Calculate `Delta_(r ) H^(0)` for the following reaction at `25^(@)C`:
`Fe_(3) O_(4) (s) + CO(g) rarr 3FeO(s) + CO_(2) (g)`
`DeltaH_(f)^(@) (kJ//mol) - 1118 " " -110.5 " " -272 " " -393.5`

Options

A. `-263kJ`
B. 54 kJ
C. 19 kJ
D. `-50 kJ`

Answer

To calculate the standard enthalpy change (\( \Delta_r H^{\circ} \)) for the reaction:

\[ \text{Fe}_3\text{O}_4 (s) + \text{CO} (g) \rightarrow 3\text{FeO} (s) + \text{CO}_2 (g) \]

we will use the enthalpy of formation values provided for each substance involved in the reaction.

### Step-by-Step Solution:

1. **Identify the enthalpy of formation values**:
- \( \Delta H_f^{\circ} \) for \( \text{Fe}_3\text{O}_4 (s) = -1118 \, \text{kJ/mol} \)
- \( \Delta H_f^{\circ} \) for \( \text{CO} (g) = -110.5 \, \text{kJ/mol} \)
- \( \Delta H_f^{\circ} \) for \( \text{FeO} (s) = -272 \, \text{kJ/mol} \)
- \( \Delta H_f^{\circ} \) for \( \text{CO}_2 (g) = -393.5 \, \text{kJ/mol} \)

2. **Write the formula for the enthalpy change of the reaction**:
The enthalpy change for the reaction can be calculated using the formula:
\[ \Delta_r H^{\circ} = \sum (\Delta H_f^{\circ} \text{ of products}) - \sum (\Delta H_f^{\circ} \text{ of reactants}) \]

3. **Calculate the total enthalpy of formation for products**:
- For \( 3 \text{FeO} \):
\[ 3 \times (-272) = -816 \, \text{kJ} \]
- For \( \text{CO}_2 \):
\[ 1 \times (-393.5) = -393.5 \, \text{kJ} \]
- Total for products:
\[ -816 + (-393.5) = -1209.5 \, \text{kJ} \]

4. **Calculate the total enthalpy of formation for reactants**:
- For \( \text{Fe}_3\text{O}_4 \):
\[ 1 \times (-1118) = -1118 \, \text{kJ} \]
- For \( \text{CO} \):
\[ 1 \times (-110.5) = -110.5 \, \text{kJ} \]
- Total for reactants:
\[ -1118 + (-110.5) = -1228.5 \, \text{kJ} \]

5. **Calculate \( \Delta_r H^{\circ} \)**:
\[ \Delta_r H^{\circ} = (-1209.5) - (-1228.5) = -1209.5 + 1228.5 = 19 \, \text{kJ} \]

### Final Answer:
\[ \Delta_r H^{\circ} = 19 \, \text{kJ} \]