Question
For the reaction
$${N_2}{O_5}\left( g \right) \to 2N{O_2}\left( g \right) + \frac{1}{2}\,{O_2}\left( g \right)$$ the value of rate of disappearance of $${N_2}{O_5}$$ is given as $$6.25 \times {10^{ - 3}}\,mol\,{L^{ - 1}}{s^{ - 1}}.$$ The rate of formation of $$N{O_2}$$ and $${O_2}$$ is given respectively as :
A.
$$\eqalign{
& 6.25 \times {10^{ - 3}}\,mol\,{L^{ - 1}}{s^{ - 1}}\,{\text{and}} \cr
& 6.25 \times {10^{ - 3}}\,mol\,{L^{ - 1}}{s^{ - 1}} \cr} $$
B.
$$\eqalign{
& 1.25 \times {10^{ - 2}}\,mol\,{L^{ - 1}}{s^{ - 1}}\,{\text{and}} \cr
& 3.125 \times {10^{ - 3}}\,mol\,{L^{ - 1}}{s^{ - 1}} \cr} $$
C.
$$\eqalign{
& 6.25 \times {10^{ - 3}}\,mol\,{L^{ - 1}}{s^{ - 1}}\,{\text{and}} \cr
& 3.125 \times {10^{ - 3}}\,mol\,{L^{ - 1}}{s^{ - 1}} \cr} $$
D.
$$\eqalign{
& 1.25 \times {10^{ - 2}}mol\,{L^{ - 1}}{s^{ - 1}}\,{\text{and}} \cr
& 6.25 \times {10^{ - 3}}\,mol\,{L^{ - 1}}{s^{ - 1}} \cr} $$
Answer :
$$\eqalign{
& 1.25 \times {10^{ - 2}}\,mol\,{L^{ - 1}}{s^{ - 1}}\,{\text{and}} \cr
& 3.125 \times {10^{ - 3}}\,mol\,{L^{ - 1}}{s^{ - 1}} \cr} $$
Solution :
$$\eqalign{
& {N_2}{O_5}\left( g \right) \to 2\,N{O_2}\left( g \right) + \frac{1}{2}{O_2}\left( g \right) \cr
& - \frac{d}{{dt}}\left[ {{N_2}{O_5}} \right] = + \frac{1}{2}\frac{d}{{dt}}\left[ {N{O_2}} \right] = 2\frac{d}{{dt}}\left[ {{O_2}} \right] \cr
& \frac{d}{{dt}}\left[ {N{O_2}} \right] = 1.25 \times {10^{ - 2}}\,mol\,{L^{ - 1}}{s^{ - 1}}\,{\text{and}} \cr
& \frac{d}{{dt}}\left[ {{O_2}} \right] = 3.125 \times {10^{ - 3}}\,mol\,{L^{ - 1}}{s^{ - 1}} \cr} $$