Q26Short answer
Use the following data to calculate ∆lattice H for NaBr. V –1 ∆sub H for sodium metal = 108.4 kJ mol –1 Ionization enthalpy of sodium = 496 kJ mol –1 Electron gain enthalpy of bromine = – 325 kJ mol Bond dissociation enthalpy of bromine = 192 kJ mol–1 V –1 ∆f H for NaBr (s) = – 360.1 kJ mol
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+735.5 kJ mol