Chapter 16, Problem 74E

The Ostwald process for the commercial production of nitric acid involves three steps:

$\begin{array}{l}4{\text{NH}}_{\text{3}}\left(g\right)+5{\text{O}}_{\text{2}}\left(g\right)\underset{{825}^{\circ }\text{C}}{\overset{\text{Pt}}{\to }}4\text{NO}\left(g\right)+6{\text{H}}_{\text{2}}\text{O}\left(g\right)\\ 2\text{NO}\left(g\right)+{\text{O}}_{\text{2}}\left(g\right)\to 2{\text{NO}}_{\text{2}}\left(g\right)\\ 3{\text{NO}}_{\text{2}}\left(g\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\to 2{\text{HNO}}_{\text{3}}\left(l\right)+\text{NO}\left(g\right)\end{array}$

a. Calculate ∆H°, ∆S°,∆G° and K (at 298 K) for each of the three steps in the Ostwald process (see Appendix 4).

b. Calculate the equilibrium constant for the first step at 825°C, assuming ∆H° and ∆S° do not depend on temperature.

c. Is there a thermodynamic reason for the high temperature in the first step, assuming standard conditions?

Interpretation Introduction

Interpretation: The values of $\Delta H^{\text{ο}},\Delta S^{\text{ο}},\Delta G^{\text{ο}}$ and $K$ at $298\text{K}$ for each of the three steps in the Ostwald process are to be calculated. The equilibrium constant for the first step at $825°\text{C}$ is to be calculated. The independency of $\Delta H^{\text{ο}}$ and $\Delta S^{\text{ο}}$ on the temperature is to be assumed. The reason for the high temperature in the first step is to be stated.

Concept introduction: The term $\Delta G^{\text{ο}}$ is a thermodynamic function. The superscript on this function represents its standard form. The term $\Delta$ represents the change. This function is known as the standard Gibb’s free energy change. It correlates the enthalpy and entropy of the system in a mathematical formula.

Explanation of Solution

Given

The reactions are given as,

$\begin{array}{l}4{\text{NH}}_{\text{3}}\left(g\right)+5{\text{O}}_2\left(g\right)\underset{825°\text{C}}{\overset{\text{Pt}}{\to }}4\text{NO}\left(g\right)+{\text{H}}_2\text{O}\left(g\right)\\ 4\text{NO}\left(g\right)+{\text{O}}_2\left(g\right)\to 2{\text{NO}}_2\left(g\right)\\ {\text{3NO}}_2\left(g\right)+{\text{H}}_2\text{O}\left(l\right)\to 2{\text{HNO}}_3\left(l\right)+\text{NO}\left(g\right)\end{array}$

The standard values of $\Delta H_{\text{f}}^{\text{ο}},\Delta S_{\text{f}}^{\text{ο}}$ and $\Delta G_{\text{f}}^{\text{ο}}$ given in appendix 4 are as follows,

$\begin{array}{cccc}\text{Substance}& \Delta H_{\text{f}}^{\text{ο}}\left(\text{kJ/mol}\right)& \Delta S_{\text{f}}^{\text{ο}}\left(\text{J/K}\cdot \text{mol}\right)& \Delta G_{\text{f}}^{\text{ο}}\left(\text{kJ/mol}\right)\\ {\text{NH}}_{\text{3}}\left(g\right)& -46& -17& 193\\ {\text{O}}_2\left(g\right)& 0& 0& 205\\ \text{NO}\left(g\right)& 90& 87& 211\\ {\text{H}}_2\text{O}\left(g\right)& -242& -229& 189\\ {\text{NO}}_2\left(g\right)& 34& 52& 240\\ {\text{H}}_2\text{O}\left(l\right)& -286& -237& 70\\ {\text{HNO}}_3\left(l\right)& -174& -81& 156\end{array}$

The value of standard enthalpy change $\Delta H^{\text{ο}}$ of the given reaction is calculated by the formula,

$\Delta H^{\text{ο}}={{\displaystyle \sum n_{\text{p}}\Delta H_{\text{f}}^{\text{ο}}}}_{\left(\text{products}\right)}-{{\displaystyle \sum n_{\text{r}}\Delta H_{\text{f}}^{\text{ο}}}}_{\left(\text{reactants}\right)}$

Where,

• $\Delta H_{\text{f}}^{\text{ο}}{}_{\left(\text{reactants}\right)}$ are the standard enthalpy of formation for the reactants.
• $\Delta H_{\text{f}}^{\text{ο}}{}_{\left(\text{products}\right)}$ are the standard enthalpy of formation for the products.
• $n_{\text{p}}$ is the number of products molecule.
• $n_{\text{r}}$ is the number of reactants molecule.
• ${\displaystyle \sum }$ is the symbol of summation.

