To calculate the standard enthalpy change H for the reaction, we need to consider the enthalpy changes of the reactants and products. The enthalpy change of neutralization of HCl aq with NaOH aq is given as -55.9 kJ/mol. This value represents the enthalpy change for the formation of 1 mol of water H2O from the reaction of NaOH aq and HCl aq . The enthalpy of formation of NaCl s is given as -411.2 kJ/mol. However, the reaction produces NaCl aq , not NaCl s . To find the enthalpy of formation for NaCl aq , we need to consider the enthalpy change associated with dissolving NaCl s in water. The dissolution of NaCl s in water is an endothermic process, meaning it requires energy. The enthalpy change for this process is approximately +3.9 kJ/mol. Now, we can calculate the enthalpy of formation for NaCl aq by adding the enthalpy of formation of NaCl s and the enthalpy change for dissolving NaCl s in water:Hf NaCl aq = Hf NaCl s + Hdissolution NaCl Hf NaCl aq = -411.2 kJ/mol + 3.9 kJ/molHf NaCl aq = -407.3 kJ/molNow we can calculate the standard enthalpy change H for the reaction using the enthalpy changes of the reactants and products:H = [Hf NaCl aq + Hf H2O l ] - [Hf NaOH aq + Hf HCl aq ]Since the enthalpy change of neutralization is given as -55.9 kJ/mol for the formation of 1 mol of water H2O from the reaction of NaOH aq and HCl aq , we can write:H = [-407.3 kJ/mol + Hf H2O l ] - -55.9 kJ/mol We are given the enthalpy of vaporization of water H2O as 40.7 kJ/mol. The enthalpy of formation of liquid water is -285.8 kJ/mol. Now we can plug in the values:H = [-407.3 kJ/mol + -285.8 kJ/mol ] - -55.9 kJ/mol H = -693.1 kJ/mol - -55.9 kJ/mol H = -637.2 kJ/molTherefore, the standard enthalpy change H for the reaction NaOH aq + HCl aq -> NaCl aq + H2O l is -637.2 kJ/mol.