Question

Use the following data to calculate Δ_lattice H^Θ for NaBr.

Δ_sub H^Θ for 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^Θ for NaBr (s) = – 360.1 kJ mol^–1

Answer 1

Solution is.........