If `H_(2)(g) Then heat of atomization of `H_(2)` is

Options

A. 52 kcals
B. 104 kcals
C. 20 kcals
D. none of the above

Answer

To find the heat of atomization of \( H_2 \), we need to understand the process involved in breaking down the \( H_2 \) molecule into individual hydrogen atoms.

### Step-by-Step Solution:

1. **Understanding the Reaction**:
The reaction given is:
\[ H_2(g) \rightarrow 2H(g) \]
This indicates that one molecule of hydrogen gas (\( H_2 \)) dissociates into two individual hydrogen atoms (\( H \)).

2. **Heat of Reaction**:
The heat change associated with this reaction is given as \( \Delta H = 104 \, \text{kcal} \). This value represents the total energy required to break the bond in one mole of \( H_2 \) to form two moles of \( H \).

3. **Heat of Atomization Definition**:
The heat of atomization is defined as the enthalpy change when one mole of a compound is converted into its individual atoms. In this case, since \( H_2 \) produces two moles of \( H \), the heat of atomization for one mole of \( H_2 \) is half of the total heat change.

4. **Calculating Heat of Atomization**:
To find the heat of atomization for one mole of \( H_2 \):
\[ \text{Heat of Atomization} = \frac{\Delta H}{2} = \frac{104 \, \text{kcal}}{2} = 52 \, \text{kcal} \]

5. **Final Answer**:
Therefore, the heat of atomization of \( H_2 \) is:
\[ \text{Heat of Atomization} = 52 \, \text{kcal} \]

### Summary:
The heat of atomization of \( H_2 \) is \( 52 \, \text{kcal} \).