Chapter 5, Problem 5.142AP

Interpretation Introduction

Interpretation: The comparison of energy required to recycle one mole of copper with that required to recover $\text{Cu}$ from $\text{CuO}$ is to be stated.

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

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

To determine: The comparison of energy required to recycle one mole of copper with that required to recover $\text{Cu}$ from $\text{CuO}$.

Answer to Problem 5.142AP

Solution

The energy required to recycle one mole of copper is less than that needed to recover copper from copper oxide.

Explanation of Solution

Explanation

Given

The melting point of copper is $1084.5°\text{C}$.

The molar heat capacity of copper $\left(c_{\text{P,}\text{Cu}}\right)$ is $24.5\text{J}/\left(\text{mol}\cdot °\text{C}\right)$.

The molar enthalpy of fusion of copper is $\left(\Delta H_{\text{fus}}\right)$ is $13.0\text{kJ}/\text{mol}$.

The molar enthalpy of formation of copper oxide $\left(\Delta H_{\text{f,}\text{CuO}\left(s\right)}^{\text{ο}}\right)$ is $-155\text{kJ}/\text{mol}$.

The balanced chemical equation for preparation of copper from $\text{CuO}$ is,

$\text{CuO}\left(s\right)+\text{CO}\left(g\right)\to \text{Cu}\left(s\right)+{\text{CO}}_2\left(g\right)$ (1)

The following information are required to solve this problem.

$\begin{array}{c}\Delta H_{\text{f,}\text{CO}\left(g\right)}^{\text{ο}}=-110.525\text{kJ}/\text{mol}\\ \Delta H_{\text{f,}{\text{CO}}_2\left(g\right)}^{\text{ο}}=-393.5\text{kJ}/\text{mol}\\ \Delta H_{\text{f,}\text{Cu}\left(\text{s}\right)}^{\text{ο}}=0\text{kJ}/\text{mol}\end{array}$

The above stated reaction is balanced. The energy required to recover copper from copper oxide is calculated by standard enthalpy change.

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 $\text{Cu}\left(s\right)$ is $1$.
• Number of moles of product ${\text{CO}}_2\left(g\right)$ is $1$.
• Number of moles of reactant $\text{CO}\left(g\right)$ is $1$.
• Number of moles of reactant $\text{CuO}\left(s\right)$ is $1$.

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,

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

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

$\begin{array}{c}{\displaystyle \sum n\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Products}\right)}=1\text{mol}\times -393.5\text{kJ}/\text{mol}+1\text{mol}\times 0\text{kJ}/\text{mol}\\ =-393.5\text{kJ}\end{array}$

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,

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

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

$\begin{array}{c}{\displaystyle \sum m\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Reactants}\right)}=1\text{mol}\times -155\text{kJ}/\text{mol}+1\text{mol}\times -110.525\text{kJ}/\text{mol}\\ =-265.52\text{kJ}\end{array}$

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).

$\begin{array}{c}\Delta H_{\text{balanced rxn}}^{\text{ο}}=-393.5\text{kJ}-\left(-265.52\text{kJ}\right)\\ =-393.5\text{kJ+}265.52\text{kJ}\\ =-127.98\text{kJ}\end{array}$

The above calculated value of $\Delta H_{\text{balanced rxn}}^{\text{ο}}$ is the enthalpy change for one mole. The recovery of copper from copper oxide evolved $\underset{\_}{127.98kJ}$ energy.

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

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

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

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

Where,

• $n$ is number of moles of copper.
• $c_{\text{P,}\text{Cu}}$ is molar heat capacity of copper.
• $\Delta T$ 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 copper.
• $T_1$ is initial temperature that is room temperature.

The melting point of copper that is final temperature $\left(T_2\right)$ is $1084.5°\text{C}$. 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).

$\begin{array}{c}\Delta T=1084.5°\text{C}-25°\text{C}\\ =1059.5°\text{C}\end{array}$

The molar heat capacity of copper $\left(c_{\text{P,}\text{Cu}}\right)$ is $24.5\text{J}/\left(\text{mol}\cdot °\text{C}\right)$. To change the units of molar heat capacity in $\text{kJ}/\left(\text{mol}\cdot °\text{C}\right)$ use conversion factor.

$\begin{array}{l}1\text{J}={10}^{-3}\text{kJ}\\ 24.5\text{J}/\left(\text{mol}\cdot °\text{C}\right)=24.5\times {10}^{-3}\text{kJ}/\left(\text{mol}\cdot °\text{C}\right)\end{array}$

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

$\begin{array}{c}q=1\text{mol}\times 24.5\times {10}^{-3}\text{kJ}/\left(\text{mol}\cdot °\text{C}\right)\times 1084.5°\text{C}\\ =26.57\text{kJ}\end{array}$

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

$\begin{array}{c}{T}_{\text{energy}}=13.0\text{kJ}/\text{mol}+26.57\text{kJ}\\ =39.\text{57}\text{kJ}\end{array}$

Thus, the total energy required to recycle one mole of copper metal is $\underset{\_}{39.57kJ}$.

Hence, the energy required to recycle one mole of copper is less than that needed to recover copper from copper oxide.

Conclusion

The energy required to recycle one mole of copper is less than that needed to recover copper from copper oxide