For $N_{2} + 3 H_{2} \to 2 N H_{3},$ rates of disappearance of $N_{2}$ and $H_{2}$ and rate of appearance of respectively $N H_{3}$ are a, b and c, then

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$a = b > c$

$a > b > c$

$a < c < b$

$a = b = c$

answer is B.

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Detailed Solution

For the given reaction $\frac{- d [N_{2}]}{d t} = \frac{- 1}{3} \frac{d [H_{2}]}{d t}$
$= \frac{1}{2} \frac{d [N H_{3}]}{d t}; a = \frac{b}{3} = \frac{c}{2}; ∴ a < c < b$

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