The net enthalpy change of a reaction is the amount of energy required to break all the bonds in reactant molecules minus amount of energy required to form all the bonds in the product molecules.What will be the enthalpy change for the following reaction.
`H_(2)(g) + Br_(2)(g) to 2HBr(g)`
Given that bond energy of `H_(2), Br_(2)` and HBr is 435 kJ `mol^(-1)`, 192 kJ `mol^(-1)` and `368 kJ mol^(-1)` respectively. (Keep your answer in magnitude and in KJ/mol)

Explanation

To find the enthalpy change for the reaction \( H_2(g) + Br_2(g) \rightarrow 2HBr(g) \), we will follow these steps:

### Step 1: Identify the bond energies
We are given the bond energies:
- Bond energy of \( H_2 \) = 435 kJ/mol
- Bond energy of \( Br_2 \) = 192 kJ/mol
- Bond energy of \( HBr \) = 368 kJ/mol

### Step 2: Calculate the total bond energy of the reactants
The reactants are \( H_2 \) and \( Br_2 \). The total bond energy for the reactants is calculated by adding the bond energies of each reactant:
\[ \text{Total bond energy of reactants} = \text{Bond energy of } H_2 + \text{Bond energy of } Br_2 \]
\[ = 435 \, \text{kJ/mol} + 192 \, \text{kJ/mol} = 627 \, \text{kJ/mol} \]

### Step 3: Calculate the total bond energy of the products
The products are \( 2HBr \). Since there are two moles of \( HBr \), we need to multiply the bond energy of \( HBr \) by 2:
\[ \text{Total bond energy of products} = 2 \times \text{Bond energy of } HBr \]
\[ = 2 \times 368 \, \text{kJ/mol} = 736 \, \text{kJ/mol} \]

### Step 4: Calculate the net enthalpy change
The net enthalpy change (\( \Delta H \)) for the reaction is given by the formula:
\[ \Delta H = \text{Total bond energy of reactants} - \text{Total bond energy of products} \]
Substituting the values we calculated:
\[ \Delta H = 627 \, \text{kJ/mol} - 736 \, \text{kJ/mol} = -109 \, \text{kJ/mol} \]

### Final Answer
The enthalpy change for the reaction is \( -109 \, \text{kJ/mol} \).

---