To calculate the standard enthalpy change for this reaction, we need to use the following equation:q = mcTwhere q is the heat absorbed or released, m is the mass of the solution, c is the specific heat capacity, and T is the change in temperature.First, we need to find the moles of HCl and NaOH in the solution:moles of HCl = volume concentration = 50.0 mL 0.200 mol/L = 0.0100 molmoles of NaOH = volume concentration = 50.0 mL 0.200 mol/L = 0.0100 molSince the moles of HCl and NaOH are equal, the reaction will go to completion, and all of the HCl and NaOH will react to form NaCl and H2O.The balanced chemical equation for this reaction is:HCl aq + NaOH aq NaCl aq + H2O l Now we need to find the enthalpy change for this reaction. The standard enthalpy change for the reaction between HCl and NaOH is -55.9 kJ/mol.H = -55.9 kJ/molSince we have 0.0100 mol of HCl and NaOH reacting, we can find the total enthalpy change for the reaction:H_total = H moles = -55.9 kJ/mol 0.0100 mol = -0.559 kJNow we can find the mass of the solution:mass = volume density = 50.0 mL + 50.0 mL 1.00 g/mL = 100.0 gWe can now use the q = mcT equation to find the change in temperature:q = H_total = -0.559 kJ = -559 J since 1 kJ = 1000 J m = 100.0 gc = 4.18 J/ gC -559 J = 100.0 g 4.18 J/ gC TNow we can solve for T:T = -559 J / 100.0 g 4.18 J/ gC = -1.34CSince the temperature change is negative, the reaction is exothermic, and the solution will heat up. The standard enthalpy change for this reaction is -0.559 kJ.