Chapter 22, Problem 22.47E

Interpretation Introduction

Interpretation:

The validation corresponding to the fact that data of carbon monoxide coverage and pressure follows a Langmuir isotherm is to be stated. The equilibrium constant for the adsorption process is to be stated.

Concept introduction:

The isotherm model that describes the adsorption process, if an adsorbate acts as an ideal gas at the isothermal situation is known as Langmuir adsorption isotherm model. In the given isothermal situation, the partial pressure of the adsorbate is related to the volume of the adsorbate which is adsorbed on the solid adsorbent.

Answer to Problem 22.47E

The occurrence of the linear straight line between $\text{1/Coverage}$ and $\text{1/Concentration}$ in the graph proves that the given data of carbon monoxide coverage and pressure follows a Langmuir isotherm.

The equilibrium constant for the adsorption process is $6.94\times {10}^{11}$.

Explanation of Solution

The given temperature is $\text{450}\text{K}$.

The given data of carbon monoxide coverage and pressure in torr is mentioned in the table below.

Coverage	Pressure (torr)
$0.25$ $0.45$ $0.65$ $0.72$ $0.78$ $0.82$	$1.5\times {10}^{-8}$ $7.0\times {10}^{-8}$ $3.5\times {10}^{-7}$ $8.7\times {10}^{-7}$ $1.9\times {10}^{-6}$ $6.8\times {10}^{-6}$

The concentration of carbon monoxide is calculated by the formula given below.

$\begin{array}{rll}PV& =& nRT\\ \raisebox{1ex}{$n$}\!\left/ \!\raisebox{-1ex}{$V$}\right.& =& \raisebox{1ex}{$P$}\!\left/ \!\raisebox{-1ex}{$RT$}\right.\end{array}$

Where,

• $P$ is the pressure of the gas.

• $V$ is the volume of the gas.

• $n$ is the number of moles.

• $R$ is the universal gas constant.

• $T$ is the temperature of the gas.

• $\raisebox{1ex}{$n$}\!\left/ \!\raisebox{-1ex}{$V$}\right.$ equals to the concentration of carbon monoxide.

The value of universal gas constant is $62.364\text{liter}\cdot \text{torr}{\text{K}}^{-1}\cdot \text{mol}$.

Pressure	Coverage	$\text{1/Coverage}$	$\raisebox{1ex}{$n$}\!\left/ \!\raisebox{-1ex}{$V$}\right.=\raisebox{1ex}{$P$}\!\left/ \!\raisebox{-1ex}{$RT$}\right.$	$\text{1/Concentration}$
$1.5\times {10}^{-8}$	$0.25$	$4$	$\frac{1.5\times {10}^{-8}}{\text{450}\times 62.364}=5.34496\times {10}^{-13}$	$\begin{array}{l}\frac{1}{5.34496\times {10}^{-13}}\\ =1.87092\times {10}^{12}\end{array}$
$7.0\times {10}^{-8}$	$0.45$	$2.222222$	$\frac{7.0\times {10}^{-8}}{\text{450}\times 62.364}=2.49432\times {10}^{-12}$	$\begin{array}{l}\frac{1}{2.49432\times {10}^{-12}}\\ =4.00911\times {10}^{11}\end{array}$
$3.5\times {10}^{-7}$	$0.65$	$1.538462$	$\frac{3.5\times {10}^{-7}}{\text{450}\times 62.364}=1.24716\times {10}^{-11}$	$\begin{array}{l}\frac{1}{1.24716\times {10}^{-11}}\\ =8.0182286\times {10}^{10}\end{array}$
$8.7\times {10}^{-7}$	$0.72$	$1.388889$	$\frac{8.7\times {10}^{-7}}{\text{450}\times 62.364}=3.1008\times {10}^{-11}$	$\begin{array}{l}\frac{1}{3.1008\times {10}^{-11}}\\ =3.225724\times {10}^{10}\end{array}$
$1.9\times {10}^{-6}$	$0.78$	$1.282051$	$\frac{1.9\times {10}^{-6}}{\text{450}\times 62.364}=6.77029\times {10}^{-11}$	$\begin{array}{l}\frac{1}{6.77029\times {10}^{-11}}\\ =1.477042\times {10}^{10}\end{array}$
$6.8\times {10}^{-6}$	$0.82$	$1.2195122$	$\frac{6.8\times {10}^{-6}}{\text{450}\times 62.364}=2.42305\times {10}^{-10}$	$\begin{array}{l}\frac{1}{2.42305\times {10}^{-10}}\\ =4.127029\times {10}^9\end{array}$

The graph between $\text{1/Coverage}$ and $\text{1/Concentration}$ is plotted as follows.

Figure 1

As, the plot between $\text{1/Coverage}$ and $\text{1/Concentration}$ is showing linear straight line, thus, it proves that the given data of carbon monoxide coverage and pressure follows a Langmuir isotherm.

The value of slope is predicted as $1.44116\times {10}^{-12}$.

The value of equilibrium constant is calculated by the expression given below.

$K=\frac{1}{\text{Slope}}$

Substitute the value of slope in the above expression.

$\begin{array}{rll}K& =& \frac{1}{1.44116\times {10}^{-12}}\\ & =& 6.94\times {10}^{11}\end{array}$

Therefore, the value of equilibrium constant is $6.94\times {10}^{11}$.

Conclusion

The given data of carbon monoxide coverage and pressure follows a Langmuir isotherm.

The equilibrium constant for the adsorption process is $6.94\times {10}^{11}$.