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 find the enthalpy change (ΔH_r) for the reaction:

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

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

### Step 1: Write down the enthalpy of formation values
- \( \Delta H_f^\circ \) for \( CO(g) = -110 \, \text{kJ/mol} \)
- \( \Delta H_f^\circ \) for \( CO_2(g) = -393 \, \text{kJ/mol} \)
- \( \Delta H_f^\circ \) for \( N_2O(g) = +81 \, \text{kJ/mol} \)
- \( \Delta H_f^\circ \) for \( N_2O_4(g) = +10 \, \text{kJ/mol} \)

### Step 2: Write the formula for ΔH_r
The change in enthalpy for the reaction can be calculated using the formula:

\[ \Delta H_r = \sum (\Delta H_f^\circ \text{ of products}) - \sum (\Delta H_f^\circ \text{ of reactants}) \]

### Step 3: Calculate the enthalpy of formation for products
The products of the reaction are \( N_2O(g) \) and \( 3CO_2(g) \).

\[ \Delta H_f^\circ \text{ of products} = \Delta H_f^\circ (N_2O) + 3 \times \Delta H_f^\circ (CO_2) \]
\[ = 81 + 3 \times (-393) \]
\[ = 81 - 1179 = -1098 \, \text{kJ} \]

### Step 4: Calculate the enthalpy of formation for reactants
The reactants of the reaction are \( N_2O_4(g) \) and \( 3CO(g) \).

\[ \Delta H_f^\circ \text{ of reactants} = \Delta H_f^\circ (N_2O_4) + 3 \times \Delta H_f^\circ (CO) \]
\[ = 10 + 3 \times (-110) \]
\[ = 10 - 330 = -320 \, \text{kJ} \]

### Step 5: Substitute values into the ΔH_r formula
Now substituting the values into the ΔH_r formula:

\[ \Delta H_r = (-1098) - (-320) \]
\[ = -1098 + 320 = -778 \, \text{kJ} \]

### Final Answer
Thus, the enthalpy change for the reaction is:

\[ \Delta H_r = -778 \, \text{kJ/mol} \]