Interrelating the Two Activity Coefficients in a Binary Mixture
The activity coefficient for species 1 in a binary mixture can be represented by
\ln \gamma_{1}=a x_{2}^{2}+b x_{2}^{3}+c x_{2}^{4}where a, b, and c are concentration-independent parameters. What is the expression for \ln \gamma_{2} in terms of these same parameters?
For a binary mixture at constant temperature and pressure we have, from Eq. 8.2-20 or 9.3-17,
x_{1}\left(\frac{\partial \bar{G}_{1}}{\partial x_{1}}\right)_{T, P}+x_{2}\left(\frac{\partial \bar{G}_{2}}{\partial x_{1}}\right)_{T, P}=0 (8.2-20)
x_{1}\left(\frac{\partial \ln \gamma_{1}}{\partial x_{1}}\right)_{T, P}+x_{2}\left(\frac{\partial \ln \gamma_{2}}{\partial x_{1}}\right)_{T, P}=0 (9.3-17)
x_{1} \frac{\partial \ln \gamma_{1}}{\partial x_{2}}+x_{2} \frac{\partial \ln \gamma_{2}}{\partial x_{2}}=0
Since x_{1}=1-x_{2} \text { and } \ln \gamma_{1}=a x_{2}^{2}+b x_{2}^{3}+c x_{2}^{4}, we have
\begin{aligned}\frac{\partial \ln \gamma_{2}}{\partial x_{2}} &=-\frac{x_{1}}{x_{2}} \frac{\partial \ln \gamma_{1}}{\partial x_{2}} \\&=-\frac{\left(1-x_{2}\right)}{x_{2}}\left(2 a x_{2}+3 b x_{2}^{2}+4 c x_{2}^{3}\right) \\&=-2 a+(2 a-3 b) x_{2}+(3 b-4 c) x_{2}^{2}+4 c x_{2}^{3}\end{aligned}
Now, by definition, \gamma_{2}\left(x_{2}=1\right)=1, \text { and } \ln \gamma_{2}\left(x_{2}=1\right)=0. Therefore,
\begin{aligned}\int_{x_{2}=1}^{x_{2}} \frac{\partial \ln \gamma_{2}}{\partial x_{2}} d x_{2}=& \ln \gamma_{2}\left(x_{2}\right)-\ln \gamma_{2}\left(x_{2}=1\right)=\ln \gamma_{2}\left(x_{2}\right) \\=& \int_{x_{2}=1}^{x_{2}}\left[-2 a+(2 a-3 b) x_{2}+(3 b-4 c) x_{2}^{2}+4 c x_{2}^{3}\right] d x_{2} \\=&-2 a\left(x_{2}-1\right)+\frac{(2 a-3 b)}{2}\left(x_{2}^{2}-1\right)+\frac{(3 b-4 c)}{3}\left(x_{2}^{3}-1\right) \\&+\frac{4 c}{4}\left(x_{2}^{4}-1\right)\end{aligned}
Again using x_{1}+x_{2}=1, \text { or } x_{2}=1-x_{1}, yields
\begin{aligned}\ln \gamma_{2}=&-2 a\left(1-x_{1}-1\right) \\&+\frac{1}{2}(2 a-3 b)\left(1-2 x_{1}+x_{1}^{2}-1\right) \\&+\frac{1}{3}(3 b-4 c)\left(1-3 x_{1}+3 x_{1}^{2}-x_{1}^{3}-1\right) \\&+c\left(1-4 x_{1}+6 x_{1}^{2}-4 x_{1}^{3}+x_{1}^{4}-1\right) \\\ln \gamma_{2}=&\left(a+\frac{3 b}{2}+2 c\right) x_{1}^{2}-\left(b+\frac{8}{3} c\right) x_{1}^{3}+c x_{1}^{4}\end{aligned}