Question
For the reaction, $${N_2} + 3{H_2} \to 2N{H_3},$$ if $$\frac{{d\left[ {N{H_3}} \right]}}{{dt}} = 2 \times {10^{ - 4}}mol\,{L^{ - 1}}{s^{ - 1}},$$ the value of $$\frac{{ - d\left[ {{H_2}} \right]}}{{dt}}$$ would be
A.
$$3 \times {10^{ - 4}}mol\,{L^{ - 1}}{s^{ - 1}}$$
B.
$$4 \times {10^{ - 4}}mol\,{L^{ - 1}}{s^{ - 1}}$$
C.
$$6 \times {10^{ - 4}}mol\,{L^{ - 1}}{s^{ - 1}}$$
D.
$$1 \times {10^{ - 4}}mol\,{L^{ - 1}}{s^{ - 1}}$$
Answer :
$$3 \times {10^{ - 4}}mol\,{L^{ - 1}}{s^{ - 1}}$$
Solution :
$$\eqalign{
& {\text{For the reaction,}} \cr
& {N_2} + 3{H_2} \to 2N{H_3} \cr} $$
$${\text{Rate}} = - \frac{{d\left[ {{N_2}} \right]}}{{dt}} = - \frac{1}{3}\frac{{d\left[ {{H_2}} \right]}}{{dt}}$$ $$ = + \frac{1}{2}\frac{{d\left[ {N{H_3}} \right]}}{{dt}}$$
$$\eqalign{
& {\text{or}}\,\, - \frac{1}{3}\frac{{d\left[ {{H_2}} \right]}}{{dt}} = + \frac{1}{2}\frac{{d\left[ {N{H_3}} \right]}}{{dt}} \cr
& - \frac{{d\left[ {{H_2}} \right]}}{{dt}} = \frac{3}{2} \times 2 \times {10^{ - 4}}mol\,{L^{ - 1}}{s^{ - 1}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 3 \times {10^{ - 4}}mol\,{L^{ - 1}}{s^{ - 1}} \cr} $$