Enthalpy change for the reaction,
`4H(g) to 2H_2(g)` is -869.6 kJ
The dissociation energy of H - H bond is

Options

A. `-869.6 kJ`
B. ` + 434.8 kJ`
C. ` +217.4 kJ`
D. `-434.8 kJ`

Explanation

To find the dissociation energy of the H-H bond from the given reaction and enthalpy change, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reaction**:
The given reaction is:
\[ 4H(g) \rightarrow 2H_2(g) \]
The enthalpy change (ΔH) for this reaction is given as -869.6 kJ.

2. **Invert the Reaction**:
To find the dissociation energy, we need to consider the reverse of the given reaction:
\[ 2H_2(g) \rightarrow 4H(g) \]
When we invert the reaction, the sign of ΔH changes. Therefore, the new ΔH for the reverse reaction will be:
\[ ΔH = +869.6 \text{ kJ} \]

3. **Relate ΔH to Bond Energy**:
The reaction shows that breaking 2 moles of H-H bonds produces 4 moles of H atoms. The energy required to break these bonds can be expressed as:
\[ ΔH = 2 \times \text{(Bond Energy of H-H)} \]
Let the bond energy of H-H be represented as \( BE \). Thus, we can write:
\[ 869.6 \text{ kJ} = 2 \times BE \]

4. **Solve for Bond Energy**:
To find the bond energy (BE), we rearrange the equation:
\[ BE = \frac{869.6 \text{ kJ}}{2} \]
Calculating this gives:
\[ BE = 434.8 \text{ kJ} \]

5. **Conclusion**:
The dissociation energy of the H-H bond is:
\[ \text{Bond Energy of H-H} = 434.8 \text{ kJ} \]

### Final Answer:
The dissociation energy of the H-H bond is **434.8 kJ**.

---