Chapter 17, Problem 17.53QP

Interpretation Introduction

Interpretation: The lowest temperature, 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 lowest temperature, at which the given reaction is spontaneous.

Answer to Problem 17.53QP

Solution

The lowest temperature, at which the given reaction is spontaneous is to be $\underset{\_}{985.05K}$.

Explanation of Solution

Explanation

The given reaction is,

${\text{H}}_{\text{2}}\text{O}\left(g\right)+\text{C}\left(s\right)\to {\text{H}}_{\text{2}}\left(g\right)+\text{CO}\left(g\right)$ (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 $\text{CO}\left(g\right)$ $\left(\Delta H_{\text{f,}\text{CO}\left(g\right)}^{\text{ο}}\right)$ is $-110.5\text{kJ}/\text{mol}$.

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

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

The standard molar entropy of $\text{CO}\left(g\right)$ $\left(\Delta S_{\text{CO}\left(g\right)}^{\text{ο}}\right)$ is $197.6{\text{Jmol}}^{-1}{\text{K}}^{-1}$.

The standard molar entropy of $\text{C}\left(s\right)$ $\left(\Delta S_{\text{C}\left(s\right)}^{\text{ο}}\right)$ is $5.7{\text{Jmol}}^{-1}{\text{K}}^{-1}$.

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 ${\text{H}}_2\left(g\right)$ $\left(S_{{\text{H}}_2\left(g\right)}^{\text{ο}}\right)$ is $130.6{\text{Jmol}}^{-1}{\text{K}}^{-1}$.

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 ${\text{H}}_2\left(g\right)$ is $1$.
• Number of moles of product $\text{CO}\left(g\right)$ is $1$.
• Number of moles of reactant ${\text{H}}_2\text{O}\left(g\right)$ is $1$.
• Number of moles of reactant $\text{C}\left(s\right)$ 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,

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

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

$\begin{array}{c}{\displaystyle \sum n\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Products}\right)}=1\text{mol}\times -110.5\text{kJ}/\text{mol}+1\text{mol}\times 0.0\text{kJ}/\text{mol}\\ =-110.5\text{kJ}\end{array}$

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

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

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

$\begin{array}{c}{\displaystyle \sum m\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Reactants}\right)}=1\text{mol}\times -241.8\text{kJ}/\text{mol}+1\text{mol}\times 0\text{kJ}/\text{mol}\\ =-241.8\text{kJ}\end{array}$

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).

$\begin{array}{c}\Delta H_{\text{rxn}}^{\text{ο}}=-110.5\text{kJ}-\left(-241.8\text{kJ}\right)\\ =-110.5\text{kJ}+241.8\text{kJ}\\ =131.3\text{kJ}\end{array}$

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

$\begin{array}{c}1\text{kJ}={10}^3\text{J}\\ 131.3\text{kJ}=131.3\times {10}^3\text{J}\end{array}$

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 $131.3\times {10}^3\text{J}$.

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,

${\displaystyle \sum n_{\text{products}}\times }{S}^{\text{ο}}\left(\text{products}\right)=1\text{mol}\times S_{\text{CO}\left(g\right)}^{\text{ο}}+1\text{mol}\times S_{{\text{H}}_2\left(g\right)}^{\text{ο}}$ (6)

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

$\begin{array}{c}{\displaystyle \sum n_{\text{products}}\times }{S}^{\text{ο}}\left(\text{products}\right)=1\text{mol}\times 197.6{\text{Jmol}}^{-1}{\text{K}}^{-1}+1\text{mol}\times 130.6{\text{Jmol}}^{-1}{\text{K}}^{-1}\\ =328.2{\text{JK}}^{-1}\end{array}$

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

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

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

$\begin{array}{c}{\displaystyle \sum m_{\text{reactants}}\times }{S}^{\text{ο}}\left(\text{reactants}\right)=1\text{mol}\times 188.8{\text{Jmol}}^{-1}{\text{K}}^{-1}+1\text{mol}\times 5.7{\text{Jmol}}^{-1}{\text{K}}^{-1}\\ =194.5{\text{JK}}^{-1}\end{array}$

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).

$\begin{array}{c}\Delta S_{\text{rxn}}^{\text{ο}}=328.2{\text{JK}}^{-1}-194.5{\text{JK}}^{-1}\\ =133.7{\text{JK}}^{-1}\end{array}$

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).

$\begin{array}{c}0>131.3\times {10}^3\text{J}-T\times 133.7{\text{JK}}^{-1}\\ T\times 133.7{\text{JK}}^{-1}>131.3\times {10}^3\text{J}\\ T>\frac{131.3\times {10}^3\text{J}}{133.7{\text{JK}}^{-1}}\\ T>985.04\text{K}\end{array}$

Thus, the lowest temperature, at which the given reaction is spontaneous is to be $\underset{\_}{985.05K}$.

Conclusion

The lowest temperature, at which the given reaction is spontaneous is to be $\underset{\_}{985.05K}$