Question

The lattice enthalpy of an ionic compound is the enthalpy when one mole of an ionic compound present in its gaseous state, dissociates into its ions. It is impossible to determine it directly by experiment. Suggest and explain an indirect method to measure lattice enthalpy of \[\ce{NaCl(s)}\].

Answer 1

The enthalpy change that occurs when one mole of an ionic compound dissociates into its ions in a gaseous state is called the lattice enthalpy of an ionic compound.

\[\ce{Ba^{+} + Cl- (s) -> Na+ (g) + Cl- (g)}\]

\[\ce{Δ_{lattice} H^Θ -> + 788 kJ mol^{-1}}\]

Let us now calculate the lattice enthalpy of \[\ce{Na+ Cl- (s)}\] by following steps given below:

1. \[\ce{Na (s) -> Na (g)}\], sublimation of sodium metal, Δ_subH^Θ = 108.4 KJmol-1

2. \[\ce{Na (g) → Na^{+} (g) + e- (g)}\] the ionization sodium atoms, ionization enthalpy.

Δ_iH^Θ = 496 kJ/mol 

3. \[\ce{1/2 Cl2 (g) -> 2Cl (g)}\], the dissociation of chlorine, the reaction enthalpy is half the bond dissociation enthalpy.

\[\ce{1/2}\] Δbond H^Θ = 121 kJ/mol

4. \[\ce{Cl (g) + e- -> Cl- (g)}\] electron gained by chlorine atoms. The electron gain enthalpy,

Δ_egH^Θ = – 348.6 kJ/mol.

5. \[\ce{Na+ (g) + Cl- (g) -> Na+ Cl- (s)}\]