The standard heat of formation of `NO_(2)(g)` and `N_(2)O_(4)(g) ` are `8.0` and `4.0 kcal mol ^(-1)` respectively. The heat of dimerisation of `NO_(2)` in kcal is`

Options

A. `-12 kcal`
B. `12 kcal`
C. `4 kcal`
D. `16 kcal`

Answer

To find the heat of dimerization of \( NO_2(g) \), we can use the standard heats of formation provided for \( NO_2(g) \) and \( N_2O_4(g) \).

### Step-by-Step Solution:

1. **Write the Dimerization Reaction:**
The dimerization of \( NO_2 \) can be represented as:
\[ 2 NO_2(g) \rightarrow N_2O_4(g) \]

2. **Identify the Standard Heats of Formation:**
From the question, we have:
- Standard heat of formation of \( NO_2(g) \): \( \Delta H_f^\circ (NO_2) = 8.0 \, \text{kcal/mol} \)
- Standard heat of formation of \( N_2O_4(g) \): \( \Delta H_f^\circ (N_2O_4) = 4.0 \, \text{kcal/mol} \)

3. **Use the Formula for Heat of Dimerization:**
The heat of dimerization (\( \Delta H_{dimerization} \)) can be calculated using the formula:
\[ \Delta H_{dimerization} = \Delta H_f^\circ (products) - \Delta H_f^\circ (reactants) \]

4. **Substitute the Values:**
For the reaction \( 2 NO_2 \rightarrow N_2O_4 \):
- Products: \( N_2O_4 \) (1 mole)
- Reactants: \( 2 NO_2 \) (2 moles)

Thus, substituting the values:
\[ \Delta H_{dimerization} = \Delta H_f^\circ (N_2O_4) - 2 \times \Delta H_f^\circ (NO_2) \]
\[ \Delta H_{dimerization} = 4.0 \, \text{kcal/mol} - 2 \times (8.0 \, \text{kcal/mol}) \]
\[ \Delta H_{dimerization} = 4.0 \, \text{kcal/mol} - 16.0 \, \text{kcal/mol} \]
\[ \Delta H_{dimerization} = 4.0 \, \text{kcal/mol} - 16.0 \, \text{kcal/mol} = -12.0 \, \text{kcal/mol} \]

5. **Final Answer:**
The heat of dimerization of \( NO_2(g) \) is:
\[ \Delta H_{dimerization} = -12.0 \, \text{kcal/mol} \]