To determine the value of [tex]\(\Delta H^{\circ}\)[/tex] for the reaction [tex]\(2 \ \text{CuO(s)} \rightarrow \text{Cu}_2\text{O(s)} + \frac{1}{2} \text{O}_2\text{(g)}\)[/tex], we can use the given reactions and Hess's Law. Hess's Law states that if a reaction is carried out in a series of steps, the [tex]\(\Delta H\)[/tex] for the overall reaction is the sum of the [tex]\(\Delta H\)[/tex] values for each step.
Given reactions:
1. [tex]\(\text{Cu(s)} + \frac{1}{2} \text{O}_2\text{(g)} \rightarrow \text{CuO(s)}, \quad \Delta H^{\circ} = -156 \ \text{kJ}\)[/tex]
2. [tex]\(\text{2Cu(s)} + \frac{1}{2} \text{O}_2\text{(g)} \rightarrow \text{Cu}_2\text{O(s)}, \quad \Delta H^{\circ} = -170 \ \text{kJ}\)[/tex]
We need to find [tex]\(\Delta H^{\circ}\)[/tex] for:
[tex]\[2 \ \text{CuO(s)} \rightarrow \text{Cu}_2\text{O(s)} + \frac{1}{2} \text{O}_2\text{(g)}\][/tex]
Step 1: Reverse the first given reaction to form [tex]\(\text{CuO(s)}\)[/tex] as a reactant:
[tex]\[\text{CuO(s)} \rightarrow \text{Cu(s)} + \frac{1}{2} \text{O}_2\text{(g)}, \quad \Delta H^{\circ} = +156 \ \text{kJ}\][/tex]
Step 2: Multiply the reversed reaction by 2 to match the stoichiometry of [tex]\(\text{CuO(s)}\)[/tex] in the desired reaction:
[tex]\[2 \ \text{CuO(s)} \rightarrow 2 \ \text{Cu(s)} + \text{O}_2\text{(g)}, \quad \Delta H^{\circ} = 2 \times 156 \ \text{kJ} = +312 \ \text{kJ}\][/tex]
Step 3: Combine the modified first reaction with the second given reaction. The second reaction does not need to be altered:
[tex]\[2 \ \text{Cu(s)} + \frac{1}{2} \text{O}_2\text{(g)} \rightarrow \text{Cu}_2\text{O(s)}, \quad \Delta H^{\circ} = -170 \ \text{kJ}\][/tex]
Adding these reactions together, we get:
[tex]\[
2 \ \text{CuO(s)} \rightarrow 2 \ \text{Cu(s)} + \text{O}_2\text{(g)}, \quad \Delta H^{\circ} = +312 \ \text{kJ} \\
2 \ \text{Cu(s)} + \frac{1}{2} \text{O}_2\text{(g)} \rightarrow \text{Cu}_2\text{O(s)}, \quad \Delta H^{\circ} = -170 \ \text{kJ}
\][/tex]
When combined:
[tex]\[
2 \ \text{CuO(s)} + 2 \ \text{Cu(s)} + \text{O}_2\text{(g)} \rightarrow 2 \ \text{Cu(s)} + \text{Cu}_2\text{O(s)} + \frac{1}{2} \ \text{O}_2\text{(g)}, \quad \Delta H^{\circ} = +312 \ \text{kJ} - 170 \ \text{kJ}
\][/tex]
Simplify the equation by canceling out common terms:
[tex]\[
2 \ \text{CuO(s)} \rightarrow \text{Cu}_2\text{O(s)} + \frac{1}{2} \text{O}_2\text{(g)}, \quad \Delta H^{\circ} = 312 \ \text{kJ} - 170 \ \text{kJ} = +142 \ \text{kJ}
\][/tex]
Therefore, the value of [tex]\(\Delta H^{\circ}\)[/tex] for the reaction [tex]\(2 \ \text{CuO(s)} \rightarrow \text{Cu}_2\text{O(s)} + \frac{1}{2} \text{O}_2\text{(g)}\)[/tex] is [tex]\(+142 \ \text{kJ}\)[/tex].
The correct answer is [tex]\( A. \ 142 \)[/tex].
