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)

Calculate Delta_(r)H^(@) for the reaction, H_(2)(g) + 1/2O_(2)(g) to

1.	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» Solution :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`.