The amount of energy when millon atoms of iodine are completely converted into `I^(-)` ions in the vapour state according to the equation, `I(g) +e^(-

Question

The amount of energy when millon atoms of iodine are completely converted into `I^(-)` ions in the vapour state according to the equation, `I(g) +e^(-

The amount of energy when millon atoms of iodine are completely converted into `I^(-)` ions in the vapour state according to the equation, `I(g) +e^(-)(g) rarr I^(-)(g) is 5.0 xx 10^(-13)J`.

Monjur Alahi

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2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

The electron gain enthalpy of iodine is equal to the energy released when 1mole of iodine atoms in vapour state are converted into `I^(-)` ions.
`= (5.0 xx 10^(-13) xx 6.023 xx 10^(23))/(10^(6)) =- 30.1 xx 10^(4)J =- 301 kJ`.
Electron gain enthalpy of iodine in `eV` per atom `= (-301)/(96.5) =- 3.12`.

4 views Answered 2023-02-05 15:29