Question

In a constant-pressure calorimeter, 50.0 mL of 0.320 M Ba(OH)2 was added to 50.0 mL of 0.640 M HCl. The reaction caused the temperature of the solution to rise from 23.76 °C to 28.12 °C. If the solution has the same density and specific heat as water (1.00 g/mL and 4.184 J/g·°C, 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

The reaction is:

Ba(OH)2 + 2HCl → BaCl2 + 2H2O

Based on stoichiometry:

1 mole of Ba(OH)2 reacts with 2 moles of HCl to form 2 moles of H2O

Based on the given data:

# moles of Ba(OH)2 present = 0.050 L * 0.320 moles/L = 0.016 moles

# moles of HCl present = 0.050 L * 0.640 moles/L = 0.032 moles

Thus, in accordance with the 1:2 (Ba(OH)2:HCl) stoichiometry, all reactants would combine to form 0.032 moles of H2O

Now,

total volume of solution = 50 + 50 = 100 ml

Since the density = density of water = 1 g/ml

Total mass, m = 100 ml* 1 g.ml-1 = 100 g

The enthalpy change or heat of the reaction is the energy released (Q),

Q = -mc(T2-T1) = 100 * 4.18*(28.12-23.76) = -1822.48 J

Therefore,  enthalpy change per mole of H2O =

= -1822.48 J/0.0320 moles = -56952.5 J/mol

Ans: ΔH/mole of H2O = -56.95 kJ/mol