Calculate `DeltaH` for the reaction
`H_(2)(g)+1//2O_(2)(g)toH_(2)O(g)`
given the bond energies of H-H and O=O bonds and O-H bond are 433 kJ `mol^(-1)`, 492 kJ `mol^(-1) and 464kJ" "mol^(-1)`.

Explanation

`H_(2)(g) +(1)/(2)O_(2)(g) rarr H_(2)O(g)`
`Delta_(f)H = BE` of product `-BE` of reactant
`= (BE of H_(2) +(1)/(2)BE of O_(2)) - (2BE of O-H)`
`= 433 +(1)/(2) xx492 -2 xx 464`
`= 433 +246 - 928 =- 249 kJ`