For the first reaction the representation in the above form is written as,

$\begin{array}{c}\Delta H^{\text{ο}}={{\displaystyle \sum n_{\text{p}}\Delta H_{\text{f}}^{\text{ο}}}}_{\left(\text{products}\right)}-{{\displaystyle \sum n_{\text{r}}\Delta H_{\text{f}}^{\text{ο}}}}_{\left(\text{reactants}\right)}\\ \Delta H^{\text{ο}}=\left[4\Delta H_{\text{f}}^{\text{ο}}{}_{\text{NO}\left(g\right)}+6\Delta H_{\text{f}}^{\text{ο}}{}_{{\text{H}}_2\text{O}\left(g\right)}\right]-\left[4\Delta H_{\text{f}}^{\text{ο}}{}_{{\text{NH}}_3\left(g\right)}+5\Delta H_{\text{f}}^{\text{ο}}{}_{{\text{O}}_2\left(g\right)}\right]\end{array}$

Substitute the value of standard enthalpy of formations in the above equation.

$\begin{array}{l}\Delta H^{\text{ο}}=\left[4\Delta H_{\text{f}}^{\text{ο}}{}_{\text{NO}\left(g\right)}+6\Delta H_{\text{f}}^{\text{ο}}{}_{{\text{H}}_2\text{O}\left(g\right)}\right]-\left[4\Delta H_{\text{f}}^{\text{ο}}{}_{{\text{NH}}_3\left(g\right)}+5\Delta H_{\text{f}}^{\text{ο}}{}_{{\text{O}}_2\left(g\right)}\right]\\ \Delta H^{\text{ο}}=\left[\begin{array}{c}\left[4\text{mol}\left(90\text{kJ/mol}\right)+6\text{mol}\left(-243\text{kJ/mol}\right)\right]-\\ \left[4\text{mol}\times \left(-46\text{kJ/mol}\right)+5\text{mol}\times \left(0\text{kJ/mol}\right)\right]\end{array}\right]\\ \Delta H^{\text{ο}}=\left(-1092\text{kJ}\right)+\left(184\text{kJ}\right)\\ \Delta H^{\text{ο}}=\underset{\_}{-908kJ}\end{array}$

For the second reaction the value of standard enthalpy of formation is calculated as,

$\begin{array}{l}\Delta H^{\text{ο}}=\left[2\Delta H_{\text{f}}^{\text{ο}}{}_{{\text{NO}}_2\left(g\right)}\right]-\left[2\Delta H_{\text{f}}^{\text{ο}}{}_{\text{NO}\left(g\right)}+\Delta H_{\text{f}}^{\text{ο}}{}_{{\text{O}}_2\left(g\right)}\right]\\ \Delta H^{\text{ο}}=\left[2\text{mol}\left(34\text{kJ/mol}\right)\right]-\left[\left(2\text{mol}\left(90\text{kJ/mol}\right)\right)+1\text{mol}\left(0\text{kJ/mol}\right)\right]\\ \Delta H^{\text{ο}}=\left(68\text{kJ}\right)-\left(180\text{kJ}\right)\\ \Delta H^{\text{ο}}=\underset{\_}{-112kJ}\end{array}$

For the third reaction the value of standard enthalpy of formation is calculated as,

