Calculate the standard enthalpy change for the reaction in which 50.0 mL of 1.0 M HCl is mixed with 50.0 mL of 1.0 M NaOH in a constant pressure calorimeter using the following data: the temperature of the HCl and NaOH solutions are both initially 25.0 °C, and the temperature of the mixture increases to 32.0 °C. The specific heat of the solution is 4.18 J/(g·°C) and the density of the solution is 1.00 g/mL. The reaction is: HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)

Question

Calculate the standard enthalpy change for the reaction in which 50.0 mL of 1.0 M HCl is mixed with 50.0 mL of 1.0 M NaOH in a constant pressure calorimeter using the following data: the temperature of the HCl and NaOH solutions are both initially 25.0 °C, and the temperature of the mixture increases to 32.0 °C. The specific heat of the s

Answer 1

To calculate the standard enthalpy change for the reaction, we first need to determine the heat  q  released or absorbed during the reaction. We can do this using the formula:q = mcTwhere m is the mass of the solution, c is the specific heat capacity, and T is the change in temperature.First, let's find the mass of the solution. Since the density of the solution is 1.00 g/mL, and we have 50.0 mL of HCl and 50.0 mL of NaOH, the total mass of the solution is:mass =  50.0 mL + 50.0 mL   1.00 g/mL = 100.0 gNext, we can calculate the change in temperature:T = T_final - T_initial = 32.0 C - 25.0 C = 7.0 CNow we can calculate the heat  q  using the formula:q = mcT =  100.0 g    4.18 J/ gC     7.0 C  = 2926 JSince the temperature of the solution increased, the reaction is exothermic, and the heat is released. Therefore, q = -2926 J.Now we need to find the moles of the limiting reactant. In this case, both HCl and NaOH have the same concentration and volume, so they are present in equal amounts:moles of HCl = moles of NaOH =  1.0 M    50.0 mL    1 L/1000 mL  = 0.050 molFinally, we can calculate the standard enthalpy change  H  for the reaction by dividing the heat  q  by the moles of the limiting reactant:H = q / moles = -2926 J / 0.050 mol = -58520 J/molSince we typically express enthalpy changes in kJ/mol, we can convert this value:H = -58.52 kJ/molSo, the standard enthalpy change for the reaction is -58.52 kJ/mol.