Calculate `Delta_(r)H^(@)` for the reaction,
`H_(2)(g) + 1/2O_(2)(g) to H_(2)O(g)`
Given that bond enthalpies of H-H bond , O = O bond and O - H bond are 433 kJ `mol^(-1)`, `492 kJ "mol"^(-1)` and `464 kJ "mol"^(-1)`

Answer

`-249 kJ`

Explanation

For the reaction :
`H -H(g) + 1/2(O = O) (g) to 2O - H(g)`
`DeltaH` = (Bond enthalpy of H - H) + `1/2`(Bond enthalpy of O = O) - 2(Bond enthalpy of O - H)
` = 433 + 1/2(492) - 2(464) = 433 + 246 - 928 = -249 kJ`.