The amount of energy released when one million atoms of iodine are completely converted into `I^(-)` ions in the vapour state according to the equation I(g) `+ e^(-) to I^(-) ( g) " is " 4.9 xx 10^(-13) J`
Calculate the electron affinity of iodine in (i) `Kj//mol` and (ii) in eV per atom.

Answer

296 `kj// mol` and 3.06 ev`//`atom

Explanation

Required energy `=0.1 xx 3.36 xx 23.06 =7.748 kcal`
Electron affinity `=6.023 xx 10^(23) xx 4.9 xx 10^(-13) xx 10^(-6)`
`=29 .5 xx 10^(4) xx 10^(-3) kJ =295 kJ mol^(-1)`
`= 295// 96.49 eV //atom =3.06 eV //atom`