`DeltaH^(0)` for a reaction `F_(2) + 2HCl rarr 2HF + Cl_(2)` is given as `-352.8 kJ. Delta H_(f)^(0)` for HF is `-268.3 KJ mol^(-1)`, then `Delta H_(f)^(0)` of HCl would be

Options

A. `-22kJ mol^(-1)`
B. `880kJ mol^(-1)`
C. `-91.9 kJ mol^(-1)`
D. `-183.8 kJ mol^(-1)`

Answer

C

Explanation

`Delta H= 2 Delta H_(f_(HF)) - 2 Delta H_(f_(HCl)) -352.8 = 2 xx (-268.3) - 2 Delta H_(f_(HCl))`
`Delta H_(f_(HCl)) = - 91.9` KJ /mol