The amount of energy released when `10^(12)` atoms of `Cl` vapours are converted to `Cl^(ɵ)` ions, according to the equation :
`Cl_((g))+e^(-) rarr Cl_((g))^(ɵ)` is `58 xx 10^(-10)J`
Calculate the `Delta_(eg)H^(ɵ)` of `Cl` atom in `kJ mol^(-1)` and `eV "atom"^(-1)`.

Explanation

For `10^10 ` atom `Cl_(g) + e^(-) to Cl^(-) , ` energy released ` = 57.86 xx 10^(-10) J`
For 1 atom ` rArr ` energy released ` = 57.86 xx 10^(-10) xx 10^(-10) J ` , energy released ` = 57.86 xx 10^(-20) J`
For 1 mole i.e. for ` 6.023 xx 10^23` atom energy released ` = 57.86 xx 10^(-20) xx 6.023 xx 10^23 J = 348.49 ` kJ/mole
energy released in eV ` = 57.86 xx N_A // 1.6 xx 10^(-19) xx 217.81 xx 10^22 eV `