The overall energy involved in the formation of CsCl from Cs(s) and Cl2(g) is -443 kJ/mol. Given the following information:
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.
Lattice Energy Of Cscl
For the best answers, search on this site https://shorturl.im/awsRx
Well, let’s think about it. When you talk about lattice energy, you talk about the energy needed to essentially COMPLETELY break apart an ionic bond (ionic solid –> separate gaseous ions). So, really, the more energy you need to pull these ionic solids apart, the more stable they were originally (or, the lower in energy they originally were). Therefore, if a compound has a higher lattice energy, it is more stable. Hope this helps! =D
The substance that has a higher lattice energy is the more stable substance as it will maintain the solid state longer. So X is the more stable substance as it will need a lot more energy to vaporize it.
The above answer is wrong but the above steps are probably correct.
The magnitude of U = -667 Kj / mol.
wow. this is hard…