Question

In the Haber's process of ammonia manufacture,

$N_2(g)+3H_2(g)\to 2N{H}_3(g)$

The rate of appearance of $N{H}_3$ is

$\frac{d[N{H}_3]}{dt}=2\times 10^{-4}mol{L}^{-}se{c}^{-1}$

The rate of disppearance of $N_2$ and $H_2$ will be

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

• A. A
• B. B
• C. C
• D. D

Answer 1

Answer: (B)

Rate $=+\frac{1}{2}\frac{d[N{H}_3]}{dt}=\frac{d[N_2]}{dt}=-\frac{1}{3}\frac{d[H_2]}{dt}$

$-\frac{d[H_2]}{dt}=\frac{3}{2}\times \frac{d[N{H}_3]}{dt}$

$=\frac{3}{2}\times 2\times 10^{-4}=3\times 10^{-4}mol{L}^{-1}se{c}^{-1}$

$-\frac{d[N_2]}{dt}=\frac{1}{2}\frac{d[N{H}_3]}{dt}=\frac{1}{2}\times 2\times 10^{-4}$

$=1\times 10^{-4}mol{L}^{-1}se{c}^{-1}$