The standard bond dissociation energies of gaseous hydrogen ($ ext{H}_2$), chlorine ($ ext{Cl}_2$)

Question

The standard bond dissociation energies of gaseous hydrogen ($	ext{H}_2$), chlorine ($	ext{Cl}_2$), and hydrogen chloride ($	ext{HCl}$) molecules are $104	ext{ kcal mol}^{-1}$, $58	ext{ kcal mol}^{-1}$, and $103	ext{ kcal mol}^{-1}$, respectively. Calculate the standard enthalpy of formation of gaseous hydrogen chloride ($\Delta H^\circ_f[	ext{HCl}(g)]$ in $	ext{kcal mol}^{-1}$). (Include the algebraic sign; if exothermic, report as a negative number)

Professor PC Thomas Classes · Answer

The formation reaction for one mole of hydrogen chloride gas from its elements in their standard states is: $\frac{1}{2}	ext{H}_2(g) + \frac{1}{2}	ext{Cl}_2(g) ightarrow 	ext{HCl}(g)$. In terms of bond dissociation energies (since bond breaking is endothermic and bond formation is exothermic), the enthalpy of formation is expressed as: $\Delta H^\circ_f[	ext{HCl}(g)] = \sum 	ext{BE}(	ext{reactants}) - \sum 	ext{BE}(	ext{products}) = \frac{1}{2}	ext{BE}(	ext{H-H}) + \frac{1}{2}	ext{BE}(	ext{Cl-Cl}) - 	ext{BE}(	ext{H-Cl})$. Substituting the given bond energies: $\Delta H^\circ_f[	ext{HCl}(g)] = \frac{1}{2}(104) + \frac{1}{2}(58) - 103 = 52 + 29 - 103 = 81 - 103 = -22	ext{ kcal mol}^{-1}$.