Explanation

(b)
$2N_{2}O_{5} \rightarrow 4NO_{2} + O_{2}$
Rate of decomposition of $N_{2}O_{5} = - \frac{1}{2}\frac{k\left\lbrack N_{2}O_{5} \right\rbrack}{dt}$
Rate of formation of $NO_{2} = \frac{1}{4}\frac{k\lbrack NO_{2}\rbrack}{dt}$
$\therefore\frac{rate\ of\ decompsition\ of\ N_{2}O_{5}}{rate\ of\ formation\ of\ NO_{2}} = \frac{\frac{1}{2}k\frac{\lbrack N_{2}O_{5}\rbrack}{dt}}{\frac{1}{4}k\frac{\lbrack NO_{2}\rbrack}{dt}}$
$or\ \ \ \ \ \frac{1}{2}k\frac{\lbrack N_{2}O_{5}\rbrack}{dt} \times \frac{4}{1}\frac{dt}{k\lbrack NO_{2}\rbrack}$
$= \frac{4}{2} = \frac{2}{1} = 2:1$