To calculate the bond energy for the O-H bond in a water molecule, we need to consider the balanced chemical equation for the combustion of water:H2 + 1/2 O2 H2OThe enthalpy change for this reaction is -286 kJ/mol. This means that 286 kJ of energy is released when 1 mole of water is formed from its elements. We also know that the O=O bond energy is 498 kJ/mol.In this reaction, one O=O bond is broken and two O-H bonds are formed. We can represent the enthalpy change for the reaction as:H = Energy of bonds broken - Energy of bonds formed-286 kJ/mol = 1/2 498 kJ/mol - 2 O-H bond energy Now, we can solve for the O-H bond energy:-286 kJ/mol + 2 O-H bond energy = 1/2 498 kJ/mol2 O-H bond energy = 498 kJ/mol / 2 + 286 kJ/mol2 O-H bond energy = 249 kJ/mol + 286 kJ/mol2 O-H bond energy = 535 kJ/molO-H bond energy = 535 kJ/mol / 2O-H bond energy = 267.5 kJ/molSo, the bond energy for the O-H bond in a water molecule is approximately 267.5 kJ/mol.