Question

In a reaction $N_2\left(g\right)+3H_2\left(g\right)\to {2NH}_3\left(g\right)$ the rate of appearance of ${NH}_3$ is $2.5\times {10}^{-4}mol{L}^{-1}{sec}^{-1}$
The rate of reaction and rate of disappearance of $H_2$ will be:

$($ in ${molL}^{-1}{sec}^{-1})$

• A. $3.75\times {10}^{-4},1.25\times {10}^{-4}$
• B. $1.25\times {10}^{-4},2.5\times {10}^{-4}$
• C. $1.25\times {10}^{-4},3.5\times {10}^{-4}$
• D. $5.0\times {10}^{-4},3.75\times {10}^{-4}$

Answer 1

The correct option is A $3.75\times {10}^{-4},1.25\times {10}^{-4}$

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

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

$\therefore$ Rate of disappearance of $H_2=\frac{3}{2}\frac{d\left[N{H}_3\right]}{dt}$

$=\frac{3}{2}\times 2.5\times {10}^{-4}=3.75\times {10}^{-4}mol{l}^{-1}{s}^{-1}$

Thus, Rate of reaction$=\frac{1}{2}\frac{H_2}{dt}=\frac{3.75\times {10}^{-4}}{3}=1.25\times {10}^{-4}$