Question
Consider the reaction $${N_2}\left( g \right) + 3{H_2}\left( g \right) \to 2N{H_3}\left( g \right)$$ The equality relationship between $$\frac{{d\left[ {N{H_3}} \right]}}{{dt}}$$ and $$ - \frac{{d\left[ {{H_2}} \right]}}{{dt}}$$ is
A.
$$ + \frac{{d\left[ {N{H_3}} \right]}}{{dt}} = - \frac{2}{3}\frac{{d\left[ {{H_2}} \right]}}{{dt}}$$
B.
$$ + \frac{{d\left[ {N{H_3}} \right]}}{{dt}} = - \frac{3}{2}\frac{{d\left[ {{H_2}} \right]}}{{dt}}$$
C.
$$\frac{{d\left[ {N{H_3}} \right]}}{{dt}} = - \frac{{d\left[ {{H_2}} \right]}}{{dt}}$$
D.
$$\frac{{d\left[ {N{H_3}} \right]}}{{dt}} = - \frac{1}{3}\frac{{d\left[ {{H_2}} \right]}}{{dt}}$$
Answer :
$$ + \frac{{d\left[ {N{H_3}} \right]}}{{dt}} = - \frac{2}{3}\frac{{d\left[ {{H_2}} \right]}}{{dt}}$$
Solution :
If we write rate of reaction in terms of concentration of $${N{H_3}}$$ and $${H_2},$$ then
Rate of reaction $$ = \frac{1}{2}\frac{{d\left[ {N{H_3}} \right]}}{{dt}} = - \frac{1}{3}\frac{{d\left[ {{H_2}} \right]}}{{dt}}$$
So, $$\frac{{d\left[ {N{H_3}} \right]}}{{dt}} = - \frac{2}{3}\frac{{d\left[ {{H_2}} \right]}}{{dt}}$$