Chapter 5, Problem 5.1IA

(a)

Interpretation Introduction

Interpretation: The activity coefficients of both the components in a mixture of iodoethane $\left(\text{I}\right)$ and ethyl ethanoate $\left(\text{E}\right)$ at $\text{50}\text{°C}$ have to be calculated on the basis of Raoult’s law.

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.

Answer to Problem 5.1IA

The activity coefficients of ethyl ethanoate calculated on the basis of Raoult’s law are shown in the table below.

$x_{\text{E}}=1-x_{\text{I}}$	${\gamma }_{\text{E}}=a_{\text{E}}/x_{\text{E}}$
1	1
0.9421	1.007505
0.8905	1.010608
0.8082	1.021168
0.7647	1.029929
0.6282	1.066769
0.4522	1.137652
0.3651	1.200958
0.1747	1.359818
0.0907	1.501313
0	0

The activity coefficients of iodoethane calculated on the basis of Raoult’s law are shown in the table below.

$x_{\text{I}}$	${\gamma }_{\text{I}}=a_{\text{I}}/x_{\text{I}}$
0	0
0.0579	1.36718
0.1095	1.362498
0.1918	1.294589
0.2353	1.267212
0.3718	1.182701
0.5478	1.101799
0.6349	1.065633
0.8253	1.017791
0.9093	1.003589
1	1

Explanation of Solution

The data for the vapour pressure of mixture of iodoethane $\left(\text{I}\right)$ and ethyl ethanoate $\left(\text{E}\right)$ is given as,

$x_{\text{I}}$	$p_{\text{I}}/\text{kPa}$	$p_{\text{E}}/\text{kPa}$
0	0	37.38
0.0579	3.73	35.48
0.1095	7.03	33.64
0.1918	11.7	30.85
0.2353	14.05	29.44
0.3718	20.72	25.05
0.5478	28.44	19.23
0.6349	31.88	16.39
0.8253	39.58	8.88
0.9093	43.00	5.09
1.000	47.12	0

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

    $x_{\text{I}}+x_{\text{E}}=1$

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

$x_{\text{I}}$	$x_{\text{E}}=1-x_{\text{I}}$	$p_{\text{I}}/\text{kPa}$	$p_{\text{E}}/\text{kPa}$
0	1	0	37.38
0.0579	0.9421	3.73	35.48
0.1095	0.8905	7.03	33.64
0.1918	0.8082	11.7	30.85
0.2353	0.7647	14.05	29.44
0.3718	0.6282	20.72	25.05
0.5478	0.4522	28.44	19.23
0.6349	0.3651	31.88	16.39
0.8253	0.1747	39.58	8.88
0.9093	0.0907	43.00	5.09
1.000	0	47.12	0

According to Raoult’s law, for ideal solutions,

The activity of ethyl ethanoate $\left(a_{\text{E}}\right)$ is expressed as,

    $a_{\text{E}}=\frac{p_{\text{E}}}{p_{\text{E}}^{*}}$                                                                                                        (1)

The activity coefficient of ethyl ethanoate $\left({\gamma }_{\text{E}}\right)$ is given by the expression,

    ${\gamma }_{\text{E}}=\frac{a_{\text{E}}}{x_{\text{E}}}$                                                                                                         (2)

Where

• $p_{\text{E}}$ is the vapour pressure of ethyl ethanoate in the solution.
• $p_{\text{E}}^{*}$ is the vapour pressure of the pure ethyl ethanoate
• $x_{\text{E}}$ is the mole fraction of ethyl ethanoate in the solution.

The vapour pressure of pure ethyl ethanoate $\left(p_{\text{E}}^{*}\right)$ is $\text{37}\text{.38}\text{kPa}$.

Mole fraction of ethyl ethanoate $\left(x_{\text{E}}\right)$ is $0.9421$.

The vapour pressure of ethyl ethanoate in the solution $\left(p_{\text{E}}\right)$ for mole fraction $\left(x_{\text{E}}\right)$ of $0.9421$ is $\text{35}\text{.48}\text{kPa}$.

Substitute the values of $p_{\text{E}}$ and $p_{\text{E}}^{*}$ in equation (1).

    $\begin{array}{c}{a}_{\text{E}}=\frac{\text{35}\text{.48}\text{kPa}}{\text{37}\text{.38}\text{kPa}}\\ =0.949\end{array}$

Substitute the values of $a_{\text{E}}$ and $x_{\text{E}}$ in equation (2).

