Chapter 18, Problem 18.57QP

(a)

Interpretation Introduction

Interpretation: The reactions only at low temperatures; those that are spontaneous only at high temperatures and those that are spontaneous at all temperatures are to be identified.

Concept introduction: A process is spontaneous at all temperature when the values of,

$\begin{array}{l}\Delta H_{\text{rxn}}^{\text{ο}}=\text{Negative}\\ \Delta S_{\text{rxn}}^{\text{ο}}=\text{Positive}\end{array}$

A process is spontaneous at low temperature when the values of,

$\begin{array}{l}\Delta H_{\text{rxn}}^{\text{ο}}=\text{Negative}\\ \Delta S_{\text{rxn}}^{\text{ο}}=\text{Negative}\end{array}$

A process is spontaneous at high temperature when the values of,

$\begin{array}{l}\Delta H_{\text{rxn}}^{\text{ο}}=\text{Positive}\\ \Delta S_{\text{rxn}}^{\text{ο}}=\text{Positive}\end{array}$

To determine: If the given reaction is spontaneous at low temperatures; at high temperatures or at all temperatures.

Answer to Problem 18.57QP

Solution

The given reaction is spontaneous at low temperature.

Explanation of Solution

Explanation

The given reaction is,

$2\text{NO}\left(g\right)+{\text{O}}_{\text{2}}\left(g\right)\to 2{\text{NO}}_2\left(g\right)$ (1)

The enthalpy of formation of $\text{NO}\left(g\right)$ $\left(\Delta H_{\text{f,}\text{NO}\left(g\right)}^{\text{ο}}\right)$ is $90.3\text{kJ}/\text{mol}$.

The enthalpy of formation of ${\text{NO}}_2\left(g\right)$ $\left(\Delta H_{\text{f,}{\text{NO}}_2\left(g\right)}^{\text{ο}}\right)$ is $33.2\text{kJ}/\text{mol}$.

The enthalpy of formation of ${\text{O}}_2\left(g\right)$ $\left(\Delta H_{\text{f,}{\text{O}}_2\left(g\right)}^{\text{ο}}\right)$ is $0.0\text{kJ}/\text{mol}$.

The standard molar entropy of $\text{NO}\left(g\right)$ $\left(\Delta S_{\text{f,}\text{NO}\left(g\right)}^{\text{ο}}\right)$ is $210.7{\text{Jmol}}^{-1}{\text{K}}^{-1}$.

The standard molar entropy of ${\text{NO}}_2\left(g\right)$ $\left(\Delta S_{\text{f,}{\text{NO}}_2\left(g\right)}^{\text{ο}}\right)$ is $240.0{\text{Jmol}}^{-1}{\text{K}}^{-1}$

The standard molar entropy of ${\text{O}}_2\left(g\right)$ $\left(\Delta S_{\text{f,}{\text{O}}_2\left(g\right)}^{\text{ο}}\right)$ is $\text{205}\text{.0}{\text{Jmol}}^{-1}{\text{K}}^{-1}$

A process is spontaneous at all temperature when the values of,

$\begin{array}{l}\Delta H_{\text{rxn}}^{\text{ο}}=\text{Negative}\\ \Delta S_{\text{rxn}}^{\text{ο}}=\text{Positive}\end{array}$ (2)

A process is spontaneous at low temperature when the values of,

$\begin{array}{l}\Delta H_{\text{rxn}}^{\text{ο}}=\text{Negative}\\ \Delta S_{\text{rxn}}^{\text{ο}}=\text{Negative}\end{array}$ (3)

A process is spontaneous at high temperature when the values of,

$\begin{array}{l}\Delta H_{\text{rxn}}^{\text{ο}}=\text{Positive}\\ \Delta S_{\text{rxn}}^{\text{ο}}=\text{Positive}\end{array}$ (4)

The enthalpy change for a reaction $\left(\Delta H_{\text{rxn}}^{\text{ο}}\right)$ is calculated by the formula,

$\Delta H_{\text{rxn}}^{\text{ο}}={\displaystyle \sum n\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Products}\right)}-{\displaystyle \sum m\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Reactants}\right)}$ (5)

Where,

• $\Delta H_{\text{f}}^{\text{ο}}\left(\text{Products}\right)$ is standard enthalpy of formation of the products.
• $\Delta H_{\text{f}}^{\text{ο}}\left(\text{Reactants}\right)$ is of standard enthalpy of formation of the reactants.
• $n$ is number of moles of products.
• $m$ is number of moles of reactants.

