Example 15

Bronze Age metallurgists were accomplished practical chemists who unknowingly used coupled reactions to obtain metals from their ores. Realizing that different ores of the same metal required different treatments, they heated copper oxide ore in the presence of charcoal (carbon) to obtain copper metal, whereas they pumped air into the reaction system if the ore was copper sulfide. Assume that a particular copper ore consists of pure cuprous oxide (Cu₂O). Using the $\Delta G_{\text{f}}^{°}$ values given for each, calculate

1. ΔG° and K_p for the decomposition of Cu₂O to metallic copper and oxygen gas $\left[\Delta G_{\text{f}}^{°}\left({\text{Cu}}_2\text{O}\right)=-146.0\text{ kJ/mol}\right]\text{.}$
2. ΔG° and K_p for the reaction of Cu₂O with carbon to produce metallic copper and carbon monoxide $\left[\Delta G_{\text{f}}^{°}\left(\text{CO}\right)=-137.2\text{ kJ/mol}\right]\text{.}$

Given: reactants and products, $\Delta G_{\text{f}}^{°}$ values for Cu₂O and CO, and temperature

Asked for: ΔG° and K_p for the formation of metallic copper from Cu₂O in the absence and presence of carbon

Strategy:

A Write the balanced equilibrium equation for each reaction. Using the “products minus reactants” rule, calculate ΔG° for the reaction.

B Substitute appropriate values into Equation 18.36 to obtain K_p.

Solution:

1. A The chemical equation for the decomposition of cuprous oxide is as follows:

   $${\text{Cu}}_2\text{O(s)}\rightleftharpoons 2\text{Cu(s)}+\frac{1}{2}{\text{O}}_2\text{(g)}$$

   The substances on the right side of this equation are pure elements in their standard states, so their $\Delta G_{\text{f}}^{°}$ values are zero. ΔG° for the reaction is therefore

   $$\begin{array}{cc}\hfill \Delta G°& =\left[2\Delta G_{\text{f}}^{°}\text{(Cu)}+\frac{1}{2}\Delta G_{\text{f}}^{°}({\text{O}}_2)\right]-\Delta G_{\text{f}}^{°}({\text{Cu}}_2\text{O})\hfill \\ \hfill & =\left[(2\cancel{\text{mol}})(0\text{ kJ/}\cancel{\text{mol}})+\left(\frac{1}{2}\cancel{\text{mol}}\right)(0\text{ kJ/}\cancel{\text{mol}})\right]-[(1\cancel{\text{mol}})(-146.0\text{ kJ/}\cancel{\text{mol}})]\hfill \\ \hfill & =146.0\text{ kJ}\hfill \end{array}$$

   B Rearranging and substituting the appropriate values into Equation 18.36,

   $$\begin{array}{cc}\hfill \text{ln }{K}_{\text{p}}& =-\frac{\Delta G°}{RT}=-\frac{(146.0\cancel{\text{kJ}})(1000\cancel{\text{J}}\text{/}\cancel{\text{kJ}})}{(8.314\cancel{\text{J}}\text{/}\cancel{\text{K}})(298.15\cancel{\text{K}})}=-58.90\hfill \\ \hfill K_{\text{p}}& =2.6\times {10}^{-26}\hfill \end{array}$$

   This is a very small number, indicating that Cu₂O does not spontaneously decompose to a significant extent at room temperature.

2. A The O₂ produced in the decomposition of Cu₂O can react with carbon to form CO:

   $$\frac{1}{2}{\text{O}}_2\text{(g)}+\text{C(s)}\rightleftharpoons \text{CO(g)}$$

   Because ΔG° for this reaction is equal to $\Delta G_{\text{f}}^{°}$ for CO (−137.2 kJ/mol), it is energetically more feasible to produce metallic copper from cuprous oxide by coupling the two reactions:

   $$\begin{array}{cccc}\hfill {\text{Cu}}_2\text{O(s)}& \rightleftharpoons 2\text{Cu(s)}+\cancel{\frac{1}{2}{\text{O}}_2\text{(g)}}\hfill & \Delta G°& =146.0\text{ kJ/mol}\hfill \\ \hfill \cancel{\frac{1}{2}{\text{O}}_2\text{(g)}}+\text{C(s)}& \rightleftharpoons \text{ CO(g)}\hfill & \Delta G°& =-137.2\text{ kJ/mol}\hfill \\ \hfill {\text{Cu}}_2\text{O(s)}+\text{C(s)}& \rightleftharpoons 2\text{Cu(s)}+\text{CO(g)}\hfill & \Delta G_{\text{f}}^{°}& =8.8\text{ kJ/mol}\hfill \end{array}$$

   B We can find the corresponding value of K_p:

   $$\begin{array}{cc}\hfill \text{ln }{K}_{\text{p}}& =-\frac{\Delta G°}{RT}=-\frac{(8.8\cancel{\text{kJ}})(1000\cancel{\text{J}}\text{/}\cancel{\text{kJ}})}{(8.314\cancel{\text{J}}\text{/}\cancel{\text{K}})(298.15\cancel{\text{K}})}=-3.6\hfill \\ \hfill K_{\text{p}}& =0.03\hfill \end{array}$$

   Although this value is still less than 1, indicating that reactants are favored over products at room temperature, it is about 24 orders of magnitude greater than K_p for the production of copper metal in the absence of carbon. Because both ΔH° and ΔS° are positive for this reaction, it becomes thermodynamically feasible at slightly elevated temperatures (greater than about 80°C). At temperatures of a few hundred degrees Celsius, the reaction occurs spontaneously, proceeding smoothly and rapidly to the right as written and producing metallic copper and carbon monoxide from cuprous oxide and carbon.

Exercise

Use the $\Delta G_{\text{f}}^{°}$ values given to calculate ΔG° and K_p for each reaction.

1. the decomposition of cuprous sulfide to copper metal and elemental sulfur [$\Delta G_{\text{f}}^{°}\left({\text{Cu}}_{\text{2}}\text{S}\right)=-\text{86}.\text{2 kJ/mol}]$
2. the reaction of cuprous sulfide with oxygen gas to produce sulfur dioxide and copper metal [$\Delta G_{\text{f}}^{°}$ [SO₂(g)] = −300.1 kJ/mol]

Answer:

1. ΔG° = 86.2 kJ/mol; K_p = 7.90 × 10⁻¹⁶
2. ΔG° = −213.9 kJ/mol; K_p = 2.99 × 10³⁷