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Ionization Enthalpy

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Question

First and second ionization energies of magnesium are $6.64 e V$ and $13.36 e V$ respectively. The amount of energy in J needed to convert all the atoms of magnesium into $M g^{2 +}$ ions present in $12 m g$ of magnesium vapour will be? $[G i v e n : 1 e V = 96.5 k J / m o l]$ .

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Solution

$12 m g M g = 0.5 m m o l M g$ as the molar mass is 24 g/mol.
$M g (g) \to M g^{2 +} (g) + 2 e^{-}$
$Δ H = I E_{1} + I E_{2} = 6.64 + 13.36 = 20 e V = 20 \times 96.5 k J / m o l$
Energy required to convert $0.5 m m o l$ $M g_{(g)}$ to $M g_{(g)}^{2 +}$ will be, $0.5 \times 20 \times 96.5 J = 965 J$

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