Import Question JSON

$\text{Hint: } \Delta\text{H}_f = \sum \text{BE}_R - \sum \text{BE}_P$ $\text{Step 1: Write the balanced equation for the formation of HCl}$ $\text{H}_2 + \text{Cl}_2 \rightarrow 2 \text{ HCl}$ $\text{Step 2: Apply the relationship between enthalpy of formation and bond energies}$ $\text{Using the equation: } \Delta\text{H}_f = \sum \text{Bond Energies (reactants)} - \sum \text{Bond Energies (products)}$ $\text{For the given reaction:}$ $\Delta\text{H}_f = \text{BE}_{\text{H-H}} + \text{BE}_{\text{Cl-Cl}} - 2 \times \text{BE}_{\text{H-Cl}}$ $\text{Rearranging to find BE}_{\text{H-Cl}}\text{:}$ $\text{BE}_{\text{H-Cl}} = \frac{\text{BE}_{\text{H-H}} + \text{BE}_{\text{Cl-Cl}} - \Delta\text{H}_f}{2}$ $\text{Step 3: Substitute the given values}$ $\text{BE}_{\text{H-Cl}} = \frac{430 \text{ kJ mol}^{-1} + 240 \text{ kJ mol}^{-1} - (-90 \text{ kJ mol}^{-1} \times 2)}{2}$ $\text{BE}_{\text{H-Cl}} = \frac{430 \text{ kJ mol}^{-1} + 240 \text{ kJ mol}^{-1} + 180 \text{ kJ mol}^{-1}}{2}$ $\text{BE}_{\text{H-Cl}} = \frac{850 \text{ kJ mol}^{-1}}{2} = 425 \text{ kJ mol}^{-1}$ $\text{Therefore, the bond enthalpy of HCl is } 425 \text{ kJ mol}^{-1}$ $\text{Note: There is a calculation error in the solution image. The value of } \Delta\text{H}_f \times 2 = -90 \times 2 = -180 \text{ kJ mol}^{-1}\text{, which when moved to the right side of the equation becomes } +180 \text{ kJ mol}^{-1}\text{. The numerator should be } 430 + 240 + 180 = 850 \text{ kJ mol}^{-1}\text{, giving a bond enthalpy of } 425 \text{ kJ mol}^{-1}\text{.}$