In a constant-pressure calorimeter of negligible heat capacity, 25 mL of 1.00 M CaCl2 is mixed with 25 mL of 2.00 M KF, resulting in solid CaF2 precipitating out of the solution.

Question

In a constant-pressure calorimeter of negligible heat capacity, 25 mL of 1.00 M CaCl₂ is mixed with 25 mL of 2.00 M KF, resulting in solid CaF₂ precipitating out of the solution. During this process, the temperature of the water rises from 25.0°C to 26.7°C. Assume the specific heat capacity of the solution is 4.184 J/°C•g and the density of the solution is 1.00 g/mL. Calculate the enthalpy of precipitation in kJ per mole of CaF₂ precipitated.

Also, what is the system and what is the surroundings in this case?

Answer 1

q = mC∆T

m = mass = 25 ml CaCl2 + 25 ml KF = 50 ml x 1.00 g/ml = 50 g

C = 4.184 J/g-deg

∆T = 1.7 deg

q = (50 g)(4.184 J/g-deg)(1.7 deg) = 355.6 J

Looking at the chemical reaction: CaCl₂(aq) + 2KF(aq) ==> CaF₂(s) + 2KCl(aq)

moles CaCl2 present = 0.025 L x 1.00 mol/L = 0.025 mol CaCl2

moles KF = 0.025 L x 2.00 mol/L = 0.05 mol KF

Neither reactant is limiting as they are both present in stoichiometric quantities. Thus, moles CaF2 formed will be 0.025 moles CaF2 formed.

∆Hppt = 355.6 J/0.025 moles = 14,224 J/mole = 14.2 kJ/mole