Chapter 5, Problem 5.2IA

Interpretation Introduction

Interpretation: The vapour pressure/composition curve for a mixture of benzene $\left(\text{B}\right)$ and ethanoic acid $\left(\text{E}\right)$ at $50\text{°C}$ for the given vapour pressure data has to be plotted.  The activity and activity coefficients of both components on the basis of Raoult’s law have to be deduced.  Taking B as a solute, its activity and activity coefficients have to be deduced on the basis of Henry’s law.  The excess Gibbs energy of the mixture has to be evaluated over the given composition range.

Concept introduction: The Raoult’s law states that the ratio of the partial pressure of each component to the partial pressure in the vapour pressure when it is present as a pure liquid is equal to its composition in the liquid phase.  Henry’s law states that the amount the partial vapour pressure is directly proportional to the amount of substance dissolved.  It involves an empirical constant $K_{\text{H}}$ called Henry’s constant.

Answer to Problem 5.2IA

The vapour pressure/composition curve for a mixture of benzene and ethanoic acid for the given vapour pressure data is plotted as below,

The activity and activity coefficient of ethanoic acid on the basis of Raoult’s law is shown in the table below.

$x_{\text{E}}$	$a_{\text{E}}$	${\gamma }_{\text{E}}$
0.016	0.016	1
0.0439	0.0439	1
0.0835	0.0835	1
0.1138	0.1138	1
0.1714	0.1714	1
0.2973	0.2973	1
0.3696	0.3696	1
0.5834	0.5834	1
0.6604	0.6604	1
0.8437	0.8437	1
0.9931	0.9931	1

The activity and activity coefficients of benzene on the basis of Raoult’s law is shown in the table below,

$x_{\text{B}}$	$a_{\text{B}}$	${\gamma }_{\text{B}}$
0.984	0.984	1
0.9561	0.9561	1
0.9165	0.9165	1
0.8862	0.8862	1
0.8286	0.8286	1
0.7027	0.7027	1
0.6304	0.6304	1
0.4166	0.4166	1
0.3396	0.3396	1
0.1563	0.1563	1
0.0069	0.0069	1

The activity and activity coefficients of benzene on the basis of Henry’s law are shown in the table below,

$x_{\text{B}}$	$a_{\text{B}}$	${\gamma }_{\text{B}}$
0.984	0.730838	0.742721
0.9561	0.710116	0.742721
0.9165	0.680704	0.742721
0.8862	0.6582	0.742721
0.8286	0.615419	0.742721
0.7027	0.52191	0.742721
0.6304	0.468212	0.742721
0.4166	0.309418	0.742721
0.3396	0.252228	0.742721
0.1563	0.116087	0.742721
0.0069	0.005125	0.742721

The excess Gibbs energy is $\underset{\_}{-766.3Jmo{l}^{-1}}$.

Explanation of Solution

The vapour pressure data for a mixture of benzene $\left(\text{B}\right)$ and ethanoic acid $\left(\text{E}\right)$ at $50\text{°C}$ is given as,

$x_{\text{E}}$	$p_{\text{E}}/\text{kPa}$	$p_{\text{B}}/\text{kPa}$
0.016	0.484	35.05
0.0439	0.967	34.29
0.0835	1.535	33.28
0.1138	1.89	32.64
0.1714	2.45	30.9
0.2973	3.31	28.16
0.3696	3.83	26.08
0.5834	4.84	20.42
0.6604	5.36	18.01
0.8437	6.76	10
0.9931	7.29	0.47

The total mole fraction of the components is $1$.

    $x_{\text{B}}+x_{\text{E}}=1$

The mole fraction of benzene $\left(x_{\text{B}}\right)$ in the solution is calculated in the table as,

$x_{\text{E}}$	$x_{\text{B}}=1-x_{\text{E}}$
0.016	0.984
0.0439	0.9561
0.0835	0.9165
0.1138	0.8862
0.1714	0.8286
0.2973	0.7027
0.3696	0.6304
0.5834	0.4166
0.6604	0.3396
0.8437	0.1563
0.9931	0.0069

The total pressure $\left(p\right)$ of the solution is,

    $p=p_{\text{E}}+p_{\text{B}}$                                                                                                   (1)

Where,

• $p_{\text{E}}$ is the vapour pressure of ethanoic acid.
• $p_{\text{B}}$ is the vapour pressure of benzene.

