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

Explanation

To find the heat of dimerization of \( NO_2(g) \), we can use the standard heats of formation provided in the question. The dimerization reaction can be represented as follows:

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

### Step 1: Write the formula for heat of dimerization
The heat of dimerization can be calculated using the formula:

\[ \Delta H_{dimerization} = \Delta H_f (products) - \Delta H_f (reactants) \]

### Step 2: Identify the heats of formation
From the question, we have:
- Heat of formation of \( N_2O_4(g) = 2.0 \, \text{kcal/mol} \)
- Heat of formation of \( NO_2(g) = 8.0 \, \text{kcal/mol} \)

### Step 3: Substitute the values into the formula
In our reaction, we have:
- Products: 1 mole of \( N_2O_4(g) \)
- Reactants: 2 moles of \( NO_2(g) \)

Thus, we can substitute the values into the formula:

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

Substituting the values:

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

### Step 4: Calculate the heat of dimerization
Now, calculate the right-hand side:

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

### Conclusion
The heat of dimerization of \( NO_2(g) \) is:

\[ \Delta H_{dimerization} = -14.0 \, \text{kcal} \]

This negative sign indicates that the reaction is exothermic, meaning energy is released during the dimerization process.

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