / \mathrm{x}_{2}\right) \, \mathrm{dx} \mathrm{x}_{2}\]

Also \(\mathrm{x}_{1}+\mathrm{x}_{2}=1\) so that \(\mathrm{dx}_{1}+\mathrm{dx}_{2}=0\)

[3] From equation (h) for a binary liquid mixture at fixed \(\mathrm{T}\) and \(\mathrm{p}\),

\[\begin{array}{r}
\left(1-x_{2}\right) \, \frac{d \ln \left(f_{1}\right)}{d x_{2}}+x_{2} \, \frac{d \ln \left(f_{2}\right)}{d x_{2}}=0 \\
\frac{d \ln \left(f_{1}\right)}{d x_{2}}-x_{2} \, \frac{d \ln \left(f_{1}\right)}{d x_{2}}+x_{2} \, \frac{d \ln \left(f_{2}\right)}{d x_{2}}=0
\end{array}\]

We divide by \(x_{2}\) and rearrange the equation.

\[\frac{\mathrm{d} \ln \left(\mathrm{f}_{1}\right)}{\mathrm{dx} \mathrm{x}_{2}}-\frac{\mathrm{d} \ln \left(\mathrm{f}_{2}\right)}{\mathrm{dx}_{2}}=\frac{1}{\mathrm{x}_{2}} \, \frac{\mathrm{d} \ln \left(\mathrm{f}_{1}\right)}{\mathrm{dx} \mathrm{x}_{2}}\]

Or,

\[\frac{\mathrm{d} \ln \left(\mathrm{f}_{1} / \mathrm{f}_{2}\right)}{\mathrm{dx} \mathrm{x}_{2}}=\frac{1}{\mathrm{x}_{2}} \, \frac{\mathrm{d} \ln \left(\mathrm{f}_{1}\right)}{\mathrm{dx_{2 }}}\]

[4] From equations (e) and (j),

\[\ln \left(f_{2}\right)=\sum_{k=1}^{k=\infty} \alpha_{k} \, x_{2}^{\lambda(k)}-\sum_{k=1}^{k=\infty} \frac{\alpha_{k} \, \lambda_{k} \, x_{2}^{\lambda(k)-1}}{\lambda_{k}-1}-I\]

But at \(x_{2} = 1, f_{2} = 1\). Then,

\[0=\sum_{\mathrm{k}=1}^{\mathrm{k}=\infty} \alpha_{\mathrm{k}}-\sum_{\mathrm{k}=1}^{\mathrm{k}=\infty} \frac{\alpha_{\mathrm{k}} \, \lambda_{\mathrm{k}}}{\lambda_{\mathrm{k}}-1}-\mathrm{I}\]

[5] J. N. Bronsted and P. Colmart, Z. Phys. Chem.,1934,A168, 381 ( as quoted in reference 1).