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Standard XII

Chemistry

Electron Gain Enthalpy

The electron ...

Question

The electron affinity of chlorine is -3.61 eV. IP value of the same element is 12.967 eV. $3 \times 10^{23}$ atoms of chlorine are converted to $C l^{-}$ ions and the energy liberated is used to ionize chlorine atoms to $C l^{+}$ ions. How many chlorine atoms could be ionized in the above process?

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Solution

(i) $C l + - C \to - C l$ $Δ H = - 3.61$ ev./atom.
So for $3 \times 10^{23}$ such reaction energy released $= 3.61 \times 3 \times 10^{23}$
(ii) $C l ⟶ - C l + - e$ $Δ H = 12.967$ .
$\Rightarrow$ No. of chlorine atoms ionized in (ii)
$= \frac{3.61 \times 3 \times 10^{23}}{12.967} = 0.835 \times 10^{23}$
So, total number of ionized atoms $= 0.835 \times 10^{23}$ atoms.

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The electron affinity of chlorine is -3.61 eV. IP value of the same element is 12.967 eV. 3×1023 atoms of chlorine are converted to Cl− ions and the energy liberated is used to ionize chlorine atoms to Cl+ ions. How many chlorine atoms could be ionized in the above process?

(i) Cl+−C→−Cl ΔH=−3.61 ev./atom.So for 3×1023 such reaction energy released =3.61×3×1023(ii) Cl⟶−Cl+−e ΔH=12.967.⇒ No. of chlorine atoms ionized in (ii)=3.61×3×102312.967=0.835×1023So, total number of ionized atoms =0.835×1023 atoms.

The electron affinity of chlorine is -3.61 eV. IP value of the same element is 12.967 eV. $3 \times 10^{23}$ atoms of chlorine are converted to $C l^{-}$ ions and the energy liberated is used to ionize chlorine atoms to $C l^{+}$ ions. How many chlorine atoms could be ionized in the above process?