According to Dalton’s law of partial pressures, the partial pressure of benzene $\left(p_{\text{B}}\text{'}\right)$ is given by the formula,

    $p_{\text{B}}\text{'}=x_{\text{B}}p$                                                                                                      (2)

Where

• $x_{\text{B}}$ is the mole fraction of benzene.

Substitute the value of $p$ from equation (1) in equation (2).

    $p_{\text{B}}\text{'}=x_{\text{B}}\left(p_{\text{E}}+p_{\text{B}}\right)$                                                                                        (3)

The mole fraction of benzene $\left(x_{\text{B}}\right)$ is $0.984$.

The vapour pressure of ethanoic acid $\left(p_{\text{E}}\right)$ is $\text{0}\text{.484}\text{kPa}$.

The vapour pressure of benzene $\left(p_{\text{B}}\right)$ is $\text{35}\text{.05}\text{kPa}$.

Substitute the value of $x_{\text{B}}$, $p_{\text{E}}$ and $p_{\text{B}}$ in equation(3).

    $\begin{array}{c}{p}_{\text{B}}\text{'}=0.984\left(\text{0}\text{.484}\text{kPa+35}\text{.05}\text{kPa}\right)\\ =34.965\text{kPa}\end{array}$

Similarly, the values of $p_B\text{'}$ and $p$ are calculated for all values of $x_{\text{B}}$ in the table as shown,

$x_{\text{B}}=1-x_{\text{E}}$	$p/\text{kPa}$	$p_{\text{B}}\text{'}/\text{kPa}$
0.984	35.534	34.96546
0.9561	35.257	33.70922
0.9165	34.815	31.90795
0.8862	34.53	30.60049
0.8286	33.35	27.63381
0.7027	31.47	22.11397
0.6304	29.91	18.85526
0.4166	25.26	10.52332
0.3396	23.37	7.936452
0.1563	16.76	2.619588
0.0069	7.76	0.053544

The graph between mole fraction of benzene $\left(x_{\text{B}}\right)$ and the partial pressure of benzene $\left(p_{\text{B}}\text{'}\right)$ is plotted as,

According to Raoult’s law the relation between the partial pressure of benzene is given by the expression,

    $p_{\text{B}}\text{'}=x_{\text{B}}{p}_{\text{B}}^{*}$

Where,

• $p_{\text{B}}^{*}$ is the vapour pressure of the pure substance.

As the graph between the partial pressure of benzene and mole fraction of benzene is a straight line, hence, it follows Raoult’s law.

According to Raoult’s law the relation between the partial pressure of ethanoic acid is given by the expression,

    $p_{\text{E}}\text{'}=x_{\text{E}}{p}_{\text{E}}^{*}$                                                                                                     (4)

Where,

• $p_{\text{E}}^{*}$ is the vapour pressure of the pure substance.
• $x_{\text{E}}$ is the mole fraction of ethanoic acid.
• $p_{\text{E}}\text{'}$ is the partial pressure of ethanoic acid.

The vapour pressure of pure ethanoic acid $\left(p_{\text{E}}^{*}\right)$ at $50\text{°C}$ is $\text{7}\text{.615}\text{kPa}$.

The mole fraction of ethanoic acid $\left(x_{\text{E}}\right)$ is $0.0160$.

Substitute the value of $p_{\text{E}}^{*}$ and $x_{\text{E}}$ in equation (4).

    $\begin{array}{c}{p}_{\text{E}}\text{'}=0.016\times 7.615\text{kPa}\\ =\text{0}\text{.1218}\text{kPa}\end{array}$

The mole fraction of ethanoic acid in vapour $\left(y_{\text{E}}\right)$ is given by the formula,

    $y_{\text{E}}=\frac{p_{\text{E}}\text{'}}{p}$                                                                                                        (5)

The partial pressure of ethanoic acid $\left(p_{\text{E}}\text{'}\right)$ is $\text{0}\text{.1218}\text{kPa}$.

