Question

When 26.5 mL of 0.500 M H₂SO₄ is added to 26.5 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).

Answer in kJ/molH₂O

Answer 1

Total volume of water = 26.5 + 26.5 = 53.0 ml

Density of water = 1.00 g/ml

So, mass of water = volume*density = 53.0*1.00 = 53.0 g

mass = m = 53.0 g

specific heat = c = 4.184 J/g oC

Temperature rise = T = 30.17-23.50 = 6.67 oC

heat of reaction = H = -mcT

= -53.0*4.184*6.67 = -1479 J = -1.479 kJ

Moles of H2SO4 = concentration*volume = 0.500*0.0265 = 0.013 mol

Moles of KOH = 1.00*0.0265 = 0.026 mol

So, 0.013 mol of H2SO4 reacts with 0.026 mol of KOH to give 0.026 mol of H2O

H = -1.479/0.026 = -56.9 kJ/mol