Chapter 19, Problem 19C.3P

(a)

Interpretation Introduction

Interpretation:

The fraction of uncovered sites under the given condition is approximately $1/\alpha p_J$ has to be derived.

Concept Introduction:

According to the Langmuir isotherm for a surface-catalyzed unimolecular reaction, the rate is

  Rate $=k_r\theta =\frac{k_r\alpha p}{1+\alpha p}$

Where, $\alpha =\frac{k_{ad}}{k_d}$

Explanation of Solution

A surface-catalyzed unimolecular reaction is one in which an adsorbed molecule undergoes decomposition on a surface.  The given scenario in the question is a surface-catalyzed unimolecular reaction.  Hence, the Langmuir isotherm can be applied to the given scenario.

According to the question, when a gas J adsorbs very strongly and its pressure is $p_J$, the fraction of uncovered sites will be

  $\begin{array}{c}\theta =\frac{\alpha p_J}{1+\alpha p_J}\\ \\ 1-\theta =\frac{1}{1+\alpha p{J}_{}}\end{array}$

Where,

$\theta =$ Fraction of covered sites

$1-\theta =$ Fraction of uncovered sites

For strongly adsorbed molecule, $k_{ad}\gg k_d$

  $\alpha p_J\gg 1$

So, 1 can be neglected in comparison to $\alpha p$ in the denominator.  Then the fraction of uncovered sites becomes

  $1-\theta =\frac{1}{\alpha p_J}$

(b)

Interpretation Introduction

Interpretation:

The rate of the reaction as given in the question has to be derived.

Concept Introduction:

According to the Langmuir isotherm for a surface-catalyzed unimolecular reaction, the rate is

  Rate $=k_r\theta =\frac{k_r\alpha p}{1+\alpha p}$

Where, $\alpha =\frac{k_{ad}}{k_d}$

Explanation of Solution

According to the question, when a gas J adsorbs very strongly and its pressure is $p_J$, the fraction of uncovered sites will be

  $\begin{array}{c}\theta =\frac{\alpha p_J}{1+\alpha p_J}\\ \\ 1-\theta =\frac{1}{1+\alpha p{J}_{}}\end{array}$

Where,

$\theta =$ Fraction of covered sites

$1-\theta =$ Fraction of uncovered sites

For strongly adsorbed molecule, $k_{ad}\gg k_d$

  $\alpha p_J\gg 1$

So, 1 can be neglected in comparison to $\alpha p$ in the denominator.  Then the fraction of uncovered sites becomes

  $1-\theta =\frac{1}{\alpha p_J}$

As the reaction rate is proportional to the pressure of ammonia and the fraction of uncovered sites by the strongly adsorbed hydrogen product, the rate of the reaction can be written as

  $\frac{d{p}_{N{H}_3}}{dt}=-k_c{p}_{N{H}_3}\left(1-\theta \right)=-\frac{k_c{p}_{N{H}_3}}{1+\alpha p_{H_2}}$

Hydrogen is strongly adsorbed on the surface. So 1 can be neglected in the denominator. Then the above expression becomes

  $\frac{d{p}_{N{H}_3}}{dt}=-\frac{k_c{p}_{N{H}_3}}{\alpha p_{H_2}}\simeq -k_c\frac{p_{N{H}_3}}{p_{H_2}}$

(c)

Interpretation Introduction

Interpretation:

The given equation in the question has to integrated by using the initial condition that at $t=0$ only ammonia is present at pressure $p^0$.

Explanation of Solution

As the reaction rate is proportional to the pressure of ammonia and the fraction of uncovered sites by the strongly adsorbed hydrogen product, the rate of the reaction can be written as

  $\frac{d{p}_{N{H}_3}}{dt}=-k_c{p}_{N{H}_3}\left(1-\theta \right)=-\frac{k_c{p}_{N{H}_3}}{1+\alpha p_{H_2}}$

Hydrogen is strongly adsorbed on the surface. So 1 can be neglected in the denominator. Then the above expression becomes

  $\frac{d{p}_{N{H}_3}}{dt}=-\frac{k_c{p}_{N{H}_3}}{\alpha p_{H_2}}\simeq -k_c\frac{p_{N{H}_3}}{p_{H_2}}$

