Question

In the Haber's process of ammonia manufacture,

$N_{2}\left(\right.g\left.\right)+3H_{2}\left(\right.g\left.\right) \rightarrow 2NH_{3}\left(\right.g\left.\right)$

The rate of appearance of $NH_{3}$ is

$\frac{d \left[\right. N H_{3} \left]\right.}{d t}=2\times 10^{- 4}molL^{-}sec^{- 1}$

The rate of disppearance of $N_{2}$ and $H_{2}$ will be

Rate in terms of $H_2$ $(mol \,L^{-1}\,sec^{-1})$	Rate in terms of $N_2$ $(mol \,L^{-1}\,sec^{-1})$
$3 \times 10^{-4}$	$2 \times 10^{-4}$
$3 \times 10^{-4}$	$1 \times 10^{-4}$
$1 \times 10^{-4}$	$3 \times 10^{-4}$
$2 \times 10^{-4}$	$2 \times 10^{-4}$

• A. A
• B. B
• C. C
• D. D

Answer 1

Answer: (B)

Rate $=+\frac{1}{2}\frac{d \left[\right. N H_{3} \left]\right.}{d t}=\frac{d \left[\right. N_{2} \left]\right.}{d t}=-\frac{1}{3}\frac{d \left[\right. H_{2} \left]\right.}{d t}$

$-\frac{d \left[\right. H_{2} \left]\right.}{d t}=\frac{3}{2}\times \frac{d \left[\right. N H_{3} \left]\right.}{d t}$

$=\frac{3}{2}\times 2\times 10^{- 4}=3\times 10^{- 4}molL^{- 1}sec^{- 1}$

$-\frac{d \left[\right. N_{2} \left]\right.}{d t}=\frac{1}{2}\frac{d \left[\right. N H_{3} \left]\right.}{d t}=\frac{1}{2}\times 2\times 10^{- 4}$

$=1\times 10^{- 4}molL^{- 1}sec^{- 1}$