The bond energies of `H--H` , `Br--Br` and `H--Br` are `433,, 192` and `364KJmol^(-1)` respectively. The `DeltaH^(@)` for the reaction
`H_(2)(g)+Br_(2)(g)rarr2HBr(g)` is

Options

A. `-261 KJ`
B. `+103 KJ`
C. `+261 KJ`
D. `-103 KJ`

Explanation

To find the enthalpy change (ΔH°) for the reaction:

\[ H_2(g) + Br_2(g) \rightarrow 2HBr(g) \]

we will use the bond energies provided:

- Bond energy of \( H-H \) = 433 kJ/mol
- Bond energy of \( Br-Br \) = 192 kJ/mol
- Bond energy of \( H-Br \) = 364 kJ/mol

### Step-by-Step Solution:

1. **Identify Bonds Broken and Formed:**
- In the reactants, we have:
- 1 mole of \( H-H \) bond
- 1 mole of \( Br-Br \) bond
- In the products, we have:
- 2 moles of \( H-Br \) bonds

2. **Calculate Energy Required to Break Bonds:**
- Energy required to break 1 mole of \( H-H \) bond = +433 kJ
- Energy required to break 1 mole of \( Br-Br \) bond = +192 kJ
- Total energy required to break bonds = \( 433 + 192 = 625 \) kJ

3. **Calculate Energy Released in Forming Bonds:**
- Energy released when forming 2 moles of \( H-Br \) bonds = \( 2 \times 364 = 728 \) kJ

4. **Calculate ΔH°:**
- The enthalpy change (ΔH°) for the reaction can be calculated using the formula:
\[ ΔH° = \text{Energy of bonds broken} - \text{Energy of bonds formed} \]
- Substituting the values:
\[ ΔH° = 625 \, \text{kJ} - 728 \, \text{kJ} = -103 \, \text{kJ} \]

5. **Conclusion:**
- The ΔH° for the reaction \( H_2(g) + Br_2(g) \rightarrow 2HBr(g) \) is \(-103 \, \text{kJ}\).

### Final Answer:
\[ ΔH° = -103 \, \text{kJ} \]