Question

A 150.0 mL sample of a 1.50 M solution of CuSO4
is 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 ΔHrxn for this reaction in units of kJ / mol of copper (II) hydroxide.
Assume the solutions is dilute enough that the specific heat and density of the solution is the same as that
of water.

Answer 1

mols CuSO4 = M x L = 1.50 x 0.150 = 0.225
mols KOH = 3.00 x 0.150 = 0.450
specific heat solns = specific heat H2O = 4.18 J/K*C

CuSO4 + 2KOH = Cu(OH)2 + 2H2O
q = mass solutions x specific heat solns x (Tfinal-Tinitial) + Ccal*deltat T
q = 300g x 4.18 x (31.3-25.2) + 24.2*(31.3-25.2)
dHrxn in J/mol= q/0.225 mol CuSO4
Then convert to kJ/mol

Answer 2

34 J

Answer 3

thank you dr bob XDXD