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Use the following data to calculate Δlattice HΘfor NaBr.
Δsub HΘfor 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
Δf HΘfor NaBr (s) = – 360.1 kJ mol–1
Sublimation of the metal(ΔsubHΘ) →Ionization of the metal (ΔiHΘ) →Dissociation of the non-metal (ΔdissHΘ) →Gain of electrons by the non-metal(ΔegHΘ)
Δf HΘ =Δsub HΘ+Δi HΘ+1/2 dissHΘ + ΔegHΘ+ ΔlattticeHΘ
To calculate the lattice enthalpy of NaBr,
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)
enthalpy of formation Δf HΘ = processes (i) + (ii) + (ii) + (iv)
Δlattice HΘ = Δf HΘ – ΔsubHΘ – ΔiHΘ – 1/2 ΔdissHΘ –ΔegHΘ
= -735.5kjmol-1.
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