`Mg(s)+2HCl (aq) to MgCl_(2) (aq) +H_(2)(g) triangle_(r) H^(@)=-467"kJ/mole"`
`MgO(s)+2HCl(aq) to MgCl_(2) (aq)+H_(2)O(l), triangle_(r)H^(@)=-151"kJ/mole"`
According to the information and given the fact that for water`triangle_(f) H^(@) "=-286 kJ/mole"`, what is the `triangle_(f)H^(@)" for "Mg0 (s)?`

Options

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

Answer

To find the standard enthalpy of formation (ΔHf°) for magnesium oxide (MgO), we can use the given reactions and their enthalpy changes. Here’s a step-by-step solution:

### Step 1: Write down the given reactions and their enthalpy changes.

1. **Reaction 1**:
\[ \text{Mg(s)} + 2 \text{HCl(aq)} \rightarrow \text{MgCl}_2\text{(aq)} + \text{H}_2\text{(g)} \quad \Delta H_r° = -467 \text{ kJ/mol} \]

2. **Reaction 2**:
\[ \text{MgO(s)} + 2 \text{HCl(aq)} \rightarrow \text{MgCl}_2\text{(aq)} + \text{H}_2\text{O(l)} \quad \Delta H_r° = -151 \text{ kJ/mol} \]

### Step 2: Reverse Reaction 2.

When we reverse Reaction 2, the enthalpy change will change its sign:

\[ \text{MgCl}_2\text{(aq)} + \text{H}_2\text{O(l)} \rightarrow \text{MgO(s)} + 2 \text{HCl(aq)} \quad \Delta H_r° = +151 \text{ kJ/mol} \]

### Step 3: Add Reaction 1 and the reversed Reaction 2.

Now, we can add the two reactions together:

\[ \text{Mg(s)} + 2 \text{HCl(aq)} + \text{MgCl}_2\text{(aq)} + \text{H}_2\text{O(l)} \rightarrow \text{MgCl}_2\text{(aq)} + \text{H}_2\text{(g)} + \text{MgO(s)} + 2 \text{HCl(aq)} \]

After canceling out the common terms on both sides, we get:

\[ \text{Mg(s)} + \text{H}_2\text{O(l)} \rightarrow \text{MgO(s)} + \text{H}_2\text{(g)} \]

### Step 4: Calculate the overall enthalpy change.

The overall enthalpy change for this reaction is the sum of the enthalpy changes of the two reactions:

\[ \Delta H_r° = -467 \text{ kJ/mol} + 151 \text{ kJ/mol} = -316 \text{ kJ/mol} \]

### Step 5: Relate the enthalpy change to the formation of MgO.

The reaction we derived can be expressed in terms of the enthalpy of formation:

\[ \Delta H_r° = \Delta H_f°(\text{MgO}) - \Delta H_f°(\text{H}_2\text{O}) - \Delta H_f°(\text{Mg}) \]

Since the standard enthalpy of formation for elements in their standard state (like Mg and H2) is zero, we have:

\[ -316 \text{ kJ/mol} = \Delta H_f°(\text{MgO}) - (-286 \text{ kJ/mol}) \]

### Step 6: Solve for ΔHf°(MgO).

Rearranging the equation gives:

\[ \Delta H_f°(\text{MgO}) = -316 \text{ kJ/mol} + 286 \text{ kJ/mol} \]
\[ \Delta H_f°(\text{MgO}) = -602 \text{ kJ/mol} \]

### Final Answer:
The standard enthalpy of formation for magnesium oxide (MgO) is:
\[ \Delta H_f°(\text{MgO}) = -602 \text{ kJ/mol} \]

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