In chemical equation (1) the,

• Number of moles of product ${\text{NO}}_2\left(g\right)$ is $2$.
• Number of moles of reactant $\text{NO}\left(g\right)$ is $2$.
• Number of moles of reactant ${\text{O}}_2\left(g\right)$ is $1$.

The ${\displaystyle \sum n\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Products}\right)}$ for the chemical reaction (1) is expressed as,

${\displaystyle \sum n\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Products}\right)}=2\text{mol}\times \Delta H_{\text{f,}{\text{NO}}_2\left(g\right)}^{\text{ο}}$ (6)

Substitute the value of $\Delta H_{\text{f,}{\text{NO}}_2\left(g\right)}^{\text{ο}}$ in equation (6).

$\begin{array}{c}{\displaystyle \sum n\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Products}\right)}=2\text{mol}\times 33.2\text{kJ}/\text{mol}\\ =66.4\text{kJ}\end{array}$

The ${\displaystyle \sum m\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Reactants}\right)}$ for the balanced chemical reaction (1) is expressed as,

${\displaystyle \sum m\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Reactants}\right)}=2\text{mol}\times \Delta H_{\text{f,}\text{NO}\left(g\right)}^{\text{ο}}+1\text{mol}\times \Delta H_{\text{f,}{\text{O}}_{\text{2}}\left(g\right)}^{\text{ο}}$ (7)

Substitute the values of $\Delta H_{\text{f,}\text{NO}\left(g\right)}^{\text{ο}}$ and $\Delta H_{\text{f,}{\text{O}}_{\text{2}}\left(g\right)}^{\text{ο}}$ in equation (7).

$\begin{array}{c}{\displaystyle \sum m\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Reactants}\right)}=2\text{mol}\times 90.3\text{kJ}/\text{mol}+1\text{mol}\times 0\text{kJ}/\text{mol}\\ =180.6\text{kJ}\end{array}$

Substitute the values of ${\displaystyle \sum n\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Products}\right)}$ and ${\displaystyle \sum m\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Reactants}\right)}$ in equation (5).

$\begin{array}{c}\Delta H_{\text{rxn}}^{\text{ο}}=66.4\text{kJ}-180.6\text{kJ}\\ =-114.2\text{kJ}\end{array}$

The standard entropy change $\left(\Delta S^{\text{ο}}\right)$ is calculated by the formula,

$\Delta S_{\text{rxn}}^{\text{ο}}={\displaystyle \sum n_{\text{products}}\times }{S}^{\text{ο}}\left(\text{products}\right)-{\displaystyle \sum m_{\text{reactants}}\times }{S}^{\text{ο}}\left(\text{reactants}\right)$ (8)

Where,

• $n_{\text{products}}$ is the number of moles of products.
• $m_{\text{reactants}}$ is the number of moles of reactants.
• $S^{\text{ο}}\left(\text{products}\right)$ is the standard molar entropy of products.
• $S^{\text{ο}}\left(\text{reactants}\right)$ is the standard molar entropy of reactants.

The ${\displaystyle \sum n_{\text{products}}\times }{S}^{\text{ο}}\left(\text{products}\right)$ is expressed as,

${\displaystyle \sum n_{\text{products}}\times }{S}^{\text{ο}}\left(\text{products}\right)=2\text{mol}\times S_{{\text{NO}}_2\left(g\right)}^{\text{ο}}$ (9)

Substitute the value of $S_{{\text{NO}}_2\left(g\right)}^{\text{ο}}$ in equation (9).

$\begin{array}{c}{\displaystyle \sum n_{\text{products}}\times }{S}^{\text{ο}}\left(\text{products}\right)=2\text{mol}\times 240.0{\text{Jmol}}^{-1}{\text{K}}^{-1}\\ =480{\text{JK}}^{-1}\end{array}$

The ${\displaystyle \sum m_{\text{reactants}}\times }{S}^{\text{ο}}\left(\text{reactants}\right)$ is expressed as,

${\displaystyle \sum m_{\text{reactants}}\times }{S}^{\text{ο}}\left(\text{reactants}\right)=\left(2\text{mol}\times S_{\text{NO}\left(g\right)}^{\text{ο}}\right)+\left(1\text{mol}\times S_{{\text{O}}_{\text{2}}\left(g\right)}^{\text{ο}}\right)$ (10)

Substitute the values of $S_{\text{NO}\left(g\right)}^{\text{ο}}$ and $S_{{\text{O}}_{\text{2}}\left(g\right)}^{\text{ο}}$ in equation (10).

