Chapter 13.13, Problem 24P

Interpretation Introduction

Interpretation:

Does the mixture form a miscibility gap? If so, how does your value is compared with the experimental value at this temperature?

Concept Introduction:

The expression of solubility parameters for methanol (1) and hexane (2) is,

${\delta }_1=\sqrt{\frac{\Delta {\underset{\_}{U}}_1^{vap}}{{\underset{\_}{V}}_1}}$

${\delta }_2=\sqrt{\frac{\Delta {\underset{\_}{U}}_2^{vap}}{{\underset{\_}{V}}_2}}$

Here, molar volume of methanol (1) and hexane (2) is ${\underset{\_}{V}}_1\text{}\text{and}{\underset{\_}{V}}_2$, change in molar internal energy of vaporization of methanol (1) and hexane (2) is $\Delta U_1^{vap}$ and $\Delta U_2^{vap}$ respectively.

The expression of change in molar internal energy of vaporization is,

$\Delta {\underset{\_}{U}}^{vap}=\Delta {\underset{\_}{H}}^{vap}-RT$

Here, change in molar enthalpy of vaporization is $\Delta {\underset{\_}{H}}^{vap}$, gas constant is R, and temperature is T.

The expression of natural logarithmic activity coefficient of component methanol is,

$\ln \left({\gamma }_1\right)=M_{12}{\left(1+\frac{M_{12}{x}_1}{M_{21}{x}_2}\right)}^{-2}$

Here, mole fraction of component methanol is $x_1$ and mole fraction of hexane is $x_2$.

The expression of natural logarithmic activity coefficient of component hexane is,

$\ln \left({\gamma }_2\right)=M_{21}{\left(1+\frac{M_{21}{x}_2}{M_{12}{x}_1}\right)}^{-2}$

The phase equilibrium relationship for phase 1 and 2 is,

$x_1^{\alpha }{\gamma }_1^{\alpha }=x_1^{\beta }{\gamma }_1^{\beta }$

$\left(1-x_1^{\alpha }\right){\gamma }_2^{\alpha }=\left(1-x_1^{\beta }\right){\gamma }_2^{\beta }$

Here, mole fraction of component 1 in phase $\alpha$ is $x_1^{\alpha }$, mole fraction of component 1 in phase $\beta$ is $x_1^{\beta }$, activity coefficient for the component 1 in the phase $\alpha$ is ${\gamma }_1^{\beta }$, activity coefficient for the component 2 in the phase $\alpha$ is ${\gamma }_2^{\alpha }$, activity coefficient for the component 2 in the phase $\beta$ is ${\gamma }_2^{\beta }$, and activity coefficient for the component 1 in the phase $\alpha$ is ${\gamma }_1^{\alpha }$.

Explanation of Solution

Write the expression of solubility parameters for methanol (1) and hexane (2).

${\delta }_1=\sqrt{\frac{\Delta {\underset{\_}{U}}_1^{vap}}{{\underset{\_}{V}}_1}}$        (1)

${\delta }_2=\sqrt{\frac{\Delta {\underset{\_}{U}}_2^{vap}}{{\underset{\_}{V}}_2}}$        (2)

Write the expression of change in molar internal energy of vaporization.

$\Delta {\underset{\_}{U}}^{vap}=\Delta {\underset{\_}{H}}^{vap}-RT$        (3)

Rewrite Equations (1) and (2) using Equation (3).

${\delta }_1=\sqrt{\frac{\Delta {\underset{\_}{H}}_1{}^{vap}-RT}{{\underset{\_}{V}}_1}}$        (4)

${\delta }_2=\sqrt{\frac{\Delta {\underset{\_}{H}}_2{}^{vap}-RT}{{\underset{\_}{V}}_2}}$        (5)

Refer Appendix C.1, “Critical point, Enthalpy of phase change, and liquid molar volume”; obtain the following properties at 300 K for methanol (1) and n-hexane (2).

