Chapter 5, Problem 5.141AP

Interpretation Introduction

Interpretation: The validation of statement that energy required to melt recycled iron less than that needed to reduce the iron in ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ to the free metal is to be stated.

Concept introduction: The total required heat energy $\left(T_{\text{energy}}\right)$ to melt one mole of iron is the summation of heat energy needed to melt $\left(q\right)$ and heat energy needed for fusion $\left(\Delta H_{\text{fus}}\right)$ of one mole of iron.

$T_{\text{energy}}=q+\Delta H_{\text{fus}}$

To determine: The validation of statement that energy required to melt recycled iron less than that needed to reduce the iron in ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ to the free metal.

Answer to Problem 5.141AP

The energy required to melt recycled iron is less than that needed to reduce the iron in ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ to the free metal.

Explanation of Solution

Explanation

Given:

The melting point of iron is $1538°\text{C}$.

The molar heat capacity of iron $\left(c_{\text{P,}\text{Fe}}\right)$ is $25.1\text{J}/\left(\text{mol}\cdot °\text{C}\right)$.

The molar enthalpy of fusion of iron is $\Delta H_{\text{fus}}$ is $19.4\text{kJ}/\text{mol}$.

The balanced chemical equation for making iron from ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ is,

$2{\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(s\right)+3\text{C}\left(s\right)\to 4\text{Fe}\left(s\right)+3{\text{CO}}_2\left(g\right)$ (1)

The following information are required to solve this problem.

$\begin{array}{c}\Delta H_{\text{f,}{\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(\text{s}\right)}^{\text{ο}}=-824.2\text{kJ}/\text{mol}\\ \Delta H_{\text{f,}{\text{CO}}_2\left(g\right)}^{\text{ο}}=-393.5\text{kJ}/\text{mol}\\ \Delta H_{\text{f,}\text{Fe}\left(\text{s}\right)}^{\text{ο}}=0\text{kJ}/\text{mol}\\ \Delta H_{\text{f,}\text{C}\left(\text{s}\right)}^{\text{ο}}=0\text{kJ}/\text{mol}\end{array}$

The above stated reaction is balanced.

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

$\Delta H_{\text{balanced rxn}}^{\text{ο}}={\displaystyle \sum n\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Products}\right)}-{\displaystyle \sum m\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Reactants}\right)}$ (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 the balanced chemical equation (1) the,

• Number of moles of product $\text{Fe}\left(s\right)$ is $4$.
• Number of moles of product ${\text{CO}}_2\left(g\right)$ is $3$.
• Number of moles of reactant $\text{C}\left(s\right)$ is $3$.
• Number of moles of reactant ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(s\right)$ is $2$.

The ${\displaystyle \sum n\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Products}\right)}$ for the balanced chemical reaction is calculated by the formula,

${\displaystyle \sum n\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Products}\right)}=3\text{mol}\times \Delta H_{\text{f,}{\text{CO}}_2\left(g\right)}^{\text{ο}}+4\text{mol}\times \Delta H_{\text{f,}\text{Fe}\left(\text{s}\right)}^{\text{ο}}$ (3)

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

$\begin{array}{c}{\displaystyle \sum n\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Products}\right)}=3\text{mol}\times -393.5\text{kJ}/\text{mol}+4\text{mol}\times 0\text{kJ}/\text{mol}\\ =-1180.5\text{kJ}\end{array}$

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

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

Substitute the values of $\Delta H_{\text{f,}{\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(\text{s}\right)}^{\text{ο}}$ and $\Delta H_{\text{f,}\text{C}\left(\text{s}\right)}^{\text{ο}}$ in equation (4).

$\begin{array}{c}{\displaystyle \sum m\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Reactants}\right)}=2\text{mol}\times -824.2\text{kJ}/\text{mol}+3\text{mol}\times 0\text{kJ}/\text{mol}\\ =-1,648.4\text{kJ}\end{array}$

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).

