The enthalpy of formation of `CO(g), CO_(2)(g),N_(2)O(g)` and `N_(2)O_(4)(g)` is `-110,-393,+81` and 10 kJ / mol respectively. For the reaction `N_(2)O_(4)(g)+3CO(g)rarr N_(2)O(g)+3CO_(2)(g). Delta H_(r )` is

Options

A. `-212`
B. `+212`
C. `+778`
D. `-778`

Answer

To solve the problem, we will calculate the change in enthalpy (ΔH) for the reaction given the enthalpy of formation values for the reactants and products.

### Step-by-Step Solution:

1. **Identify the Reaction**:
The reaction we are considering is:
\[ N_2O_4(g) + 3CO(g) \rightarrow N_2O(g) + 3CO_2(g) \]

2. **List the Enthalpy of Formation Values**:
- ΔH_f for \( CO(g) = -110 \, \text{kJ/mol} \)
- ΔH_f for \( CO_2(g) = -393 \, \text{kJ/mol} \)
- ΔH_f for \( N_2O(g) = +81 \, \text{kJ/mol} \)
- ΔH_f for \( N_2O_4(g) = +10 \, \text{kJ/mol} \)

3. **Write the Formula for ΔH of the Reaction**:
The change in enthalpy for the reaction (ΔH_r) can be calculated using the formula:
\[ \Delta H_r = \sum (\Delta H_f \text{ of products}) - \sum (\Delta H_f \text{ of reactants}) \]

4. **Calculate the Enthalpy of Products**:
- For \( N_2O(g) \): \( \Delta H_f = +81 \, \text{kJ/mol} \)
- For \( 3CO_2(g) \): \( 3 \times (-393) = -1179 \, \text{kJ} \)
- Total for products:
\[ \Delta H_f \text{ (products)} = 81 + (-1179) = 81 - 1179 = -1098 \, \text{kJ} \]

5. **Calculate the Enthalpy of Reactants**:
- For \( N_2O_4(g) \): \( \Delta H_f = +10 \, \text{kJ/mol} \)
- For \( 3CO(g) \): \( 3 \times (-110) = -330 \, \text{kJ} \)
- Total for reactants:
\[ \Delta H_f \text{ (reactants)} = 10 + (-330) = 10 - 330 = -320 \, \text{kJ} \]

6. **Calculate ΔH_r**:
Now substituting the values into the ΔH_r formula:
\[ \Delta H_r = (-1098) - (-320) = -1098 + 320 = -778 \, \text{kJ} \]

7. **Final Answer**:
The change in enthalpy for the reaction is:
\[ \Delta H_r = -778 \, \text{kJ} \]