Question
For the reaction $${N_2} + 3{H_2} \to 2N{H_3},$$ how are the rate of reaction expressions inter-related $$\frac{{d\left[ {{H_2}} \right]}}{{dt}}$$ and $$\frac{{d\left[ {N{H_3}} \right]}}{{dt}}?$$
A.
$$ - \frac{1}{3}\frac{{d\left[ {{H_2}} \right]}}{{dt}} = + \frac{1}{2}\frac{{d\left[ {N{H_3}} \right]}}{{dt}}$$
B.
$$ - \frac{1}{2}\frac{{d\left[ {{H_2}} \right]}}{{dt}} = + \frac{1}{3}\frac{{d\left[ {N{H_3}} \right]}}{{dt}}$$
C.
$$ + \frac{1}{2}\frac{{d\left[ {{H_2}} \right]}}{{dt}} = - \frac{1}{3}\frac{{d\left[ {N{H_3}} \right]}}{{dt}}$$
D.
$$ + \frac{1}{3}\frac{{d\left[ {{H_2}} \right]}}{{dt}} = - \frac{1}{2}\frac{{d\left[ {N{H_3}} \right]}}{{dt}}$$
Answer :
$$ - \frac{1}{3}\frac{{d\left[ {{H_2}} \right]}}{{dt}} = + \frac{1}{2}\frac{{d\left[ {N{H_3}} \right]}}{{dt}}$$
Solution :
$${H_2}$$ decreases three times as fast as that of $${N_2}$$ while $$N{H_3}$$ increases twice as fast as that of $${N_2}$$ decreases.
Hence, Rate $$ = - \frac{1}{3}\frac{{d\left[ {{H_2}} \right]}}{{dt}} = + \frac{1}{2}\frac{{d\left[ {N{H_3}} \right]}}{{dt}}$$