The total pressure of the mixture $\left(p\right)$ when mole fraction of ethanoic acid is $0.016$ is $\text{35}\text{.534}\text{kPa}$

Substitute the value of $p_{\text{E}}\text{'}$ and $p$ in equation (5).

    $\begin{array}{c}{y}_{\text{E}}=\frac{\text{0}\text{.1218}\text{kPa}}{\text{35}\text{.534}\text{kPA}}\\ =0.00342\end{array}$

The activity of ethanoic acid $\left(a_{\text{E}}\right)$ is given as,

    $a_{\text{E}}=\frac{y_{\text{E}}p}{p_{\text{E}}^{*}}$                                                                                                       (6)

Substitute the values of $y_{\text{E}}$, $p$, and $p_{\text{E}}^{*}$ in equation (6).

    $\begin{array}{c}{a}_{\text{E}}=\frac{0.00342\times 35.534\text{kPa}}{7.615\text{kPa}}\\ =0.0159\end{array}$

The activity coefficient $\left({\gamma }_{\text{E}}\right)$ of ethanoic acid is given as,

    ${\gamma }_{\text{E}}=\frac{a_{\text{E}}}{x_{\text{E}}}$                                                                                                         (7)

Substitute the value of $a_{\text{E}}$ and $x_{\text{E}}$ in equation (7).

    $\begin{array}{c}{\gamma }_{\text{E}}=\frac{0.0159}{0.016}\\ =0.99\end{array}$

Similarly, the values of $a_{\text{E}}$ and ${\gamma }_{\text{E}}$ are calculated for all values of $x_{\text{E}}$ in the table as shown,

$x_{\text{E}}$	$p_{\text{E}}\text{'}/\text{kPa}$	$y_{\text{E}}$	$a_{\text{E}}$	${\gamma }_{\text{E}}$
0.016	0.12184	0.003429	0.016	1
0.0439	0.334299	0.009482	0.0439	1
0.0835	0.635853	0.018264	0.0835	1
0.1138	0.866587	0.025097	0.1138	1
0.1714	1.305211	0.039137	0.1714	1
0.2973	2.26394	0.07194	0.2973	1
0.3696	2.814504	0.094099	0.3696	1
0.5834	4.442591	0.175875	0.5834	1
0.6604	5.028946	0.215188	0.6604	1
0.8437	6.424776	0.38334	0.8437	1
0.9931	7.562457	0.974543	0.9931	1

According to Raoult’s law the relation between the partial pressure of benzene is given by the expression,

    $p_{\text{B}}\text{'}=x_{\text{B}}{p}_{\text{B}}^{*}$                                                                                                     (8)

Where,

• $p_{\text{B}}^{*}$ is the vapour pressure of the pure substance.
• $x_{\text{B}}$ is the mole fraction of benzene.
• $p_{\text{B}}\text{'}$ is the partial pressure of benzene.

The vapour pressure of pure benzene $\left(p_{\text{B}}^{*}\right)$ at $50\text{°C}$ is $\text{34}\text{.66}\text{kPa}$.

The mole fraction of benzene $\left(x_{\text{B}}\right)$ is $0.984$.

Substitute the value of $p_{\text{B}}^{*}$ and $x_{\text{B}}$ in equation (8).

    $\begin{array}{c}{p}_{\text{B}}\text{'}=0.984\times 33.66\text{kPa}\\ \text{=33}\text{.121kPa}\end{array}$

The mole fraction of benzene in vapour $\left(y_{\text{B}}\right)$ is given by the formula,

    $y_{\text{B}}=\frac{p_{\text{B}}\text{'}}{p}$                                                                                                       (9)

The partial pressure of benzene $\left(p_{\text{B}}\text{'}\right)$ is $\text{33}\text{.121}\text{kPa}$.

The total pressure of the mixture $\left(p\right)$ when mole fraction of benzene is $0.984$ is $\text{35}\text{.534}\text{kPa}$.

Substitute the value of $p_{\text{B}}\text{'}$ and $p$ in equation (9).

