Question

In the reaction $N_2+3H_2\rightleftharpoons 2N{H}_3$, $\left(\frac{dx}{dt}\right)=1\times {10}^2\left[N_2\right][H_2{]}^3-1\times {10}^{-3}[N{H}_3]$ and at same instant if $\frac{\left[N_2\right]{\left[H_2\right]}^2}{{\left[N{H}_3\right]}^2}={10}^{-5}{M}^2$ then at this instant value of $\left(\frac{dx}{dt}\right)$ is:

• A. $0$
• B. $1\times {10}^5$
• C. $1\times {10}^{-5}$
• D. $1\times {10}^{-3}$

Answer 1

The correct option is A $0$

$N_2+{3H}_2\rightleftharpoons 2{NH}_3$

now, $\frac{\left[N_2\right]{\left[H_2\right]}^3}{{\left[{NH}_3\right]}^2}={10}^{-5}$

$\Rightarrow \left[N_2\right]{\left[H_2\right]}^3={10}^{-5}{\left[{NH}_3\right]}^2$

now, $\frac{dx}{dt}=1\times {10}^2\left[N_2\right]{\left[H_2\right]}^3-{10}^{-3}{\left[{NH}_3\right]}^2$

$=1\times {10}^2\times {10}^{-5}{\left[{NH}_3\right]}^2-{10}^{-3}{\left[{NH}_3\right]}^2$

$={10}^{-3}{\left[{NH}_3\right]}^2-{10}^{-3}{\left[{NH}_3\right]}^2$

$=0$

Correct answer $0$ (A).