In the reaction $2 \mathrm{~N}_{2} \mathrm{O}_{5(\mathrm{~g})} \longrightarrow 4…
In the reaction $2 \mathrm{~N}_{2} \mathrm{O}_{5(\mathrm{~g})} \longrightarrow 4 \mathrm{NO}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})}$ the rate of formation of $\mathrm{NO}_{2(\mathrm{~g})}$ and $\mathrm{O}_{2(\mathrm{~g})}$ are in
the ratio of
$1: 4$
$1: 1$
$6: 1$
$4: 1$
Solution
By Differential Rate law;
$\frac{-1}{2} \frac{d\left(N_{2} O_{5}\right)}{d t}=\frac{1}{4} \frac{d\left(N O_{2}\right)}{d t}=\frac{d\left(O_{2}\right)}{d t}$
d $\frac{\text { Rate of formation of } \mathrm{NO}_{2}}{\text { Ratc of formation of } \mathrm{O}_{2}}=\frac{d\left(\mathrm{NO}_{2}\right) / d t}{d\left(\mathrm{O}_{2} / d t\right.}=4: 1$