$\begin{array}{l}\Delta {\underset{\_}{H}}_1{}^{vap}=35.21\frac{\text{kJ}}{\text{mol}}\\ \Delta {\underset{\_}{H}}_2{}^{vap}=28.85\frac{\text{kJ}}{\text{mol}}\\ {\underset{\_}{V}}_1=40.73\frac{{\text{cm}}^{\text{3}}}{\text{mol}}\\ {\underset{\_}{V}}_2=131.59\frac{{\text{cm}}^{\text{3}}}{\text{mol}}\end{array}$

Substitute $8.314\frac{\text{J}}{\text{K}\cdot \text{mol}}$ for R, 300 K for T, $35.21\frac{\text{kJ}}{\text{mol}}$ for $\Delta {\underset{\_}{H}}_1{}^{vap}$, and $40.73\frac{{\text{cm}}^{\text{3}}}{\text{mol}}$ for ${\underset{\_}{V}}_1$ in Equation (4).

$\begin{array}{c}{\delta }_1=\sqrt{\frac{35.21\frac{\text{kJ}}{\text{mol}}-\left(8.314\frac{\text{J}}{\text{K}\cdot \text{mol}}\right)300\text{K}}{40.73\frac{{\text{cm}}^{\text{3}}}{\text{mol}}}}\\ =\sqrt{\frac{35.21\frac{\text{kJ}\times \text{1000}\frac{\text{J}}{\text{kJ}}}{\text{mol}}-\left(8.314\frac{\text{J}}{\text{K}\cdot \text{mol}}\right)300\text{K}}{40.73\frac{{\text{cm}}^{\text{3}}}{\text{mol}}}}\\ =\sqrt{\frac{32715.8\frac{\text{J}}{\text{mol}}}{40.73\frac{{\text{cm}}^{\text{3}}}{\text{mol}}}}\\ =\sqrt{803.23\frac{\text{J}}{{\text{cm}}^3}\times \frac{\text{Pa}\cdot {\text{m}}^3}{\text{1}\text{J}}\times \frac{{\left(\text{100}\text{cm}\right)}^3}{{\text{m}}^{\text{3}}}}\\ =28341.42{\left(\text{Pa}\right)}^{0.5}\end{array}$

Substitute $8.314\frac{\text{J}}{\text{K}\cdot \text{mol}}$ for R, 300 K for T, $28.85\frac{\text{kJ}}{\text{mol}}$ for $\Delta {\underset{\_}{H}}_2{}^{vap}$, and $131.59\frac{{\text{cm}}^{\text{3}}}{\text{mol}}$ for ${\underset{\_}{V}}_2$ in Equation (5).

$\begin{array}{c}{\delta }_2=\sqrt{\frac{28.85\frac{\text{kJ}}{\text{mol}}-\left(8.314\frac{\text{J}}{\text{K}\cdot \text{mol}}\right)300\text{K}}{131.59\frac{{\text{cm}}^{\text{3}}}{\text{mol}}}}\\ =\sqrt{\frac{28.85\frac{\text{kJ}\times \text{1000}\frac{\text{J}}{\text{kJ}}}{\text{mol}}-\left(8.314\frac{\text{J}}{\text{K}\cdot \text{mol}}\right)300\text{K}}{131.59\frac{{\text{cm}}^{\text{3}}}{\text{mol}}}}\\ =\sqrt{200.287\frac{\text{J}}{{\text{cm}}^3}\times \frac{\text{Pa}\cdot {\text{m}}^3}{\text{1}\text{J}}\times \frac{{\left(\text{100}\text{cm}\right)}^3}{{\text{m}}^{\text{3}}}}\\ =14152.3{\left(\text{Pa}\right)}^{0.5}\end{array}$

Write the expression of the parameter $M_{12}$.

