Molecular masses of N 2 and N H 3 are x 2 and x 1 , respectively. In the reaction: N 2 + 3 H 2 → 2N H 3 , their equivalent weights are y 2 and y 1 . Then ( y 2 − y1 ) is

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ChemistryRedox ReactionsEquivalent Weight and n-factorEasy2 minQB

ChemistryEasysingle choice

Molecular masses of $N_{2}$ and $N H_{3}$ are $x_{2}$ and $x_{1}$ , respectively. In the reaction: $N_{2}$ + $3 H_{2}$ $\to$ $2N H_{3}$ , their equivalent weights are $y_{2}$ and $y_{1}$ . Then ( $y_{2 - y1}$ ) is

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Solution:

$y_{2}$ = $x_{2} /6$ (6 $e^{-}$ per $N_{2}$ ) and $y_{1}$ = $x_{1} /3$ (N changes by 3 units). $y_{2}$ - $y_{1}$ = $x_{2} /6$ - $x_{1} /3$ = ( $x_{2}$ - $2 x_{1}) /6$ . Hence (a).

Stream:JEESubject:ChemistryTopic:Redox ReactionsSubtopic:Equivalent Weight and n-factor

⏱ 2mℹ️ Source: QB

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