Chapter 18, Problem 18.47QP

Interpretation Introduction

Interpretation: The value of $\Delta G^{\text{ο}}$ for the reduction of iron ore for given thermodynamic values at $25°\text{C}$ is to be calculated.

Concept introduction: The free energy change $\left(\Delta G^{\text{ο}}\right)$ is calculated by the formula,

$\Delta G^{\text{ο}}=\Delta H_{\text{rxn}}^{\text{ο}}-T\Delta S_{\text{rxn}}^{\text{ο}}$

To determine: The value of $\Delta G^{\text{ο}}$ for the reduction of iron ore for given thermodynamic values at $25°\text{C}$.

Answer to Problem 18.47QP

Solution

The value of $\Delta G^{\text{ο}}$ for the reduction of iron ore is $\underset{\_}{56.54kJ}$.

Explanation of Solution

Explanation

Given

The enthalpy change for formation $\left(\Delta H_{\text{f}}^{\text{ο}}\right)$ of ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(s\right)$ is $-824.2\text{kJ}/\text{mol}$.

The enthalpy change for formation $\left(\Delta H_{\text{f}}^{\text{ο}}\right)$ of $\text{Fe}\left(s\right)$ is $0\text{kJ}/\text{mol}$.

The enthalpy change for formation $\left(\Delta H_{\text{f}}^{\text{ο}}\right)$ of ${\text{H}}_{\text{2}}\left(g\right)$ is $0\text{kJ}/\text{mol}$.

The enthalpy change for formation $\left(\Delta H_{\text{f}}^{\text{ο}}\right)$ of ${\text{H}}_{\text{2}}\text{O}\left(g\right)$ is $-241.8\text{kJ}/\text{mol}$.

The standard molar entropy $\left(S^{\text{ο}}\right)$ of ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(s\right)$ is $87.4{\text{Jmol}}^{-1}{\text{K}}^{-1}$.

The standard molar entropy $\left(S^{\text{ο}}\right)$ of $\text{Fe}\left(s\right)$ is $27.3{\text{Jmol}}^{-1}{\text{K}}^{-1}$.

The standard molar entropy $\left(S^{\text{ο}}\right)$ of ${\text{H}}_{\text{2}}\left(g\right)$ is $130.6{\text{Jmol}}^{-1}{\text{K}}^{-1}$.

The standard molar entropy $\left(S^{\text{ο}}\right)$ of ${\text{H}}_{\text{2}}\text{O}\left(g\right)$ is $188.8{\text{Jmol}}^{-1}{\text{K}}^{-1}$.

The reaction at $25°\text{C}$ temperature $\left(T\right)$ is,

${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(s\right)+3{\text{H}}_{\text{2}}\left(g\right)\to 2\text{Fe}\left(s\right)+3{\text{H}}_{\text{2}}\text{O}\left(g\right)$ (1)

The free energy change $\left(\Delta G^{\text{ο}}\right)$ is calculated by the formula,

$\Delta G^{\text{ο}}=\Delta H_{\text{rxn}}^{\text{ο}}-T\Delta S_{\text{rxn}}^{\text{ο}}$ (2)

The equation (1) is balanced reaction.

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

$\Delta H_{\text{rxn}}^{\text{ο}}={\displaystyle \sum n\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Products}\right)}-{\displaystyle \sum m\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Reactants}\right)}$ (3)

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 the balanced chemical equation (1) the,

• Number of moles of product $\text{Fe}\left(s\right)$ is $2$.
• Number of moles of product ${\text{H}}_{\text{2}}\text{O}\left(g\right)$ is $3$.
• Number of moles of reactant ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(s\right)$ is $1$.
• Number of moles of reactant ${\text{H}}_{\text{2}}\left(g\right)$ is $3$.

