Question

In a constant-pressure calorimeter, 60.0 mL of 0.320 M Ba(OH)2 was added to 60.0 mL of 0.640 M HCl. The reaction caused the temperature of the solution to rise from 21.70 °C to 26.06 °C. If the solution has the same density and specific heat as water (1.00 g/mL and 4.184 J/g·K, respectively), what is ΔH for this reaction (per mole of H2O produced)? Assume that the total volume is the sum of the individual volumes.

Answer 1

Equation for the reaction

          Ba(OH)2 + 2 HCl → BaCl2 + 2 H2O

The barium hydroxide and hydrochloric acid are present in stoichiometric amounts, so either can be considered the limiting reactant:

Given- 60.0 mL of 0.320 M Ba(OH)2

             60.0 mL of 0.640 M HCl

Mol of H₂O =  

          (60ml/1000 L) x (0.640 mol/L HCl) x (2 mol H2O / 2 mol HCl) = 0.0384 mol H2O

Now we can calculate heat gained by solution

            q = mxs xt

               = (60.0g +60.0g ) x (4.186 J/g·°C x (26.06 - 21.70)°C

               = 2186.976 J

         ΔH for per mole of H2O = (2186.976 J) / (0.0384 mol H2O) =56952.5 J

                                           = 56.9525 kJ