To calculate the standard enthalpy change for the neutralization of HCl and NaOH, we first need to write the balanced chemical equation for the reaction:HCl aq + NaOH aq NaCl aq + H2O l The standard enthalpy change for this reaction, H, can be found in literature or reference books. For the neutralization of a strong acid HCl and a strong base NaOH , the standard enthalpy change is approximately -57.3 kJ/mol.Next, we need to determine the number of moles of HCl and NaOH in the given solutions. Since both solutions have a concentration of 1M and a volume of 50mL, we can calculate the moles as follows:moles = concentration volume in liters moles of HCl = 1M 0.050L = 0.050 molmoles of NaOH = 1M 0.050L = 0.050 molSince the moles of HCl and NaOH are equal, the reaction goes to completion, and all of the acid and base are neutralized.Now we can calculate the enthalpy change for the reaction:H = moles standard enthalpy change per moleH = 0.050 mol -57.3 kJ/mol = -2.865 kJThe standard enthalpy change for the neutralization of 50mL of 1M HCl solution with 50mL of 1M NaOH solution at 25C is approximately -2.865 kJ.