The chemical equation for the standard enthalpy of formation of HCl is: $\frac{1}{2}\text{H}_2\text{(g)} + \frac{1}{2}\text{Cl}_2\text{(g)} \rightarrow \text{HCl(g)}$

The enthalpy of formation ($\Delta_f H$) is related to bond enthalpies by the equation: $\Delta_f H = \sum \text{B.E.(reactants)} - \sum \text{B.E.(products)}$ $\Delta_f H = \left[ \frac{1}{2}\text{B.E.}(\text{H}-\text{H}) + \frac{1}{2}\text{B.E.}(\text{Cl}-\text{Cl}) \right] - \text{B.E.}(\text{H}-\text{Cl})$

Given: $\Delta_f H = -90 \text{ kJ mol}^{-1}$ $\text{B.E.}(\text{H}-\text{H}) = 430 \text{ kJ mol}^{-1}$ $\text{B.E.}(\text{Cl}-\text{Cl}) = 240 \text{ kJ mol}^{-1}$

Substituting the values: $-90 = \left[ \frac{1}{2}(430) + \frac{1}{2}(240) \right] - \text{B.E.}(\text{H}-\text{Cl})$ $-90 = (215 + 120) - \text{B.E.}(\text{H}-\text{Cl})$ $-90 = 335 - \text{B.E.}(\text{H}-\text{Cl})$ $\text{B.E.}(\text{H}-\text{Cl}) = 335 + 90 = 425 \text{ kJ mol}^{-1}$