Homnom - 2-11-2012 at 06:59
OK so I'm getting really irritated by these equations for forming lattice energies. The equations I am using are
Lelect = - (Z+Z-A|e|2) divided by 4(pi)ε0 x R
Lrep = (Z+Z-A|e|2) divided by 4(pi)ε0 x n x R
and finally, L = Lrep + Lelect
I'm supposed to calculate L for the first 4 alkali halides then the average affinity of chlorine. problem is I'm getting stupid values.
For example, for LiCl, I'm using the values: R = 2.57; A = 1.74756; n = 5; e as below; E as below
so Lelect = (1 x 1 x 1.74756 x (1.602177 x 10-19)2 divided by 4 x pi x 8.85429 x 10-12 x 2.57
x10-10 = 1.57 x 10-18) x (NA / 1000) (to convert to kj) = -944.7 Kj mol-1
and Lrep = (1 x 1 x 1.74756 x (1.602177 x 10-19)2 divided by 4 x pi x 8.85429 x 10-12 x 2.57
x10-10 = 1.57 x 10-18x 5) x (NA / 1000) = 188.9
added together this = -755.8, which is nowhere near the value it should be near 861 and is the value I should have got for NaCl (I got -738.2) Can
someone see where I am going wrong???
[Edited on 2-11-2012 by Homnom]
DJF90 - 2-11-2012 at 07:55
For LiCl, the Borne exponent is not n=5! LiCl has n=7 (average of the two closed shell ions, Li+ is [He] = 5, Cl- is [Ar] = 9). Sodium chloride has
n=8 (Na+ is [Ne] = 7). The first four alkali metal chlorides have the NaCl structure; I suspect they're not asking for CsCl because it has its own
structure, with a different madelung constant (A) associated with it. I think the n-value is the only problem with your working, so give it a go and
see if you get the right answer. If not, I'll have a look and see if I can spot anything else troublesome.
Any chance you're a first year chemistry undergraduate 
?
Homnom - 2-11-2012 at 08:46
Thanks a lot! I was stupidly just using the Borne exponent for the cation only which messed up all my answers . still not spot on but for NaCl -753 kJ
mol-1 is as good as I'm going to get with their data and my own rounding I suspect. They did actually ask for Cs, I just assumed that was the fourth
one as I don't remember the table that well yet haha (they did give me the constant though).
You're correct in your assumption, greetings from Cardiff uni 
DJF90 - 2-11-2012 at 09:21
Glad it helped. The answer will probably not match exactly an experimental value as there are certain approximations/assumptions involved in the
derivation of the equation. You're going to have fun over the next couple of years, I assure you...
Homnom - 2-11-2012 at 09:38
Haha thanks again, that last sentence did sound ominous though.
DJF90 - 2-11-2012 at 11:29
Been there (not there, but you know what I mean), done that. I suspect we'll be hearing from you again