QuestionApril 23, 2025

Show your work, using Hess's Law to solve for Delta H for the combustion of Magnesium. Mg(s)+1/2O_(2)(g)arrow MgO(s) square

Show your work, using Hess's Law to solve for Delta H for the combustion of Magnesium. Mg(s)+1/2O_(2)(g)arrow MgO(s) square
Show your work, using Hess's Law to solve for Delta H for the combustion of Magnesium. Mg(s)+1/2O_(2)(g)arrow MgO(s) square

Solution
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Answer

\Delta H = -44.3 \text{ kJ/mol} Explanation 1. Identify Given Reactions Use the following reactions with known \Delta H values: 1. Mg(s) + 2HCl(aq) \rightarrow MgCl_2(aq) + H_2(g), \Delta H = -467.1 \text{ kJ/mol} 2. H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l), \Delta H = -285.8 \text{ kJ/mol} 3. MgCl_2(aq) + H_2O(l) \rightarrow MgO(s) + 2HCl(aq), \Delta H = -125.6 \text{ kJ/mol} 2. Apply Hess's Law Combine reactions to form the target reaction: Mg(s) + \frac{1}{2}O_2(g) \rightarrow MgO(s). 3. Calculate \Delta H for Target Reaction Add the \Delta H values of the combined reactions: - Reverse reaction 1: MgCl_2(aq) + H_2(g) \rightarrow Mg(s) + 2HCl(aq), \Delta H = +467.1 \text{ kJ/mol} - Reaction 2 remains unchanged. - Reaction 3 remains unchanged. Sum: \Delta H = (+467.1) + (-285.8) + (-125.6) = -44.3 \text{ kJ/mol}

Explanation

1. Identify Given Reactions<br /> Use the following reactions with known $\Delta H$ values:<br />1. $Mg(s) + 2HCl(aq) \rightarrow MgCl_2(aq) + H_2(g)$, $\Delta H = -467.1 \text{ kJ/mol}$<br />2. $H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l)$, $\Delta H = -285.8 \text{ kJ/mol}$<br />3. $MgCl_2(aq) + H_2O(l) \rightarrow MgO(s) + 2HCl(aq)$, $\Delta H = -125.6 \text{ kJ/mol}$<br /><br />2. Apply Hess's Law<br /> Combine reactions to form the target reaction: $Mg(s) + \frac{1}{2}O_2(g) \rightarrow MgO(s)$.<br /><br />3. Calculate $\Delta H$ for Target Reaction<br /> Add the $\Delta H$ values of the combined reactions:<br />- Reverse reaction 1: $MgCl_2(aq) + H_2(g) \rightarrow Mg(s) + 2HCl(aq)$, $\Delta H = +467.1 \text{ kJ/mol}$<br />- Reaction 2 remains unchanged.<br />- Reaction 3 remains unchanged.<br /><br /> Sum: $\Delta H = (+467.1) + (-285.8) + (-125.6) = -44.3 \text{ kJ/mol}$
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