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

Options

A. `-44 kcal cdot mol^(-1)`
B. `-4.4 kcal cdot mol^(-1)`
C. `-22 kcal cdot mol^(-1)`
D. `-220 kcal cdot mol^(-1)`

Answer

C

Explanation

Reaction : `(1)/(2) H_(2) +(1)/(2) Cl_(2) rarr HCl`
`Delta H = Sigma B.E.("Reactants")- Sigma B.E. ("Products")`
`Delta H=B.E. (HCl)-[(1)/(2) B.E. (H_(2)) +(1)/(2) B.E. (Cl_(2))]`
`Delta H =-103 -[(1)/(2) xx (-104) +(1)/(2) xx (-58)]`
`=-103 -(52-29) =-22 kcal`