$\begin{array}{l}\Delta H^{\text{ο}}=\left[2\Delta H_{\text{f}}^{\text{ο}}{}_{{\text{HNO}}_3\left(g\right)}+\Delta H_{\text{f}}^{\text{ο}}{}_{\text{NO}\left(g\right)}\right]-\left[3\Delta H_{\text{f}}^{\text{ο}}{}_{{\text{NO}}_2\left(g\right)}+5\Delta H_{\text{f}}^{\text{ο}}{}_{{\text{HO}}_2\left(l\right)}\right]\\ \Delta H^{\text{ο}}=\left[\begin{array}{c}\left[2\text{mol}\left(-174\text{kJ/mol}\right)+1\text{mol}\left(-90\text{kJ/mol}\right)\right]-\\ \left[3\text{mol}\times \left(34\text{kJ/mol}\right)+1\text{mol}\times \left(-286\text{kJ/mol}\right)\right]\end{array}\right]\\ \Delta H^{\text{ο}}=\left(-258\text{kJ}\right)+\left(184\text{kJ}\right)\\ \Delta H^{\text{ο}}=\underset{\_}{-74kJ}\end{array}$

The calculation of standard entropy change for both the reactions is given below.

The value of standard entropy change $\Delta S^{\text{ο}}$ of the given reaction is calculated by the formula,

$\Delta S^{\text{ο}}={{\displaystyle \sum n_{\text{p}}\Delta S_{\text{f}}^{\text{ο}}}}_{\left(\text{products}\right)}-{{\displaystyle \sum n_{\text{r}}\Delta S_{\text{f}}^{\text{ο}}}}_{\left(\text{reactants}\right)}$

Where,

• $\Delta S_{\text{f}}^{\text{ο}}{}_{\left(\text{reactants}\right)}$ are the standard entropy of formation for the reactants.
• $\Delta S_{\text{f}}^{\text{ο}}{}_{\left(\text{products}\right)}$ are the standard entropy of formation for the products.
• $n_{\text{p}}$ is the number of products molecule.
• $n_{\text{r}}$ is the number of reactants molecule.
• ${\displaystyle \sum }$ is the symbol of summation.

For the first reaction the representation in the above form is written as,

$\begin{array}{c}\Delta S^{\text{ο}}={{\displaystyle \sum n_{\text{p}}\Delta S_{\text{f}}^{\text{ο}}}}_{\left(\text{products}\right)}-{{\displaystyle \sum n_{\text{r}}\Delta S_{\text{f}}^{\text{ο}}}}_{\left(\text{reactants}\right)}\\ \Delta S^{\text{ο}}=\left[4\Delta S_{\text{f}}^{\text{ο}}{}_{\text{NO}\left(g\right)}+6\Delta S_{\text{f}}^{\text{ο}}{}_{{\text{H}}_2\text{O}\left(g\right)}\right]-\left[4\Delta S_{\text{f}}^{\text{ο}}{}_{{\text{NH}}_3\left(g\right)}+5\Delta S_{\text{f}}^{\text{ο}}{}_{{\text{O}}_2\left(g\right)}\right]\end{array}$

Substitute the values of standard entropy of formations in the above equation.

$\begin{array}{c}\Delta S^{\text{ο}}=\left[4\Delta S_{\text{f}}^{\text{ο}}{}_{\text{NO}\left(g\right)}+6\Delta S_{\text{f}}^{\text{ο}}{}_{{\text{H}}_2\text{O}\left(g\right)}\right]-\left[4\Delta S_{\text{f}}^{\text{ο}}{}_{{\text{NH}}_3\left(g\right)}+5\Delta S_{\text{f}}^{\text{ο}}{}_{{\text{O}}_2\left(g\right)}\right]\\ \Delta S^{\text{ο}}=\left[\begin{array}{l}\left[4\text{mol}\left(211\text{J/K}\cdot \text{mol}\right)+6\text{mol}\left(189\text{J/K}\cdot \text{mol}\right)\right]-\\ \left[4\text{mol}\times \left(193\text{J/K}\cdot \text{mol}\right)+5\text{mol}\times \left(205\text{J/K}\cdot \text{mol}\right)\right]\end{array}\right]\\ \Delta S^{\text{ο}}=\left(1978\text{J/K}\cdot \text{mol}\right)-\left(1797\text{J/K}\cdot \text{mol}\right)\\ \Delta S^{\text{ο}}=\underset{\_}{181J/K}\end{array}$

