The enthalpy of precipitation of calcium fluoride is -13.974 kJ/mol.
In order to solve this, first we need to write the balanced chemical reaction equation of the reaction in question:
CaCl₂(aq) + 2KF(aq) → CaF₂(s) + 2KCl(aq)
In order to calculate the enthalpy of precipitation (ΔH), we need the amount of heat released (Q) and the number of moles of CaF₂ (n):
ΔH = Q/n
To calculate the number of moles, we can use the molarity (c = 1.00 M) of the calcium chloride solution and its volume (V = 25 mL = 0.025 L):
c = n/V ⇒ n = c*V
n = 1.00 M * 0.025 L = 0.025 mol
We calculate the amount of heat released (Q) using the following equation:
Q = (t₁ - t₂) * C * m
t₁ - initial temperature (25.0 ⁰C)
t₂ - final temperature (26.7 ⁰C)
C - specific heat capacity (4.184 J/⁰Cg)
m - the mass of the solution
We need the mass of the solution, which we can calculate using the density (d = 1.00 g/mL) and the volume (V = 25 mL + 25 mL = 50 mL) of the solution:
d = m/V ⇒ m = d*V
m = 1.00 g/mL * 50 mL = 50 g
Q = (25.0 ⁰C - 26.7 ⁰C) * 4.184 J/°Cg * 50 g
Q = -349.4 J
ΔH = -349.4 J / 0.025 mol
ΔH = -13974 J/mol = -13.974 kJ/mol
You can learn more about enthalpy here:
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