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.

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Infinity Learn · Accepted Answer

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 $$3.6(96.45)kJ$$                                                                                $$X=1.66$$×$$1023$$

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.

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