The `Delta_(eg)H^(ɵ)` of `Br` is `3.4 eV`. How much energy in kcal is released when `0.8 g` of `Br (g)` is completely converted to `Br^(ɵ) (g)` ions. `(1 eV = 23.06 kcal mol^(1-))`
b. The energy released when `10^(7)` atoms of `I (g)` is converted to `I^(ɵ) (g)` ions, is `5 xx 10^(-13) J`. Calculate `Delta_(eg)H^(ɵ)` of `I (g)` in (i) `eV "atom"^(-1)` and (ii) `kJ mol^(-1)`.

Explanation

a. Moles of `Br = ("Mass")/("Atomic mass") = (0.8)/(80) = 10^(-2)`
Energy released `= 10^(-2) xx 3.4 eV xx 23.06`
`= 0.784 kcal`
b. i. `Delta_(eg)H^(ɵ) = (6.023 xx 10^(23) xx 5 xx 10^(-13))/(10^(7)) = 30.115 xx 10^(3)`
`= 30115 kJ mol^(-1)`
ii. `96.49 kJ mol^(-1) = eV "atom"^(-1)`
`:. Delta_(eg)H^(ɵ) = (30115//96.49) eV "atom"^(-1)`
`= 312.104 eV "atom"^(-1)`