To calculate the standard enthalpy of formation of water using Hess's law, we need to manipulate the given reactions to match the desired reaction: H2 g + 1/2O2 g -> H2O g .First, we need to convert the second reaction from liquid water to gaseous water. To do this, we need to consider the enthalpy of vaporization of water:H2O l -> H2O g Hvap = 44 kJ/molNow, we can add the enthalpy of vaporization to the second reaction to get the formation of gaseous water:H2 g + 1/2O2 g -> H2O g Hrxn = -286 kJ/mol + 44 kJ/mol = -242 kJ/molNow, we can compare the first reaction to the desired reaction. The first reaction produces 2 moles of gaseous water, while the desired reaction produces only 1 mole. Therefore, we need to divide the first reaction by 2:1/2 2H2 g + O2 g -> 2H2O g Hrxn = 1/2 -484 kJ/mol = -242 kJ/molNow, both reactions are equivalent:H2 g + 1/2O2 g -> H2O g Hrxn = -242 kJ/molThus, the standard enthalpy of formation of water gaseous is -242 kJ/mol.