Find the value of `Delta_(f) H^(@)` for the reaction
`N_(2) O_(4) (g) + 3 CO(g) rarr N_(2) O (g) + 3 CO_(2) (g)`
Standard enthalpies of formation of `CO(g), CO_(2) (g), N_(2) O (g)`, and `N_(2) O_(4) (g)` are `- 110, - 393, 81`, and `9.7 kJ mol^(-1)`, respectively.
Strategy : The standard enthalpy change of a reaction is equal to the sum of the standard molar enthalpie of formation of the products each multiplied by its stiochiometric coefficient in the balanced equation, minus the corresponding sum of the standard molar enthalpies of formation of the reactants

Explanation

To find the standard enthalpy change (Δ_f H^(@)) for the reaction:

\[ N_2O_4(g) + 3 CO(g) \rightarrow N_2O(g) + 3 CO_2(g) \]

we will use the formula for the standard enthalpy change of a reaction:

\[ \Delta H_{reaction} = \sum (\Delta H_f^@ \text{ of products}) - \sum (\Delta H_f^@ \text{ of reactants}) \]

### Step-by-Step Solution:

1. **Identify the standard enthalpies of formation (ΔH_f^@)**:
- For \( N_2O_4(g) \): \( 9.7 \, \text{kJ/mol} \)
- For \( CO(g) \): \( -110 \, \text{kJ/mol} \)
- For \( N_2O(g) \): \( 81 \, \text{kJ/mol} \)
- For \( CO_2(g) \): \( -393 \, \text{kJ/mol} \)

2. **Write the equation for ΔH of the reaction**:
\[ \Delta H_{reaction} = \left[ \Delta H_f^@ (N_2O) + 3 \cdot \Delta H_f^@ (CO_2) \right] - \left[ \Delta H_f^@ (N_2O_4) + 3 \cdot \Delta H_f^@ (CO) \right] \]

3. **Substitute the values into the equation**:
\[ \Delta H_{reaction} = \left[ 81 + 3 \cdot (-393) \right] - \left[ 9.7 + 3 \cdot (-110) \right] \]

4. **Calculate the enthalpy of formation for the products**:
- For \( N_2O \): \( 81 \, \text{kJ/mol} \)
- For \( 3 \cdot CO_2 \): \( 3 \cdot (-393) = -1179 \, \text{kJ/mol} \)
- Total for products: \( 81 - 1179 = -1098 \, \text{kJ/mol} \)

5. **Calculate the enthalpy of formation for the reactants**:
- For \( N_2O_4 \): \( 9.7 \, \text{kJ/mol} \)
- For \( 3 \cdot CO \): \( 3 \cdot (-110) = -330 \, \text{kJ/mol} \)
- Total for reactants: \( 9.7 - 330 = -320.3 \, \text{kJ/mol} \)

6. **Combine the results to find ΔH of the reaction**:
\[ \Delta H_{reaction} = (-1098) - (-320.3) = -1098 + 320.3 = -777.7 \, \text{kJ/mol} \]

### Final Answer:
\[ \Delta H_{reaction} = -777.7 \, \text{kJ/mol} \]