For the second reaction the representation in the above form is written as,

$\begin{array}{c}\Delta S^{\text{ο}}={{\displaystyle \sum n_{\text{p}}\Delta S_{\text{f}}^{\text{ο}}}}_{\left(\text{products}\right)}-{{\displaystyle \sum n_{\text{r}}\Delta S_{\text{f}}^{\text{ο}}}}_{\left(\text{reactants}\right)}\\ \Delta S^{\text{ο}}=\left[2\Delta S_{\text{f}}^{\text{ο}}{}_{{\text{NO}}_2\left(g\right)}\right]-\left[2\Delta S_{\text{f}}^{\text{ο}}{}_{\text{NO}\left(g\right)}+\Delta S_{\text{f}}^{\text{ο}}{}_{{\text{O}}_2\left(g\right)}\right]\end{array}$

Substitute the values of standard entropy of formations in the above equation.

$\begin{array}{l}\Delta S^{\text{ο}}=\left[2\Delta S_{\text{f}}^{\text{ο}}{}_{{\text{NO}}_2\left(g\right)}\right]-\left[2\Delta S_{\text{f}}^{\text{ο}}{}_{\text{NO}\left(g\right)}+\Delta S_{\text{f}}^{\text{ο}}{}_{{\text{O}}_2\left(g\right)}\right]\\ \Delta S^{\text{ο}}=\left[2\text{mol}\left(240\text{J/K}\cdot \text{mol}\right)\right]-\left[2\text{mol}\left(211\text{J/K}\cdot \text{mol}\right)+1\text{mol}\left(205\text{J/K}\cdot \text{mol}\right)\right]\\ \Delta S^{\text{ο}}=\left[\left(480\text{J/K}\right)\right]-\left[\left(422\text{J/K}\right)+\left(205\text{J/K}\right)\right]\\ \Delta S^{\text{ο}}=\underset{\_}{-147J/K}\end{array}$

For the third reaction the representation in the above form is written as,

$\begin{array}{l}\Delta S^{\text{ο}}={{\displaystyle \sum n_{\text{p}}\Delta S_{\text{f}}^{\text{ο}}}}_{\left(\text{products}\right)}-{{\displaystyle \sum n_{\text{r}}\Delta S_{\text{f}}^{\text{ο}}}}_{\left(\text{reactants}\right)}\\ \Delta S^{\text{ο}}=\left[2\Delta S_{\text{f}}^{\text{ο}}{}_{{\text{HNO}}_3\left(l\right)}+\Delta S_{\text{f}}^{\text{ο}}{}_{\text{NO}\left(g\right)}\right]-\left[3\Delta S_{\text{f}}^{\text{ο}}{}_{{\text{NO}}_2\left(g\right)}+\Delta S_{\text{f}}^{\text{ο}}{}_{{\text{HO}}_2\left(l\right)}\right]\end{array}$

Substitute the values of standard entropy of formations in the above equation.

$\begin{array}{l}\Delta S^{\text{ο}}=\left[2\Delta S_{\text{f}}^{\text{ο}}{}_{{\text{HNO}}_3\left(l\right)}+\Delta S_{\text{f}}^{\text{ο}}{}_{\text{NO}\left(g\right)}\right]-\left[3\Delta S_{\text{f}}^{\text{ο}}{}_{{\text{NO}}_2\left(g\right)}+\Delta S_{\text{f}}^{\text{ο}}{}_{{\text{HO}}_2\left(l\right)}\right]\\ \Delta S^{\text{ο}}=\left[\begin{array}{l}\left[2\text{mol}\left(156\text{J/K}\cdot \text{mol}\right)+1\text{mol}\left(211\text{J/K}\cdot \text{mol}\right)\right]-\\ \left[3\text{mol}\left(240\text{J/K}\cdot \text{mol}\right)+1\text{mol}\left(70\text{J/K}\cdot \text{mol}\right)\right]\end{array}\right]\\ \Delta S^{\text{ο}}=\left[\left(523\text{J/K}\right)\right]-\left[\left(790\text{J/K}\right)\right]\\ \Delta S^{\text{ο}}=\underset{\_}{-267J/K}\end{array}$

