Chapter 5, Problem 5A.4P

Interpretation Introduction

Interpretation:

The partial pressures of methylbenzene (A) and butanone have to be calculated.  The calculated partial pressures have to be plotted against the mole fraction of the liquid mixture.  Henry’s law constant for both the components has to be calculated.

Concept introduction:

Henry’s law states that the partial vapour pressure is directly proportional to the amount of substance dissolved.  It involves an empirical constant $K_{\text{H}}$ called Henry’s law constant.

Answer to Problem 5A.4P

The partial pressures of methylbenzene (A) are calculated in the table given below.

$y_{\text{A}}$	$x_{\text{A}}$	$p/\text{kPa}$	$p_{\text{A}}/\text{kPa}$
0	0	36.066	0
0.041	0.0898	34.121	1.398961
0.1154	0.2476	30.9	3.56586
0.1762	0.3577	28.626	5.043901
0.2772	0.5194	25.239	6.996251
0.3393	0.6036	23.402	7.940299
0.445	0.7188	20.6984	9.210788
0.5435	0.8019	18.592	10.10475
0.7284	0.9105	15.496	11.28729
1	1	12.295	12.295

The partial pressures of butanone are calculated in the table given below.

$y_{\text{B}}$	$x_{\text{B}}$	$p/\text{kPa}$	$p_{\text{B}}/\text{kPa}$
1	1	36.066	36.066
0.959	0.9102	34.121	32.72204
0.8846	0.7524	30.9	27.33414
0.8238	0.6423	28.626	23.5821
0.7228	0.4806	25.239	18.24275
0.6607	0.3964	23.402	15.4617
0.555	0.2812	20.6984	11.48761
0.4565	0.1981	18.592	8.487248
0.2716	0.0895	15.496	4.208714
0	0	12.295	0

The partial pressure of methylbenzene (A) is plotted against the mole fraction of the liquid mixture in the graph shown below.

Figure 1

The partial pressure of butanone is plotted against the mole fraction of the liquid mixture in the graph shown below.

Figure 2

The value of Henry’s law constant for methylbenzene (A) is $\underset{\_}{15.578kPa}$.  The value of Henry’s law constant for butanone is $\underset{\_}{47.025kPa}$.

Explanation of Solution

The data for the mole fraction of methylbenzene (A) in the liquid phase, $x_{\text{A}}$, vapour phase, $y_{\text{A}}$ and the total pressure, $p$ is given below.

$x_{\text{A}}$	$y_{\text{A}}$	$p/\text{kPa}$
0	0	36.066
0.0898	0.041	34.121
0.2476	0.1154	30.9
0.3577	0.1762	28.626
0.5194	0.2772	25.239
0.6036	0.3393	23.402
0.7188	0.445	20.6984
0.8019	0.5435	18.592
0.9105	0.7284	15.496
1	1	12.295

According to Dalton’s law of partial pressures, the partial pressure of methylbenzene (A) $\left(p_{\text{A}}\right)$ is calculated by the formula given below.

    $p_{\text{A}}=y_{\text{A}}p$                                                                                                       (1)

Where,

• $y_{\text{A}}$ is the mole fraction of methylbenzene (A) in the vapour phase.
• $p$ is the total vapour pressure.

The total pressure $\left(p\right)$ at $y_{\text{A}}=0.0410$ is $\text{34}\text{.121}\text{kPa}$.

Substitute the value of $p$ and $y_{\text{A}}$ in equation (1).

    $\begin{array}{c}{p}_{\text{A}}=0.0410\times \text{34}\text{.121}\text{kPa}\\ =\underset{\_}{1.398961kPa}\end{array}$

Therefore, the partial pressure of methylbenzene (A) $\left(p_{\text{A}}\right)$ is calculated for all vapour phase composition $\left(y_{\text{A}}\right)$ in the table below.

$y_{\text{A}}$	$x_{\text{A}}$	$p/\text{kPa}$	$p_{\text{A}}/\text{kPa}$
0	0	36.066	0
0.041	0.0898	34.121	1.398961
0.1154	0.2476	30.9	3.56586
0.1762	0.3577	28.626	5.043901
0.2772	0.5194	25.239	6.996251
0.3393	0.6036	23.402	7.940299
0.445	0.7188	20.6984	9.210788
0.5435	0.8019	18.592	10.10475
0.7284	0.9105	15.496	11.28729
1	1	12.295	12.295

The graph between the mole fraction in the liquid phase, $x_{\text{A}}$ and the partial pressure, $p_{\text{A}}$ is given below.

