Chapter 5, Problem 5.19P

(a)

Interpretation Introduction

Interpretation:

The equilibrium constant of dissociation $N{H}_{4}Cl$ has to be calculated.

Concept Introduction:

The equilibrium constant is given by,

  $\begin{array}{l}AB\stackrel{}{\to }C+D\\ \\ K=\frac{[C][D]}{[AB]}\end{array}$

Explanation of Solution

Given,

The dissociation vapour of $N{H}_{4}Cl$ at $427°C$ is $608kPa$ and at $459°C$ is $1115kPa$.

The balanced dissociation reaction of $N{H}_{4}Cl$ is,

  $N{H}_{4}Cl{}_{(s|)}\stackrel{}{\to }N{H}_{3(g)}+HC{l}_{(g)}$

From the above reaction the equilibrium constant is,

    $Kp=\frac{P_{N{H}_3}{P}_{HCl}}{P_{N{H}_{4}Cl}}$

Solidi $N{H}_{4}Cl$ is taking as 1 so the equilibrium constant is,

  $\begin{array}{l}Kp=P_{N{H}_3}{P}_{HCl}\\ Kp=\frac{P_{N{H}_3}}{P^o}\frac{P_{HCl}}{P^o}=\frac{P_{N{H}_3}{}^2}{P^o}=\frac{1}{4}\times {\left(\frac{P}{P^o}\right)}^2\end{array}$

The equilibrium constantat $427°C:$

    $K=\frac{1}{4}\times {\left(\frac{608KPa}{100KPa}\right)}^2=9.24160$

The equilibrium constantat $459°C:$

    $K=\frac{1}{4}\times {\left(\frac{1115KPa}{100KPa}\right)}^2=31.08063$

(b)

Interpretation Introduction

Interpretation:

The standard reaction Gibbs energy of dissociation $N{H}_{4}Cl$ has to be calculated.

Concept Introduction:

The equilibrium constant is given by,

  $\begin{array}{l}AB\stackrel{}{\to }C+D\\ \\ K=\frac{[C][D]}{[AB]}\end{array}$

Explanation of Solution

Given,

The dissociation vapour of $N{H}_{4}Cl$ at $427°C$ is $608kPa$ and at $459°C$ is $1115kPa$.

The balanced dissociation reaction of $N{H}_{4}Cl$ is,

  $N{H}_{4}Cl{}_{(s|)}\stackrel{}{\to }N{H}_{3(g)}+HC{l}_{(g)}$

From the above reaction the equilibrium constant is,

    $Kp=\frac{P_{N{H}_3}{P}_{HCl}}{P_{N{H}_{4}Cl}}$

Solid $N{H}_{4}Cl$ is taking as 1 so the equilibrium constant is,

  $\begin{array}{l}Kp=P_{N{H}_3}{P}_{HCl}\\ Kp=\frac{P_{N{H}_3}}{P^o}\frac{P_{HCl}}{P^o}=\frac{P_{N{H}_3}{}^2}{P^o}=\frac{1}{4}\times {\left(\frac{P}{P^o}\right)}^2\end{array}$

The equilibrium constantat $427°C:$

    $K=\frac{1}{4}\times {\left(\frac{608KPa}{100KPa}\right)}^2=9.24160$

The equilibrium constantat $459°C:$

    $K=\frac{1}{4}\times {\left(\frac{1115KPa}{100KPa}\right)}^2=31.08063$

Gibbs energy of dissociation $N{H}_{4}Cl$ at $427°C:$

    $\begin{array}{l}\Delta G=-RT\ln K\\ =-12.945KJmo{l}^{-1}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The standard reaction enthalpy of dissociation $N{H}_{4}Cl$ has to be calculated.

Concept Introduction:

The equilibrium constant is given by,

  $\begin{array}{l}AB\stackrel{}{\to }C+D\\ \\ K=\frac{[C][D]}{[AB]}\end{array}$

Explanation of Solution

Given,

The dissociation vapour of $N{H}_{4}Cl$ at $427°C$ is $608kPa$ and at $459°C$ is $1115kPa$.

