Question
Consider the reaction, $$2{N_2}{O_5} \to 4N{O_2} + {O_2}$$
In the reaction $$N{O_2}$$ is being formed at the rate of $$0.0125\,mol\,{L^{ - 1}}\,{s^{ - 1}}.$$ What is the rate of reaction at this time?
A.
$$0.0018\,mol\,{L^{ - 1}}\,{s^{ - 1}}$$
B.
$$0.0031\,mol\,{L^{ - 1}}\,{s^{ - 1}}$$
C.
$$0.0041\,mol\,{L^{ - 1}}\,{s^{ - 1}}$$
D.
$$0.050\,mol\,{L^{ - 1}}\,{s^{ - 1}}$$
Answer :
$$0.0031\,mol\,{L^{ - 1}}\,{s^{ - 1}}$$
Solution :
$$\eqalign{
& {\text{Rate}} = \frac{1}{4}\frac{{d\left[ {N{O_2}} \right]}}{{dt}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{4} \times 0.0125 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\, = 0.0031\,mol\,{L^{ - 1}}\,{s^{ - 1}} \cr} $$