Figure 1

In the above graph, when $x_{\text{A}}$ is $0.0898$, $p_{\text{A}}$ is $\text{1}\text{.398961}\text{kPa}$.

According to Henry’s law, the partial pressure of methylbenzene (A)  is given as,

  $p_{\text{A}}=x_{\text{A}}{K}_{\text{H}}$                                                                                             (2)

Where,

• $p_{\text{A}}$ is the partial pressure of methylbenzene (A).
• $K_{\text{H}}$ is Henry’s law constant.
• $x_{\text{A}}$ is the mole fraction of methylbenzene (A) in the liquid phase.

Substitute the value of $x_{\text{A}}$ and $p_{\text{A}}$ in equation (2).

    $\begin{array}{c}\text{1}\text{.398961}\text{kPa}=0.0898\times K_{\text{H}}\\ K_{\text{H}}=\frac{\text{1}\text{.398961}\text{kPa}}{0.0898}\\ =\underset{\_}{15.578kPa}\end{array}$

Therefore, the value of Henry’s law constant for methylbenzene (A) is $\underset{\_}{15.578kPa}$.

The total mole fraction in the vapour phase equal to $1$.

    $\begin{array}{c}{y}_{\text{A}}+y_{\text{B}}=1\\ y_{\text{B}}=1-y_{\text{A}}\end{array}$                                                                                              (3)

Where,

• $y_{\text{B}}$ is the mole fraction of butanone in the vapour phase.

The mole fraction of methylbenzene (A) in the vapour phase $\left(y_{\text{A}}\right)$ is $0.0410$.

Substitute the value of $y_{\text{A}}$ in equation (3).

    $\begin{array}{c}{y}_{\text{B}}=1-0.0410\\ =0.959\end{array}$

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

    $p_{\text{B}}=y_{\text{B}}p$                                                                                                        (4)

Where,

• $y_{\text{B}}$ is the mole fraction of butanone (B) in the vapour phase.
• $p$ is the total vapour pressure.

The total pressure $\left(p\right)$ at $y_{\text{B}}=0.959$ is $\text{34}\text{.121}\text{kPa}$.

Substitute the value of $p$ and $y_{\text{B}}$ in equation (4).

    $\begin{array}{c}{p}_{\text{B}}=0.959\times \text{34}\text{.121}\text{kPa}\\ =\underset{\_}{32.722kPa}\end{array}$

Therefore, the partial pressure of butanone (B) $\left(p_{\text{B}}\right)$ is calculated for all vapour phase composition $\left(y_{\text{B}}\right)$ in the table below.

$y_{\text{B}}$	$x_{\text{B}}$	$p/\text{kPa}$	$p_{\text{B}}/\text{kPa}$
1	1	36.066	36.066
0.959	0.9102	34.121	32.72204
0.8846	0.7524	30.9	27.33414
0.8238	0.6423	28.626	23.5821
0.7228	0.4806	25.239	18.24275
0.6607	0.3964	23.402	15.4617
0.555	0.2812	20.6984	11.48761
0.4565	0.1981	18.592	8.487248
0.2716	0.0895	15.496	4.208714
0	0	12.295	0

The graph between the mole fraction in the liquid phase, $x_{\text{B}}$ and the partial pressure, $p_{\text{B}}$ is given below.

Figure 2

In the above graph, when $x_{\text{B}}$ is $0.0895$, $p_{\text{B}}$ is $4.208714\text{kPa}$.

According to Henry’s law, the partial pressure of butanone (B) is given as,

  $p_{\text{B}}=x_{\text{B}}{K}_{\text{H}}$                                                                                              (5)

Where,

• $p_{\text{B}}$ is the partial pressure of butanone (B).
• $K_{\text{H}}$ is Henry’s law constant.
• $x_{\text{B}}$ is the mole fraction of butanone (B) in the liquid phase.

Substitute the value of $x_{\text{B}}$ and $p_{\text{B}}$ in equation (5).

    $\begin{array}{c}4.208714\text{kPa}\text{kPa}=0.0895\times K_{\text{H}}\\ K_{\text{H}}=\frac{4.208714\text{kPa}}{0.0895}\\ =\underset{\_}{47.025kPa}\end{array}$

Therefore, the value of Henry’s law constant for butanone is $\underset{\_}{47.025kPa}$.