The standard enthalpy change for the neutralization reaction between hydrochloric acid HCl and sodium hydroxide NaOH at 25 can be found using the heat of formation values for the reactants and products.H = Hf products - Hf reactants For the given reaction:HCl aq + NaOH aq NaCl aq + H2O l The standard enthalpy of formation Hf values at 25 are as follows:HCl aq : -167.2 kJ/molNaOH aq : -469.15 kJ/molNaCl aq : -407.27 kJ/molH2O l : -285.83 kJ/molNow, we can calculate the standard enthalpy change H for the reaction: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/molTherefore, the standard enthalpy change for the neutralization reaction between hydrochloric acid and sodium hydroxide at 25 is -56.75 kJ/mol.