    $\begin{array}{c}{\gamma }_{\text{E}}=\frac{0.949}{0.9421}\\ =1.007\end{array}$

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,

$x_{\text{E}}=1-x_{\text{I}}$	$p_{\text{E}}/\text{kPa}$	$a_{\text{E}}=p_{\text{E}}/p_{\text{E}}^{*}$	${\gamma }_{\text{E}}=a_{\text{E}}/x_{\text{E}}$
1	37.38	1	1
0.9421	35.48	0.949171	1.007505
0.8905	33.64	0.899946	1.010608
0.8082	30.85	0.825308	1.021168
0.7647	29.44	0.787587	1.029929
0.6282	25.05	0.670144	1.066769
0.4522	19.23	0.514446	1.137652
0.3651	16.39	0.43847	1.200958
0.1747	8.88	0.23756	1.359818
0.0907	5.09	0.136169	1.501313
0	0	0	0

The activity of iodoethane $\left(a_{\text{I}}\right)$ is expressed as,

    $a_{\text{I}}=\frac{p_{\text{I}}}{p_{\text{I}}^{*}}$                                                                                                        (3)

The activity coefficient of iodoethane $\left({\gamma }_{\text{I}}\right)$ is given by the expression,

    ${\gamma }_{\text{I}}=\frac{a_{\text{I}}}{x_{\text{I}}}$                                                                                                         (4)

Where

• $p_{\text{I}}$ is the vapour pressure of iodoethane in the solution.
• $p_{\text{I}}^{*}$ is the vapour pressure of the pure iodoethane.
• $x_{\text{I}}$ is the mole fraction of iodoethane in the solution.

The vapour pressure of pure iodoethane $\left(p_{\text{I}}^{*}\right)$ is $\text{47}\text{.12}\text{kPa}$.

Mole fraction of iodoethane $\left(x_{\text{I}}\right)$ is $0.9093$.

The vapour pressure of iodoethane in the solution $\left(p_{\text{I}}\right)$ for mole fraction $\left(x_{\text{I}}\right)$ of $0.9093$ is $\text{43}\text{.00}\text{kPa}$.

Substitute the values of $p_{\text{I}}$ and $p_{\text{I}}^{*}$ in equation (3).

    $\begin{array}{c}{a}_{\text{I}}=\frac{\text{43}\text{.00}\text{kPa}}{\text{47}\text{.12}\text{kPa}}\\ =0.912\end{array}$

Substitute the values of $a_{\text{I}}$ and $x_{\text{I}}$ in equation (4).

    $\begin{array}{c}{\gamma }_{\text{I}}=\frac{0.912}{0.9093}\\ =1.002\end{array}$

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

$x_{\text{I}}$	$p_{\text{I}}/\text{kPa}$	$a_{\text{I}}=p_{\text{I}}/p_{\text{I}}^{*}$	${\gamma }_{\text{I}}=a_{\text{I}}/x_{\text{I}}$
0	0	0	0
0.0579	3.73	0.07916	1.36718
0.1095	7.03	0.149194	1.362498
0.1918	11.7	0.248302	1.294589
0.2353	14.05	0.298175	1.267212
0.3718	20.72	0.439728	1.182701
0.5478	28.44	0.603565	1.101799
0.6349	31.88	0.67657	1.065633
0.8253	39.58	0.839983	1.017791
0.9093	43	0.912564	1.003589
1	47.12	1	1

(b)

Interpretation Introduction

Interpretation: The activity coefficients of both the components in a mixture of iodoethane $\left(\text{I}\right)$ and ethyl ethanoate $\left(\text{E}\right)$ at $\text{50}\text{°C}$ have to be calculated on the basis of Henry’s law.

Concept introduction: 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 5.1IA

The activity coefficients of ethyl ethanoate calculated on the basis of Henry’s law are shown in the table below.

$x_{\text{E}}=1-x_{\text{I}}$	${\gamma }_{\text{E}}=a_{\text{E}}/x_{\text{E}}$
1	0.970657
0.9421	0.977942
0.8905	0.980954
0.8082	0.991203
0.7647	0.999708
0.6282	1.035467
0.4522	1.10427
0.3651	1.165718
0.1747	1.319917
0.0907	1.45726
0	0

The activity coefficients of iodoethane calculated on the basis of Henry’s law are shown in the table below.

$x_{\text{I}}$	${\gamma }_{\text{I}}=a_{\text{I}}/x_{\text{I}}$
0	0
0.0579	1.160746
0.1095	1.156773
0.1918	1.099118
0.2353	1.075874
0.3718	1.004124
0.5478	0.935437
0.6349	0.904732
0.8253	0.864114
0.9093	0.852056
1	0.849009

Explanation of Solution

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

  $p_{\text{I}}=x_{\text{I}}{K}_{\text{H}}$

Where

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

The data for the vapour pressure of mixture of iodoethane $\left(\text{I}\right)$ and ethyl ethanoate $\left(\text{E}\right)$ is given as,

$x_{\text{I}}$	$p_{\text{I}}/\text{kPa}$
0	0
0.0579	3.73
0.1095	7.03
0.1918	11.7
0.2353	14.05
0.3718	20.72
0.5478	28.44
0.6349	31.88
0.8253	39.58
0.9093	43.00
1.000	47.12

Graph between $p_{\text{I}}$ and $x_{\text{I}}$ is plotted below.

From the graph the value of $K_{\text{H}}$ is observed as $55.5\text{kPa}$ .

