Enthalpy change of the reaction
`2H(g) to H_(2)(g)` is `-104 kcal`
The `H-H` bond dissociation energy is

Options

A. `104 kcal`
B. `-104 kcal`
C. `-52 kcal`
D. `+52 kcal`

Explanation

To find the bond dissociation energy of the H-H bond, we can analyze the given reaction and the enthalpy change associated with it.

### Step-by-Step Solution:

1. **Understanding the Reaction**:
The reaction given is:
\[ 2H(g) \rightarrow H_2(g) \]
This indicates that two hydrogen atoms in the gaseous state combine to form one molecule of hydrogen gas (H2).

2. **Identifying the Enthalpy Change**:
The enthalpy change (ΔH) for this reaction is given as:
\[ \Delta H = -104 \text{ kcal} \]
This negative sign indicates that the reaction is exothermic, meaning energy is released when the bond is formed.

3. **Understanding Bond Dissociation Energy**:
The bond dissociation energy is defined as the energy required to break a bond in a molecule. In this case, it refers to the energy needed to break the H-H bond in H2 to form two hydrogen atoms.

4. **Reversing the Reaction**:
If we reverse the reaction, we get:
\[ H_2(g) \rightarrow 2H(g) \]
For this reverse reaction, the enthalpy change will be the opposite of the forward reaction. Therefore:
\[ \Delta H = +104 \text{ kcal} \]
This positive value indicates that this is the energy required to break the H-H bond.

5. **Conclusion**:
Thus, the bond dissociation energy of the H-H bond is:
\[ \text{Bond Dissociation Energy} = 104 \text{ kcal} \]

### Final Answer:
The bond dissociation energy of the H-H bond is **104 kcal**.

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