Q. For the reaction
$N_{2}+3H_{2} \rightarrow 2NH_{3}$
The rate of disappearance of hydrogen is $0.3\times 10^{- 4} \, Ms^{- 1}$ . The rate of formation of ammonia is

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A

$-0.2\times 10^{- 4} \, Ms^{- 1}$

B

$0.2\times 10^{- 4}Ms^{- 1}$

C

$0.1\times 10^{- 4} \, Ms^{- 1}$

D

$0.3\times 10^{- 4} \, Ms^{- 1}$

Solution:

$N_{2}+3H_{2} \rightarrow 2NH_{3}$
$\frac{d \left[H_{2}\right]}{dt}=-0.3\times 10^{- 4}Ms^{- 1}$
$Rate=-\frac{1}{3}\frac{d \left[H_{2}\right]}{dt}=+\frac{1}{2}\frac{d \left[NH_{3}\right]}{dt}$
$=\frac{d \left[NH_{3}\right]}{dt}=-\frac{2}{3}\frac{d \left[H_{2}\right]}{dt}$
$=-\frac{2}{3}\times \left(\right.-0.3\times \left(10\right)^{- 4}\left.\right)$
$\text{=} \, \text{0} \text{.2} \, \text{\times } \, \text{10}^{- \text{4}} \, \text{Ms}^{- \text{1}}$

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