To calculate the standard enthalpy change for the neutralization reaction, we can use the formula: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 write the balanced chemical equation for the neutralization reaction:HCl aq + NaOH aq NaCl aq + H2O l Since both solutions have the same concentration and volume, the reaction will go to completion, and all of the acid and base will react.Next, we need to find the mass of the solution. Since the density of the solutions is 1.00 g/mL, we can find the mass by multiplying the density by the total volume of the solution:mass = density volumemass = 1.00 g/mL 25.0 mL + 25.0 mL mass = 1.00 g/mL 50.0 mLmass = 50.0 gNow, we need to find the heat released during the reaction. The heat released can be calculated using the heat of neutralization, which is the heat released when one mole of water is formed from the reaction of an acid and a base. The heat of neutralization for strong acids and bases, like HCl and NaOH, is approximately -57.1 kJ/mol.Since we have 1.0 M solutions and 25.0 mL of each, we can find the moles of HCl and NaOH:moles = concentration volumemoles = 1.0 mol/L 0.025 Lmoles = 0.025 molNow, we can find the heat released during the reaction:q = moles heat of neutralizationq = 0.025 mol -57.1 kJ/molq = -1.4275 kJFinally, we can find the change in temperature using the formula q = mcT:T = q / mc T = -1.4275 kJ / 50.0 g 4.18 J/gC T = -1.4275 10^3 J / 50.0 g 4.18 J/gC T = -6.82CSince the temperature change is negative, the temperature of the solution decreases during the reaction. The standard enthalpy change for the neutralization of 25.0 mL of 1.0 M HCl with 25.0 mL of 1.0 M NaOH is -1.4275 kJ.