To calculate the standard enthalpy change for the reaction between hydrochloric acid HCl and sodium hydroxide NaOH , we can use the following equation:H = -q_p / nwhere H is the standard enthalpy change, q_p is the heat released at constant pressure, and n is the number of moles of the limiting reactant.First, we need to determine the limiting reactant. The balanced chemical equation for the reaction between HCl and NaOH is:HCl aq + NaOH aq NaCl aq + H2O l From the balanced equation, we can see that the mole ratio between HCl and NaOH is 1:1.Now, let's calculate the moles of each reactant:Moles of HCl = volume in L concentration in mol/L = 25 mL 1 L/1000 mL 2.0 M = 0.050 molMoles of NaOH = volume in L concentration in mol/L = 50 mL 1 L/1000 mL 1.0 M = 0.050 molSince the moles of HCl and NaOH are equal, the reaction will go to completion, and there is no limiting reactant.Next, we need to find the heat released at constant pressure q_p . We can use the heat of neutralization for the reaction between a strong acid and a strong base, which is approximately -57.3 kJ/mol.Now, we can calculate the standard enthalpy change H :H = -q_p / n = - -57.3 kJ/mol / 0.050 mol = 1146 kJ/molThe standard enthalpy change for the reaction between hydrochloric acid and sodium hydroxide in aqueous solutions is approximately 1146 kJ/mol.