$\begin{array}{c}{\displaystyle \sum m_{\text{reactants}}\times }{S}^{\text{ο}}\left(\text{reactants}\right)=\left(2\text{mol}\times 210.7{\text{Jmol}}^{-1}{\text{K}}^{-1}\right)+\left(1\text{mol}\times 205.0{\text{Jmol}}^{-1}{\text{K}}^{-1}\right)\\ =421.4{\text{JK}}^{-1}+205.0{\text{JK}}^{-1}\\ =626.4{\text{JK}}^{-1}\end{array}$

Substitute the values of ${\displaystyle \sum m_{\text{reactants}}\times }{S}^{\text{ο}}\left(\text{reactants}\right)$ and ${\displaystyle \sum n_{\text{products}}\times }{S}^{\text{ο}}\left(\text{products}\right)$ in equation (8).

$\begin{array}{c}\Delta S_{\text{rxn}}^{\text{ο}}=480{\text{JK}}^{-1}-626.4{\text{JK}}^{-1}\\ =-146.4{\text{JK}}^{-1}\end{array}$

The values of $\Delta H_{\text{rxn}}^{\text{ο}}$ and $\Delta S_{\text{rxn}}^{\text{ο}}$ satisfies the equation (3).

Hence the given reaction is spontaneous at low temperature.

(b)

Interpretation Introduction

To determine: If the given reaction is spontaneous at low temperatures; at high temperatures or at all temperatures.

Answer to Problem 18.57QP

Solution

The given reaction is spontaneous at low temperature.

Explanation of Solution

Explanation

The given reaction is,

$2{\text{NH}}_3\left(g\right)+2{\text{O}}_{\text{2}}\to {\text{N}}_2\text{O}\left(g\right)+3{\text{H}}_{\text{2}}\text{O}$ (11)

The enthalpy of formation of ${\text{NH}}_3\left(g\right)$ $\left(\Delta H_{\text{f,}{\text{NH}}_3\left(g\right)}^{\text{ο}}\right)$ is $-46.1\text{kJ}/\text{mol}$.

The enthalpy of formation of ${\text{N}}_2\text{O}\left(g\right)$ $\left(\Delta H_{\text{f,}{\text{N}}_2\text{O}\left(g\right)}^{\text{ο}}\right)$ is $82.1\text{kJ}/\text{mol}$.

The enthalpy of formation of ${\text{H}}_2\text{O}\left(g\right)$ $\left(\Delta H_{\text{f,}{\text{H}}_2\text{O}\left(g\right)}^{\text{ο}}\right)$ is $-241.8\text{kJ}/\text{mol}$.

The enthalpy of formation of ${\text{O}}_2\left(g\right)$ $\left(\Delta H_{\text{f,}{\text{O}}_2\left(g\right)}^{\text{ο}}\right)$ is $0.0\text{kJ}/\text{mol}$.

The standard molar entropy of ${\text{NH}}_3\left(g\right)$ $\left(S_{{\text{NH}}_3\left(g\right)}^{\text{ο}}\right)$ is $192.5{\text{Jmol}}^{-1}{\text{K}}^{-1}$.

The standard molar entropy of ${\text{N}}_2\text{O}\left(g\right)$ $\left(S_{{\text{N}}_2\text{O}\left(g\right)}^{\text{ο}}\right)$ is $219.9{\text{Jmol}}^{-1}{\text{K}}^{-1}$.

The standard molar entropy of ${\text{H}}_2\text{O}\left(g\right)$ $\left(S_{{\text{H}}_2\text{O}\left(g\right)}^{\text{ο}}\right)$ is $188.8{\text{Jmol}}^{-1}{\text{K}}^{-1}$

The standard molar entropy of ${\text{O}}_2\left(g\right)$ $\left(S_{{\text{O}}_2\left(g\right)}^{\text{ο}}\right)$ is $\text{205}\text{.0}{\text{Jmol}}^{-1}{\text{K}}^{-1}$

In chemical equation (11) the,

• Number of moles of product ${\text{N}}_2\text{O}\left(g\right)$ is $1$.
• Number of moles of product ${\text{H}}_2\text{O}\left(g\right)$ is $3$.
• Number of moles of reactant ${\text{NH}}_3\left(g\right)$ is $2$.
• Number of moles of reactant ${\text{O}}_2\left(g\right)$ is $2$.