The value of standard Gibb’s free energy $\Delta G^{\text{ο}}$ of the given reaction is calculated by the formula,

$\Delta G^{\text{ο}}={{\displaystyle \sum n_{\text{p}}\Delta G_{\text{f}}^{\text{ο}}}}_{\left(\text{products}\right)}-{{\displaystyle \sum n_{\text{r}}\Delta G_{\text{f}}^{\text{ο}}}}_{\left(\text{reactants}\right)}$

Where,

• $\Delta G_{\text{f}}^{\text{ο}}{}_{\left(\text{reactants}\right)}$ are the standard free energy of formation for the reactants.
• $\Delta G_{\text{f}}^{\text{ο}}{}_{\left(\text{products}\right)}$ are the standard free energy of formation for the products.
• $n_{\text{p}}$ is the number of products molecule.
• $n_{\text{r}}$ is the number of reactants molecule.
• ${\displaystyle \sum }$ is the symbol of summation.

For the first reaction the representation in the above form is written as,

$\begin{array}{c}\Delta G^{\text{ο}}={{\displaystyle \sum n_{\text{p}}\Delta G_{\text{f}}^{\text{ο}}}}_{\left(\text{products}\right)}-{{\displaystyle \sum n_{\text{r}}\Delta G_{\text{f}}^{\text{ο}}}}_{\left(\text{reactants}\right)}\\ \Delta G^{\text{ο}}=\left[4\Delta G_{\text{f}}^{\text{ο}}{}_{\text{NO}\left(g\right)}+6\Delta G_{\text{f}}^{\text{ο}}{}_{{\text{H}}_2\text{O}\left(g\right)}\right]-\left[4\Delta G_{\text{f}}^{\text{ο}}{}_{{\text{NH}}_3\left(g\right)}+5\Delta G_{\text{f}}^{\text{ο}}{}_{{\text{O}}_2\left(g\right)}\right]\end{array}$

Substitute the values of Gibb’s energy of formations in the above equation.

$\begin{array}{l}\Delta G^{\text{ο}}=\left[4\Delta G_{\text{f}}^{\text{ο}}{}_{\text{NO}\left(g\right)}+6\Delta G_{\text{f}}^{\text{ο}}{}_{{\text{H}}_2\text{O}\left(g\right)}\right]-\left[4\Delta G_{\text{f}}^{\text{ο}}{}_{{\text{NH}}_3\left(g\right)}+5\Delta G_{\text{f}}^{\text{ο}}{}_{{\text{O}}_2\left(g\right)}\right]\\ \Delta G^{\text{ο}}=\left[\begin{array}{c}\left[4\text{mol}\left(87\text{kJ/mol}\right)+6\text{mol}\left(-229\text{kJ/mol}\right)\right]-\\ \left[4\text{mol}\times \left(-17\text{kJ/mol}\right)+5\text{mol}\times \left(0\text{kJ/mol}\right)\right]\end{array}\right]\\ \Delta G^{\text{ο}}=\left(-1026\text{kJ}\right)+\left(68\text{kJ}\right)\\ \Delta G^{\text{ο}}=\underset{\_}{-958kJ}\end{array}$

For the second reaction the representation in the above form is written as,

$\begin{array}{c}\Delta G^{\text{ο}}={{\displaystyle \sum n_{\text{p}}\Delta G_{\text{f}}^{\text{ο}}}}_{\left(\text{products}\right)}-{{\displaystyle \sum n_{\text{r}}\Delta G_{\text{f}}^{\text{ο}}}}_{\left(\text{reactants}\right)}\\ \Delta G^{\text{ο}}=\left[2\Delta G_{\text{f}}^{\text{ο}}{}_{{\text{NO}}_2\left(g\right)}\right]-\left[2\Delta G_{\text{f}}^{\text{ο}}{}_{\text{NO}\left(g\right)}+\Delta G_{\text{f}}^{\text{ο}}{}_{{\text{O}}_2\left(g\right)}\right]\end{array}$

Substitute the values of Gibb’s energy of formations in the above equation.

