To calculate the standard enthalpy change of the oxidation of hydrogen gas H2 to form water H2O , we can use the following equation:H_reaction = H_products - H_reactantsThe reaction can be written as:H2 g + 1/2 O2 g H2O l The standard enthalpy change of the reaction H_reaction is the difference between the standard enthalpy of formation of water H_products and the bond dissociation enthalpy of the H-H bond H_reactants .H_products = -285.8 kJ/mol standard enthalpy of formation of water Since the bond dissociation enthalpy of the H-H bond is given as 436 kJ/mol, we need to consider that the reaction involves breaking one mole of H-H bonds and forming two moles of O-H bonds. The standard enthalpy change for breaking one mole of O=O bonds is also needed. The bond dissociation enthalpy of the O=O bond is 498 kJ/mol. Since we only need half a mole of O2 for the reaction, we will use half of this value.H_reactants = 436 kJ/mol H-H bond + 1/2 * 498 kJ/mol O=O bond = 436 + 249 = 685 kJ/molNow, we can calculate the standard enthalpy change of the reaction:H_reaction = H_products - H_reactants = -285.8 kJ/mol - 685 kJ/mol = -970.8 kJ/molThe standard enthalpy change of the oxidation of hydrogen gas H2 to form water H2O at 298 K is -970.8 kJ/mol.