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)`

Answer

`-269kJ mol^(-1)`

Explanation

Aim `: (1)/(2) H_(2)+(1)/(2) F_(2) rarr HF, Delta_(r) H =?`
`Delta_(r)H= Sigma B.E.` ( Reactants) `- Sigma`B.E. ( Products)
`=(1)/(2)B.E. ( H_(2)) + (1)/(2) B.E. (F_(2))- B.E. (HF) = (1)/(2 ) xx434 +(1)/(2) xx 158 - 565 = - 269 kJ mol^(-1)`