    $\begin{array}{c}{y}_{\text{B}}=\frac{33.121\text{kPa}}{\text{35}\text{.534}\text{kPA}}\\ =0.932\end{array}$

The activity of benzene $\left(a_{\text{B}}\right)$ is given as,

    $a_{\text{B}}=\frac{y_{\text{B}}p}{p_{\text{B}}^{*}}$                                                                                                     (10)

Substitute the values of $y_{\text{B}}$, $p$, and $p_{\text{B}}^{*}$ in equation (10).

    $\begin{array}{c}{a}_{\text{B}}=\frac{0.932\times 35.534\text{kPa}}{33.66\text{kPa}}\\ =0.984\end{array}$

The activity coefficient $\left({\gamma }_{\text{B}}\right)$ of benzene is given as,

    ${\gamma }_{\text{B}}=\frac{a_{\text{B}}}{x_{\text{B}}}$                                                                                                       (11)

Substitute the value of $a_{\text{B}}$ and $x_{\text{B}}$ in equation (11).

    $\begin{array}{c}{\gamma }_{\text{B}}=\frac{0.984}{0.984}\\ =1\end{array}$

Similarly, the values of $a_{\text{B}}$ and ${\gamma }_{\text{B}}$ are calculated for all values of $x_{\text{B}}$ in the table as shown,

$x_{\text{B}}$	$p_{\text{B}}\text{'}/\text{kPa}$	$y_{\text{B}}$	$a_{\text{B}}$	${\gamma }_{\text{B}}$
0.984	33.12144	0.932106	0.984	1
0.9561	32.18233	0.912793	0.9561	1
0.9165	30.84939	0.886095	0.9165	1
0.8862	29.82949	0.863872	0.8862	1
0.8286	27.89068	0.836302	0.8286	1
0.7027	23.65288	0.751601	0.7027	1
0.6304	21.21926	0.709437	0.6304	1
0.4166	14.02276	0.555137	0.4166	1
0.3396	11.43094	0.489129	0.3396	1
0.1563	5.261058	0.313906	0.1563	1
0.0069	0.232254	0.02993	0.0069	1

According to Henry’s law the partial pressure of benzene $\left(p_{\text{B}}\text{'}\right)$ is given as,

  $p_{\text{B}}\text{'}=x_{\text{B}}{K}_{\text{H}}$                                                                                                  (12)

Where

• $p_{\text{B}}\text{'}$ is the vapour pressure of benzene in the solution.
• $K_{\text{H}}$ is Henry’s constant.
• $x_{\text{B}}$ is the mole fraction of benzene in the solution.

From the graph plotted between the partial pressure of benzene and mole fraction of benzene,

When mole fraction of benzene $\left(x_{\text{B}}\right)$ is $0.4$, the partial pressure of benzene $\left(p_{\text{B}}\text{'}\right)$ is $\text{10}\text{kPa}$.

Substitute the value of $x_{\text{B}}$ and $p_{\text{B}}\text{'}$ in equation (12).

    $\begin{array}{c}10\text{kPa}=0.4\times K_{\text{H}}\\ K_{\text{H}}=25\text{kPa}\end{array}$

The mole fraction of benzene in the vapour phase is given as,

    $y_{\text{B}}=\frac{K_{\text{H}}{x}_{\text{B}}}{p}$                                                                                                    (13)

The mole fraction of benzene $\left(x_{\text{B}}\right)$ is $0.984$.

The total pressure of the mixture $\left(p\right)$ when mole fraction of benzene is $0.984$ is $\text{35}\text{.534}\text{kPa}$.

Substitute the value of $K_{\text{H}}$, $x_{\text{B}}$, and $p$ in the equation (8).

    $\begin{array}{c}{y}_{\text{B}}=\frac{25\text{kPa}\times 0.984}{\text{35}\text{.534}\text{kPa}}\\ =0.6922\end{array}$

The activity of benzene $\left(a_{\text{B}}\right)$ is given as,

    $a_{\text{B}}=\frac{y_{\text{B}}p}{p_{\text{B}}^{*}}$                                                                                                     (10)

Substitute the values of $y_{\text{B}}$, $p$, and $p_{\text{B}}^{*}$ in equation (10).

