Na( s) ⟶ Na+(g) ; Δ H=495.8 kJ/molElectron gain enthalpy:
21 Br2(ℓ)+e− ⟶ Br−(g); Δ H=325 kJ / molLattice energy:
Na+(g)+Br−(g) ⟶ NaBr( s); Δ H=−728.4 kJ/ molFormation enthalpy:
Na(s)+21 Br2(ℓ) ⟶ NaBr(s); Δ H=?
Δ H=495.8−325−728.4
Δ H=−557.6 kJ=−5576 × 10−1 kJ