Q. 26

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Answer :

o Lattice enthalpy of an ionic compound is the enthalpy change which occurs when one mole of an ionic compound dissociates into its constituent ions in the gaseous state. Under standard conditions.

Formation of an ionic compound in the gaseous state involves several processes like :

Sublimation of the metalsubHΘ) Ionization of the metal (ΔiHΘ) Dissociation of the non-metal dissHΘ) Gain of electrons by the non-metal(ΔegHΘ)

The enthalpy of formation is related to these enthalpies of these processes by the relation:

Δf HΘ = ΔsubHΘ + ΔiHΘ + 1/2 ΔdissHΘ +ΔegHΘ + ΔlatticeHΘ

o To calculate the lattice enthalpy of NaBr, we have to consider a set of reactions for the process:

Na(s) Na(g) ; ΔsubHΘ =108.4 kJ mol–1 (i)

Na Na+ + e- ; ΔiHΘ = 496 kJ mol–1 (ii)

1/2 Br2 Br ; 1/2 ΔdissHΘ = ( ) = 96 kJ mol–1 (iii)

Br + e- Br- ;ΔegHΘ= – 325 kJ mol–1 (iv)

And enthalpy of formation Δf HΘfor NaBr (s) = – 360.1 kJ mol–1

Enthalpies in different steps involved in the formation of NaBr (s):

Hence,

enthalpy of formation Δf HΘ = processes (i) + (ii) + (ii) + (iv)

Δf HΘ = ΔsubHΘ + ΔiHΘ + 1/2 ΔdissHΘ +ΔegHΘ + ΔlatticeHΘ

Therefore, Δlattice HΘ = Δf HΘ - ΔsubHΘ - ΔiHΘ - 1/2 ΔdissHΘ -ΔegHΘ

= (–360.1 - 108.4 -96 – 496 +325) kJ mol–1

= - 735.5 kJ mol–1

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