Question

If the rate of formation of $N{H}_3=2\times {10}^{-2}M/s$. Then calculate the Rate of disapperance of $H_2$ in the following reaction.
$N_2+3H_2\to 2N{H}_3$

• A. $3\times {10}^{-2}M/s$
• B. $4\times {10}^{-2}M/s$
• C. $2\times {10}^{-2}M/s$
• D. None of these

Answer 1

The correct option is A $3\times {10}^{-2}M/s$

The given reaction is :-

$N_2+3H_2⟶2N{H}_3$

So, the rate of the reaction can be expressed as:

$r=\frac{decreaseinconcentrationof{N}_2}{time}=\frac{-d\left[N_2\right]}{dt}$ $-\left(i\right)$

or, $r=\frac{-1}{3}\times \frac{d\left[H_2\right]}{dt}$ $-\left(ii\right)$ or $r=\frac{1}{2}\frac{d\left[N{H}_3\right]}{dt}$ $-\left(iii\right)$

Now, from (ii) & (iii)

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

Since, $\frac{d\left[N{H}_3\right]}{dt}=2\times {10}^{-2}M/s$ (given)

$\Rightarrow$ $\frac{-d\left[H_2\right]}{dt}=\frac{3}{2}\times 2\times {10}^{-2}M/s$

=$3\times {10}^{-2}M/s$