When a mole of magnesium bromide is prepared from 1 mole of magnesium and 1 mole of liquid bromine, 524 kJ of energy is released. The heat of sublimation of Mg metal is `148 "kJ mol"^(-1)`. The heat of dissociation of bromine gas into atoms is `193 "kJ mol"^(-1)`. The heat of vaporization of liquid bromine is 31 kJ `"mol"^(-1)`. The ionisation energy of magnesium is 2187 kJ `mol^(-1)` and the electron affinity of bromine is `- 662 "kJ mol"^(-1)`. Calculate the lattice energy of magnesium bromide.

Explanation

Given :
`Mg(s) + Br_2(l) to MgBr_(2)(s) " " DeltaH_f^0=-524 "KJ mol"^(-1)`
Sublimation :
`Mg(s) to Mg(g) " " DeltaH_1^0=+148 "KJ mol"^(-1)`
Ionisation : `Mg(g) to Mg^(2+)(g) +2e^(-) " " DeltaH_2^0 =2187 "KJ mol"^(-1)`
Vapourisaton :
`Br_2(l) to Br_(2)(g) " " DeltaH_3^0=+31 "KJ mol"^(-1)`
Dissociation :
`Br_2(g) to 2Br(g) " " DeltaH_4^0=+193 "KJ mol"^(-1)`
Electron affinity :
`Br(g) +e^(-) to Br^(-) (g) " " DeltaH_5^0=-331 "KJ mol"^(-1)`

`DeltaH_f=DeltaH_1 + DeltaH_2+DeltaH_3+DeltaH_4+2DeltaH_5+u`
-524=148+2187+31+193+(2 x -331) + u
-524 =1897 + u
u=-524 - 1897
`u=-2421 "kJ mol"^(-1)`