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Q.

For $N_{2} + 3 H_{2} ⇌ 2 {NH}_{3}$ rate of disappearance of $H_{2}$ is $0.01$ $M . \min^{- 1}$ The amount of ${NH}_{3}$ formed at that instant will be

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a

$0.17 g$

b

$0.1132 g$

c

$- 0.02 mol$

d

Zero

answer is C.

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Detailed Solution

$\frac{1}{3} d \frac{{dH}_{2}}{dt} = \frac{1}{2} \frac{{NH}_{3} d}{dt} \begin{aligned} \frac{2}{3} \frac{{dH}_{2}}{dt} = \frac{d ({NH}_{3})}{dt} \\ \frac{2}{3} \times 0 .01 = \frac{d ({NH}_{3})}{dt (17)} \\ \frac{d ({NH}_{3})}{dt} = \frac{2}{3} \times 0 .01 \times 17 \\ = 0.1132 \end{aligned}$

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