The Student Room Group
Reply 1
there is 2 moles of na so double the enthalpy of *whatever* when your doing the calculation. On the diagram if it was atmoisation energy it would be x2 with just one arrow. This is hard to explain but can you understand?
Reply 2
It's a bit difficult to explain without drawing the thing but I'll try...

You'd have Na2O (s) at the bottom.

First line up - the reverse of ∆f. Thus you have 2Na (s) + 1/2 O2 (g) as those are the elements in their standard states.

Second line up - you make 2Na (g) + 1/2 O2 (g), as you have atomised the 2Na (s) in line 1 into 2Na (g). Because the ∆atm applies for molar quantities, you have to multiply your ∆atm of Na by two as you have 2 moles of Na.

Third line up - you should atomise your O2. Remember ∆atm is making 1 mole of gaseous atoms; on line 2 you only have 1/2 mol of oxygen. So you'd produce O (g) from 1/2 O2 (g), which was your ∆atm for oxygen.

Fourth line up - you should ionise your Na first; to get from 2Na (g) to 2Na+ (g) you needed to ionise 2 moles of Na. Thus you multiply ∆IE of sodium by two.

Fifth line - this is where you add one electron to O (g) to get O- (g). This is your first electron affinity for oxygen, and because it is exothermic your fifth line should go down.

Last line - this is where you get the final O2- (g) with your second electron affinity; as this is endothermic your line should come back up. And this should leave you with 2Na+ (g) + O2- (g), which you can then draw a line back down to the bottom line for your lattice enthalpy.

Hope that wasn't too confusing!
Reply 3
It's similar whether it's the anion or cation which is more than one. The enthalpy changes that involve the metal cation are the enthalpy change of atomisation of it and ionisation energies of it. So, you just have to multiply those values accordingly with the number of metal cations in the compound.
For example, comparing between NaCl and Na2O..
For NaCl, it's just Na (s) -> Na (g) for the atomisation part.
For Na2O, (since there are 2 Na), it'll be 2 Na (s) -> 2 Na (g). Hence, you multiply the enthalpy change of atomisation of Na by 2.

Hope this helped.

Edit: oops, didn't see excalibur post before me..haha
Reply 4
Cheers Excalibur, you really made it simple