Question
For the reaction $$2N{H_3} \to {N_2} + 3{H_2},$$ if $$ - \frac{{d\left[ {N{H_3}} \right]}}{{dt}} = {k_1}\left[ {N{H_3}} \right],$$ $$\frac{{d\left[ {{N_2}} \right]}}{{dt}} = {k_2}\left[ {N{H_3}} \right],$$ $$\frac{{d\left[ {{H_2}} \right]}}{{dt}} = {k_3}\left[ {N{H_3}} \right]$$ then the relation between $${k_1},{k_2}$$ and $${k_3}$$ is
A.
$${k_1} = {k_2} = {k_3}$$
B.
$${k_1} = 3{k_2} = 2{k_3}$$
C.
$$1.5{k_1} = 3{k_2} = {k_3}$$
D.
$$2{k_1} = {k_2} = 3{k_3}$$
Answer :
$$1.5{k_1} = 3{k_2} = {k_3}$$
Solution :
$$2N{H_3} \to {N_2} + 3{H_2}$$
$${\text{Rate}} = - \frac{1}{2}\frac{{d\left[ {N{H_3}} \right]}}{{dt}} = \frac{{d\left[ {{N_2}} \right]}}{{dt}}$$ $$ = \frac{1}{3}\frac{{d\left[ {{H_2}} \right]}}{{dt}}$$
$$\frac{1}{2}{k_1}\left[ {N{H_3}} \right] = {k_2}\left[ {N{H_3}} \right] = \frac{1}{3}{k_3}\left[ {N{H_3}} \right];$$ $$1.5{k_1} = 3{k_2} = {k_3}$$