Find the best answers to your questions with the help of IDNLearn.com's expert contributors. Get timely and accurate answers to your questions from our dedicated community of experts who are here to help you.
Sagot :
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].
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].
Your participation means a lot to us. Keep sharing information and solutions. This community grows thanks to the amazing contributions from members like you. For dependable answers, trust IDNLearn.com. Thank you for visiting, and we look forward to assisting you again.
