To calculate the standard enthalpy change H for the neutralization reaction, we first need to determine the balanced chemical equation for the reaction between HCl and NaOH:HCl aq + NaOH aq NaCl aq + H2O l Next, we need to determine the limiting reactant. To do this, we can calculate the moles of each reactant:moles of HCl = volume of HCl concentration of HCl = 50.0 mL 0.100 mol/L = 5.00 mmolmoles of NaOH = volume of NaOH concentration of NaOH = 75.0 mL 0.0750 mol/L = 5.625 mmolSince there are fewer moles of HCl 5.00 mmol than NaOH 5.625 mmol , HCl is the limiting reactant.Now, we can use the stoichiometry of the balanced chemical equation to determine the moles of water produced:moles of H2O = moles of HCl = 5.00 mmolThe standard enthalpy change for the neutralization reaction between a strong acid and a strong base is approximately -55.8 kJ/mol of water produced. Therefore, we can calculate the standard enthalpy change for this reaction:H = moles of H2O -55.8 kJ/mol = 5.00 mmol -55.8 kJ/mol = -279 kJ/molSo, the standard enthalpy change for the neutralization reaction of 50.0 mL of 0.100 M HCl with 75.0 mL of 0.0750 M NaOH at a constant pressure of 1 atm and a temperature of 25C is approximately -279 kJ/mol.