To calculate the standard enthalpy of formation of water H2O , we can use the given information and Hess's Law. Hess's Law states that the total enthalpy change for a reaction is the same, regardless of the path taken.First, let's write the balanced chemical equations for the combustion reactions:1. Combustion of hydrogen gas: H2 g + 1/2 O2 g H2O l H = -286 kJ/mol2. Combustion of carbon: C s + O2 g CO2 g H = -394 kJ/mol3. Combustion of methane gas: CH4 g + 2 O2 g CO2 g + 2 H2O l H = -890 kJ/molWe can use equations 2 and 3 to find the enthalpy change for the formation of water from its elements. First, we need to eliminate CO2 from the equations. To do this, we can reverse equation 2 and add it to equation 3:Reverse equation 2: CO2 g C s + O2 g H = +394 kJ/molAdd reversed equation 2 to equation 3: CH4 g + 2 O2 g + CO2 g CO2 g + 2 H2O l + C s + O2 g H = -890 kJ/mol + 394 kJ/mol = -496 kJ/molNow, we can simplify the equation: CH4 g + O2 g 2 H2O l + C s H = -496 kJ/molNow, we need to eliminate CH4 and C from the equation. To do this, we can use the fact that the standard enthalpy of formation of hydrogen gas H2 and carbon C is 0 kJ/mol. We can write the formation reaction for methane as: C s + 2 H2 g CH4 g H = 0 kJ/molNow, we can add this equation to the previous one: CH4 g + O2 g + C s + 2 H2 g 2 H2O l + C s + 2 H2 g H = -496 kJ/mol + 0 kJ/mol = -496 kJ/molSimplify the equation: CH4 g + O2 g 2 H2O l H = -496 kJ/molNow, we can divide the equation by 2 to find the enthalpy change for the formation of 1 mole of water: 1/2 CH4 g + 1/2 O2 g H2O l H = -496 kJ/mol 2 = -248 kJ/molTherefore, the standard enthalpy of formation of water H2O is -248 kJ/mol.