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IP1 and IP2 o...

Question

$I P_{1} a n d I P_{2}$ of Mg are 178 and 348 K. cal $m o l^{- 1} .$ The enthalpy required for the reaction $M g \to M g^{2 +} + 2 e^{-}$ is:

+ 170

K.cal

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+ 526

K.cal

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- 170

K.cal

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- 526

K.cal

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Solution

The correct option is B $+ 526$ K.cal
$I P_{1} a n d I P_{2}$ of $M g$ are $1784$ and $4348 K .$ cal $m o l^{- 1} .$

The enthalpy required for the reaction $M g \to M g^{2 +} + 2 e^{-}$ is:

$I P_{1} + I P_{2}$ i.e. $178 + 348 = 526$ kCal.

So, the correct option is $B$

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