How many Cl atoms can you ionize in the process of Cl → Cl+ + e– by the energy liberated for the process Cl + e– → Cl– for one Avogadro number of atoms? Given IP=13 eV and EA=3.60 eV.

EA of Chlorine is 3.6 ev/atomThus ΔHeg of Cl= -3.6 ev/atom⇒ Clg+e→Cl-g, ∆Heg=-3.696.45kJThis is the energy released by the conversion of avagadro's number of Clg→Cl-gIP of Chlorine is 13ev/atomThus ∆HIE=+13.6 eV⇒ Clg→Cl+g+e, ∆HIE=+13.696.45kJThis is the energy required for the convertion of Avagadro's number of Clg→Cl+gAvailable energy is the energy released by the conversion of NAatoms of Clg→Cl-g which is 3.6(96.45)kJ .NA atoms Clg→Cl-g release 3.6(96.45)kJ = available energyNA atoms Clg→Cl+g require 13(96.45)kJX atoms Clg→Cl+g require 3.6(96.45)kJ                                                                                X=1.66×1023