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/atom

Thus ΔHeg of Cl= -3.6 ev/atom

⇒ $\mathrm{Cl}\left(\mathrm{g}\right)+\mathrm{e}\to {\mathrm{Cl}}^{-}\left(\mathrm{g}\right), ∆{\mathrm{H}}_{\mathrm{eg}}=-3.6\left(96.45\right)\mathrm{kJ}$

This is the energy released by the conversion of avagadro's number of $\mathrm{Cl}\left(\mathrm{g}\right)\to {\mathrm{Cl}}^{-}\left(\mathrm{g}\right)$

IP of Chlorine is 13ev/atom

Thus $∆{\mathrm{H}}_{\mathrm{IE}}=+13.6 \mathrm{eV}$

⇒ $\mathrm{Cl}\left(\mathrm{g}\right)\to {\mathrm{Cl}}^{+}\left(\mathrm{g}\right)+\mathrm{e}, ∆{\mathrm{H}}_{\mathrm{IE}}=+13.6\left(96.45\right)\mathrm{kJ}$

This is the energy required for the convertion of Avagadro's number of $\mathrm{Cl}\left(\mathrm{g}\right)\to {\mathrm{Cl}}^{+}\left(\mathrm{g}\right)$

Available energy is the energy released by the conversion of ${\mathrm{N}}_{\mathrm{A}}$atoms of $\mathrm{Cl}\left(\mathrm{g}\right)\to {\mathrm{Cl}}^{-}\left(\mathrm{g}\right)$ which is 3.6(96.45)kJ .

${\mathrm{N}}_{\mathrm{A} }$atoms $\left[\mathrm{Cl}\left(\mathrm{g}\right)\to {\mathrm{Cl}}^{-}\left(\mathrm{g}\right)\right]$ release 3.6(96.45)kJ = available energy

${\mathrm{N}}_{\mathrm{A}}$ atoms $\left[\mathrm{Cl}\left(\mathrm{g}\right)\to {\mathrm{Cl}}^{+}\left(\mathrm{g}\right)\right]$ require 13(96.45)kJ

X atoms $\left[\mathrm{Cl}\left(\mathrm{g}\right)\to {\mathrm{Cl}}^{+}\left(\mathrm{g}\right)\right]$require 3.6(96.45)kJ

                                                                                $\mathrm{X}=1.66\times {10}^{23}$