$\begin{array}{l}\Delta G^{\text{ο}}=\left[2\Delta G_{\text{f}}^{\text{ο}}{}_{{\text{NO}}_2\left(g\right)}\right]-\left[2\Delta G_{\text{f}}^{\text{ο}}{}_{\text{NO}\left(g\right)}+\Delta G_{\text{f}}^{\text{ο}}{}_{{\text{O}}_2\left(g\right)}\right]\\ \Delta G^{\text{ο}}=\left[2\text{mol}\left(52\text{kJ/mol}\right)\right]-\left[\left(2\text{mol}\left(87\text{kJ/mol}\right)\right)+1\text{mol}\left(0\text{kJ/mol}\right)\right]\\ \Delta G^{\text{ο}}=\left(104\text{kJ}\right)-\left(174\text{kJ}\right)\\ \Delta G^{\text{ο}}=\underset{\_}{-70kJ}\end{array}$

For the third reaction the representation in the above form is written as,

$\begin{array}{l}\Delta G^{\text{ο}}={{\displaystyle \sum n_{\text{p}}\Delta G_{\text{f}}^{\text{ο}}}}_{\left(\text{products}\right)}-{{\displaystyle \sum n_{\text{r}}\Delta G_{\text{f}}^{\text{ο}}}}_{\left(\text{reactants}\right)}\\ \Delta G^{\text{ο}}=\left[2\Delta G_{\text{f}}^{\text{ο}}{}_{{\text{HNO}}_3\left(l\right)}+\Delta G_{\text{f}}^{\text{ο}}{}_{\text{NO}\left(g\right)}\right]-\left[3\Delta G_{\text{f}}^{\text{ο}}{}_{{\text{NO}}_2\left(g\right)}+\Delta G_{\text{f}}^{\text{ο}}{}_{{\text{HO}}_2\left(l\right)}\right]\end{array}$

Substitute the values of Gibb’s energy of formations in the above equation.

$\begin{array}{l}\Delta G^{\text{ο}}=\left[2\Delta G_{\text{f}}^{\text{ο}}{}_{{\text{HNO}}_3\left(g\right)}+\Delta G_{\text{f}}^{\text{ο}}{}_{\text{NO}\left(g\right)}\right]-\left[3\Delta G_{\text{f}}^{\text{ο}}{}_{{\text{NO}}_2\left(g\right)}+5\Delta G_{\text{f}}^{\text{ο}}{}_{{\text{HO}}_2\left(g\right)}\right]\\ \Delta G^{\text{ο}}=\left[\begin{array}{c}\left[2\text{mol}\left(-81\text{kJ/mol}\right)+1\text{mol}\left(87\text{kJ/mol}\right)\right]-\\ \left[3\text{mol}\times \left(52\text{kJ/mol}\right)+1\text{mol}\times \left(-237\text{kJ/mol}\right)\right]\end{array}\right]\\ \Delta G^{\text{ο}}=\left(-75\text{kJ}\right)-\left(-81\text{kJ}\right)\\ \Delta G^{\text{ο}}=\underset{\_}{+6kJ}\end{array}$

The value of rate constant, $K$ is calculated by the formula,

$\begin{array}{c}\Delta G^{\text{ο}}=-\text{R}T\ln K\\ \ln K=\frac{-\Delta G^{\text{ο}}}{\text{R}T}\\ K=e^{\frac{-\Delta G^{\text{ο}}}{\text{R}T}}\end{array}$

Where,

• $\text{R}$ is gas constant.
• $T$ is the temperature in Kelvin.
• $e^{\frac{-\Delta G^{\text{ο}}}{\text{R}T}}$ is the exponential term have the power of $-\Delta G^{\text{ο}}/\text{R}T$.

For the first reaction substitute the values in the above formula.

$\begin{array}{l}K=e^{\frac{-\Delta G^{\text{ο}}}{\text{R}T}}\\ K=e^{-\left[\frac{\left(-958\times {10}^3\text{J/mol}\right)}{8.314\text{J/mol}\cdot \text{K}\times \text{298}\text{K}}\right]}\\ K={\text{e}}^{\left(\text{387}\right)}\\ K=\underset{\_}{1.18\times 10^{168}}\end{array}$

For the second reaction substitute the values in the above formula.

