Chapter 17, Problem 17.54QP

Interpretation Introduction

Interpretation: The temperature range, at which the given reaction is spontaneous, is to be calculated.

Concept introduction: The spontaneous reaction is only possible when the value of $\Delta G_{\text{rxn}}^{\text{ο}}$ is negative or less than zero.

To determine: The temperature range, at which the given reaction is spontaneous.

Answer to Problem 17.54QP

Solution

The temperature range, at which the given reaction is spontaneous is to be greater than $\underset{\_}{542.92K}$.

Explanation of Solution

Explanation

The given reaction is,

${\text{SO}}_2\left(g\right)+2{\text{H}}_{\text{2}}\text{S}\left(g\right)\to \frac{3}{8}{\text{S}}_8\left(s\right)+2{\text{H}}_{\text{2}}\text{O}\left(g\right)$ (1)

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 ${\text{SO}}_2\left(g\right)$ $\left(\Delta H_{\text{f,}{\text{SO}}_2\left(g\right)}^{\text{ο}}\right)$ is $-296.8\text{kJ}/\text{mol}$.

The enthalpy of formation of ${\text{S}}_8\left(s\right)$ $\left(\Delta H_{\text{f,}{\text{S}}_8\left(s\right)}^{\text{ο}}\right)$ is $0.0\text{kJ}/\text{mol}$.

The enthalpy of formation of ${\text{H}}_2\text{S}\left(g\right)$ $\left(\Delta H_{\text{f,}{\text{H}}_2\text{S}\left(g\right)}^{\text{ο}}\right)$ is $-20.17\text{kJ}/\text{mol}$.

The standard molar entropy of ${\text{SO}}_2\left(g\right)$ $\left(\Delta S_{{\text{SO}}_2\left(g\right)}^{\text{ο}}\right)$ is $248.2{\text{Jmol}}^{-1}{\text{K}}^{-1}$.

The standard molar entropy of ${\text{S}}_8\left(s\right)$ $\left(\Delta S_{{\text{S}}_8\left(s\right)}^{\text{ο}}\right)$ is $32.1{\text{Jmol}}^{-1}{\text{K}}^{-1}$.

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 ${\text{H}}_2\text{S}\left(g\right)$ $\left(S_{{\text{H}}_2\text{S}\left(g\right)}^{\text{ο}}\right)$ is $205.6{\text{Jmol}}^{-1}{\text{K}}^{-1}$.

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

$\Delta H_{\text{rxn}}^{\text{ο}}={\displaystyle \sum n\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Products}\right)}-{\displaystyle \sum m\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Reactants}\right)}$ (2)

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 ${\text{S}}_8\left(s\right)$ is $\frac{3}{8}$.
• Number of moles of product ${\text{H}}_2\text{O}\left(g\right)$ is $2$.
• Number of moles of reactant ${\text{H}}_2\text{S}\left(g\right)$ is $2$.
• Number of moles of reactant ${\text{SO}}_2\left(g\right)$ 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,

${\displaystyle \sum n\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Products}\right)}=\frac{3}{8}\text{mol}\times \Delta H_{\text{f,}{\text{S}}_8\left(s\right)}^{\text{ο}}+2\text{mol}\times \Delta H_{\text{f,}{\text{H}}_2\text{O}\left(g\right)}^{\text{ο}}$ (3)

Substitute the values of $\Delta H_{\text{f,}{\text{S}}_8\left(s\right)}^{\text{ο}}$ and $\Delta H_{\text{f,}{\text{H}}_2\text{O}\left(g\right)}^{\text{ο}}$ in equation (3).

$\begin{array}{c}{\displaystyle \sum n\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Products}\right)}=\frac{3}{8}\text{mol}\times 0.0\text{kJ}/\text{mol}+2\text{mol}\times -241.8\text{kJ}/\text{mol}\\ =-483.6\text{kJ}\end{array}$

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

${\displaystyle \sum m\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Reactants}\right)}=2\text{mol}\times \Delta H_{\text{f,}{\text{H}}_2\text{S}\left(g\right)}^{\text{ο}}+1\text{mol}\times \Delta H_{\text{f,}{\text{SO}}_2\left(g\right)}^{\text{ο}}$ (4)

Substitute the values of $\Delta H_{\text{f,}{\text{H}}_2\text{S}\left(g\right)}^{\text{ο}}$ and $\Delta H_{\text{f,}{\text{SO}}_2\left(g\right)}^{\text{ο}}$ in equation (4).

$\begin{array}{c}{\displaystyle \sum m\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Reactants}\right)}=2\text{mol}\times -20.17\text{kJ}/\text{mol}+1\text{mol}\times -296.8\text{kJ}/\text{mol}\\ =-337.14\text{kJ}\end{array}$

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 (2).

