The amount of energy when a million atoms of iodine are completely converted into $I^{-}$ ions in the vapor state, according to the equation $I (g) + e^{-} (g) \to I^{-} (g)$ , is $5.0 \times 1 0^{- 13}$ J.
Calculate the magnitude of the electron gain enthalpy of iodine in terms of kJ mol⁻¹.

Question

The amount of energy when a million atoms of iodine are completely converted into $I^{-}$ ions in the vapor state, according to the equation $I (g) + e^{-} (g) \to I^{-} (g)$ , is $5.0 \times 1 0^{- 13}$ J.

Calculate the magnitude of the electron gain enthalpy of iodine in terms of kJ mol⁻¹.

TestHub · Accepted Answer

The electron gain enthalpy of iodine is equal to the energy released when 1 mole of iodine atoms in vapor state are converted into $I^{-}$ ions.

$\frac{5 \times 1 0 ^{- 13} \times 6.023 \times 1 0 ^{23}}{1 0 ^{6}}$

$= 30.1 \times 1 0^{4} J = 301 K J$

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