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

${\displaystyle \sum n\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Products}\right)}=2\text{mol}\times \Delta H_{\text{f,}\text{Fe}\left(s\right)}^{\text{ο}}+3\text{mol}\times \Delta H_{\text{f,}{\text{H}}_{\text{2}}\text{O}\left(g\right)}^{\text{ο}}$ (4)

Where,

• $\Delta H_{\text{f,}\text{Fe}\left(s\right)}^{\text{ο}}$ is the enthalpy change for formation $\left(\Delta H_{\text{f}}^{\text{ο}}\right)$ of $\text{Fe}\left(s\right)$
• $\Delta H_{\text{f,}{\text{H}}_{\text{2}}\text{O}\left(g\right)}^{\text{ο}}$ is the enthalpy change for formation $\left(\Delta H_{\text{f}}^{\text{ο}}\right)$ of ${\text{H}}_{\text{2}}\text{O}\left(g\right)$.

Substitute the values of $\Delta H_{\text{f,}\text{Fe}\left(s\right)}^{\text{ο}}$ and $\Delta H_{\text{f,}{\text{H}}_{\text{2}}\text{O}\left(g\right)}^{\text{ο}}$ in equation (4).

$\begin{array}{c}{\displaystyle \sum n\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Products}\right)}=2\text{mol}\times 0\text{kJ}/\text{mol}+3\text{mol}\times -241.8\text{kJ}/\text{mol}\\ =-\text{725}\text{.4}\text{kJ}\end{array}$

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

${\displaystyle \sum m\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Reactants}\right)}=1\text{mol}\times \Delta H_{\text{f,}{\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(s\right)}^{\text{ο}}+3\text{mol}\times \Delta H_{\text{f,}{\text{H}}_{\text{2}}\left(g\right)}^{\text{ο}}$ (5)

Where,

• $\Delta H_{\text{f,}{\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(s\right)}^{\text{ο}}$ is the enthalpy change for formation $\left(\Delta H_{\text{f}}^{\text{ο}}\right)$ of ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(s\right)$.
• $\Delta H_{\text{f,}{\text{H}}_{\text{2}}\left(g\right)}^{\text{ο}}$ is the enthalpy change for formation $\left(\Delta H_{\text{f}}^{\text{ο}}\right)$ of ${\text{H}}_{\text{2}}\left(g\right)$.

Substitute the values of $\Delta H_{\text{f,}{\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(s\right)}^{\text{ο}}$ and $\Delta H_{\text{f,}{\text{H}}_{\text{2}}\left(g\right)}^{\text{ο}}$ in equation (5).

$\begin{array}{c}{\displaystyle \sum m\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Reactants}\right)}=1\text{mol}\times -824.2\text{kJ}/\text{mol}+3\text{mol}\times 0\text{kJ}/\text{mol}\\ =-824.2\text{kJ}\end{array}$

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 (3).

$\begin{array}{c}\Delta H_{\text{rxn}}^{\text{ο}}=\text{725}\text{.4}\text{kJ}-\left(-824.2\text{kJ}\right)\\ =-\text{725}\text{.4}\text{kJ}+824.2\text{kJ}\\ =98.8\text{kJ}\end{array}$

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

$\Delta S_{\text{rxn}}^{\text{ο}}={\displaystyle \sum n_{\text{products}}\times }{S}^{\text{ο}}\left(\text{products}\right)-{\displaystyle \sum m_{\text{reactants}}\times }{S}^{\text{ο}}\left(\text{reactants}\right)$ (6)

Where,

• $n_{\text{products}}$ is the number of moles of products.
• $m_{\text{reactants}}$ is the number of moles of 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.

In the balanced chemical equation (1) the,

• Number of moles of product $\text{Fe}\left(s\right)$ is $2$.
• Number of moles of product ${\text{H}}_{\text{2}}\text{O}\left(g\right)$ is $3$.
• Number of moles of reactant ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(s\right)$ is $1$.
• Number of moles of reactant ${\text{H}}_{\text{2}}\left(g\right)$ is $3$.

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

${\displaystyle \sum n_{\text{products}}\times }{S}^{\text{ο}}\left(\text{products}\right)=2\text{mol}\times S_{\text{Fe}\left(s\right)}^{\text{ο}}+3\text{mol}\times S_{{\text{H}}_{\text{2}}\text{O}\left(g\right)}^{\text{ο}}$ (7)

Where,

• $S_{\text{Fe}\left(s\right)}^{\text{ο}}$ is the standard molar entropy of $\text{Fe}\left(s\right)$.
• $S_{{\text{H}}_{\text{2}}\text{O}\left(g\right)}^{\text{ο}}$ is the standard molar entropy of ${\text{H}}_{\text{2}}\text{O}\left(g\right)$.