The decomposition of ammonia can be written as

  $N{H}_3\left(g\right)\to \frac{1}{2}{N}_2+\frac{3}{2}{H}_2$

At $t=0$, $p_{N{H}_3\left(g\right)}=p^0$ and at $t=t$, $p_{N{H}_3\left(g\right)}=p$

Then, $p_{H_2}=\frac{3}{2}\left\{p^0-p\right\}$

The rate of the reaction can be modified like

  $-\frac{dp}{dt}=\frac{k_{c}p}{\frac{3}{2}\alpha \left\{p^0-p\right\}}=\frac{kp}{p^0-p},k=\frac{2k_c}{3\alpha }$

The integration of the above equation results

  $\begin{array}{l}{\displaystyle \underset{p^0}{\overset{p}{\int }}\left(1-\frac{p^0}{p}\right)}dp=k{\displaystyle \underset{0}{\overset{t}{\int }}dt}\\ \\ \boxed{\frac{p-p^0}{t}=k+\frac{p^0}{t}\ln \frac{p}{p^0}}\end{array}$

(d)

Interpretation Introduction

Interpretation:

The form of a suitable straight-line plot to determine $k_c$ has to be derived.

Explanation of Solution

As the reaction rate is proportional to the pressure of ammonia and the fraction of uncovered sites by the strongly adsorbed hydrogen product, the rate of the reaction can be written as

  $\frac{d{p}_{N{H}_3}}{dt}=-k_c{p}_{N{H}_3}\left(1-\theta \right)=-\frac{k_c{p}_{N{H}_3}}{1+\alpha p_{H_2}}$

Hydrogen is strongly adsorbed on the surface. So 1 can be neglected in the denominator. Then the above expression becomes

  $\frac{d{p}_{N{H}_3}}{dt}=-\frac{k_c{p}_{N{H}_3}}{\alpha p_{H_2}}\simeq -k_c\frac{p_{N{H}_3}}{p_{H_2}}$

The decomposition of ammonia can be written as

  $N{H}_3\left(g\right)\to \frac{1}{2}{N}_2+\frac{3}{2}{H}_2$

At $t=0$, $p_{N{H}_3\left(g\right)}=p^0$ and at $t=t$, $p_{N{H}_3\left(g\right)}=p$

Then, $p_{H_2}=\frac{3}{2}\left\{p^0-p\right\}$

The rate of the reaction can be modified like

  $-\frac{dp}{dt}=\frac{k_{c}p}{\frac{3}{2}\alpha \left\{p^0-p\right\}}=\frac{kp}{p^0-p},k=\frac{2k_c}{3\alpha }$

The integration of the above equation results

  $\begin{array}{l}{\displaystyle \underset{p^0}{\overset{p}{\int }}\left(1-\frac{p^0}{p}\right)}dp=k{\displaystyle \underset{0}{\overset{t}{\int }}dt}\\ \\ \boxed{\frac{p-p^0}{t}=k+\frac{p^0}{t}\ln \frac{p}{p^0}}\end{array}$

Let, $A=\frac{p^0}{t}\ln \frac{p}{p^0}andB=\frac{p-p^0}{t}$

Then, the above equation can be written in the form

  $B=k+A$

Hence, a plot of B against A should give a straight line with the intercept k. In other words, the difference $B-A$ should be a constant, k.

The required table for the plot:

  $\underset{\_}{\begin{array}{cccccccc}t/s& 0& 30& 60& 100& 160& 200& 250\\ p/kPa& 13.3& 11.7& 11.2& 10.7& 10.3& 9.9& 9.6\\ B/\left(kPa{s}^{-1}\right)& & -0.0533& -0.035& -0.026& -0.0188& -0.017& -0.015\\ A/\left(kPa{s}^{-1}\right)& & -0.0568& -0.0381& -0.0289& -0.0213& -0.0196& -0.0173\\ \left(B-A\right)/\left(kPa{s}^{-1}\right)& & 0.00349& 0.00309& 0.00293& 0.00250& 0.00263& 0.00254\end{array}}$

Therefore, the data fit the rate law and takin average of the values of $B-A$, the constant came to be $k_c=0.00286kPa{s}^{-1}$.