The Best Answer for Can anyone help with lattice energy?
You're just playing, aren't you Danny?
I'm sure you know about Hess's Law and Born-Haber cycles...
Cs(s) + ½Cl2(g) --> CsCl(s) .....ΔHf(CsCl(s)) = -443 kJ mol‾¹
Because enthalpy is a state function, it is invariant on the route by which that system state is reached. Hence, the overall reaction can be broken down into steps, each of which are experimentally favourable in the determination of their respective ΔH values. Thus:
Cs(s) + ½Cl2(g) --> Cs(g) + ½Cl2(g) .....ΔHsub(Cs) = +76 kJ mol‾¹
Cs(g) + ½Cl2(g) --> Cs(g) + Cl(g) .....BD(½Cl2) = +121 kJ mol‾¹
Cs(g) + Cl(g) --> Cs+(g) + e‾ + + Cl(g) .....IE1(Cs) = +376 kJ mol‾¹
Cs+(g) + e‾ + + Cl(g) --> Cs+(g) + Cl‾(g) .....EA(Cl) = -349 kJ mol‾¹
Cs+(g) + Cl‾(g --> CsCl(s) .....–Ulatt(CsCl) = unknown
ΔHf(CsCl(s)) = ΔHsub(Cs) + BD(½Cl2) + IE1(Cs) + EA(Cl) – Ulatt(CsCl)
==> –Ulatt(CsCl) = ΔHf(CsCl(s)) - ΔHsub(Cs) - BD(½Cl2) - IE1(Cs) - EA(Cl)
= (-443) - (76) - (121) - (376) - (-349)
= -667kJ mol‾¹
The lattice energy of CsCl is calculated to be 667 kJ mol‾¹.