$M_{12}=\frac{{\underset{\_}{V}}_1}{RT}{\left({\delta }_1-{\delta }_2\right)}^2$        (6)

Substitute $8.314\frac{\text{J}}{\text{K}\cdot \text{mol}}$ for R, 300 K for T, $28341.42{\left(\text{Pa}\right)}^{0.5}$ for ${\delta }_1$, $14152.3{\left(\text{Pa}\right)}^{0.5}$ for ${\delta }_2$, and $40.73\frac{{\text{cm}}^{\text{3}}}{\text{mol}}$ for ${\underset{\_}{V}}_1$ in Equation (6).

$\begin{array}{c}{M}_{12}=\frac{40.73\frac{{\text{cm}}^{\text{3}}}{\text{mol}}}{\left(8.314\frac{\text{J}}{\text{K}\cdot \text{mol}}\right)300\text{K}}{\left(28341.42{\left(\text{Pa}\right)}^{0.5}-14152.3{\left(\text{Pa}\right)}^{0.5}\right)}^2\\ =3.288\end{array}$

Write the expression of the parameter $M_{21}$.

$M_{21}=\frac{{\underset{\_}{V}}_2}{RT}{\left({\delta }_1-{\delta }_2\right)}^2$        (7)

Substitute $8.314\frac{\text{J}}{\text{K}\cdot \text{mol}}$ for R, 300 K for T, $28341.42{\left(\text{Pa}\right)}^{0.5}$ for ${\delta }_1$, $14152.3{\left(\text{Pa}\right)}^{0.5}$ for ${\delta }_2$, and $131.59\frac{{\text{cm}}^{\text{3}}}{\text{mol}}$ for ${\underset{\_}{V}}_2$ in Equation (7).

$\begin{array}{c}{M}_{21}=\frac{131.59\frac{{\text{cm}}^{\text{3}}}{\text{mol}}}{\left(8.314\frac{\text{J}}{\text{K}\cdot \text{mol}}\right)300\text{K}}{\left(28341.42{\left(\text{Pa}\right)}^{0.5}-14152.3{\left(\text{Pa}\right)}^{0.5}\right)}^2\\ =10.622\end{array}$

Write the expression of natural logarithmic activity coefficient of component methanol.

$\ln \left({\gamma }_1\right)=M_{12}{\left(1+\frac{M_{12}{x}_1}{M_{21}{x}_2}\right)}^{-2}$        (8)

Write the expression of natural logarithmic activity coefficient of component hexane.

$\ln \left({\gamma }_2\right)=M_{21}{\left(1+\frac{M_{21}{x}_2}{M_{12}{x}_1}\right)}^{-2}$        (9)

Write the phase equilibrium relationship for phase 1 and 2.

$x_1^{\alpha }{\gamma }_1^{\alpha }=x_1^{\beta }{\gamma }_1^{\beta }$

$\left(1-x_1^{\alpha }\right){\gamma }_2^{\alpha }=\left(1-x_1^{\beta }\right){\gamma }_2^{\beta }$        (10)

Write the expression of the molar Gibbs free energy of mixing.

$\begin{array}{l}\Delta {\underset{\_}{G}}_{mix}={\underset{\_}{G}}^E+RT\left[x_1\ln \left(x_1\right)+x_2\ln \left(x_2\right)\right]\\ \frac{\Delta {\underset{\_}{G}}_{mix}}{RT}=\frac{{\underset{\_}{G}}^E}{RT}+\left[x_1\ln \left(x_1\right)+x_2\ln \left(x_2\right)\right]\end{array}$        (11)

If we plot the graph of $\frac{\Delta {\underset{\_}{G}}_{mix}}{RT}$ as a function of $x_1$, we can obtain the double tangency at $x_1=0.05$ and $x_2=0.99$.

By using this approach, obtain the nearest values of $x_1^{\alpha }\text{and}{x}_1^{\beta }$ as 0.041 and 0.99997 respectively.

When the SH approach does, a miscibility gap is predicted. That is the gap over a bigger range is experimentally observed. Actually, it predicts an immiscible liquid system.