To calculate the bond dissociation energy of the O-H bond in a water molecule, we first need to determine the enthalpy changes for the formation of the bonds in the reactants and products.For the given reaction:2H2 g + O2 g -> 2H2O l Enthalpy change for the reaction, H = -285.8 kJ/molIn a water molecule, there are two O-H bonds. Therefore, we need to consider the energy required to break both O-H bonds in one water molecule.Let's denote the bond dissociation energy of the O-H bond as BDE O-H .The enthalpy change for the formation of the bonds in the reactants is the sum of the bond dissociation energies of the H-H bonds in the hydrogen molecules and the O=O bond in the oxygen molecule.Enthalpy change for reactants = 2 * BDE H-H + BDE O=O The enthalpy change for the formation of the bonds in the products is the sum of the bond dissociation energies of the O-H bonds in the water molecules.Enthalpy change for products = 4 * BDE O-H According to Hess's Law, the enthalpy change for the reaction is the difference between the enthalpy change for the products and the enthalpy change for the reactants.H = Enthalpy change for products - Enthalpy change for reactants-285.8 kJ/mol = 4 * BDE O-H - 2 * BDE H-H + BDE O=O We need the values of BDE H-H and BDE O=O to solve for BDE O-H .BDE H-H = 436 kJ/mol average value for H-H bond BDE O=O = 498 kJ/mol average value for O=O bond Now, we can plug these values into the equation:-285.8 kJ/mol = 4 * BDE O-H - 2 * 436 kJ/mol + 498 kJ/mol -285.8 kJ/mol = 4 * BDE O-H - 1370 kJ/molNow, solve for BDE O-H :4 * BDE O-H = -285.8 kJ/mol + 1370 kJ/mol4 * BDE O-H = 1084.2 kJ/molBDE O-H = 1084.2 kJ/mol / 4BDE O-H = 271.05 kJ/molThe bond dissociation energy of the O-H bond in a water molecule is approximately 271.05 kJ/mol.