To calculate the bond dissociation energy of the O-H bond in a water molecule, we need to consider the enthalpy change for the formation of water and the average bond energy of an O-H bond.The enthalpy change for the formation of water is given as -285.8 kJ/mol. This means that 285.8 kJ of energy is released when one mole of water is formed from its constituent elements H2 and O2 in their standard states.The average bond energy of an O-H bond is given as 463 kJ/mol. Since there are two O-H bonds in a water molecule, the total energy required to break both O-H bonds is 2 463 kJ/mol = 926 kJ/mol.Now, we can calculate the bond dissociation energy of the O-H bond in a water molecule using the following equation:Bond dissociation energy O-H = Total energy required to break O-H bonds - Enthalpy change for the formation of waterBond dissociation energy O-H = 926 kJ/mol - -285.8 kJ/mol Bond dissociation energy O-H = 926 kJ/mol + 285.8 kJ/molBond dissociation energy O-H = 1211.8 kJ/molHowever, this value is for breaking both O-H bonds in a water molecule. To find the bond dissociation energy for a single O-H bond, we need to divide this value by 2:Bond dissociation energy O-H = 1211.8 kJ/mol 2Bond dissociation energy O-H 605.9 kJ/molTherefore, the bond dissociation energy of the O-H bond in a water molecule is approximately 605.9 kJ/mol.