Chapter 13.12, Problem 4E

For an ethylene glycol n-butyl ether (1) + water (2) system at 340 K with 80% by mass water, determine if the system is one stable liquid phase or two stable liquid phases at equilibrium. If the latter, provide the mass fraction of the co-existing phases and the amount of each phase.

Interpretation Introduction

Interpretation:

Find out the stable phase of the system and mass fraction of the co-existing phases.

Concept introduction:

Mass to mole conversion is given as follows:

$n_i=\frac{m_i}{\text{M}{\text{.W}}_i}$

Here, mass of the component is $m_i$ and molecular weight of the component is $M{W}_i$.

Molecular weight of ethylene glycol n-butyl ether is $118.17\text{g}/\text{mol}$.

Molecular weight of water is $18.01\text{g}/\text{mol}$.

The composition of this mixture in terms of percentage is given as follows:

$x_i=\frac{n_i}{n_i+n_w}\times 100\%$

Here, mole of the component is $n_i$ and mole of the water is $n_w$.

The overall material balance.

$\begin{array}{l}N=L^{\alpha }+L^{\beta }\\ n_1+n_2=L^{\alpha }+L^{\beta }\end{array}$

Here, overall number of moles is N, and moles in phases $\alpha \text{and}\beta$ is $L^{\alpha }\text{and}{L}^{\beta }$ respectively.

The ethylene glycol n-butyl ether material balance.

$\begin{array}{l}{x}_{1}N=x_1^{\alpha }{L}^{\alpha }+x_1^{\beta }{L}^{\beta }\\ x_1\left(n_1+n_2\right)=x_1^{\alpha }{L}^{\alpha }+x_1^{\beta }{L}^{\beta }\end{array}$

Explanation of Solution

Given information:

Mass of the ethylene glycol -butyl ether is $m_{ethyleneglycoln-butylether}=0.20\text{g}$

Mass of water is $m_{water}=0.80\text{g}$

Mass to moles conversion for ethylene glycol n-butyl ether is given as follows,

$n_{ethyleneglycoln-butylether}=\frac{m_{ethyleneglycoln-butylether}}{\text{M}{\text{.W}}_{ethyleneglycoln-butylether}}$        (1)

Substitute 0.20 g for $m_{ethyleneglycoln-butylether}$ and $118.17\text{g}/\text{mol}$ for $\text{M}{\text{.W}}_{ethyleneglycoln-butylether}$ in equation (1).

$\begin{array}{c}{n}_{ethyleneglycoln-butylether}=\frac{0.20\text{g}}{118.17\text{g}/\text{mol}}\\ =0.0016\text{mol}\end{array}$

Mass to moles conversion for water is given as follows,

$n_{water}=\frac{m_{water}}{\text{M}{\text{.W}}_{water}}$        (2)

Substitute 0.80 g for $m_{water}$ and $18.01\text{g}/\text{mol}$ for $\text{M}{\text{.W}}_{water}$ in equation (2).

$\begin{array}{c}{n}_{water}=\frac{0.80\text{g}}{18.01\text{g}/\text{mol}}\\ =0.044\text{mol}\end{array}$

Calculate the composition of this mixture in terms of percentage.

$x_{ethyleneglycoln-butylether}=\frac{n_{ethyleneglycoln-butylether}}{n_{ethyleneglycoln-butylether}+n_{water}}\times 100\%$        (3)

Substitute 0.0016 mole for $n_{ethyleneglycoln-butylether}$ and 0.044 mole for $n_{water}$ in equation (3).

$\begin{array}{c}{x}_{ethyleneglycoln-butylether}=\frac{0.0016\text{mole}}{0.0016\text{mole}+0.044\text{mole}}\times 100\%\\ =3.67\%\end{array}$

Thus the composition of this mixture in terms of moles is then 3.67%.

Refer Figure 13-4 “Liquid-Liquid equilibrium of the ethylene glycol n-butyl ether + water system at 1 bar” at 340 K and the calculated composition mixture at 0.036703, The mixture is

Two stable liquid phase.

Thus, the given composition lies inside of the two phase region for the given temperature, so the given system is Two stable liquid phase.

Calculate the mass fraction of the liquid phase $\left(\alpha \right)$ :

Refer Figure 13-4, “Liquid-Liquid equilibrium of the ethylene glycol n-butyl ether + water system at 1 bar”,

In phase diagram, move horizontally towards left at 340 K to find out the liquid phase of the composition ($\alpha$) for ethylene glycol n-butyl ether and water are $x_1^{\alpha }=0.01$ and $x_2^{\alpha }=0.99$

Thus, the mass fraction for phase $\alpha$ is $\underset{\_}{0.01}$ and $\underset{\_}{0.99}$ respectively.

Calculate the mass fraction of the liquid phase $\left(\beta \right)$ :

Refer Figure 13-4, “Liquid-Liquid equilibrium of the ethylene glycol n-butyl ether + water system at 1 bar”,

In phase diagram, move horizontally towards left at 340 K to find out the liquid phase of the composition ($\beta$) for ethylene glycol n-butyl ether and water are $x_1^{\alpha }=0.18$ and $x_2^{\alpha }=0.82$

Thus, the mass fraction for phase $\alpha$ is $\underset{\_}{0.18}$ and $\underset{\_}{0.82}$ respectively.

