Using the chart below, calculate enthalpy change when fluorite (CaF 2 ) dissolve
ID: 508407 • Letter: U
Question
Using the chart below, calculate enthalpy change when fluorite (CaF2) dissolves in water in the standard state (STAP). Calculate the Gibbs free energy change of the reaction. Predict how the solubility of fluorite will change with T. Calculate the solubility of Fluroite in water at 25° C using these calculations. Express in moles and in mg/l. Write the formula for the equation that you would use to calculate how "K" would changge with the changes in T. Calculate the solubility of Fluorite at T= 10° C and 30° C. Show your work.
Species
G°f(kcal/mole)
H°(kcal/mole)
Ca2+
-132.3
-129.74
F2-
-66.4
-79.5
CaF2
-279
-291.5
Species
G°f(kcal/mole)
H°(kcal/mole)
Ca2+
-132.3
-129.74
F2-
-66.4
-79.5
CaF2
-279
-291.5
Explanation / Answer
CaF2 Ca2+ + 2 F-
Go = Go products - Go reactants
= [-132 +2x(-66.4)] - -279 = 14.2 kCal/mole = 14200 cal/mol
Go = -2.303RT logK
At 250C
Go = -2.303 x 1.987 cal K-1 mol-1 x 298 K log K
log K = - 14200 cal/mol / (2.303 x 1.987 cal K-1 mol-1 x 298 K)
log K = - 10.41
K = 3.9 x 10-11
K = Ksp
Ksp = 4S3
4S3 = 3.9 x 10-11
S = 33.9 x 10-11/4
S = 2.1 x 10-4
At 100C
Go = -2.303 x 1.987 cal K-1 mol-1 x 298 K log K
log K = - 14200 cal/mol / (2.303 x 1.987 cal K-1 mol-1 x 283 K)
log K = - 10.96
K = 1.1 x 10-11
K = Ksp
Ksp = 4S3
4S3 = 1.1 x 10-11
S = 31.1 x 10-11/4
S = 1.4 x 10-4
At 300C
Go = -2.303 x 1.987 cal K-1 mol-1 x 303 K log K
log K = - 14200 cal/mol / (2.303 x 1.987 cal K-1 mol-1 x 298 K)
log K = - 10.24
K = 5.8 x 10-11
K = Ksp
Ksp = 4S3
4S3 = 5.8 x 10-11
S = 35.8 x 10-11/4
S = 2.4 x 10-4
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