Calculate the lattice energy of Na2O for the following data

Calculate the lattice energy of Na2O for the following data:

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

Electron affenity of O(g) 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 2 Na(s) + 1/2 O2(g)---->Na2O(s)

Solution

k .. we want Na2O so we can start with 2Na (s)+ 1/2O2(g) but Na needs to be in gaseous state so

1. 2Na (s)+ 1/2O2(g) => 2Na (g)+ 1/2O2(g) dH: (109*2) KJ/mol since we have 2 moles of Na

next we need to break the O2 bond to create O atoms
2. 2Na (s)+ 1/2O2(g) => 2Na (g)+ O(g) dH: (499/2) KJ/mol since we have 1/2 moles of O2

next we need to ionize Na
3. 2Na (s)+ 1/2O2(g) => 2Na+ (g) + 2e- + O (g) dH: (495*2) KJ/mol since we have 2 moles of Na

next O will accept the 2e- (electron affinity of O for 2e-)
4. 2Na+ (g) + 2e- + O (g) => 2Na+ + O^(2-) (g) dH: (603)KJ/mol

reaction 5 is our lattice energy reaction
5. 2Na+ + O^(2-) (g) => Na2O (s) dH: unknown (to find)

since we know that enthalpy is a state function, instead of doing all these steps we could do the reaction in one step and have the same enthalpy for net reaction.
2 Na (s) + O2 (g) => Na2O (s) dH: -416 kj/mol

so -416 = 1 +2 +3 +4 +5
therefore,

5 = -416-(1 +2 +3 +4) viz. lattice energy

lattice energy= -416-((109*2)+(499/2)+(495*2)+(603))
= -2 476.5 KJ/mol

this is the answer for question 6 HW19 .. lol .. ;)