Question

In a catalytic conversion of $N_2$ to $NH_3$ 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 are no side reaction, the rate of the reaction as expressed in terms of hydrogen is $\left(in\, mol\, L^{-1} s^{-1}\right)$

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

Answer 1

Answer: (A)

$\frac{-d\left(N_{2}\right)}{dt}=-\frac{1}{3}\frac{d\left(H_{2}\right)}{dt}=\frac{1}{2}\frac{d\left(NH_{3}\right)}{dt}$
$=\frac{d\left(H_{2}\right)}{dt}=\frac{3}{2}\frac{d\left(NH_{3}\right)}{dt}=\frac{3}{2}\times40\times10^{-3}$
$=60\times10^{-3}$