Calculate the enthalpy of formation of water, given that the bond energies of `H-H, O=O` and `O-H` bond are `433 kJ mol^(-1), 492 kJ mol^(-1)`, and `464 kJ mol^(-1)`, respectively.

Answer

`-249kJ mol^(-1)`

Explanation

Aim `: H_(2)+(1)/(2)O_(2) rarr H_(2)O,Delta_(f) H^(@) = ? `or `H-H+(1)/(2) O = O rarr H-O-H ,DeltaH = ?`
`433 + (1)/(2) xx 492 ` `2 xx 464`
`ubrace(" =679kJ " )_("Energy absorbed")` `ubrace(" =928kJ " )_("Energy released")`
`:.`Net released `= 928 - 679 = 249 kJ mol^(-1) , i.e., Delta _(f) H^(@) = - 249 kJ mol^(-1)`
Alternatively , `DeltaH = B.E. (H_(2))+ (1)/(2) B.E. (O_(2))- 2B.E. ( O-H)`
`=433 + (1)/(2) xx 492 - 2 xx 464 = - 249 kJ mol^(-1)`