To find the standard enthalpy of formation (\( \Delta_f H^\circ \)) for \( \text{MgO}(s) \), we can use Hess's law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for the individual steps. We will use the given reactions and their enthalpy changes to derive the value for \( \Delta_f H^\circ \) for \( \text{MgO}(s) \).
### Step-by-Step Solution:
1. **Write the given reactions and their enthalpy changes:**
- Reaction 1:
\[
\text{Mg}(s) + 2 \text{HCl}(aq) \rightarrow \text{MgCl}_2(aq) + \text{H}_2(g), \quad \Delta_r H^\circ = -467 \, \text{kJ/mol}
\]
- Reaction 2:
\[
\text{MgO}(s) + 2 \text{HCl}(aq) \rightarrow \text{MgCl}_2(aq) + \text{H}_2O(l), \quad \Delta_r H^\circ = -151 \, \text{kJ/mol}
\]
- Enthalpy of formation of water:
\[
\text{H}_2(g) + \frac{1}{2} \text{O}_2(g) \rightarrow \text{H}_2O(l), \quad \Delta_f H^\circ = -26 \, \text{kJ/mol}
\]
2. **Identify the target reaction for the formation of \( \text{MgO}(s) \):**
\[
\text{Mg}(s) + \frac{1}{2} \text{O}_2(g) \rightarrow \text{MgO}(s)
\]
3. **Use Hess's Law to combine the reactions:**
- We need to manipulate the reactions to arrive at the formation of \( \text{MgO}(s) \).
- From Reaction 2, we can reverse it to express it in terms of \( \text{MgO}(s) \):
\[
\text{MgCl}_2(aq) + \text{H}_2O(l) \rightarrow \text{MgO}(s) + 2 \text{HCl}(aq), \quad \Delta_r H^\circ = +151 \, \text{kJ/mol}
\]
4. **Add the modified Reaction 2 to Reaction 1:**
- Now, we add Reaction 1 and the reversed Reaction 2:
\[
\text{Mg}(s) + 2 \text{HCl}(aq) \rightarrow \text{MgCl}_2(aq) + \text{H}_2(g) \quad (-467 \, \text{kJ/mol})
\]
\[
\text{MgCl}_2(aq) + \text{H}_2O(l) \rightarrow \text{MgO}(s) + 2 \text{HCl}(aq) \quad (+151 \, \text{kJ/mol})
\]
- The \( \text{MgCl}_2(aq) \) and \( 2 \text{HCl}(aq) \) cancel out, leading to:
\[
\text{Mg}(s) + \frac{1}{2} \text{O}_2(g) \rightarrow \text{MgO}(s)
\]
5. **Calculate the total enthalpy change:**
- The total enthalpy change for the formation of \( \text{MgO}(s) \) is:
\[
\Delta_f H^\circ = (-467 \, \text{kJ/mol}) + (+151 \, \text{kJ/mol}) - (-26 \, \text{kJ/mol})
\]
- Simplifying this gives:
\[
\Delta_f H^\circ = -467 + 151 + 26 = -290 \, \text{kJ/mol}
\]
### Final Answer:
The standard enthalpy of formation for \( \text{MgO}(s) \) is:
\[
\Delta_f H^\circ = -290 \, \text{kJ/mol}
\]
