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

Expert verified
  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

01

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:

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

02

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 \(86.2\;{\rm{kJ}}/{\rm{mol}}\).

03

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 \( - 300.2\;{\rm{kJ}}/{\rm{mol}}\).

04

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 \({\rm{Cu}}\).

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Most popular questions from this chapter

In the laboratory, hydrogen chloride \({\bf{(HCl(g))}}\) and ammonia \(\left( {{\bf{N}}{{\bf{H}}_{\bf{3}}}{\bf{(g)}}} \right)\)often escape from bottles of their solutions and react to form the ammonium chloride\(\left( {{\bf{N}}{{\bf{H}}_{\bf{4}}}{\bf{Cl(s)}}} \right)\), the white glaze often seen on glassware. Assuming that the number of moles of each gas that escapes into the room is the same, what is the maximum partial pressure of \({\bf{HCl}}\) and \({\bf{N}}{{\bf{H}}_{\bf{3}}}\)in the laboratory at room temperature? (Hint: The partial pressures will be equal and are at their maximum value when at equilibrium.)

Many plastic materials are organic polymers that contain carbon and hydrogen. The oxidation of these plastics in air to form carbon dioxide and water is a spontaneous process; however, plastic materials tend to persist in the environment. Explain.

Without doing a numerical calculation, determine which of the following will reduce the free energy change for the reaction, that is, make it less positive or more negative, when the temperature is increased. Explain.

(a) \({{\bf{N}}_{\bf{2}}}{\bf{(g) + 3}}{{\bf{H}}_{\bf{2}}}{\bf{(g)}} \to {\bf{2N}}{{\bf{H}}_{\bf{3}}}{\bf{(g)}}\)

(b) \({\bf{HCl(g) + N}}{{\bf{H}}_{\bf{3}}}{\bf{(g)}} \to {\bf{N}}{{\bf{H}}_{\bf{4}}}{\bf{Cl(s)}}\)

(c) \({\left( {{\bf{N}}{{\bf{H}}_{\bf{4}}}} \right)_{\bf{2}}}{\bf{C}}{{\bf{r}}_{\bf{2}}}{{\bf{O}}_{\bf{7}}}{\bf{(s)}} \to {\bf{C}}{{\bf{r}}_{\bf{2}}}{{\bf{O}}_{\bf{3}}}{\bf{(s) + 4}}{{\bf{H}}_{\bf{2}}}{\bf{O(g) + }}{{\bf{N}}_{\bf{2}}}{\bf{(g)}}\)

(d) \({\bf{2Fe(s) + 3}}{{\bf{O}}_{\bf{2}}}{\bf{(g)}} \to {\bf{F}}{{\bf{e}}_{\bf{2}}}{{\bf{O}}_{\bf{3}}}{\bf{(s)}}\)

Calculate ΔG° using

(a) free energies of formation and

(b) enthalpies of formation and entropies(Appendix G). Do the results indicate the reaction to be spontaneous or nonspontaneous at 25 °C?

\({{\bf{C}}_{\bf{2}}}{{\bf{H}}_{\bf{4}}}{\bf{(g)}} \to {{\bf{H}}_{\bf{2}}}{\bf{(g) + }}{{\bf{C}}_{\bf{2}}}{{\bf{H}}_{\bf{4}}}{\bf{(g)}}\)

Use the information in Appendix G to estimate the boiling point of CS2.

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