The ${\displaystyle \sum n\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Products}\right)}$ for the chemical reaction (11) is expressed as,

${\displaystyle \sum n\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Products}\right)}=1\text{mol}\times \Delta H_{\text{f,}{\text{N}}_2\text{O}\left(g\right)}^{\text{ο}}+3\text{mol}\times \Delta H_{\text{f,}{\text{H}}_2\text{O}\left(g\right)}^{\text{ο}}$ (12)

Substitute the values of $\Delta H_{\text{f,}{\text{N}}_2\text{O}\left(g\right)}^{\text{ο}}$ and $\Delta H_{\text{f,}{\text{H}}_2\text{O}\left(g\right)}^{\text{ο}}$ in equation (12).

$\begin{array}{c}{\displaystyle \sum n\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Products}\right)}=1\text{mol}\times 82.1\text{kJ}/\text{mol}+3\text{mol}\times -241.8\text{kJ}/\text{mol}\\ =-643.3\text{kJ}\end{array}$

The ${\displaystyle \sum m\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Reactants}\right)}$ for the balanced chemical reaction (11) is expressed as,

${\displaystyle \sum m\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Reactants}\right)}=2\text{mol}\times \Delta H_{\text{f,}{\text{NH}}_3\left(g\right)}^{\text{ο}}+2\text{mol}\times \Delta H_{\text{f,}{\text{O}}_{\text{2}}\left(g\right)}^{\text{ο}}$ (13)

Substitute the values of $\Delta H_{\text{f,}{\text{NH}}_3\left(g\right)}^{\text{ο}}$ and $\Delta H_{\text{f,}{\text{O}}_{\text{2}}\left(g\right)}^{\text{ο}}$ in equation (13).

$\begin{array}{c}{\displaystyle \sum m\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Reactants}\right)}=2\text{mol}\times -46.1\text{kJ}/\text{mol}+1\text{mol}\times 0\text{kJ}/\text{mol}\\ =-92.2\text{kJ}\end{array}$

Substitute the values of ${\displaystyle \sum n\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Products}\right)}$ and ${\displaystyle \sum m\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Reactants}\right)}$ in equation (5).

$\begin{array}{c}\Delta H_{\text{rxn}}^{\text{ο}}=-643.3\text{kJ}-\left(-92.2\text{kJ}\right)\\ =-551.3\text{kJ}\end{array}$

The ${\displaystyle \sum n_{\text{products}}\times }{S}^{\text{ο}}\left(\text{products}\right)$ is expressed as,

${\displaystyle \sum n_{\text{products}}\times }{S}^{\text{ο}}\left(\text{products}\right)=1\text{mol}\times S_{{\text{N}}_2\text{O}\left(g\right)}^{\text{ο}}+3\text{mol}\times S_{{\text{H}}_2\text{O}\left(g\right)}^{\text{ο}}$ (14)

Substitute the values of $S_{{\text{N}}_2\text{O}\left(g\right)}^{\text{ο}}$ and $S_{{\text{H}}_2\text{O}\left(g\right)}^{\text{ο}}$ in equation (14).

$\begin{array}{c}{\displaystyle \sum n_{\text{products}}\times }{S}^{\text{ο}}\left(\text{products}\right)=1\text{mol}\times 219.9{\text{Jmol}}^{-1}{\text{K}}^{-1}+3\text{mol}\times 188.8{\text{Jmol}}^{-1}{\text{K}}^{-1}\\ =786.3{\text{JK}}^{-1}\end{array}$

The ${\displaystyle \sum m_{\text{reactants}}\times }{S}^{\text{ο}}\left(\text{reactants}\right)$ is expressed as,

${\displaystyle \sum m_{\text{reactants}}\times }{S}^{\text{ο}}\left(\text{reactants}\right)=2\text{mol}\times S_{{\text{NH}}_3\left(g\right)}^{\text{ο}}+2\text{mol}\times \Delta S_{{\text{O}}_{\text{2}}\left(g\right)}^{\text{ο}}$ (15)

Substitute the values of $S_{{\text{NH}}_3\left(g\right)}^{\text{ο}}$ and $S_{{\text{O}}_{\text{2}}\left(g\right)}^{\text{ο}}$ in equation (15).

