To calculate the standard enthalpy change for the neutralization reaction, we can use the following equation:H = -n * H_neutralizationwhere H is the standard enthalpy change, n is the number of moles of water formed, and H_neutralization is the standard enthalpy change of neutralization which is -57.32 kJ/mol for the reaction between HCl and NaOH .First, we need to determine the number of moles of HCl and NaOH in the reaction. We can use the equation:moles = volume L * concentration M For HCl:moles_HCl = 0.050 L * 2 M = 0.1 molesFor NaOH:moles_NaOH = 0.050 L * 2 M = 0.1 molesSince the moles of HCl and NaOH are equal, the reaction goes to completion and all of the acid and base will react to form water. The balanced chemical equation for the reaction is:HCl aq + NaOH aq NaCl aq + HO l The number of moles of water formed n is equal to the moles of HCl or NaOH, which is 0.1 moles.Now we can calculate the standard enthalpy change:H = -n * H_neutralizationH = -0.1 moles * -57.32 kJ/mol H = 5.732 kJThe standard enthalpy change for the neutralization reaction of 50 mL of 2M hydrochloric acid and 50 mL of 2M sodium hydroxide at 25C is 5.732 kJ.