Question

12. What is the molar enthalpy of reaction of barium hydroxide in the following experiment?

(we assume the volume of calorimeter water is the two solutions added together! We also
assume the specific heat of the solution is the same as the specific heat of water)
Barium hydroxide solution
Volume: 50.0 mL
Concentration: 2.00 mol/L
Initial Temperature: 21.0°C
Hydrochloric acid
Volume: 200.0 mL
Concentration: 1.00 mol/L
Initial Temperature: 21.0°C
Calori
Final Temperature of the mixture: 32.5°C
a. Using data from above, calculate the molar enthalpy of neutralization for
hydrochloric acid.
b. Write the thermochemical equation for the reaction.
c. Use your thermochemical equation to determine the molar enthalpy of
neutralization for barium hydroxide.

Answer 1

a. First, calculate the heat absorbed or released (q) by the solution using q = mcΔT. Here, m is the mass of the solution (250 g), c is the specific heat capacity of water (4.18 J/g°C), and ΔT is the change in temperature (32.5°C - 21.0°C = 11.5°C).

So, q = 250 g * 4.18 J/g°C * 11.5°C = 12035 Joules.

The reaction is exothermic, so q = -12035 J.

The moles of HCl is volume (L) * concentration (mol/L) = 0.2 L * 1 mol/L = 0.2 mol.

The molar enthalpy of neutralization for hydrochloric acid is q/n = -12035 J / 0.2 mol = -60175 J/mol or -60.175 kJ/mol.

b. Thermochemical equation for the reaction is:

Ba(OH)2 + 2HCl → BaCl2 + 2H2O, ΔH = -60.175 kJ/mol (for HCl)

c. The moles of Ba(OH)2 is volume (L) * concentration (mol/L) = 0.05 L * 2 mol/L = 0.1 mol.

The molar enthalpy of neutralization for barium hydroxide is q/n = -12035 J / 0.1 mol = -120350 J/mol or -120.35 kJ/mol.