To calculate the standard enthalpy change for the reaction between HCl and NaOH, we can use the following equation:H = -q / nwhere H is the standard enthalpy change, q is the heat released during the reaction, and n is the number of moles of the limiting reactant.First, let's write the balanced chemical equation for the reaction:HCl aq + NaOH aq NaCl aq + H2O l Next, we need to determine the limiting reactant. Since both HCl and NaOH have the same concentration 1.0 M and the same volume 50.0 mL , they will react in a 1:1 ratio, and neither is in excess.Now, let's calculate the number of moles of the limiting reactant either HCl or NaOH :moles = Molarity Volume in L moles = 1.0 M 0.050 L = 0.050 molesThe standard enthalpy change of neutralization for a strong acid and a strong base, like HCl and NaOH, is approximately -57.1 kJ/mol. This value represents the heat released q when 1 mole of H2O is formed.Now we can calculate the heat released q during the reaction:q = moles H_neutralizationq = 0.050 moles -57.1 kJ/mol = -2.855 kJFinally, we can calculate the standard enthalpy change H for the reaction:H = -q / nH = - -2.855 kJ / 0.050 moles = 57.1 kJ/molThe standard enthalpy change for the reaction between HCl and NaOH to form NaCl and H2O when 50.0 mL of 1.0 M HCl reacts with 50.0 mL of 1.0 M NaOH at 25C is 57.1 kJ/mol.