The bond dissociation energy BDE of a molecule is the energy required to break a specific bond in the molecule. In the case of water H2O , we are interested in the BDE of the H-O bond. To calculate this, we need to know the enthalpy of formation of water and the enthalpy of formation of the individual atoms H and O .1. Enthalpy of formation of water H2O : Hf H2O = -285.8 kJ/mol2. Enthalpy of formation of hydrogen H2 and oxygen O2 molecules:Hf H2 = 0 kJ/mol since hydrogen is in its elemental form Hf O2 = 0 kJ/mol since oxygen is in its elemental form Now, we can use Hess's Law to calculate the BDE of the H-O bond. Hess's Law states that the total enthalpy change for a reaction is the same, regardless of the path taken. In this case, we can consider the following reaction:H2O g H g + OH g We can break this reaction down into two steps:Step 1: H2O g H2 g + 1/2 O2 g Step 2: H2 g + 1/2 O2 g H g + OH g Now, we can calculate the enthalpy change for each step:H1 = Hf H2 + 1/2 Hf O2 - Hf H2O = 0 + 0 - -285.8 = 285.8 kJ/molFor step 2, we need to know the BDE of the O-H bond in the hydroxyl radical OH . This value is typically given in reference tables:BDE O-H in OH = 428 kJ/molH2 = BDE O-H in OH = 428 kJ/molNow, we can calculate the total enthalpy change for the reaction:H = H1 + H2 = 285.8 kJ/mol + 428 kJ/mol = 713.8 kJ/molThus, the bond dissociation energy of the H-O bond in water H2O is approximately 713.8 kJ/mol.