`Mg(s) + 2HCl(aq) rarr Mg Cl_(2) (aq) + H_(2)(g) , Delta_(r )H^(@) = -467 kJ//mol`
`MgO(s) + 2HCl (aq) rarr MgCl_(2)(aq) + H_(2)O (l), Delta_(r )H^(@) = -151 kJ//mol W`.
According to the information, and given the fact that for water, `Delta_(f)H^(@) = -286 kJ//mol` what is the `Delta_(f)H^(@)` for MgO(s)?

Options

A. `-904 kJ//mol`
B. `-602 kJ//mol`
C. `-334 kJ//mol`
D. `-30 kJ//mol`

Explanation

To find the standard heat of formation (\( \Delta_f H^\circ \)) for \( \text{MgO}(s) \), we can use the provided thermochemical equations and the heat of formation for water. Here's a step-by-step solution:

### Step 1: Write the reactions and their enthalpy changes

1. For the first reaction:
\[ \text{Mg}(s) + 2\text{HCl}(aq) \rightarrow \text{MgCl}_2(aq) + \text{H}_2(g), \quad \Delta_r H^\circ = -467 \, \text{kJ/mol} \]

2. For the second reaction:
\[ \text{MgO}(s) + 2\text{HCl}(aq) \rightarrow \text{MgCl}_2(aq) + \text{H}_2O(l), \quad \Delta_r H^\circ = -151 \, \text{kJ/mol} \]

### Step 2: Write the heat of formation equations

Using the definition of heat of reaction:
\[ \Delta_r H^\circ = \sum \Delta_f H^\circ (\text{products}) - \sum \Delta_f H^\circ (\text{reactants}) \]

For the first reaction:
\[ -467 = \Delta_f H^\circ (\text{MgCl}_2) + \Delta_f H^\circ (\text{H}_2) - \left( \Delta_f H^\circ (\text{Mg}) + 2 \Delta_f H^\circ (\text{HCl}) \right) \]
Since \( \Delta_f H^\circ (\text{Mg}) = 0 \) and \( \Delta_f H^\circ (\text{H}_2) = 0 \):
\[ -467 = \Delta_f H^\circ (\text{MgCl}_2) - 2 \Delta_f H^\circ (\text{HCl}) \quad \text{(Equation 1)} \]

For the second reaction:
\[ -151 = \Delta_f H^\circ (\text{MgCl}_2) + \Delta_f H^\circ (\text{H}_2O) - \left( \Delta_f H^\circ (\text{MgO}) + 2 \Delta_f H^\circ (\text{HCl}) \right) \]
Rearranging gives:
\[ -151 = \Delta_f H^\circ (\text{MgCl}_2) + \Delta_f H^\circ (\text{H}_2O) - \Delta_f H^\circ (\text{MgO}) - 2 \Delta_f H^\circ (\text{HCl}) \quad \text{(Equation 2)} \]

### Step 3: Substitute known values

We know \( \Delta_f H^\circ (\text{H}_2O) = -286 \, \text{kJ/mol} \). Now we can substitute this value into Equation 2:
\[ -151 = \Delta_f H^\circ (\text{MgCl}_2) - \Delta_f H^\circ (\text{MgO}) - 2 \Delta_f H^\circ (\text{HCl}) - 286 \]
This simplifies to:
\[ \Delta_f H^\circ (\text{MgCl}_2) - \Delta_f H^\circ (\text{MgO}) - 2 \Delta_f H^\circ (\text{HCl}) = 135 \quad \text{(Equation 3)} \]

### Step 4: Solve the equations

Now we have two equations:
1. From Equation 1:
\[ \Delta_f H^\circ (\text{MgCl}_2) = -467 + 2 \Delta_f H^\circ (\text{HCl}) \]

2. Substitute this into Equation 3:
\[ -467 + 2 \Delta_f H^\circ (\text{HCl}) - \Delta_f H^\circ (\text{MgO}) - 2 \Delta_f H^\circ (\text{HCl}) = 135 \]
Simplifying gives:
\[ -467 - \Delta_f H^\circ (\text{MgO}) = 135 \]
Rearranging yields:
\[ \Delta_f H^\circ (\text{MgO}) = -467 - 135 = -602 \, \text{kJ/mol} \]

### Final Answer
\[ \Delta_f H^\circ (\text{MgO}) = -602 \, \text{kJ/mol} \]