Question

The bond dissociation energy of gaseous `H_(2),Cl_(2)` and `HCl` are `104,58` and `103 kcal mol^(-1)` respecitvely. Calculate the enthalpy of formation for `HCl` gas.

Answer 1

Objective reaction `(1)/(2)H_(2)(g)+(1)/(2)Cl_(2)(g)rarr HCl(g) , Delta H = ?`
`Delta H = sum("Bond energy")_(P)- sum ("Bond energy")_(P) = [(1)/(2)xx104+(1)/(2)xx58]-[103]=22 kcal mol^(-1)`