Question

Use the following information to calculate ?H

Answer 1

Lattice energies, U(latt), cannot be measured directly, however they can be claculated by realising:- <Enthalpy of formation> = <Sum of your other enthalpies> all of which can be determined by other means.

1. Sublimation of Mg
Mg(s) + Cl2(g) ==> Mg(g) + Cl2(g) .....ΔH°(sub) = 148 kJ molˉ¹
2. Ionisations of Mg
Mg(g) + Cl2(g) ==> Mg+(g) + e- + Cl2(g) .....IE(1) = 738 kJ molˉ¹
Mg+(g) + e- + Cl2(g) ==> Mg2+(g) + 2e- + Cl2(g) .....IE(2) = 1450 kJ molˉ¹
3. Dissociation of Cl2
Mg2+(g) + 2e- + Cl2(g) ==> Mg2+(g) + 2e- + 2Cl(g) .....BD = 243 kJ molˉ¹
4. Electron Affinities of Cl's
Mg2+(g) + 2e- + 2Cl(g) ==> Mg2+(g) + e- + Clˉ(g) + Cl(g) .....EA = -349 kJ molˉ¹
Mg2+(g) + e- + Clˉ(g) + Cl(g) ==> Mg2+(g) + 2Clˉ(g) .....EA = -349 kJ molˉ¹
5. Lattice Energy of MgCl2
Mg2+(g) + 2Clˉ(g) ==> MgCl2(s) .....U(latt) = your unknown

6. Formation of MgCl2
Mg(s) + Cl2(g) ==> MgCl2(s) .....ΔHf° = -642 kJ molˉ¹

You should notice that overall equation #6 is the same as going stepwise from #1 through to #5.

So in terms of energies (thermodynamics):
ΔHf = ΔH(sub) + IE(1) + IE(2) + BD + 2EA + U(latt)
==> U(latt) = ΔHf° - ΔH°(sub) - IE(1) - IE(2) - BD - 2EA
= -642 - 148 - 738 - 1450 - 243 + 2(349) kJ molˉ¹
= -2523 kJ molˉ¹

So the lattice engery of magnesium chloride is found to be -2495.6 kJ molˉ¹.

This above is the Born-Haber cycle, an application of Hess' Law - look up these for more information