Write the overall material balance.

$\begin{array}{l}N=L^{\alpha }+L^{\beta }\\ n_1+n_2=L^{\alpha }+L^{\beta }\end{array}$        (4)

Substitute $0.0016\text{mol}$ for $n_1$ and $0.044\text{mol}$ for $n_2$ in Equation (1).

$\begin{array}{l}0.0016\text{mol}+0.044\text{mol}=L^{\alpha }+L^{\beta }\\ 0.0456\text{mol}=L^{\alpha }+L^{\beta }\\ L^{\alpha }=0.0456-L^{\beta }\end{array}$        (5)

Write the ethylene glycol n-butyl ether material balance.

$\begin{array}{l}{x}_{1}N=x_1^{\alpha }{L}^{\alpha }+x_1^{\beta }{L}^{\beta }\\ x_1\left(n_1+n_2\right)=x_1^{\alpha }{L}^{\alpha }+x_1^{\beta }{L}^{\beta }\end{array}$        (6)

Substitute 0.0367 for $x_1$, $0.0016\text{mol}$ for $n_1$, $0.044\text{mol}$ for $n_2$, 0.01 for $x_1^{\alpha }$, and 0.18 for $x_1^{\beta }$ in Equation (3).

$\begin{array}{l}0.0367\left(0.0016\text{mol}+0.044\text{mol}\right)=\left(0.01\right)L^{\alpha }+\left(0.18\right)L^{\beta }\\ 0.001674=\left(0.01\right)L^{\alpha }+\left(0.18\right)L^{\beta }\end{array}$        (7)

Substitute equation (5) in equation (7).

$\begin{array}{l}0.001674=\left(0.01\right)\left(0.0456-L^{\beta }\right)+\left(0.18\right)\left(L^{\beta }\right)\\ 0.001674=0.000456-0.01L^{\beta }+0.18L^{\beta }\\ 0.001218=0.17L^{\beta }\\ L^{\beta }=0.007165\text{moles}\end{array}$

Substitute $0.007165\text{moles}$ for $L^{\beta }$ in equation (5).

$\begin{array}{c}{L}^{\alpha }=0.0456-0.007165\\ =0.038435\text{moles}\end{array}$

Mole to mass conversion

components	$x_i^p$	$L^p$ in  moles	$\text{M}\text{.W}$ in $\text{g}/\text{mol}$	$m_i=x_i^p{L}^p\left(\text{M}{\text{.W}}_i\right)$ in grams
components 1 (phase $\alpha$)	0.01	0.038435	118.17	0.0454
components 2 (phase $\alpha$)	0.99	0.038435	18.01	0.685
components 1 (phase $\alpha$)	0.18	0.007165	118.17	0.1524
components 2 (phase $\beta$)	0.82	0.007165	18.01	0.1058

Write the expression for the composition for phase $\alpha$ for component 1.

$x_1=\frac{m_1}{m_1+m_2}$

Substitute 0.0454 g for $m_1$ and 0.685 g for $m_2$.

$\begin{array}{c}{x}_1=\frac{0.0454\text{g}}{0.0454\text{g}+0.685\text{g}}\\ =0.06215\end{array}$

Write the expression for the composition for phase $\alpha$ for component 2.

$x_2=\frac{m_2}{m_1+m_2}$

Substitute 0.0454 g for $m_1$ and 0.685 g for $m_2$.

$\begin{array}{c}{x}_2=\frac{0.685\text{g}}{0.685\text{g+}0.0454\text{g}}\\ =0.9378\end{array}$

Write the expression for the total mass of phase $\alpha$.

$m_{\alpha }=m_1+m_2$

Substitute 0.0454 g for $m_1$ and 0.685 g for $m_2$.

$\begin{array}{c}{m}_{\alpha }=0.0454+0.685\\ =0.7304\text{g}\end{array}$

Thus, the amount in phase $\alpha$ is $\underset{\_}{0.7304\text{g}}$.

Similarly, calculating the mass of the $\beta$ phase using Table (1)

Calculate the composition for phase $\beta$ for component 1.

$x_1=\frac{m_1}{m_1+m_2}$

Substitute 0.1524 g for $m_1$ and 0.1058 g for $m_2$.

$\begin{array}{c}{x}_1=\frac{0.1524\text{g}}{0.1524\text{g}+0.1058\text{g}}\\ =0.590\end{array}$

Calculate the composition for phase $\beta$ for component 2.

$x_2=\frac{m_2}{m_1+m_2}$

Substitute 0.1524 g for $m_1$ and 0.1058 g for $m_2$.

$\begin{array}{c}{x}_2=\frac{0.1058\text{g}}{0.1058\text{g+}0.1524\text{g}}\\ =0.409\end{array}$

Calculate the total mass of phase $\beta$.

$m_{\beta }=m_1+m_2$

Substitute 0.1524 g for $m_1$ and 0.1058 g for $m_2$.

$\begin{array}{c}{m}_{\beta }=0.1524+0.1058\\ =0.258\text{g}\end{array}$

Thus, the amount in phase $\beta$ is $\underset{\_}{0.258\text{g}}$.