Use the following data to calculate ∆lattice Hᶱ for NaBr. ∆sub Hᶱ for sodium metal = 108.4 kJ/ mol
Ionization enthalpy of sodium = 496 kJ/mol
Electron gain enthalpy of bromine = – 325 kJ mol–1 Bond dissociation enthalpy of bromine = 192 kJ mol–1 ∆f Hᶱ for NaBr (s) = – 360.1 kJ/ mol
Use the following data to calculate ∆lattice Hᶱ for NaBr. ∆sub Hᶱ for sodium metal = 108.4 kJ/ mol
Ionization enthalpy of sodium = 496 kJ/mol
Electron gain enthalpy of bromine = – 325 kJ mol–1 Bond dissociation enthalpy of bromine = 192 kJ mol–1 ∆f Hᶱ for NaBr (s) = – 360.1 kJ/ mol
This is a Short Answer Type Questions as classified in NCERT Exemplar
In order to calculate the lattice enthalpy of NaBr,
(i) Na (s) →Na (g) ; ΔsubH? =108.4 kJ/mol
(ii) Na→Na+ + e- ΔiH? = 496kJ/ mol
(iii) Br→ Br, Δdiss H? = 96kJ/ mol
(iv) Br+e-Br- ΔegH? = - 325 kJmol-1
? fH? =? subH? + Δdiss H
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Chemistry NCERT Exemplar Solutions Class 11th Chapter Six 2025
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