$\begin{array}{c}{\displaystyle \sum m_{\text{reactants}}\times }{S}^{\text{ο}}\left(\text{reactants}\right)=2\text{mol}\times 192.5{\text{Jmol}}^{-1}{\text{K}}^{-1}+2\text{mol}\times 205.0{\text{Jmol}}^{-1}{\text{K}}^{-1}\\ =795{\text{JK}}^{-1}\end{array}$

Substitute the values of ${\displaystyle \sum m_{\text{reactants}}\times }{S}^{\text{ο}}\left(\text{reactants}\right)$ and ${\displaystyle \sum n_{\text{products}}\times }{S}^{\text{ο}}\left(\text{products}\right)$ in equation (8).

$\begin{array}{c}\Delta S_{\text{rxn}}^{\text{ο}}=786.3{\text{JK}}^{-1}-795{\text{JK}}^{-1}\\ =-8.7{\text{JK}}^{-1}\end{array}$

The values of $\Delta H_{\text{rxn}}^{\text{ο}}$ and $\Delta S_{\text{rxn}}^{\text{ο}}$ satisfies the equation (3).

Hence the given reaction is spontaneous at low temperature.

(c)

Interpretation Introduction

To determine: If the given reaction is spontaneous at low temperatures; at high temperatures or at all temperatures.

Answer to Problem 18.57QP

Solution

The given reaction is spontaneous at all temperature.

Explanation of Solution

Explanation

The given reaction is,

${\text{NH}}_4{\text{NO}}_{\text{3}}\left(s\right)\to 2{\text{H}}_2\text{O}\left(g\right)+{\text{N}}_{\text{2}}\text{O}\left(g\right)$ (16)

The enthalpy of formation of ${\text{NH}}_4{\text{NO}}_{\text{3}}\left(s\right)$ $\left(\Delta H_{\text{f,}{\text{NH}}_4{\text{NO}}_{\text{3}}\left(s\right)}^{\text{ο}}\right)$ is $-365.6\text{kJ}/\text{mol}$.

The enthalpy of formation of ${\text{N}}_2\text{O}\left(g\right)$ $\left(\Delta H_{\text{f,}{\text{N}}_2\text{O}\left(g\right)}^{\text{ο}}\right)$ is $82.1\text{kJ}/\text{mol}$.

The enthalpy of formation of ${\text{H}}_2\text{O}\left(g\right)$ $\left(\Delta H_{\text{f,}{\text{H}}_2\text{O}\left(g\right)}^{\text{ο}}\right)$ is $-241.8\text{kJ}/\text{mol}$.

The standard molar entropy of ${\text{NH}}_4{\text{NO}}_{\text{3}}\left(s\right)$ $\left(S_{{\text{NH}}_4{\text{NO}}_{\text{3}}\left(s\right)}^{\text{ο}}\right)$ is $151.1{\text{Jmol}}^{-1}{\text{K}}^{-1}$.

The standard molar entropy of ${\text{N}}_2\text{O}\left(g\right)$ $\left(S_{{\text{N}}_2\text{O}\left(g\right)}^{\text{ο}}\right)$ is $219.9{\text{Jmol}}^{-1}{\text{K}}^{-1}$.

The standard molar entropy of ${\text{H}}_2\text{O}\left(g\right)$ $\left(S_{{\text{H}}_2\text{O}\left(g\right)}^{\text{ο}}\right)$ is $188.8{\text{Jmol}}^{-1}{\text{K}}^{-1}$

In chemical equation (16) the,

• Number of moles of product ${\text{N}}_2\text{O}\left(g\right)$ is $1$.
• Number of moles of product ${\text{H}}_2\text{O}\left(g\right)$ is $3$.
• Number of moles of reactant ${\text{NH}}_3\left(g\right)$ is $2$.
• Number of moles of reactant ${\text{O}}_2\left(g\right)$ is $2$.

The ${\displaystyle \sum n\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Products}\right)}$ for the chemical reaction (16) is expressed as,

${\displaystyle \sum n\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Products}\right)}=1\text{mol}\times \Delta H_{\text{f,}{\text{N}}_2\text{O}\left(g\right)}^{\text{ο}}+2\text{mol}\times \Delta H_{\text{f,}{\text{H}}_2\text{O}\left(g\right)}^{\text{ο}}$ (17)

Substitute the values of $\Delta H_{\text{f,}{\text{N}}_2\text{O}\left(g\right)}^{\text{ο}}$ and $\Delta H_{\text{f,}{\text{H}}_2\text{O}\left(g\right)}^{\text{ο}}$ in equation (17).

