To calculate the standard enthalpy change of the reaction, we first need to determine the heat released or absorbed during the reaction. The reaction between HCl and NaOH is an acid-base neutralization reaction, which can be represented as: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 HCl and NaOH will react.First, let's find the moles of HCl and NaOH:moles of HCl = 0.100 M 100.0 mL 1 L / 1000 mL = 0.0100 molmoles of NaOH = 0.100 M 100.0 mL 1 L / 1000 mL = 0.0100 molNow, let's find the heat released or absorbed during the reaction. We can use the formula:q = mcTwhere q is the heat released or absorbed, m is the mass of the solution, c is the specific heat capacity, and T is the change in temperature.Since the density of the final solution is 1.00 g/mL, the mass of the 200.0 mL solution is:m = 200.0 mL 1.00 g/mL = 200.0 gWe are not given the change in temperature T in the problem, so we cannot directly calculate the heat released or absorbed q . However, we can express the standard enthalpy change of the reaction H in terms of q:H = -q / moles of reactionSince the reaction goes to completion, the moles of reaction are equal to the moles of HCl or NaOH 0.0100 mol . Therefore, the standard enthalpy change of the reaction can be expressed as:H = -q / 0.0100 molTo find the value of H, we would need the change in temperature T during the reaction. This information is not provided in the problem, so we cannot calculate the exact value of H. However, we have set up the necessary equations to solve for H if the change in temperature were provided.