For the reaction; $2 \mathrm{~N}_{2} \mathrm{O}_{5} \longrightarrow 4 \mathrm{NO}_{2}+\mathrm{O}_{2}

Question

For the reaction; $2 \mathrm{~N}_{2} \mathrm{O}_{5} \longrightarrow 4 \mathrm{NO}_{2}+\mathrm{O}_{2}$ rate and rate constant are $1.02 	imes 10^{-4}$ and $3.4 	imes 10^{-5} \mathrm{sec}^{-1}$ respectively, then concentration of $\mathrm{N}_{2} \mathrm{O}_{5}$ at that time will be\

Professor PC Thomas Classes · Answer

(b) : $2 \mathrm{~N}_{2} \mathrm{O}_{5} ightarrow 4 \mathrm{NO}_{2}+\mathrm{O}_{2}$ This is a first order reaction.\ $	herefore \quad$ rate $=k\left[\mathrm{~N}_{2} \mathrm{O}_{5}ight]$; $$ \left[\mathrm{N}_{2} \mathrm{O}_{5}ight]=	ext { rate } / k=\frac{1.02 	imes 10^{-4}}{3.4 	imes 10^{-5}}=3 $$