Question

When 23.8 mL of 0.500 M H2SO4 is added to 23.8 mL of 1.00 M KOH in a coffee-cup calorimeter at 23.50°C, the temperature rises to 30.17°C. Calculate ΔH of this reaction. (Assume that the total volume is the sum of the individual volumes and that the density and specific heat capacity of the solution are the same as for pure water.) (d for water = 1.00 g/mL; c for water = 4.184 J/g·°C.)

?: kJ/mol H2O

Answer 1

Total volume = 23.8 mL + 23.8 mL = 47.6 mL

Density of the solution = 1 g/mL (Assumed same as that of water)

Therefore, mass of the solution = Volume x Density = 47.6 g

Heat gained by the calori meter, q_reaction = mcT = 47.6 g x 4.184 J/goC x (30.17- 23.50) = 1328.4 J

Heat released during the reaction, q_{neutralization} = -1328.4 J

Moles of H₂SO₄ used = 0.5 M x (0.0238 L) = 0.0119 moles

Moles of KOH used = 1 M x (0.0238 L) = 0.0238 moles

H₂SO₄ + 2KOH K₂SO₄ + 2H₂O

Thus, one mole of H₂SO₄ produces two moles of H₂O

Therefore, 0.0119 moles of H₂SO₄ will produce 0.0119 x 2 = 0.0238 moles of H₂O

Similarly, 0.0238 moles of KOH will produce 0.0238 moles of H₂O

Therefore,

H = -q_{neutralization}/ Moles of H₂O = -1328.4 J/ 0.0238 = -55.8 kJ/mol

Enthalpy of the reaction = -55.8 kJ/mol