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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 $58 \times 10^{- 10} J$ . The magnitude of $Δ_{e g} H$ of $C l$ atom (in $k J m o l^{- 1}$ ) is

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Solution

The amount of energy released when 1 mole ( $\approx 6 \times 10^{23}$ atoms) of gaseous $C l$ are converted to $C l^{-}$ ions according to the given equation is termed as $Δ_{e g} H$ of $C l$ vapour.
$∴ Δ_{e g} H = \frac{- 58 \times 10^{- 10} J \times 6 \times 10^{23}}{10^{12}} = - 3480 J m o l^{- 1} = - 3.48 k J m o l^{- 1}$

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