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Entropy

IEI and IEII ...

Question

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

+ 170 {kcal mol}^{- 1}

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+ 526 {kcal mol}^{- 1}

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- 170 {kcal mol}^{- 1}

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- 526 {kcal mol}^{- 1}

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Solution

The correct option is B $+ 526 {kcal mol}^{- 1}$
$M g \to M g^{+} + e^{-}; Δ H_{1} = 178 kcal M g^{+} \to M g^{+ 2} + e^{-}; Δ H_{2} = 348 kcal S o, Δ H f o r M g \to M g^{2 +} + 2 e^{-} i s Δ H_{1} + Δ H_{2} = 178 + 348 = 526 kcal .$

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