Question

Write the born-haber cycle for the formation of cacl2

Answer 1

The Born-Haber's cycle of formation of CaCl₂ takes following steps:
1. Solid calcium is converted to gaseous state: (Enthalpy of atomization)
Ca_(s) --> Ca _(g) $\Delta {H^0}_a=178KJ/mol$
2. The calcium is converted to ionic form (divalent cation): (Ionization enthalpy)
Ca --> Ca⁺ + e^- , $\Delta {H^0}_{IE}=590KJ/mol$
Ca⁺ --> Ca⁺² + e^- , $\Delta {H^0}_{IE}=1145KJ/mol$
3. Atomization of chlorine molecule to chlorine atom : (Atomization enthalpy)
1/2Cl₂ --> Cl $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\Delta {H^0}_{Cl-Cl}=121KJ/mol$
For two chlorine atoms required, the atomization energy = 2 x 121 = 242 KJ/mol
4. Conversion of chlorine atom to ion : Electron affinity :
Cl + e^- --> Cl^- , $\Delta {H^0}_{ea}=-364KJ/mol$
5. Finally the two ions join together by lattice enrgy as:(here two Cl^- ions are involved)
Ca⁺² + 2Cl^- --> CaCl₂
Hence,
Lattice nergy = Heat of formation - heat of atomization-dissociation energy-(sum of ionization energies)-(sum of electron affinities)
Since, heat of formation(i,e. standard enthalpy of formation) of CaCl₂ = -796 KJ/mol
Or, Lattice energy = -796 - 178-242 - (590+1145)-(2 x-364) = -2223 KJ/mol