calculate the lattice energy of sodium oxide (Na2O) from the following data:

Ionization energy of Na(g): 495 kJ/mol

Electron affinity of O2 for 2e: 603 kJ/mol

Energy to vaporize Na(s): 109 kJ/mol

O2(g) bond energy: 499 kJ/mol

Energy change for the reaction 2Na(s)+ 1/2 O2(g)--> Na2O(s): -416 kJ/mol

3 answers

2Na(s) ==> 2Na(g) dHsub = 109*2 = ?
2Na(g) ==> 2Na^ + 2e IP = 495*2 = ?
O2(g) ==> 2O(g) B.E. = 499*1/2 = ?
O + 2e ==> O^2- EA = -603 = ?
2Na^+(g) + O^2-(g) ==> Na2O(s) Ecrystal
------------------------------------
2Na(s) + 1/2O2(g) ==> Na2O(s) = -416
Hf = 416
-416=dHsub + IP + BE + EA(-) + Ecrystal
Solve for Ecrystal.

1170

-2477