To calculate the enthalpy change ([tex]\( \Delta H \)[/tex]) for the reaction:
[tex]\[ H_2 + Br_2 \rightarrow 2HBr \][/tex]
we need to follow these steps:
### Step 1: Identify the bond energies
From the given data:
- Bond energy of [tex]\( H-H \)[/tex] (Hydrogen-Hydrogen bond): [tex]\( 432 \)[/tex] kJ/mol
- Bond energy of [tex]\( Br-Br \)[/tex] (Bromine-Bromine bond): [tex]\( 190 \)[/tex] kJ/mol
- Bond energy of [tex]\( H-Br \)[/tex] (Hydrogen-Bromine bond): [tex]\( 362 \)[/tex] kJ/mol
### Step 2: Calculate the energy required to break the bonds in the reactants
For [tex]\( H_2 \)[/tex]:
[tex]\[ 1 \times \text{bond energy of } H-H = 1 \times 432 \, \text{kJ} = 432 \, \text{kJ} \][/tex]
For [tex]\( Br_2 \)[/tex]:
[tex]\[ 1 \times \text{bond energy of } Br-Br = 1 \times 190 \, \text{kJ} = 190 \, \text{kJ} \][/tex]
Total energy required to break the bonds in the reactants:
[tex]\[ \text{Energy required for breaking bonds} = 432 \, \text{kJ} + 190 \, \text{kJ} = 622 \, \text{kJ} \][/tex]
### Step 3: Calculate the energy released by forming the bonds in the products
For [tex]\( 2HBr \)[/tex]:
[tex]\[ 2 \times \text{bond energy of } H-Br = 2 \times 362 \, \text{kJ} = 724 \, \text{kJ} \][/tex]
### Step 4: Calculate the enthalpy change ([tex]\( \Delta H \)[/tex])
The enthalpy change for the reaction is the energy required to break bonds minus the energy released by forming bonds:
[tex]\[ \Delta H = \text{Energy required for breaking bonds} - \text{Energy released by forming bonds} \][/tex]
Substituting the values calculated:
[tex]\[ \Delta H = 622 \, \text{kJ} - 724 \, \text{kJ} = -102 \, \text{kJ} \][/tex]
Thus, the enthalpy change ([tex]\( \Delta H \)[/tex]) for the reaction is:
[tex]\[ \Delta H = -102 \, \text{kJ} \][/tex]