$\begin{array}{c}\Delta H_{\text{rxn}}^{\text{ο}}=-483.6\text{kJ}-\left(-337.14\text{kJ}\right)\\ =-483.6\text{kJ}+337.14\text{kJ}\\ =-146.46\text{kJ}\end{array}$

To change the above value in joules, use conversion factor.

$\begin{array}{c}1\text{kJ}={10}^3\text{J}\\ -146.46\text{kJ}=-146.46\times {10}^3\text{J}\end{array}$

Thus the new value of $\Delta H_{\text{rxn}}^{\text{ο}}$ in joules $\left(\Delta H_{\text{rxn}}^{\text{ο}}\left(\text{J}\right)\right)$ is $-146.46\times {10}^3\text{J}$.

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

$\Delta S_{\text{rxn}}^{\text{ο}}={\displaystyle \sum n_{\text{products}}\times }{S}^{\text{ο}}\left(\text{products}\right)-{\displaystyle \sum m_{\text{reactants}}\times }{S}^{\text{ο}}\left(\text{reactants}\right)$ (5)

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,

${\displaystyle \sum n_{\text{products}}\times }{S}^{\text{ο}}\left(\text{products}\right)=\frac{3}{8}\text{mol}\times S_{{\text{S}}_8\left(s\right)}^{\text{ο}}+2\text{mol}\times S_{{\text{H}}_2\text{O}\left(g\right)}^{\text{ο}}$ (6)

Substitute the values of $S_{{\text{S}}_8\left(s\right)}^{\text{ο}}$ and $S_{{\text{H}}_2\text{O}\left(g\right)}^{\text{ο}}$ in equation (6).

$\begin{array}{c}{\displaystyle \sum n_{\text{products}}\times }{S}^{\text{ο}}\left(\text{products}\right)=\frac{3}{8}\text{mol}\times 32.1{\text{Jmol}}^{-1}{\text{K}}^{-1}+2\text{mol}\times 188.8{\text{Jmol}}^{-1}{\text{K}}^{-1}\\ =12.037{\text{JK}}^{-1}+377.6{\text{JK}}^{-1}\\ =389.637{\text{JK}}^{-1}\end{array}$

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

${\displaystyle \sum m_{\text{reactants}}\times }{S}^{\text{ο}}\left(\text{reactants}\right)=2\text{mol}\times S_{{\text{H}}_2\text{S}\left(g\right)}^{\text{ο}}+1\text{mol}\times S_{{\text{SO}}_2\left(g\right)}^{\text{ο}}$ (7)

Substitute the values of $S_{{\text{H}}_2\text{S}\left(g\right)}^{\text{ο}}$ and $S_{{\text{SO}}_2\left(g\right)}^{\text{ο}}$ in equation (7).

$\begin{array}{c}{\displaystyle \sum m_{\text{reactants}}\times }{S}^{\text{ο}}\left(\text{reactants}\right)=2\text{mol}\times 205.6{\text{Jmol}}^{-1}{\text{K}}^{-1}+1\text{mol}\times 248.2{\text{Jmol}}^{-1}{\text{K}}^{-1}\\ =411.2{\text{JK}}^{-1}+248.2{\text{JK}}^{-1}\\ =659.4{\text{JK}}^{-1}\end{array}$

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 (5).

$\begin{array}{c}\Delta S_{\text{rxn}}^{\text{ο}}=389.637{\text{JK}}^{-1}-659.4{\text{JK}}^{-1}\\ =-269.763{\text{JK}}^{-1}\end{array}$

The $\Delta G_{\text{rxn}}^{\text{ο}}$ of a reaction is calculated by the formula,

$\Delta G_{\text{rxn}}^{\text{ο}}=\Delta H_{\text{rxn}}^{\text{ο}}-T\Delta S_{\text{rxn}}^{\text{ο}}$ (8)

The spontaneous reaction is only possible when the value of $\Delta G_{\text{rxn}}^{\text{ο}}$ is negative or less than zero. Thus, the equation (8) becomes,

$0>\Delta H_{\text{rxn}}^{\text{ο}}-T\Delta S_{\text{rxn}}^{\text{ο}}$ (9)

Substitute the values of $\Delta S_{\text{rxn}}^{\text{ο}}$ and $\Delta H_{\text{rxn}}^{\text{ο}}\left(\text{J}\right)$ in equation (9).

$\begin{array}{c}0>-146.46\times {10}^3\text{J}-T\times -269.763{\text{JK}}^{-1}\\ T\times -269.763{\text{JK}}^{-1}>-146.46\times {10}^3\text{J}\\ T>\frac{-146.46\times {10}^3\text{J}}{-269.763{\text{JK}}^{-1}}\\ T>542.92\text{K}\end{array}$

Thus, the temperature range, at which the given reaction is spontaneous is to be greater than $\underset{\_}{542.92K}$.

Conclusion

The temperature range, at which the given reaction is spontaneous is to be greater than $\underset{\_}{542.92K}$