Question

11 Question (3 points) See page 247 Copper has been used for thousands of years, either as a pure metal or in alloys. It is frequently used today in the production of wires and cables. Copper can be obtained through smelting or recycling. Determine the energy associated with each of these processes in order to recycle 145 mol Cu The smelting of copper occurs by the balanced chemical equation: Cuo(4) + CO(g) – Cu(s) + CO,(g) where AH cuois -155 kJ/mol. Assume the process of recycling copper is simplified to just the melting of the solid Cu starting at 25°C. The melting point of Cu is 1084.5'C with AH - 13.0 kJ/mol and a molar heat capacity, cc, 24.5J/mol C.

Answer 1

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Given reaction:

$CuO_{(s)} + CO_{(g)} \rightarrow Cu_{(s)}+ CO_{2}_{(g)}$

We know that, Enthalpy change for the reaction is given by,

$\Delta H_{rxn}=\Delta H_{f}(products)-\Delta H_{f}(reactants)$

$\Rightarrow \Delta H_{rxn}=\left [\Delta H_{f}(Cu_{(s)})+\Delta H_{f}(CO_{2}_{(g)})\right ]- \left [ \Delta H_{f}(CO_{(g)})+\Delta H_{f}(CuO_{(s)}) \right ]$

.$\Rightarrow \Delta H_{rxn}=\left [\0\;kJ/mol+(-393.5\;kJ/mol)\right ]- \left [ -110.5\;kJ/mol+(-155\;kJ/mol ) \right ]$

.

$\Rightarrow \Delta H_{rxn}=-393.5\;kJ/mol+110.5\;kJ/mol$

.

$\mathbf{\therefore \Delta H_{rxn}=-128\;kJ/mol }$

Given that, No.of moles of Cu, $n=1.45\;mol$

$\therefore \Delta H_{rxn}=-128\;kJ/mol \times 1.45\;mol$

$\mathbf{\therefore \Delta H_{rxn}=-185.6\;kJ}$

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We know that,Total energy required for recycling of Cu = Heat required to raise the temperature of Cu from 25°C to 1084.5°C (H₁) + Heat required to melt Cu at 1084.5°C(H₂).

Heat required to raise temperature is given by,

$H_{1}=nC_{p}\Delta T\\ \;\;\;=1.45\;mol\times 24.5\;J/mol^{o}C\times (1084.5^{o}C - 25^{o}C)$

$\therefore H_{1}=1.45\;mol\times 24.5\;J/mol^{o}C\times (1059.5^{o}C)$

$H_{1}=37638.74\;J$

$\therefore H_{1}=37638.74\times 10^{-3}\;kJ$

$\mathbf{\therefore H_{1}=37.64\;kJ}$

Heat required to melt Cu is given by,

$H_{2}=\Delta H_{fus}\times n=1.45\;mol \times 13.0\;kJ/mol$

$\therefore H_{2}=18.85\;kJ$

Hence, total energy is given by,

$H_{total}=H_{1}+H_{2}=37.64\;kJ+18.85\;kJ$

$\mathbf{\therefore H_{total}=56.49\;kJ}$

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