Acommon laboratory reaction is the neutralization of an acid with a base. when 32.8 ml of 0.500 m hcl at 25.0°c is added to 66.3 ml of 0.500 m naoh at 25.0°c in a coffee cup calorimeter (with a negligible heat capacity), the temperature of the mixture rises to 28.2°c. what is the heat of reaction per mole of nacl (in kj/mol)? assume the mixture has a specific heat capacity of 4.18 j/(g·k) and that the densities of the reactant solutions are both 1.07 g/ml.

Explanation:

The given data is as follows.

  Densities of both the reactant solutions = 1.07 g/mL.

  Specific heat = 4.18

  Volume of HCl = 32.8 ml

  Volume of NaOH = 66.3 ml

The reaction will be as follows.

Since, density is mass divided by volume. Hence, calculate the mass of HCl as follows.

             Mass of HCl = Density × Volume of HCl

                                   =

                                   = 35.096 g

Similarly, mass of NaOH will be calculated as follows.

            Mass of NaOH = Density × Volume of NaOH

                                     =

                                     = 70.941 g

Therefore, total mass will be as follows.

               Total mass = Mass of HCl + Mass of NaOH

                                  = 35.096 g + 70.941 g

                                  = 106.037 g

Change in temperature will be calculated as follows.

                =

                          =

Therefore, calculate the heat of reaction as follows.

                       q =

                           =

                           = 1418.35 J

or,                        = 1.418 kJ         (as 1 kJ = 1000 J)

No. of moles of HCl = Molarity × Volume

                                 = 0.5 M × 32.8 ml

                                 = 16.4 mol

No. of moles of NaOH = Molarity × Volume

                                 = 0.5 M × 66.3 ml

                                 = 33.15 mol

So, HCl is the limiting reagent and heat of reaction produces by per mole of NaCl will be calculated as follows.

  Heat released for 1 mole of NaCl =

                                                          = 0.0864 kJ/mol

Thus, we can conclude that the heat of reaction per mole of NaCl is 0.0864 kJ/mol.