The balanced dissociation reaction of $N{H}_{4}Cl$ is,

  $N{H}_{4}Cl{}_{(s|)}\stackrel{}{\to }N{H}_{3(g)}+HC{l}_{(g)}$

From the above reaction the equilibrium constant is,

    $Kp=\frac{P_{N{H}_3}{P}_{HCl}}{P_{N{H}_{4}Cl}}$

Solidi $N{H}_{4}Cl$ is taking as 1 so the equilibrium constant is,

  $\begin{array}{l}Kp=P_{N{H}_3}{P}_{HCl}\\ Kp=\frac{P_{N{H}_3}}{P^o}\frac{P_{HCl}}{P^o}=\frac{P_{N{H}_3}{}^2}{P^o}=\frac{1}{4}\times {\left(\frac{P}{P^o}\right)}^2\end{array}$

The equilibrium constantat $427°C:$

    $K=\frac{1}{4}\times {\left(\frac{608KPa}{100KPa}\right)}^2=9.24160$

The equilibrium constantat $459°C:$

    $K=\frac{1}{4}\times {\left(\frac{1115KPa}{100KPa}\right)}^2=31.08063$

Gibbs energy of dissociation $N{H}_{4}Cl$ at $427°C:$

    $\begin{array}{l}\Delta G=-RT\ln K\\ =-12.945KJmo{l}^{-1}\end{array}$

$\Delta H$ doesn’t vary with temperature so plug the calculated values in beloe equation to get $\Delta H$.

    $\begin{array}{l}\ln K_2-\ln K_1=\frac{\Delta H^o}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)\\ \Delta H^o=161.54KJmo{l}^{-1}\end{array}$

(d)

Interpretation Introduction

Interpretation:

The standard reaction entropy of dissociation $N{H}_{4}Cl$ has to be calculated.

Concept Introduction:

The equilibrium constant is given by,

  $\begin{array}{l}AB\stackrel{}{\to }C+D\\ \\ K=\frac{[C][D]}{[AB]}\end{array}$

Explanation of Solution

Given,

The dissociation vapour of $N{H}_{4}Cl$ at $427°C$ is $608kPa$ and at $459°C$ is $1115kPa$.

The balanced dissociation reaction of $N{H}_{4}Cl$ is,

  $N{H}_{4}Cl{}_{(s|)}\stackrel{}{\to }N{H}_{3(g)}+HC{l}_{(g)}$

From the above reaction the equilibrium constant is,

    $Kp=\frac{P_{N{H}_3}{P}_{HCl}}{P_{N{H}_{4}Cl}}$

Solidi $N{H}_{4}Cl$ is taking as 1 so the equilibrium constant is,

  $\begin{array}{l}Kp=P_{N{H}_3}{P}_{HCl}\\ Kp=\frac{P_{N{H}_3}}{P^o}\frac{P_{HCl}}{P^o}=\frac{P_{N{H}_3}{}^2}{P^o}=\frac{1}{4}\times {\left(\frac{P}{P^o}\right)}^2\end{array}$

The equilibrium constantat $427°C:$

    $K=\frac{1}{4}\times {\left(\frac{608KPa}{100KPa}\right)}^2=9.24160$

The equilibrium constantat $459°C:$

    $K=\frac{1}{4}\times {\left(\frac{1115KPa}{100KPa}\right)}^2=31.08063$

Gibbs energy of dissociation $N{H}_{4}Cl$ at $427°C:$

    $\begin{array}{l}\Delta G=-RT\ln K\\ =-12.945KJmo{l}^{-1}\end{array}$

$\Delta H$ doesn’t vary with temperature so plug the calculated values in beloe equation to get $\Delta H$.

    $\begin{array}{l}\ln K_2-\ln K_1=\frac{\Delta H^o}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)\\ \Delta H^o=161.54KJmo{l}^{-1}\end{array}$

The calculated $\Delta H$ and $\Delta G$ are plugged in below equation to get $\Delta S$ of a reaction.

    $\begin{array}{l}\Delta G=\Delta H-\Delta S\\ \Delta S=249.21J{K}^{-1}mo{l}^{-1}\end{array}$