For the reaction, $$2{N_2}{O_5} \to 4N{O_2} + {O_2},$$     the rate equation can be expressed in two ways $$ - \frac{{d\left[ {{N_2}{O_5}} \right]}}{{dt}} = k\left[ {{N_2}{O_5}} \right]$$     and $$ + \frac{{d\left[ {N{O_2}} \right]}}{{dt}} = k'\left[ {{N_2}{O_5}} \right]$$     $$k$$  and $$k'$$  are related as :

Solution :
Rate of disappearance of reactant = Rate of appearance of products
$$\eqalign{ & - \frac{1}{2}\frac{{d\left[ {{N_2}{O_5}} \right]}}{{dt}} = \frac{1}{4}\frac{{d\left[ {N{O_2}} \right]}}{{dt}} \cr & \frac{1}{2}k\left[ {{N_2}{O_5}} \right] = \frac{1}{4}k'\left[ {{N_2}{O_5}} \right] \cr & \frac{k}{2} = \frac{{k'}}{4} \cr & \therefore \,\,k' = 2k \cr} $$