$\begin{array}{c}\Delta H_{\text{balanced rxn}}^{\text{ο}}=-1180.5\text{kJ}-\left(-1,648.4\text{kJ}\right)\\ =-1180.5\text{kJ+}1,648.4\text{kJ}\\ =468\text{kJ}\end{array}$

The above calculated value of $\Delta H_{\text{balanced rxn}}^{\text{ο}}$ is the enthalpy change for one mole. Divide the $\Delta H_{\text{balanced rxn}}^{\text{ο}}$ with $4$ moles of Iron produced in a balanced equation (1)

$\begin{array}{l}=\frac{468\text{kJ}}{4\text{mol}}\\ =117\text{kJ}/\text{mol}\end{array}$

Thus, the actual energy required for formation of one mole of iron is $\underset{\_}{117kJ}$.

The total required heat energy $\left(T_{\text{energy}}\right)$ to melt one mole of iron is the summation of heat energy needed to melt $\left(q\right)$ and heat energy needed for fusion $\left(\Delta H_{\text{fus}}\right)$ of one mole of iron.

$T_{\text{energy}}=q+\Delta H_{\text{fus}}$ (5)

The heat energy $\left(q\right)$ needed to melt one mole $\left(n\right)$ of iron is calculated by the formula,

$q=n\times c_{\text{P,}\text{Fe}}\times \Delta T$ (6)

Where,

• $n$ is number of moles of iron.
• $c_{\text{P,}\text{Fe}}$ is molar heat capacity of iron.
• $\Delta T$ is temperature difference between final and initial temperatures.

The temperature difference $\left(\Delta T\right)$ is calculated by the formula,

$\Delta T=T_2-T_1$ (7)

Where,

• $T_2$ is final temperature that is melting point of iron.
• $T_1$ is initial temperature that is room temperature.

The melting point of iron that is final temperature $\left(T_2\right)$ is $1538°\text{C}$. The initial temperature $\left(T_1\right)$ is assumed as $25°\text{C}$.

Substitute the values of $T_2$ and $T_1$ in equation (7).

$\begin{array}{c}\Delta T=1538°\text{C}-25°\text{C}\\ =1513°\text{C}\end{array}$

The molar heat capacity of iron $\left(c_{\text{P,}\text{Fe}}\right)$ is $25.1\text{J}/\left(\text{mol}\cdot °\text{C}\right)$. To change the units of molar heat capacity in $\text{kJ}/\left(\text{mol}\cdot °\text{C}\right)$ use conversion factor.

$\begin{array}{l}1\text{J}={10}^{-3}\text{kJ}\\ 25.1\text{J}/\left(\text{mol}\cdot °\text{C}\right)=25.1\times {10}^{-3}\text{kJ}/\left(\text{mol}\cdot °\text{C}\right)\end{array}$

Substitute the values of $n$, $c_{\text{P,}\text{Fe}\left(\text{kJ}/\left(\text{mol}\cdot °\text{C}\right)\right)}$ and $\Delta T$ in equation (6).

$\begin{array}{c}q=1\text{mol}\times 25.1\text{kJ}/\left(\text{mol}\cdot °\text{C}\right)\times {10}^{-3}\times 1513°\text{C}\\ =\text{37}\text{.976}\text{kJ}\end{array}$

Substitute the values of $\Delta H_{\text{fus}}$ and $q$ for one mole in equation (5).

$\begin{array}{c}{T}_{\text{energy}}=\text{37}\text{.976}\text{kJ}+19.4\text{kJ}\\ =\text{57}\text{.37}\text{kJ}\end{array}$

Thus, the total energy required to melt one mole of recycled iron metal is $\underset{\_}{57.37kJ}$.

Hence, the energy required to melt recycled iron is less than that needed to reduce the iron in ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ to the free metal.

Conclusion

Conclusion

The energy required to melt recycled iron is less than that needed to reduce the iron in ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ to the free metal.