If `H_(2)(g)=2H(g),DeltaH=104cal` , then heat of atomisation of hydrogen is

Options

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

Explanation

To find the heat of atomization of hydrogen given the reaction \( H_2(g) \rightarrow 2H(g) \) with \( \Delta H = 104 \, \text{cal} \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Reaction**: The reaction given is the dissociation of one mole of hydrogen gas (\( H_2 \)) into two moles of hydrogen atoms (\( H \)). This is represented as:
\[ H_2(g) \rightarrow 2H(g) \]

2. **Identify the Given Data**: We are provided with the enthalpy change (\( \Delta H \)) for this reaction, which is:
\[ \Delta H = 104 \, \text{cal} \]
This value represents the energy required to break the bond in one mole of \( H_2 \) to form two moles of hydrogen atoms.

3. **Define Heat of Atomization**: The heat of atomization is defined as the energy change when one mole of a compound is dissociated into its constituent atoms in the gaseous state. In this case, the heat of atomization of hydrogen refers to the energy required to break one mole of \( H_2 \) into two moles of \( H \).

4. **Relate the Given Data to Heat of Atomization**: Since the reaction involves the dissociation of one mole of \( H_2 \) into two moles of \( H \), the heat of atomization for hydrogen is equal to the enthalpy change for this reaction. Therefore:
\[ \text{Heat of Atomization of } H = \Delta H = 104 \, \text{cal} \]

5. **Conclusion**: The heat of atomization of hydrogen is:
\[ \text{Heat of Atomization} = 104 \, \text{cal} \]

### Final Answer:
The heat of atomization of hydrogen is **104 calories**.