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Ionization Enthalpy

IE 1 and IE 2...

Question

$I E_{1}$ and $I E_{2}$ of $M g$ are $178$ and $348 kcal$ respectively. The energy required (in kcal) for the following reaction is:
$M g \to M g^{2 +} + 2 e^{-}$
(Where $I E$ = ionization energy)

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Solution

According to given conditions
$M g \to M g^{+} + e^{-} I E_{1} = 178 kcal$
$M g^{+} \to M g^{2 +} + e^{-} I E_{2} = 348 kcal$
Therefore,
$M g \to M g^{2 +} + 2 e^{-} I E = I E_{1} + I E_{2}$
Hence, ionization energy of the given reaction is:
$I E = 178 kcal + 348 kcal$
$= 526 kcal$

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