Question

A 50.0-ml sample of a 1.00 m solution of cuso4 is mixed with 50.0 ml of 2.00 M KOH in a calorimeter. the temperature of both solutions was 20.2C before mixing and 26.3C after mixing. The heat capacity of the calorimeter is 12.1 J/k. from these data calculate ΔH for the process:

CuSO₄ (1M) +2KOH (2M)---->Cu(OH)₂ (s) + K₂SO₄ (0.5M)

Assume the specific heat and density of the solution after mixing are the same as those of pure water and that the volumes are addictive.

Answer 1

50.0-ml sample of a 1.00 m solution of cuso4 is mixed with 50.0 ml of 2.00 M KOH.

total volume of solution = (50.0 + 50.0) = 100.0 ml

mass of solution = 100.0 g as the density of solution = 1.0 g / ml

heat = m1 * s1 * dT1 + heat capacity of the calorimeter * dT1 = 100.0 * 4.184 * (26.3 - 20.2) + 12.1 * (26.3 - 20.2)

or

heat = 2552.2 + 73.81

or

heat = 2626 J = 2.626 KJ

mole of CuSO4 = 0.050 L * 1.0 mole / L = 0.050 mole.

thus

delta H for this process = - 2.626 KJ / 0.050 mole = - 52.52 KJ / mole (answer)