    $\begin{array}{c}{a}_{\text{B}}=\frac{0.6922\times 35.534\text{kPa}}{33.66\text{kPa}}\\ =0.73\end{array}$

The activity coefficient $\left({\gamma }_{\text{B}}\right)$ of benzene is given as,

    ${\gamma }_{\text{B}}=\frac{a_{\text{B}}}{x_{\text{B}}}$                                                                                                       (11)

Substitute the value of $a_{\text{B}}$ and $x_{\text{B}}$ in equation (11).

    $\begin{array}{c}{\gamma }_{\text{B}}=\frac{0.743}{0.984}\\ =0.742\end{array}$

Similarly, the values of $a_{\text{B}}$ and ${\gamma }_{\text{B}}$ are calculated for all values of $x_{\text{B}}$ in the table.

$x_{\text{B}}$	$p/\text{kPa}$	$y_{\text{B}}$	$a_{\text{B}}$	${\gamma }_{\text{B}}$
0.984	35.534	0.692295	0.730838	0.742721
0.9561	35.257	0.67795	0.710116	0.742721
0.9165	34.815	0.658121	0.680704	0.742721
0.8862	34.53	0.641616	0.6582	0.742721
0.8286	33.35	0.621139	0.615419	0.742721
0.7027	31.47	0.55823	0.52191	0.742721
0.6304	29.91	0.526914	0.468212	0.742721
0.4166	25.26	0.412312	0.309418	0.742721
0.3396	23.37	0.363286	0.252228	0.742721
0.1563	16.76	0.233144	0.116087	0.742721
0.0069	7.76	0.022229	0.005125	0.742721

The excess Gibbs energy is given as,

    $\Delta G_{\text{e}}=x_{\text{E}}RT\ln {\gamma }_{\text{E}}+x_{\text{B}}RT\ln {\gamma }_{\text{B}}$                                                                   (14)

Where,

• $R$ is the gas constant $\left(\text{8}\text{.314}\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\right)$.
• $T$ is the temperature in Kelvin.

The temperature of the mixture is $\text{50}\text{°C}$

The temperature values in degree Celsius is converted to Kelvin using the equation given below as,

    $T\left(\text{K}\right)=T\left(°\text{C}\right)+273.15$

Substitute the temperature value in the above equation as follows.

    $\begin{array}{c}T\left(\text{K}\right)=T\left(°\text{C}\right)+273.15\\ =50+273.15\\ =323.15\text{K}\end{array}$

The value of mole fraction of benzene $\left(x_{\text{B}}\right)$ is $0.984$.

The value of mole fraction of ethanoic acid $\left(x_{\text{E}}\right)$ is $0.016$.

The activity coefficient of benzene is $\left({\gamma }_{\text{B}}\right)$ is $0.7427$.

The activity coefficient of ethanoic acid $\left({\gamma }_{\text{E}}\right)$ is $1$.

Substitute the value of $R$, $T$, $x_{\text{B}}$, $x_{\text{E}}$, ${\gamma }_{\text{B}}$ and ${\gamma }_{\text{E}}$ in equation (14).

    $\begin{array}{c}\Delta G_{\text{e}}=x_{\text{E}}RT\ln {\gamma }_{\text{E}}+x_{\text{B}}RT\ln {\gamma }_{\text{B}}\\ =\left(\begin{array}{l}\text{0}\text{.016×8}\text{.314}\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{×323}\text{K}\ln \text{1+}\\ \text{0}\text{.984×8}\text{.314}\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{×323}\text{K}\text{ln}\text{0}\text{.7427}\end{array}\right)\\ =\text{0}+\text{0}\text{.984×8}\text{.314}\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{×323}\text{K×}\left(\text{-0}\text{.29}\right)\\ \text{=}-\text{766}\text{.3}\text{J}{\text{mol}}^{\text{-1}}\end{array}$

Thus the excess Gibbs energy is $\underset{\_}{-766.3Jmo{l}^{-1}}$.