CHM 501

Homework 13

Due November 8, 2001

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

Answers

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

Ca(s) + F2 CaF2(s)	Hfo = –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:

–H_f^o + S + (IP₁ + IP₂) + D + (–2EA) –E_lat = 0

Rearranging and substituting numbers:

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

Using the Born-Landé equation:

N_o = 6.022×10²³

Z⁺ = +2

Z^– = –1

e = 1.602×10^–19

M = 2.51939 (lecture notes)

= 3.14159

_o = 8.854×10^–12

r_o 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)

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

n is found from the compressability, :

= 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!