$\begin{array}{l}K=e^{\frac{-\Delta G^{\text{ο}}}{\text{R}T}}\\ K=e^{-\left[\frac{\left(-70\times {10}^3\text{J/mol}\right)}{8.314\text{J/mol}\cdot \text{K}\times \text{298}\text{K}}\right]}\\ K={\text{e}}^{\left(\text{30}\right)}\\ K=\underset{\_}{10^{13}}\end{array}$

For the third reaction substitute the values in the above formula.

$\begin{array}{l}K=e^{\frac{-\Delta G^{\text{ο}}}{\text{R}T}}\\ K=e^{-\left[\frac{\left(6\times {10}^3\text{J/mol}\right)}{8.314\text{J/mol}\cdot \text{K}\times \text{298}\text{K}}\right]}\\ K={\text{e}}^{\text{-}\left(\text{2}\text{.421}\right)}\\ K=\underset{\_}{0.088}\end{array}$

Interpretation Introduction

Interpretation: The values of $\Delta H^{\text{ο}},\Delta S^{\text{ο}},\Delta G^{\text{ο}}$ and $K$ at $298\text{K}$ for each of the three steps in the Ostwald process are to be calculated. The equilibrium constant for the first step at $825°\text{C}$ is to be calculated. The independency of $\Delta H^{\text{ο}}$ and $\Delta S^{\text{ο}}$ on the temperature is to be assumed. The reason for the high temperature in the first step is to be stated.

Concept introduction: The term $\Delta G^{\text{ο}}$ is a thermodynamic function. The superscript on this function represents its standard form. The term $\Delta$ represents the change. This function is known as the standard Gibb’s free energy change. It correlates the enthalpy and entropy of the system in a mathematical formula.

Explanation of Solution

Given

The entropy and enthalpy are independent of the temperature.

Temperature is $825°\text{C}\left(825+273=1098\text{K}\right)$.

The free energy change of the first reaction at the given temperature is calculated by the formula,

$\Delta G^{\text{ο}}=\Delta H^{\text{ο}}-T\Delta S^{\text{ο}}$

Substitute the values in the above equation.

$\begin{array}{c}\Delta G^{\text{ο}}=\Delta H^{\text{ο}}-T\Delta S^{\text{ο}}\\ =\left(-908000\text{J}\right)-\left(1098\text{K}\right)\left(181\text{J/K}\right)\\ =\left(-1106735\text{J}\right)\end{array}$

The equilibrium constant for the first step is calculated as,

$\begin{array}{l}K=e^{\frac{-\Delta G^{\text{ο}}}{\text{R}T}}\\ K=e^{-\left[\frac{\left(-1106738\text{J/mol}\right)}{8.314\text{J/mol}\cdot \text{K}\times 10\text{98}\text{K}}\right]}\\ K={\text{e}}^{\left(\text{121}\text{.23}\right)}\\ K=\underset{\_}{4.7\times 10^{52}}\end{array}$

Interpretation Introduction

Interpretation: The values of $\Delta H^{\text{ο}},\Delta S^{\text{ο}},\Delta G^{\text{ο}}$ and $K$ at $298\text{K}$ for each of the three steps in the Ostwald process are to be calculated. The equilibrium constant for the first step at $825°\text{C}$ is to be calculated. The independency of $\Delta H^{\text{ο}}$ and $\Delta S^{\text{ο}}$ on the temperature is to be assumed. The reason for the high temperature in the first step is to be stated.

Concept introduction: The term $\Delta G^{\text{ο}}$ is a thermodynamic function. The superscript on this function represents its standard form. The term $\Delta$ represents the change. This function is known as the standard Gibb’s free energy change. It correlates the enthalpy and entropy of the system in a mathematical formula.

Explanation of Solution

There is no thermodynamic reason observed in the Ostwald process for the high temperature in first step. Only kinetic reason is responsible. As at the low temperature the reaction is very slow and a side reaction is observed. The side reaction is given as,

$4{\text{NH}}_{\text{3}}+6\text{NO}\to {\text{5N}}_2+6{\text{H}}_{\text{2}}\text{O}$.

This side reaction occurs at the low temperature in the first step and it decreases the percent yield of the product. Therefore, to increase the percent yield the high temperature is to be given.