The amount of energy released when 1012 atoms of Cl vapour are converted to Cl -ions- according to the equation-Cl g - e - Cl g -is 58 - 10-10 JCalculate the - eg H -of Cl atom in eV atom -1-A- -0-036 eV atom -1B- -3-480 eV atom -1C- -0-038 eV atom -1D- -0-361 eV atom -1

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Standard XII

Chemistry

Electron Gain Enthalpy

The amount of...

Question

The amount of energy released when $10^{12}$ atoms of $C l$ (vapour) are converted to $C l^{-}$ ions, according to the equation:
$C l_{(g)} + e^{-} \to C l_{(g)}^{-} is 58 \times 10^{- 10} J$
Calculate the $Δ_{e g} H^{-}$ of $C l$ atom in ${eV atom}^{- 1}$ .

- 0.036 {eV atom}^{- 1}

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- 3.480 {eV atom}^{- 1}

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+ 0.038 {eV atom}^{- 1}

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- 0.361 {eV atom}^{- 1}

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Solution

The correct option is A $- 0.036 {eV atom}^{- 1}$
The amount of energy released when 1 mole of $C l$ is converted to $C l^{-}$ ions,
$= \frac{- 58 \times 10^{- 10} J \times 6 \times 10^{23}}{10^{12}}$
$= - 3.48 k J m o l^{- 1}$
We know that,
$1 e V a t o m^{- 1} = 96.49 k J m o l^{- 1} \frac{- 3.48}{96.49} = - 0.036 e V a t o m^{- 1}$

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The amount of energy released when 1012 atoms of Cl (vapour) are converted to Cl− ions, according to the equation: Cl(g)+e−→Cl−(g) is 58×10−10 J Calculate the ΔegH− of Cl atom in eV atom−1.

The correct option is A −0.036 eV atom−1The amount of energy released when 1 mole of Cl is converted to Cl− ions, =−58×10−10J×6×10231012 =− 3.48kJ mol−1 We know that, 1eVatom−1=96.49 kJmol−1−3.4896.49=−0.036 eVatom−1

The amount of energy released when $10^{12}$ atoms of $C l$ (vapour) are converted to $C l^{-}$ ions, according to the equation:
$C l_{(g)} + e^{-} \to C l_{(g)}^{-} is 58 \times 10^{- 10} J$
Calculate the $Δ_{e g} H^{-}$ of $C l$ atom in ${eV atom}^{- 1}$ .