Use the Born-Mayer equation and other data given below to calculate the lattice

Question

Use the Born-Mayer equation and other data given below to calculate the lattice enthalpy for calcium fluoride, CaF2, assuming it adopts the fluorite structure. r(Ca2+) = 100 pm r(F–) = 133 pm A(fluorite) = 2.52 d* = 34.5 pm

(ii) Using the value you calculated in (i), and the data below, calculate the enthalpy change upon dissolution of CaF2 in water. Listing any assumptions you make, predict whether calcium fluoride dissolves in water. H(hydration) Ca2+ = +1507 kJmol–1 H(hydration) F– = +505 kJmol–1

(iii) The radius of the Sr2+ ion is 118 pm. Would you expect SrF2 to be more or less soluble in water than CaF2? Explain your reasoning.

Explanation / Answer

i. Let's use the Born-Haber cycle since it's an easier method. This means we split the process of formation of the salt from its elements into steps, for which the enthalpy change is known. The sum of the enthalpy changes equals the enthalpy of formation of the salt.

The asked lattice energy represents one of the steps. Since we know the enthalpy of formation and the enthalpy of all other steps, we can calculate from the over all enthalpy of the process.


Steps:
1. atomization of calcium
Ca(s) + F(g) Ca(g) + F(g)
Enthalpy of this step is the sublimation enthalpy of calcium
H = 121kJ/mol

2. ionization of calcium
Ca(g) + F(g) Ca²(g) + F(g) + 2 e
enthalpy of this step is sum of first and second ionization energy
H = 590kJ/mol + 1145kJ/mol = 1735kJ/mol

3. atomization of fluorine
Ca²(g) + F(g) + 2 e Ca²(g) + F(g) + 2 e
enthalpy of this step is equal to the bond energy of the fluorine molecule:
H = 157kJ/mol

3. atomization of fluorine
Ca²(g) + F(g) + 2 e Ca²(g) + 2 F(g)
enthalpy of this step is equal to the bond energy of the fluorine molecule:
H = 157kJ/mol

4. ionization of fluorine
Ca²(g) + 2 F(g) + 2 e Ca²(g) + 2 F(g)
enthalpy of this step is equal to the electron affinity of fluorine:
H = -328 kJ/mol

5. formation of solid calcium fluoride
Ca²(g) + 2 F(g) CaF(s)
enthalpy of this step is equal to lattice energy:
H = EL

Overall:
Ca(s) + F(g) CaF(s)
with
Hf = H + H + H + H + EL
=>
EL = Hf - H - H - H - H
= -1120kJ/mol - 121kJ/mol - 1735kJ/mol - 157kJ/mol - (-328kJ/mol)
= -2805kJ/mol

ii. CaF2 + aq ------> Ca2+(aq) + 2F-(aq)

Hdissol = EL - H(hydration)Ca2+ - 2*H(hydration)F- = -2805 kJ -1507 kJ -2*505 kJ = -5322 kJ/mol

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