The activity is given as,

    $a_{\text{I}}=\frac{p_{\text{I}}}{K_{\text{H}}}$                                                                                                         (5)

The value of Henry’s constant $\left(K_{\text{H}}\right)$ is $55.5\text{kPa}$.

The mole fraction of iodoethane $\left(x_{\text{I}}\right)$ is $0.579$.

The vapour pressure of iodoethane in the solution $\left(p_{\text{I}}\right)$ for mole fraction $\left(x_{\text{I}}\right)$ of $0.0579$ is $\text{3}\text{.73}\text{kPa}$.

Substitute the values of $K_{\text{H}}$ and $p_{\text{I}}$ in equation (5).

    $\begin{array}{c}{a}_{\text{I}}=\frac{\text{3}\text{.73}\text{kPa}}{\text{55}\text{.5}\text{kPa}}\\ =0.0672\end{array}$

The activity coefficient of iodoethane $\left({\gamma }_{\text{I}}\right)$ is given by the expression,

    ${\gamma }_{\text{I}}=\frac{a_{\text{I}}}{x_{\text{I}}}$                                                                                                           (4)

Substitute the values of $a_{\text{I}}$ and $x_{\text{I}}$ in equation (4).

    $\begin{array}{c}{\gamma }_{\text{I}}=\frac{0.0672}{0.0579}\\ =1.160\end{array}$

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

$x_{\text{I}}$	$p_{\text{I}}/\text{kPa}$	$a_{\text{I}}=p_{\text{I}}/K_{\text{H}}$	${\gamma }_{\text{I}}=a_{\text{I}}/x_{\text{I}}$
0	0	0	0
0.0579	3.73	0.067207	1.160746
0.1095	7.03	0.126667	1.156773
0.1918	11.7	0.210811	1.099118
0.2353	14.05	0.253153	1.075874
0.3718	20.72	0.373333	1.004124
0.5478	28.44	0.512432	0.935437
0.6349	31.88	0.574414	0.904732
0.8253	39.58	0.713153	0.864114
0.9093	43	0.774775	0.852056
1	47.12	0.849009	0.849009

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

  $p_{\text{E}}=x_{\text{E}}{K}_{\text{H}}$

Where

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

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

    $x_{\text{I}}+x_{\text{E}}=1$

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

$x_{\text{E}}=1-x_{\text{I}}$	$p_{\text{E}}/\text{kPa}$
1	37.38
0.9421	35.48
0.8905	33.64
0.8082	30.85
0.7647	29.44
0.6282	25.05
0.4522	19.23
0.3651	16.39
0.1747	8.88
0.0907	5.09
0	0

Graph between $p_{\text{E}}$ and $x_{\text{E}}$ is plotted below.

From the graph, the value of $K_{\text{H}}$ is observed as $38.51\text{kPa}$.

The activity of ethyl ethanoate $\left(a_{\text{E}}\right)$ is expressed as,

    $a_{\text{E}}=\frac{p_{\text{E}}}{K_{\text{H}}}$                                                                                                        (6)

The activity coefficient of ethyl ethanoate $\left({\gamma }_{\text{E}}\right)$ is given by the expression,

    ${\gamma }_{\text{E}}=\frac{a_{\text{E}}}{x_{\text{E}}}$                                                                                                         (2)

Where

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

The value of Henry’s constant $\left(K_{\text{H}}\right)$ is $38.51\text{kPa}$.

Mole fraction of ethyl ethanoate $\left(x_{\text{E}}\right)$ is $0.9421$.

The vapour pressure of ethyl ethanoate in the solution $\left(p_{\text{E}}\right)$ for mole fraction $\left(x_{\text{E}}\right)$ of $0.9421$ is $\text{35}\text{.48}\text{kPa}$.

Substitute the values of $p_{\text{E}}$ and $K_{\text{H}}$ in equation (6).

    $\begin{array}{c}{a}_{\text{E}}=\frac{\text{35}\text{.48}\text{kPa}}{38.51\text{kPa}}\\ =0.9213\end{array}$

Substitute the values of $a_{\text{E}}$ and $x_{\text{E}}$ in equation (2).

    $\begin{array}{c}{\gamma }_{\text{E}}=\frac{0.9213}{0.9421}\\ =0.978\end{array}$

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,

$x_{\text{E}}=1-x_{\text{I}}$	$p_{\text{E}}/\text{kPa}$	$a_{\text{E}}=p_{\text{E}}/K_{\text{H}}$	${\gamma }_{\text{E}}=a_{\text{E}}/x_{\text{E}}$
1	37.38	0.970657	0.970657
0.9421	35.48	0.921319	0.977942
0.8905	33.64	0.873539	0.980954
0.8082	30.85	0.801091	0.991203
0.7647	29.44	0.764477	0.999708
0.6282	25.05	0.65048	1.035467
0.4522	19.23	0.499351	1.10427
0.3651	16.39	0.425604	1.165718
0.1747	8.88	0.230589	1.319917
0.0907	5.09	0.132173	1.45726
0	0	0	0