Question

Calculate the total heat change when 100 ml of 0.5 M HCl is mixed with 100 ml of 0.5 M NaOH at 25°C. Assume the heat capacity of the solution is 4.18 J/(g°C) and the density of the solution is 1 g/mL. The enthalpy of neutralization of HCl and NaOH is -57.3 kJ/mol.

Answer 1

To calculate the total heat change when 100 ml of 0.5 M HCl is mixed with 100 ml of 0.5 M NaOH, we first need to determine the number of moles of HCl and NaOH in the solution. Moles of HCl = volume  L   concentration  M Moles of HCl = 0.1 L  0.5 M = 0.05 molesMoles of NaOH = volume  L   concentration  M Moles of NaOH = 0.1 L  0.5 M = 0.05 molesSince the moles of HCl and NaOH are equal, the reaction will go to completion, and all of the HCl and NaOH will react to form water and NaCl. The enthalpy of neutralization for this reaction is -57.3 kJ/mol.Now, we can calculate the heat change  q  using the enthalpy of neutralization and the moles of HCl  or NaOH, since they are equal .q = moles  enthalpy of neutralizationq = 0.05 moles  -57.3 kJ/mol = -2.865 kJThe total heat change for this reaction is -2.865 kJ. This is an exothermic reaction, as the heat change is negative, meaning heat is released during the reaction.