Identify the bonds broken and formed in the reaction.

The given reaction is:

Bonds broken (reactants):

• One H-H bond in $\text{H}_2$
• One Br-Br bond in $\text{Br}_2$

Bonds formed (products):

• Two H-Br bonds in $2\text{HBr}$

Energy to break H-H bond = $436 \text{ kJ mol}^{-1}$
Energy to break Br-Br bond = $192 \text{ kJ mol}^{-1}$
Total energy absorbed = $436 \text{ kJ mol}^{-1} + 192 \text{ kJ mol}^{-1} = 628 \text{ kJ mol}^{-1}$

Energy released per H-Br bond formed = $368 \text{ kJ mol}^{-1}$
Total energy released for two H-Br bonds = $2 \times 368 \text{ kJ mol}^{-1} = 736 \text{ kJ mol}^{-1}$

The enthalpy change for a reaction using bond energies is given by:

This value is for the formation of 2 moles of HBr.

The heat of formation is defined for 1 mole of the compound. Since the reaction produces 2 moles of HBr, we divide the $\Delta H_{rxn}$ by 2.

The heat of formation of hydrogen bromide is $-54 \text{ kJ mol}^{-1}$.