Alright Ronneydomnick — let's do this. Step 1: Identify the bonds broken and formed in the reaction. The given reaction is: $$\text{H}_2(\text{g}) + \text{Br}_2(\text{g}) \longrightarrow 2\text{HBr}(\text{g})$$ 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 $2\text{HBr}$ Step 2: Calculate the total energy absorbed to break bonds. 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 = $436 \text{ kJ mol}^{-1} + 192 \text{ kJ mol}^{-1} = 628 \text{ kJ mol}^{-1}$ Step 3: Calculate the total energy released when bonds are formed. Energy released per H-Br bond formed = $368 \text{ kJ mol}^{-1}$ Total energy released for two H-Br bonds = $2 \times 368 \text{ kJ mol}^{-1} = 736 \text{ kJ mol}^{-1}$ Step 4: Calculate the enthalpy change for the reaction ($\Delta H_{rxn}$). The enthalpy change for a reaction using bond energies is given by: $$\Delta H_{rxn} = \sum (\text{bond energies of bonds broken}) - \sum (\text{bond energies of bonds formed})$$ $$\Delta H_{rxn} = 628 \text{ kJ mol}^{-1} - 736 \text{ kJ mol}^{-1}$$ $$\Delta H_{rxn} = -108 \text{ kJ mol}^{-1}$$ This value is for the formation of 2 moles of HBr. Step 5: Calculate the heat of formation of hydrogen bromide ($\Delta H_f$). 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. $$\Delta H_f (\text{HBr}) = \frac{\Delta H_{rxn}}{2}$$ $$\Delta H_f (\text{HBr}) = \frac{-108 \text{ kJ mol}^{-1}}{2}$$ $$\Delta H_f (\text{HBr}) = -54 \text{ kJ mol}^{-1}$$ The heat of formation of hydrogen bromide is $\boxed{-54 \text{ kJ mol}^{-1}}$. Send me the next one 📸