At standard conditions, if the change in the enthalpy for the following reaction is -109KJ`mol^(-1)`
`H_2(g) + Br_2(g) rarr 2HBr(g)`
Given that bond energy of `H_2 and Br_2` is 435kj`mol^(-1)` and 192kj`mol^(-1)`, respectively. What is the bond energy of HBr ?

Options

A. 368
B. 736
C. 518
D. 259

Answer

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

### Step 1: Write the reaction and identify the enthalpy change
The reaction is:
\[ H_2(g) + Br_2(g) \rightarrow 2 HBr(g) \]
The change in enthalpy (\( \Delta H \)) for this reaction is given as:
\[ \Delta H = -109 \, \text{kJ/mol} \]

### Step 2: List the bond energies of the reactants
The bond energies given are:
- Bond energy of \( H_2 \) = 435 kJ/mol
- Bond energy of \( Br_2 \) = 192 kJ/mol

### Step 3: Calculate the total bond energy of the reactants
The total bond energy of the reactants can be calculated by adding the bond energies of \( H_2 \) and \( Br_2 \):
\[ \text{Total bond energy of reactants} = \text{Bond energy of } H_2 + \text{Bond energy of } Br_2 \]
\[ = 435 \, \text{kJ/mol} + 192 \, \text{kJ/mol} = 627 \, \text{kJ/mol} \]

### Step 4: Set up the equation for the enthalpy change
The enthalpy change for the reaction can be expressed as:
\[ \Delta H = \text{Total bond energy of reactants} - \text{Total bond energy of products} \]
Let the bond energy of \( HBr \) be \( x \). Since 2 moles of \( HBr \) are formed, the total bond energy of products will be \( 2x \):
\[ -109 \, \text{kJ/mol} = 627 \, \text{kJ/mol} - 2x \]

### Step 5: Solve for \( x \)
Rearranging the equation gives:
\[ 2x = 627 \, \text{kJ/mol} + 109 \, \text{kJ/mol} \]
\[ 2x = 736 \, \text{kJ/mol} \]
Now, divide both sides by 2 to find \( x \):
\[ x = \frac{736 \, \text{kJ/mol}}{2} = 368 \, \text{kJ/mol} \]

### Conclusion
The bond energy of \( HBr \) is:
\[ \text{Bond energy of } HBr = 368 \, \text{kJ/mol} \]