The chemical equation for the formation of $\text{HCl(g)}$ from its constituent elements in their standard states is: $\frac{1}{2}\text{H}_2\text{(g)} + \frac{1}{2}\text{Cl}_2\text{(g)} \rightarrow \text{HCl(g)}$

The standard enthalpy of formation ($\Delta_f H^{\circ}$) can be calculated from the given bond dissociation enthalpies ($\Delta_{\text{bond}} H^{\circ}$) using the relation: $\Delta_f H^{\circ} = \sum \Delta_{\text{bond}} H^{\circ}(\text{reactants}) - \sum \Delta_{\text{bond}} H^{\circ}(\text{products})$

Substituting the given values into the equation: $\Delta_f H^{\circ}(\text{HCl}) = \left[ \frac{1}{2} \Delta_{\text{bond}} H^{\circ}(\text{H}_2) + \frac{1}{2} \Delta_{\text{bond}} H^{\circ}(\text{Cl}_2) \right] - \left[ \Delta_{\text{bond}} H^{\circ}(\text{HCl}) \right]$ $\Delta_f H^{\circ}(\text{HCl}) = \left[ \frac{1}{2}(434) + \frac{1}{2}(242) \right] - 431$ $\Delta_f H^{\circ}(\text{HCl}) = (217 + 121) - 431$ $\Delta_f H^{\circ}(\text{HCl}) = 338 - 431 = -93 \text{ kJ mol}^{-1}$