CHM 501

35. 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 34.

Answer

2 Al(s) + 3/2 O2(g) → Al2O3(s)

ΔHfo = –1675.7

2 Al(s) → 2 Al(g)

2S = 2(330.0) = 660.0

2 Al(g) → 2 Al3+(g)

2(IP1 + IP2 + IP3) = 2(5.986 + 18.828 + 28.447)96.4869 = 10278.0

3/2 O2(g) → 3 O(g)

3/2BDE = 3/2(493.6) = 740.4

3 O(g) → 3 O2–(g)

–3(EA1 + EA2) = –3(141.0 + –779.6) = 1915.8

2 Al3+(g) + 3 O2–(g) → Al2O3(s)

–Elat

For the thermochemical cycle:

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

Elat = –ΔHfo + 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.