Chemistry - Chemical Kinetics Question with Solution | TestHub

For the given reaction $2{\text{N}}_2{\text{O}}_{5(g)}\to 4{\text{NO}}_{2(g)}+{\text{O}}_{2(g)}$, the rate of reaction can be expressed as:

$\text{Rate}=-\frac{1}{2}\frac{d[{\text{N}}_2{\text{O}}_5]}{dt}=\frac{1}{4}\frac{d[{\text{NO}}_2]}{dt}=\frac{d[{\text{O}}_2]}{dt}$

Given the rate of formation of ${\text{NO}}_2$ is $\frac{d[{\text{NO}}_2]}{dt}=0.08{\text{mol dm}}^{-3}{\text{min}}^{-1}$.

The rate of disappearance of ${\text{N}}_2{\text{O}}_5$ is $-\frac{d[{\text{N}}_2{\text{O}}_5]}{dt}$.

Using the rate expression:

$-\frac{d[{\text{N}}_2{\text{O}}_5]}{dt}=\frac{2}{4}\frac{d[{\text{NO}}_2]}{dt}$

$-\frac{d[{\text{N}}_2{\text{O}}_5]}{dt}=\frac{1}{2}\times 0.08{\text{mol dm}}^{-3}{\text{min}}^{-1}=0.04{\text{M min}}^{-1}$

Answer: $0.04{\text{M min}}^{-1}$.