Substitute the values of $S_{\text{Fe}\left(s\right)}^{\text{ο}}$ and $S_{{\text{H}}_{\text{2}}\text{O}\left(g\right)}^{\text{ο}}$ in equation (7).

$\begin{array}{c}{\displaystyle \sum n_{\text{products}}\times }{S}^{\text{ο}}\left(\text{products}\right)=2\text{mol}\times 27.3{\text{Jmol}}^{-1}{\text{K}}^{-1}+3\text{mol}\times 188.8{\text{Jmol}}^{-1}{\text{K}}^{-1}\\ =54.6{\text{JK}}^{-1}+566.4{\text{JK}}^{-1}\\ =621{\text{JK}}^{-1}\end{array}$

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

${\displaystyle \sum m_{\text{reactants}}\times }{S}^{\text{ο}}\left(\text{reactants}\right)=\left(1\text{mol}\times S_{{\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(s\right)}^{\text{ο}}\right)+\left(3\text{mol}\times S_{{\text{H}}_{\text{2}}\left(g\right)}^{\text{ο}}\right)$ (8)

Where,

• $S_{{\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(s\right)}^{\text{ο}}$ is the standard molar entropy of ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(s\right)$.
• $S_{{\text{H}}_{\text{2}}\left(g\right)}^{\text{ο}}$ is the standard molar entropy of ${\text{H}}_{\text{2}}\left(g\right)$.

Substitute the values of $S_{{\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(s\right)}^{\text{ο}}$ and $S_{{\text{H}}_{\text{2}}\left(g\right)}^{\text{ο}}$ in equation (8).

$\begin{array}{c}{\displaystyle \sum m_{\text{reactants}}\times }{S}^{\text{ο}}\left(\text{reactants}\right)=\left(1\text{mol}\times 87.4{\text{Jmol}}^{-1}{\text{K}}^{-1}\right)+\left(3\text{mol}\times 130.6{\text{Jmol}}^{-1}{\text{K}}^{-1}\right)\\ =87.4{\text{JK}}^{-1}+391.8{\text{JK}}^{-1}\\ =479.2{\text{JK}}^{-1}\end{array}$

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 (6).

$\begin{array}{c}\Delta S_{\text{rxn}}^{\text{ο}}=621{\text{JK}}^{-1}-479.2{\text{JK}}^{-1}\\ =141.8{\text{JK}}^{-1}\end{array}$

To change the value of $\Delta S_{\text{rxn}}^{\text{ο}}$ in ${\text{kJK}}^{-1}$, use conversion factor.

$\begin{array}{c}1\text{J}={10}^{-3}\text{kJ}\\ 141.8{\text{JK}}^{-1}=141.8\times {10}^{-3}{\text{kJK}}^{-1}\end{array}$

To change the value of temperature $\left(T\right)$ in Kelvin $\left(\text{K}\right)$, use conversion factor.

$\begin{array}{c}0°\text{C}=\left(0+273\right)\text{K}\\ \text{25}°\text{C}=\left(25+273\right)\text{K}\\ =\text{298}\text{K}\end{array}$

Substitute the values of $\Delta S_{\text{rxn}}^{\text{ο}}$ $\left({\text{kJK}}^{-1}\right)$, $\Delta H_{\text{rxn}}^{\text{ο}}$ and $T$ $\left(\text{K}\right)$ in equation (2).

$\begin{array}{c}\Delta G^{\text{ο}}=\Delta H_{\text{rxn}}^{\text{ο}}-T\Delta S_{\text{rxn}}^{\text{ο}}\\ =98.8\text{kJ}-\left(\text{298}\text{K}\times 141.8\times {10}^{-3}{\text{kJK}}^{-1}\right)\\ =98.8\text{kJ}-42.25\text{kJ}\\ =56.54\text{kJ}\end{array}$

Thus, the value of $\Delta G^{\text{ο}}$ for the reduction of iron ore is $\underset{\_}{56.54kJ}$.

Conclusion

The value of $\Delta G^{\text{ο}}$ for the reduction of iron ore is $\underset{\_}{56.54kJ}$