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The amount of energy released when 10^(1...

The amount of energy released when `10^(12)` atoms of `Cl` vapours are converted to `Cl^(ɵ)` ions, according to the equation :
`C_((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)`.

Text Solution

Verified by Experts

The amount of energy released when `1` mol `(~~ 6.0 xx 10^(23)` atoms) of `Cl` are converted to `Cl^(ɵ)` ions according to the above equation is `Delta_(eg)H^(ɵ)` of `Cl` vapours.
`:. Delta_(eg)H^(ɵ)` of `Cl` vapours
`= (- 58 xx 10^(-10) J xx 6 xx 10^(23))/(10^(12))`
`= -348 xx 10 J mol^(-1)`
`= -3.48 kJ mol^(-1)`
We know that
`1 eV` atom^(-1) `= 96.49 kJ mol^(-1)`
Therefore, `Delta_(eg)H^(ɵ)` of `Cl` vapours in `eV` is
`(-3.48)/(96.49) = - 0.036 eV "atom"^(-1)`
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