To calculate the standard enthalpy change for the neutralization reaction, we need to know the standard enthalpy of formation for the products and reactants. The standard enthalpy of formation for HCl aq , NaOH aq , NaCl aq , and H2O l are as follows:HCl aq : -167.2 kJ/molNaOH aq : -469.15 kJ/molNaCl aq : -407.27 kJ/molH2O l : -285.83 kJ/molThe balanced equation for the reaction is:HCl aq + NaOH aq NaCl aq + H2O l Using Hess's Law, we can calculate the standard enthalpy change for the reaction as follows:H = [Hf products - Hf reactants ]H = [ -407.27 kJ/mol + -285.83 kJ/mol - -167.2 kJ/mol + -469.15 kJ/mol ]H = [ -693.1 kJ/mol - -636.35 kJ/mol ]H = -56.75 kJ/molNow, we need to determine the number of moles of HCl and NaOH in the reaction. Since the volume of both solutions is 50 mL and the concentration is 0.1 M, we can calculate the moles as follows:moles = volume concentrationmoles of HCl = 0.050 L 0.1 mol/L = 0.005 molmoles of NaOH = 0.050 L 0.1 mol/L = 0.005 molSince the reaction is 1:1, both reactants will be completely consumed in the reaction. Therefore, we can calculate the enthalpy change for the reaction using the moles and the enthalpy change per mole:H_reaction = H molesH_reaction = -56.75 kJ/mol 0.005 molH_reaction = -0.28375 kJThe standard enthalpy change for the neutralization of 50 mL of 0.1 M hydrochloric acid with 50 mL of 0.1 M sodium hydroxide at a constant pressure and a temperature of 25C is approximately -0.284 kJ.