Question 6 of 7: Deriving Activity Coefficients from a $G^E$ Correlation

Question 6: Deriving Activity Coefficients from a $G^E$ Correlation (Part C — 30 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. $g\equiv G^E/RT=x_1x_2\,(A+Bx_1^2)$, with $x_1+x_2=1$. This is a derivation question; no numbers are supplied.

Find. (a) $\ln\gamma_1,\ln\gamma_2(A,B,x_1,x_2)$; (b) $A,B$ in terms of $\ln\gamma_i^\infty$; (c) a sketch of the reduced function with the infinite-dilution limits marked.

Approach. The activity coefficient is the partial molar excess Gibbs energy, so write $nG^E/RT$ in terms of mole numbers and differentiate at constant $T$, $P$ and the other mole number. The infinite-dilution values then follow by setting the relevant mole fraction to zero, and the sketch reads those limits off the ends of the fitted curve.

1. (a) Write $ng$ in mole numbers. With $x_1=n_1/n$ and $x_2=n_2/n$: $$\frac{nG^E}{RT}=A\,\frac{n_1n_2}{n}+B\,\frac{n_1^3n_2}{n^3}.$$
2. (a) Differentiate with respect to $n_1$. Using $\partial n/\partial n_1=1$: $$\ln\gamma_1=A\Big(\frac{n_2}{n}-\frac{n_1n_2}{n^2}\Big)+B\Big(\frac{3n_1^2n_2}{n^3}-\frac{3n_1^3n_2}{n^4}\Big)=Ax_2^2+3Bx_1^2x_2^2\;\Rightarrow\;\boxed{\ln\gamma_1=x_2^2\big(A+3Bx_1^2\big)}.$$
3. (a) Differentiate with respect to $n_2$. $$\ln\gamma_2=A\Big(\frac{n_1}{n}-\frac{n_1n_2}{n^2}\Big)+B\Big(\frac{n_1^3}{n^3}-\frac{3n_1^3n_2}{n^4}\Big)=Ax_1^2+Bx_1^3(1-3x_2)\;\Rightarrow\;\boxed{\ln\gamma_2=x_1^2\big[A+Bx_1(3x_1-2)\big]}.$$ Check (summability): $x_1\ln\gamma_1+x_2\ln\gamma_2=Ax_1x_2(x_1+x_2)+Bx_1^3x_2(3x_2+1-3x_2)=x_1x_2(A+Bx_1^2)=G^E/RT$.
4. (b) Infinite-dilution limits. Species 1 is infinitely dilute when $x_1\to0$ ($x_2\to1$); species 2 when $x_2\to0$ ($x_1\to1$): $$\ln\gamma_1^\infty=\lim_{x_1\to0}x_2^2(A+3Bx_1^2)=\boxed{A},\qquad \ln\gamma_2^\infty=\lim_{x_1\to1}x_1^2\big[A+Bx_1(3x_1-2)\big]=A+B(3-2)=\boxed{A+B}.$$ Hence $A=\ln\gamma_1^\infty$ and $B=\ln\gamma_2^\infty-\ln\gamma_1^\infty$. Approach: evaluate each $\ln\gamma_i$ where that species vanishes. The squared prefactor goes to 1 and the bracket collapses to its end value. The same result comes straight from the reduced function, because $\lim_{x_1\to0}G^E/(RTx_1x_2)=\ln\gamma_1^\infty$ and $\lim_{x_1\to1}G^E/(RTx_1x_2)=\ln\gamma_2^\infty$ for any $G^E$ model.
5. (c) Sketch and graphical relationships. On a plot of $G^E/(RTx_1x_2)$ vs $x_1$, the fit $A+Bx_1^2$ is a parabola with its vertex on the left axis (zero slope at $x_1=0$). It rises toward the right axis if $B>0$ and falls if $B<0$:
   • At $x_1=0$ (left axis) the curve meets the axis at $A=\ln\gamma_1^\infty$.
   • At $x_1=1$ (right axis) the curve reaches $A+B=\ln\gamma_2^\infty$.
   • $B$ is the vertical rise between the two end points, $B=\ln\gamma_2^\infty-\ln\gamma_1^\infty$. It is not the slope at either end: the slope is $2Bx_1$, which is 0 at $x_1=0$ and $2B$ at $x_1=1$, so $B$ can also be read as half the end slope at $x_1=1$.
   Figure 6 — Sketch of the reduced excess Gibbs energy A + B x1² versus x1 (illustrative A = 0.4, B = 0.5). The end-point values are the infinite-dilution activity coefficients; the rise between them is B.
   Extrapolating the curve fitted to the plotted data out to both axes therefore gives both infinite-dilution activity coefficients. If the data plot as a horizontal line ($B=0$), the model reduces to the one-parameter symmetric Margules form, with $\ln\gamma_1^\infty=\ln\gamma_2^\infty=A$.