Calculate the enthalpyof 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 464kJ mol^(-1) respectively.

Calculate the enthalpyof formation of water, given that the bond energ

1.	Calculate the enthalpyof 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 464kJ mol^(-1) respectively.
Answer» Solution :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")` `:.`Netreleased`= 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)`