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. **Understand the Reaction**:
The dimerization of nitrogen dioxide (\( NO_2 \)) can be represented as:
\[ 2 NO_2(g) \rightarrow N_2O_4(g) \]

2. **Write the Heat of Dimerization Formula**:
The heat of dimerization (\( \Delta H_{dimerization} \)) can be calculated using the following formula:
\[ \Delta H_{dimerization} = \Delta H_f^{products} - \Delta H_f^{reactants} \]
where \( \Delta H_f \) is the standard heat of formation.

3. **Substitute the Values**:
From the problem, we know:
- The standard heat of formation of \( N_2O_4(g) \) is \( 4.0 \, \text{kcal/mol} \).
- The standard heat of formation of \( NO_2(g) \) is \( 8.0 \, \text{kcal/mol} \).

Since we have 2 moles of \( NO_2 \) as reactants, we can substitute the values into the formula:
\[ \Delta H_{dimerization} = \Delta H_f^{N_2O_4} - 2 \times \Delta H_f^{NO_2} \]
\[ \Delta H_{dimerization} = 4.0 \, \text{kcal/mol} - 2 \times 8.0 \, \text{kcal/mol} \]

4. **Calculate**:
\[ \Delta H_{dimerization} = 4.0 \, \text{kcal/mol} - 16.0 \, \text{kcal/mol} \]
\[ \Delta H_{dimerization} = -12.0 \, \text{kcal/mol} \]

5. **Interpret the Result**:
The negative sign indicates that the dimerization process is exothermic, meaning heat is released during the reaction.

### Final Answer:
The heat of dimerization of \( NO_2 \) is \( -12.0 \, \text{kcal} \).

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