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

Solution :
Key Idea Rate of disappearance of reactant = rate of appearance of product
$${\text{or}}$$  $$ - \frac{1}{{{\text{Stoichiometric coefficient of reactant}}}}$$        $$\frac{{{\text{d [reactant]}}}}{{dt}}$$
$$ = + \frac{1}{{{\text{Stoichiometric coefficient}}\,{\text{of product}}}}$$        $$\frac{{{\text{d[product]}}}}{{dt}}$$

$$\eqalign{ & {\text{For the reaction,}} \cr & {N_2}{O_5}\left( g \right) \to 2N{O_2}\left( g \right) + \frac{1}{2}{O_2}\left( g \right) \cr} $$
$$\frac{{ - d\left[ {{N_2}{O_5}} \right]}}{{dt}} = + \frac{1}{2}\frac{{d\left[ {N{O_2}} \right]}}{{dt}}$$      $$ = + \frac{{2\,d\left[ {{O_2}} \right]}}{{dt}}$$
$$\eqalign{ & \therefore \,\,\frac{{d\left[ {N{O_2}} \right]}}{{dt}} = - 2\frac{{d\left[ {{N_2}{O_5}} \right]}}{{dt}} \cr & = 2 \times 6.25 \times {10^{ - 3}}\,mol\,{L^{ - 1}}{s^{ - 1}} \cr & = 12.5 \times {10^{ - 3}}\,mol\,{L^{ - 1}}{s^{ - 1}} \cr & = 1.25 \times {10^{ - 2}}\,mol\,{L^{ - 1}}{s^{ - 1}} \cr & \frac{{d\left[ {{O_2}} \right]}}{{dt}} = - \frac{{d\left[ {{N_2}{O_5}} \right]}}{{dt}} \times \frac{1}{2} \cr & = \frac{{6.25 \times {{10}^{ - 3}}mol\,{L^{ - 1}}{s^{ - 1}}}}{2} \cr & = 3.125 \times {10^{ - 3}}\,mol\,{L^{ - 1}}{s^{ - 1}} \cr} $$