Question

Consider two solutions, the first being 50.0 mL of 1.00 M CuSO4 and the second 50.0 mL of 2.00 M KOH . When the two solutions are mixed in a constant-pressure calorimeter, a precipitate forms and the temperature of the mixture rises from 21.5 ∘C to 27.7 ∘C .

From the calorimetric data, calculate ΔH for the reaction that occurs on mixing. Assume that the calorimeter absorbs only a negligible quantity of heat, that the total volume of the solution is 100.0 mL , and that the specific heat and density of the solution after mixing are the same as that of pure water.

Answer 1

H = -51.9 kJ/mol

Explanation

The balanced reaction equation is :

CuSO₄ (aq) + 2 KOH (aq) Cu(OH)₂ (s) + K₂SO₄ (aq)

Moles CuSO₄ = (molarity) * (volume)

Moles CuSO₄ = (1.00 M) * (50.0 x 10^-3 L)

Moles CuSO₄ = 0.05 mol

moles KOH = (molarity) * (volume)

moles KOH = (2.00 M) * (50.0 x 10^-3 L)

moles KOH = 0.10 mol

moles Cu(OH)₂ formed = moles CuSO₄ consumed = (0.5) * (moles KOH consumed) = 0.05 mol

Volume of solution = 100.0 mL

mass of solution = (Volume of solution) * (density of solution)

mass of solution = (100.0 mL) * (1.00 g/mL)

mass of solution = 100.0 g

Heat absorbed by solution = (mass of solution) * (specific heat of solution) * (final temp. - initial temp.)

Heat absorbed by solution = (100.0 g) * (4.184 J/g.^oC) * (27.7 ^oC - 21.5 ^oC)

Heat absorbed by solution = 2594.08 J

Heat lost by reaction = -(Heat absorbed by solution)

Heat lost by reaction = -(2594.08 J)

Heat lost by reaction = -2594.08 J

H = (Heat lost by reaction) / (moles Cu(OH)₂ formed)

H = (-2594.08 J) / (0.05 mol)

H = -51881.6 J/mol

H = -51.9 kJ/mol