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.

Options

A. `-44` kcal
B. `-88` kcal
C. `-22` kcal
D. `-11` kcal

Answer

C

Explanation

Aim :`1/2 H_(2)+1/2 Cl_(2) rarr HCI `
`DeltaH =sum`B.E.(Reactants)-`sum` B.E (Products )
`=[1/2 B.E. (H_(2)) +1/2B.E(Cl_(2))]- B.E. (HCl)`
` [1/2xx(104)+1/2(58)]-103 `
(52+29)-103=-22kcal.