Question

An important source of copper is from the copper ore, chalcocite, a form of copper(I) sulfide. When heated, the \({\bf{C}}{{\bf{u}}_{\bf{2}}}{\bf{S}}\) decomposes to form copper and sulfur described by the following equation:

\({\bf{C}}{{\bf{u}}_{\bf{2}}}{\bf{\;S(s)}} \to {\bf{Cu(s) + S(s)}}\)

(a) Determine \({\bf{\Delta G}}_{{\bf{298}}}^{\bf{^\circ }}\)for the decomposition of \({\bf{C}}{{\bf{u}}_{\bf{2}}}{\bf{S(\;s)}}\).

(b) The reaction of sulfur with oxygen yields sulfur dioxide as the only product. Write an equation that describes this reaction, and determine\({\bf{\Delta G}}_{{\bf{298}}}^{\bf{^\circ }}\)for the process.

(c) The production of copper from chalcocite is performed by roasting the \({\bf{C}}{{\bf{u}}_{\bf{2}}}{\bf{S}}\) in air to produce the \({\bf{Cu}}\). By combining the equations from Parts (a) and (b), write the equation that describes the roasting of the chalcocite, and explain why coupling these reactions together makes for a more efficient process for the production of the copper.

Short answer

1. The value of \({\bf{\Delta G}}_{{\bf{298}}}^{\bf{0}}\) for decomposition of \({\bf{C}}{{\bf{u}}_{\bf{2}}}{\bf{S(s)}}\) is \({\bf{86}}{\bf{.2\;kJ/mol}}\).
2. The value of free energy change is \({\bf{ - 300}}{\bf{.2\;kJ/mol}}\).
3. The value of standard Gibbs free energy change is negative thus, the reaction is spontaneous. This is because this is the efficient method to produce \({\bf{Cu}}\).

Step-by-step solution

Define entropy of reaction

For a general reaction as follows:

\({\rm{A}} \to {\rm{B}} + {\rm{C}}\)

The change in Gibbs free energy for the reaction can be calculated as follows:

\(\Delta {G_{298}}^0 = \Delta G_B^0 + \Delta G_C^0 - \Delta G_A^0\)

a) Determine the decomposition reaction.

The decomposition reaction of \({\rm{C}}{{\rm{u}}_2}\;{\rm{S}}(s)\) is as follows:

\({\rm{C}}{{\rm{u}}_2}\;{\rm{S}}(s) \to 2{\rm{Cu}}(s) + {\rm{S}}(s)\)

The change in Gibbs free energy for the reaction can be calculated as follows:

\(\Delta {G_{298}}^0 = 2\Delta {G^0}{\rm{Cu}}(s) + \Delta {G^0}\;{\rm{S}}(s) - \Delta {G^0}{\rm{C}}{{\rm{u}}_2}\;{\rm{S}}(s)\)

From Appendix G:

\(\begin{array}{l}\Delta {{\rm{G}}_{{f_{{\rm{C}}{{\rm{u}}_2}\;{\rm{S}}(s)}}}} = - 86.1904\;{\rm{kJ}}/{\rm{mol}}\\\Delta {{\rm{G}}^0}{\rm{Cu}}(s) = 0\\\Delta {{\rm{G}}^0}\;{\rm{S}}(s) = 0\end{array}\)

Putting the values,

\(\begin{array}{l}\Delta {G_{298}}^0 = 0 + 0 - ( - 86.1904\;{\rm{kJ}}/{\rm{mol}})\\ = 86.1904\;{\rm{kJ}}/{\rm{mol}}\end{array}\)

Thus, the value of \(\Delta G_{298}^0\) for decomposition of \({\rm{C}}{{\rm{u}}_2}\;{\rm{S}}(s)\) is \(86.2\;{\rm{kJ}}/{\rm{mol}}\).

b) Determine the free energy of the decomposition reaction.

The reaction is as follows:

\({\rm{S}}(s) + {{\rm{O}}_2}(g) \to {\rm{S}}{{\rm{O}}_2}(g)\)

For the above reaction, change in standard free energy can be calculated as follows:

\(\Delta {G^0}_{298} = \Delta {G^0}{\rm{S}}{{\rm{O}}_2}(g) - \left( {\Delta {G^0}\;{\rm{S}}(s) + \Delta {G^0}{{\rm{O}}_2}(g)} \right)\)

From Appendix G, the value of Gibbs free energy change is as follows:

\(\begin{array}{l}\Delta {{\rm{G}}^0}\;{\rm{S}}(s) = 0\\\Delta {{\rm{G}}^0}{{\rm{O}}_2}(g) = 0\\\Delta {{\rm{G}}^0}{\rm{S}}{{\rm{O}}_2}(g) = - 300.2\;{\rm{kJ}}/{\rm{mol}}\end{array}\)

Putting the values,

\(\begin{array}{l}\Delta {G^0}_{298} = - 300.2\;{\rm{kJ}}/{\rm{mol}} - (0 + 0)\\ = - 300.2\;{\rm{kJ}}/{\rm{mol}}\end{array}\)

Thus, the value of free energy change is \( - 300.2\;{\rm{kJ}}/{\rm{mol}}\).

Determine the value of standard Gibbs free energy change

Adding reaction (1) and (2) from part (a) and (b), the overall reaction will be as follows:

\({\rm{C}}{{\rm{u}}_2}\;{\rm{S}}(s) + {{\rm{O}}_2}(g) \to 2{\rm{Cu}}(s) + {\rm{S}}{{\rm{O}}_2}(g)\)

The value of Gibbs free energy can be calculated as follows:

\(\Delta {G^0}_{298} = 2\Delta {G^0}{\rm{Cu}}(s) + \Delta {G^0}{\rm{S}}{{\rm{O}}_2}(g) - \Delta {G^0}{\rm{C}}{{\rm{u}}_2}\;{\rm{S}}(s) + \Delta {G^0}{{\rm{O}}_2}(g)\)

Putting the values,

\(\Delta G_{298}^0 = 2(0) + ( - 300.2\;{\rm{kJ}}/{\rm{mol}}) - (86.2\;{\rm{kJ}}/{\rm{mol}}) - (0) = - 386.4\;{\rm{kJ}}/{\rm{mol}}\)

Since, the value of standard Gibbs free energy change is negative thus, the reaction is spontaneous. This is because this is the efficient method to produce \({\rm{Cu}}\).