Question

In a catalytic conversion of $N_2$ to $N{H}_3$ by Haber's process, the rate of reaction was expressed as change in the concentration of ammonia per time is $40\times {10}^{-3}$ mol L${}^{-1}$ s${}^{-1}$. If there is no side reaction, the rate of the reaction as expressed in terms of hydrogen is :

(in mol L${}^{-1}$ s${}^{-1}$)

• A. $60\times {10}^{-3}$
• B. $20\times {10}^{-3}$
• C. $1.200$
• D. $10.3\times {10}^{-3}$

Answer 1

The correct option is A $60\times {10}^{-3}$

Balanced reaction is as following :

$N_2+3H_2⟶{2NH}_3$

$\frac{\Delta conc}{\Delta time}=\frac{-d\left(N_2\right)}{dt}=\frac{-1}{3}\frac{d\left[H_2\right]}{dt}=\frac{1}{2}\frac{\left[{NH}_3\right]}{dt}$

$\frac{1}{2}\frac{\left[{NH}_3\right]}{dt}=\frac{40}{2}\times {10}^{-3}mol/L.sec$

Since, we have to express it in terms of hydrogen we write :

$\frac{1}{2}\frac{\left[{NH}_3\right]}{dt}=-\frac{1}{3}\frac{d\left[H_2\right]}{dt}$

$\frac{3}{2}\frac{d\left[{NH}_3\right]}{dt}=-\frac{d\left[H_2\right]}{dt}$

$\therefore -\frac{d\left[H_2\right]}{dt}=\frac{3}{2}\times 40\times {10}^{-3}mol/L.sec=60\times {10}^{-3}mol/L.sec$

Therefore, the correct option is A.