Question

Use the following information to answer questions 5-10. A student determined the heat of neutralization of copper (II) sulfate (CuSO4) mixed with potassium hydroxide (KOH) using the procedure described part 2 of this experiment. A 150.0 mL sample of a 1.50 M solution of CuSO4 was mixed with a 150.0 mL sample of 3.00 M KOH in a coffee cup calorimeter. The temperature of both solutions and the calorimeter was 25.2 °C before mixing and 31.3 °C after mixing. The heat capacity of the calorimeter is 24.2 J/K. Calculate the heat of reaction (ΔHrxn) for this reaction in units of J/mole of copper (II) hydroxide (Cu(OH)2). Assume the solution is dilute enough that the specific heat capacity and density of the solution is the same as that of water, 4.18 J/g∙K and 1.00 g/mL respectively. Calculate the number of moles of copper (II) hydroxide (Cu(OH)2) formed during this reaction. Assume the reaction goes to 100% completion.

Answer 1

Volume of CuSO₄ = 150 mL

Volume of KOH = 150 mL

Total volume of solution = 150 + 150 = 300 mL = 0.300L

density of solution = 1.00g/mL

Mass of solution = density * volume = 1.00g/mL * 300 mL = 300 g

change in temperature = Final temperature - Initial temperature = 31.3 - 25.2 = 6.1 ^oC

Specific heat of solution = 4.18 J/g.K = 4.18 J/g.^oC

Specific heat of calorimeter = 24.2 J/K = 24.2 J/^oC

thus, heat (q) = (mass* specific heat* change in temperature)_solution + specific heat*(change in temperature)_calorimeter

q = (300g * 4.18 J/g.^oC * 6.1 ^oC) + (6.1 ^oC* 24.2 J/^oC) = 7797.02 J

but temperature increases, thus reaction is exothermic and q should be negative in sign

so, q = -7797.02 J

Now, reaction is:

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

Now, Molarity of CuSO₄ = 1.5 M

Moles of CuSO₄ = Molarity * volume = 1.5 M * 150 mL = 1.5mol/L * 0.150 L = 0.225 moles

from reaction, 1 ml of CuSO₄ GIVES 1 mol of  Cu(OH)₂

thus, 0.225 moles of CuSO₄ GIVES 0.225 mol of  Cu(OH)₂

Hence, number of moles of Cu(OH)₂ produced = 0.225 mol

Now, H_rxn = q/moles of Cu(OH)₂ =  -7797.02 J/ 0.225 mol =- 34653.42 J/mol