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

Explanation

To calculate 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 1: Write the reaction and identify bonds broken and formed
In the reaction, we break the following bonds:
- 1 \( H-H \) bond
- 1 \( Br-Br \) bond

And we form:
- 2 \( H-Br \) bonds

### Step 2: Calculate the total energy required to break the bonds
The total energy required to break the bonds is the sum of the bond energies of the bonds broken:

\[ \text{Energy required} = \text{Bond energy of } H-H + \text{Bond energy of } Br-Br \]
\[ = 433 \, \text{kJ/mol} + 192 \, \text{kJ/mol} = 625 \, \text{kJ/mol} \]

### Step 3: Calculate the total energy released in forming the bonds
The total energy released when forming the bonds is given by:

\[ \text{Energy released} = 2 \times \text{Bond energy of } H-Br \]
\[ = 2 \times 364 \, \text{kJ/mol} = 728 \, \text{kJ/mol} \]

### Step 4: Calculate ΔH for the reaction
Now, we can calculate ΔH using the formula:

\[ \Delta H = \text{Energy required} - \text{Energy released} \]
\[ \Delta H = 625 \, \text{kJ/mol} - 728 \, \text{kJ/mol} = -103 \, \text{kJ/mol} \]

### Conclusion
The enthalpy change (ΔH) for the reaction \( H_2(g) + Br_2(g) \rightarrow 2HBr(g) \) is:

\[ \Delta H = -103 \, \text{kJ/mol} \]