Use the following data to calculate Δ_lattice H^Θfor NaBr.

Δ_sub H^Θfor sodium metal = 108.4 kJ mol^–1

Ionization enthalpy of sodium = 496 kJ mol^–1

Electron gain enthalpy of bromine = – 325 kJ mol^–1

Bond dissociation enthalpy of bromine = 192 kJ mol^–1

Δ_f H^Θfor NaBr (s) = – 360.1 kJ mol^–1

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 metal(Δ_subH^Θ) →Ionization of the metal (Δ_iH^Θ) →Dissociation of the non-metal (Δ_dissH^Θ) →Gain of electrons by the non-metal(Δ_egH^Θ)

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

Δ_f H^Θ = Δ_subH^Θ + Δ_iH^Θ + 1/2 Δ_dissH^Θ +Δ_egH^Θ + Δ_latticeH^Θ

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

Na(s) → Na(g) ; Δ_subH^Θ =108.4 kJ mol^–1 (i)

Na→ Na⁺ + e^- ; Δ_iH^Θ = 496 kJ mol^–1 (ii)

1/2 Br₂→ Br ; 1/2 Δ_dissH^Θ = ( ) = 96 kJ mol^–1 (iii)

Br + e^-→ Br^- ;Δ_egH^Θ= – 325 kJ mol^–1 (iv)

And enthalpy of formation Δ_f H^Θfor NaBr (s) = – 360.1 kJ mol^–1

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

Hence,

enthalpy of formation Δ_f H^Θ = processes (i) + (ii) + (ii) + (iv)

Δ_f H^Θ = Δ_subH^Θ + Δ_iH^Θ + 1/2 Δ_dissH^Θ +Δ_egH^Θ + Δ_latticeH^Θ

Therefore, Δ_lattice H^Θ = Δ_f H^Θ - Δ_subH^Θ - Δ_iH^Θ - 1/2 Δ_dissH^Θ -Δ_egH^Θ

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

= - 735.5 kJ mol^–1