The standard heats of formation in Kcal `mol^(-1)` of `NO_(2)(g)` and `N_(2)O_(4)(g)` are 8.0 and 2.0 respectively. The heat of dimerization of `NO_(2)` in Kcal is. Given : `2NO_(2)(g)hArr N_(2)O_(4)(g)`

Options

A. `10.0`
B. `-6.0`
C. `-12.0`
D. `-14.0`

Answer

To find the heat of dimerization of \( NO_2(g) \) to form \( N_2O_4(g) \), we can use the standard heats of formation provided for both compounds. The reaction can be represented as:

\[ 2 NO_2(g) \rightleftharpoons N_2O_4(g) \]

### Step-by-Step Solution:

1. **Write the equation for the heat of dimerization:**
The heat of dimerization (\( \Delta H_{dimerization} \)) can be calculated using the standard heats of formation of the reactants and products.

\[ \Delta H_{dimerization} = \Delta H_f(N_2O_4) - 2 \Delta H_f(NO_2) \]

2. **Substitute the given values:**
From the problem, we know:
- \( \Delta H_f(NO_2) = 8.0 \, \text{kcal/mol} \)
- \( \Delta H_f(N_2O_4) = 2.0 \, \text{kcal/mol} \)

Now substituting these values into the equation:

\[ \Delta H_{dimerization} = 2.0 \, \text{kcal/mol} - 2 \times 8.0 \, \text{kcal/mol} \]

3. **Calculate the heat of dimerization:**
Now perform the calculation:

\[ \Delta H_{dimerization} = 2.0 \, \text{kcal/mol} - 16.0 \, \text{kcal/mol} \]
\[ \Delta H_{dimerization} = 2.0 - 16.0 = -14.0 \, \text{kcal/mol} \]

4. **Conclusion:**
The heat of dimerization of \( NO_2(g) \) is \( -14.0 \, \text{kcal} \).

### Final Answer:
The heat of dimerization of \( NO_2(g) \) is \( -14.0 \, \text{kcal} \).

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