$\begin{array}{c}{\displaystyle \sum n\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Products}\right)}=1\text{mol}\times 82.1\text{kJ}/\text{mol}+2\text{mol}\times -241.8\text{kJ}/\text{mol}\\ =-401.5\text{kJ}\end{array}$

The ${\displaystyle \sum m\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Reactants}\right)}$ for the chemical reaction (16) is expressed as,

${\displaystyle \sum m\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Reactants}\right)}=1\text{mol}\times \Delta H_{\text{f,}{\text{NH}}_4{\text{NO}}_{\text{3}}\left(s\right)}^{\text{ο}}$ (18)

Substitute the value of $\Delta H_{\text{f,}{\text{NH}}_4{\text{NO}}_{\text{3}}\left(s\right)}^{\text{ο}}$ in equation (18).

$\begin{array}{c}{\displaystyle \sum m\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Reactants}\right)}=1\text{mol}\times -365.6\text{kJ}/\text{mol}\\ =-365.6\text{kJ}\end{array}$

Substitute the values of ${\displaystyle \sum n\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Products}\right)}$ and ${\displaystyle \sum m\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Reactants}\right)}$ in equation (5).

$\begin{array}{c}\Delta H_{\text{rxn}}^{\text{ο}}=-401.5\text{kJ}-\left(-365.6\text{kJ}\right)\\ =-35.9\text{kJ}\end{array}$

The ${\displaystyle \sum n_{\text{products}}\times }{S}^{\text{ο}}\left(\text{products}\right)$ is expressed as,

${\displaystyle \sum n_{\text{products}}\times }{S}^{\text{ο}}\left(\text{products}\right)=1\text{mol}\times S_{{\text{N}}_2\text{O}\left(g\right)}^{\text{ο}}+2\text{mol}\times S_{{\text{H}}_2\text{O}\left(g\right)}^{\text{ο}}$ (19)

Substitute the values of $S_{{\text{N}}_2\text{O}\left(g\right)}^{\text{ο}}$ and $S_{{\text{H}}_2\text{O}\left(g\right)}^{\text{ο}}$ in equation (19).

$\begin{array}{c}{\displaystyle \sum n_{\text{products}}\times }{S}^{\text{ο}}\left(\text{products}\right)=1\text{mol}\times 219.9{\text{Jmol}}^{-1}{\text{K}}^{-1}+2\text{mol}\times 188.8{\text{Jmol}}^{-1}{\text{K}}^{-1}\\ =597.5{\text{JK}}^{-1}\end{array}$

The ${\displaystyle \sum m_{\text{reactants}}\times }{S}^{\text{ο}}\left(\text{reactants}\right)$ is expressed as,

${\displaystyle \sum m_{\text{reactants}}\times }{S}^{\text{ο}}\left(\text{reactants}\right)=1\text{mol}\times S_{{\text{NH}}_4{\text{NO}}_{\text{3}}\left(s\right)}^{\text{ο}}$ (20)

Substitute the value of $S_{{\text{NH}}_4{\text{NO}}_{\text{3}}\left(s\right)}^{\text{ο}}$ in equation (20).

$\begin{array}{c}{\displaystyle \sum m_{\text{reactants}}\times }{S}^{\text{ο}}\left(\text{reactants}\right)=1\text{mol}\times 151.1{\text{Jmol}}^{-1}{\text{K}}^{-1}\\ =151.1{\text{JK}}^{-1}\end{array}$

Substitute the values of ${\displaystyle \sum m_{\text{reactants}}\times }{S}^{\text{ο}}\left(\text{reactants}\right)$ and ${\displaystyle \sum n_{\text{products}}\times }{S}^{\text{ο}}\left(\text{products}\right)$ in equation (8).

$\begin{array}{c}\Delta S_{\text{rxn}}^{\text{ο}}=597.5{\text{JK}}^{-1}-151.1{\text{JK}}^{-1}\\ =446.4{\text{JK}}^{-1}\end{array}$

The values of $\Delta H_{\text{rxn}}^{\text{ο}}$ is negative and $\Delta S_{\text{rxn}}^{\text{ο}}$ is positive. Thus it satisfies the equation (2).

Hence the given reaction is spontaneous at all temperature.

Conclusion

1. a. The given reaction is spontaneous at low temperature.
2. b. The given reaction is spontaneous at low temperature.
3. c. The given reaction is spontaneous at all temperature