Question

In the Haber's process of ammonia manufacture, $N_2\left(g\right)+3H_2\left(g\right)⟶2N{H}_3\left(g\right)$, the rate of appearance of $N{H}_3$ is:
$\frac{d\left[N{H}_3\right]}{dt}=2\times {10}^{-4}$ $mol$ $L^{-1}{sec}^{-1}$
The rates of the reaction expressed in terms of $N_2$ and $H_2$ will be:

Rates in terms of $H_2$ $\left(mol{L}^{-1}{sec}^{-1}\right)$

Rates in terms of $N_2$ $\left(mol{L}^{-1}{sec}^{-1}\right)$

• A. $3\times {10}^{-4}2\times {10}^{-4}$
• B. $3\times {10}^{-4}1\times {10}^{-4}$
• C. $1\times {10}^{-4}3\times {10}^{-4}$
• D. $2\times {10}^{-4}2\times {10}^{-4}$

Answer 1

The correct option is B $3\times {10}^{-4}1\times {10}^{-4}$

Rate $=+\frac{1}{2}\frac{d\left[N{H}_3\right]}{dt}=\frac{-d\left[N_2\right]}{dt}=-\frac{1}{3}\frac{d\left[H_2\right]}{dt}$
$\frac{-d\left[H_2\right]}{dt}=\frac{3}{2}\times \frac{d\left[N{H}_3\right]}{dt}$
$=\frac{3}{2}\times 2\times {10}^{-4}=3\times {10}^{-4}$ $mol$ $L^{-1}{sec}^{-1}$
$\frac{-d\left[N_2\right]}{dt}=\frac{1}{2}\frac{d\left[N{H}_3\right]}{dt}=\frac{1}{2}\times 2\times {10}^{-4}$
$=1\times {10}^{-4}$ $mol$ $L^{-1}{sec}^{-1}$