Question

The amount of energy released when ${10}^{12}$ atoms of $Cl$ vapour are converted to $C{l}^{-}$ ions, according to the equation;
$Cl\left(g\right)+e^{-}\to C{l}^{-}\left(g\right)$ is $58\times {10}^{-10}J$. The magnitude of ${\Delta }_{eg}H$ of $Cl$ atom (in $kJmo{l}^{-1}$) is

Answer 1

The amount of energy released when 1 mole ($\approx 6\times {10}^{23}$ atoms) of gaseous $Cl$ are converted to $C{l}^{-}$ ions according to the given equation is termed as ${\Delta }_{eg}H$ of $Cl$ vapour.
$\therefore {\Delta }_{eg}H=\frac{-58\times {10}^{-10}J\times 6\times {10}^{23}}{{10}^{12}}=-3480Jmo{l}^{-1}=-3.48kJmo{l}^{-1}$