CHM 501

32. Estimate the lattice energy for calcium fluoride using both the Born-Landé equation and a Born-Haber cycle. Compare the two results.

Answer

Using a Born-Haber cycle to find the experimental lattice energy:

Ca(s) + F2 → CaF2(s)

ΔHfo = –1228.0 kJ/mol
(CRC handbook)

Ca(s) → Ca(g)

S = 177.8 kJ/mol
(CRC handbook)

Ca(g) → Ca2+(g)

IP1 + IP2 = (6.113 + 11.871)×96.4869 = 1735.2 kJ/mol
(DMA, Table 1.7, page 43)

F2(g) → 2 F(g)

D = 154.6 kJ/mol
(DMA, Table 2.8, page 90)

2 F(g) → 2 F(g)

–2×EA = –2×328.0 = –656.0 kJ/mol
(DMA, Table 1.8, page 48)

Ca2+(g) + 2 F → CaF2(s)

–Elat

Summing the thermodynamic quantities gives:

–ΔHfo + S + (IP1 + IP2) + D + (–2EA) –Elat = 0

Rearranging and substituting numbers:

Elat = –(–1228.0) + 177.8 + (1735.2) + 154.6 + (–656.0) = 2639.6 kJ/mol


Using the Born-Landé equation:

No = 6.022×1023

Z+ = +2

Z = –1

e = 1.602×10–19

M = 2.51939 (lecture notes)

π = 3.14159

εo = 8.854×10–12

ro is estimated using ionic radii:

r(Ca2+) = 114 pm (DMA, Table 5.8, page 225)

r(F) = 119 pm (DMA, Table 5.10, page 227)

ro ~ 114 + 119 = 233 pm = 2.33×10–-10 m

n is found from the compressability, κ:

κ = 1.23×10–11 (lecture notes)

Putting numbers into equations:

This compares to n = 8 using electron configuration estimates (see page 224 in DMA).

Finally, find the lattice energy:

The two methods compare exceedingly well!