To calculate the standard enthalpy change for the reaction between HCl and NaOH, we first need to determine the heat released during the reaction q and then divide it by the moles of the limiting reactant.The reaction between HCl and NaOH is an exothermic neutralization reaction:HCl aq + NaOH aq NaCl aq + H2O l First, let's find the limiting reactant. Since both solutions have the same volume 50 mL and concentration 0.2 M , they will react in a 1:1 ratio, and there is no limiting reactant.Next, let's find the total mass of the combined solution:Mass = Volume DensityMass = 50 mL + 50 mL 1.0 g/mLMass = 100 gNow, we need to find the change in temperature T during the reaction. Since the reaction is exothermic, the temperature of the solution will increase. Let's assume the final temperature of the solution is Tf and the initial temperature is 25C.We can use the heat released q formula:q = mass specific heat capacity TRearranging the formula to find T:T = q / mass specific heat capacity Since we don't have the value for q, we can use the heat of neutralization for a strong acid and strong base reaction, which is approximately -57.32 kJ/mol.Now, let's find the moles of HCl and NaOH:Moles = Volume ConcentrationMoles = 50 mL 0.2 MMoles = 0.01 L 0.2 mol/LMoles = 0.002 molSince the reaction is 1:1, there are 0.002 mol of both HCl and NaOH.Now, let's find the heat released q during the reaction:q = moles heat of neutralizationq = 0.002 mol -57.32 kJ/molq = -0.11464 kJNow, we can find the change in temperature T :T = q / mass specific heat capacity T = -0.11464 kJ / 100 g 4.18 J/g C T = -0.11464 10^3 J / 100 4.18 T -0.274 CSince the reaction is exothermic, the final temperature Tf will be:Tf = 25C + -0.274C 24.726CFinally, we can calculate the standard enthalpy change H for the reaction:H = q / molesH = -0.11464 kJ / 0.002 molH -57.32 kJ/molThe standard enthalpy change for the reaction between 50 mL of 0.2 M HCl and 50 mL of 0.2 M NaOH at 25C is approximately -57.32 kJ/mol.