Find the experimental lattice energy of aluminum oxide using a Born-Haber cycle. In addition to data available in your textbook, the following thermodynamic values will be of use: the second electron affinity for oxygen is –779.6 kJ/mol, the sublimation energy for Al is 330.0 kJ/mol, and the heat of formation of aluminum oxide is –1675.7 kJ/mol. Compare the experimental value to the calculated value from problem 1.

2Al(s) + 3/2O2(g)  Al2O3(s) Hfo = –1675.7
2Al(s) 2Al(g) 2S = 2(330.0) = 660.0
2Al(g) 2Al3+(g) 2(IP1 + IP2 + IP3) = 2(5.986 + 18.828 + 28.447)96.4869

= 10278.0
3/2O2(g) 3O(g) 3/2BDE = 3/2(493.6) = 740.4
3O(g) 3O2–(g) –3(EA1 + EA2) = –3(141.0 + –779.6) = 1915.8
2Al3+(g) + 3O2–(g) Al2O3(s) –Elat

For the thermochemical cycle:

–Hfo + 2S + 2(IP1 + IP2 + IP3) + 3/2BDE – 3(EA1 + EA2) – Elat = 0

Elat = –Hfo + 2S + 2(IP1 + IP2 + IP3) + 3/2BDE – 3(EA1 + EA2)

Elat = 1675.7 + 660.0 + 10278.0 + 740.4 + 1915.8 = 15269.9 kJ/mol

There is ~2.4% difference between the experimental and theoretical values, which is quite acceptable.