To calculate the standard enthalpy change for the neutralization reaction, we first need to determine the heat released during the reaction q using the formula:q = mcTwhere m is the mass of the solution, c is the specific heat capacity, and T is the temperature change.The total volume of the solution is 25 mL of HCl + 25 mL of NaOH = 50 mL. Since the density of the solution is 1.00 g/mL, the mass of the solution is:m = 50 mL * 1.00 g/mL = 50 gNow we can calculate the heat released q :q = 50 g 4.18 J/ g*K 4.5C = 940.5 JNext, we need to determine the moles of HCl and NaOH that reacted. Since both solutions have the same concentration 0.10 M and volume 25 mL , they will react in a 1:1 ratio. To find the moles of HCl and NaOH, we use the formula:moles = concentration * volume in liters moles of HCl = moles of NaOH = 0.10 mol/L 0.025 L = 0.0025 molNow we can calculate the standard enthalpy change H for the reaction by dividing the heat released q by the moles of HCl or NaOH that reacted:H = q / moles = 940.5 J / 0.0025 mol = 376200 J/molSince the standard unit for enthalpy change is kJ/mol, we can convert the value:H = 376.2 kJ/molTherefore, the standard enthalpy change for the neutralization of 25 mL of 0.10 M hydrochloric acid with 25 mL of 0.10 M sodium hydroxide is -376.2 kJ/mol the negative sign indicates that the reaction is exothermic, meaning it releases heat .