The amount of energy when 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 `4.9X10^(-13)` J.What would be the electron gain enthalpy of iodine in terms of KJ `mol^(-1)` and eV per atom?

Explanation

The amount of energy released for the conversion of ne mole `(6.02 xx 10 ^(23))` of atoms of iodine into `I ^(-)` ions can be calculated. This corresponds to electron gain enthalpy. Thus,
Amount of energy released for `1 xx 10 ^(6)` atoms of iodine
`=4.9 xx 10 ^(-13) J`
Amount of energy released for `6.02 xx 10^(23)` atoms of iodine
`= (4.9 xx 10 ^(-13))/(1 xx 10 ^(6)) xx 6. 02 xx 10 ^(23)`
` = 29 . 5 xx 10 ^(4) J = 295 kJ// mol.`
Now 1 ev/atom`= 96.3 kJ mol^(-1)`
`therefore ` Electron gain enthapy `=-295kJ mol ^(-1) `
`= (295)/(96.3)=- 3.06 eV//` atom