Enthalpy change for the reaction
`4H(g)rarr2H_2(g)` is `-869.6kJ`
The dissociation energy of `H-H` bond is:

Options

A. `=217.4KJ`
B. `-434.8KJ`
C. `-869.6KJ`
D. `+434.8KJ`

Explanation

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

### Step 1: Understand the Reaction
The reaction given is:
\[ 4H(g) \rightarrow 2H_2(g) \]
with an enthalpy change (\( \Delta H \)) of \(-869.6 \, \text{kJ}\).

### Step 2: Reverse the Reaction
To find the bond dissociation energy, we need to reverse the reaction because the enthalpy change is given for the formation of \( H_2 \) from \( H \). The reversed reaction will be:
\[ 2H_2(g) \rightarrow 4H(g) \]
When we reverse the reaction, the sign of the enthalpy change also reverses:
\[ \Delta H = +869.6 \, \text{kJ} \]

### Step 3: Calculate the Energy per Bond
In the reversed reaction, we have 2 moles of \( H_2 \) producing 4 moles of \( H \). Each \( H_2 \) molecule contains one H-H bond. Therefore, to find the energy required to break one H-H bond, we divide the total energy by the number of bonds broken.

Since there are 2 moles of \( H_2 \), we have:
\[ \text{Number of H-H bonds} = 2 \]

Now, we can calculate the bond dissociation energy:
\[ \text{Bond dissociation energy} = \frac{869.6 \, \text{kJ}}{2} = 434.8 \, \text{kJ} \]

### Final Answer
The dissociation energy of the H-H bond is:
\[ \text{Dissociation energy} = 434.8 \, \text{kJ} \]

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