To calculate the standard enthalpy change for the neutralization reaction, we first need to write the balanced chemical equation for the reaction:HCl aq + NaOH aq NaCl aq + H2O l Now, we need to determine the amount of heat released during the reaction. Since both solutions have the same concentration and volume, they will react in a 1:1 ratio, and the limiting reactant will be completely consumed.First, let's find the moles of HCl and NaOH:moles of HCl = 1.0 M * 50.0 mL * 1 L / 1000 mL = 0.050 molmoles of NaOH = 1.0 M * 50.0 mL * 1 L / 1000 mL = 0.050 molSince both reactants are present in equal amounts, the reaction will go to completion.Next, we need to calculate the heat released during the reaction. To do this, we can use the formula:q = mcTwhere q is the heat released, m is the mass of the solution, c is the specific heat capacity, and T is the change in temperature.First, let's find the mass of the final solution:mass of final solution = 50.0 mL HCl + 50.0 mL NaOH * 1 g/mL = 100.0 gNow, we need to determine the change in temperature T . Since we are not given the initial and final temperatures, we cannot directly calculate T. However, we can use the standard enthalpy change of neutralization for HCl and NaOH, which is -57.32 kJ/mol, to find the heat released.q = moles of HCl * -57.32 kJ/mol = 0.050 mol * -57.32 kJ/mol = -2.866 kJNow, we can use the formula q = mcT to find T:-2.866 kJ = 100.0 g * 4.18 J/g K * TFirst, convert kJ to J:-2866 J = 100.0 g * 4.18 J/g K * TNow, solve for T:T = -2866 J / 100.0 g * 4.18 J/g K = -6.85 KSince we are looking for the standard enthalpy change, we can use the formula:H = q / moles of HCl = -2866 J / 0.050 mol = -57320 J/molThe standard enthalpy change for the neutralization reaction of HCl and NaOH is -57.32 kJ/mol.