Find the enthalpy of formation of hydrogen flouride on the basis of following data:
Bond energy of `H-H` bond `=434 kJ mol^(-1)`
Bond energy of `F-F` bond `=158 kJ mol^(-1)`
Bond enegry of `H -F` bond `=565 kJ mol^(-1)`

Explanation

`(1)/(2) H_(2)(g) +(1)/(2) F_(2)(g) rarr HF(g)`
`Delta_(f)H = BE` of reactant -`BE` of product
`= ((1)/(2)BE of H_(2) +(1)/(2)BE of H_(2)) - (BE of HF)`
`=(1)/(2) xx 434 +(1)/(2) xx 158 - 565 = 217 +79 - 565`
`=- 269 kJ`