NO 2 required for a reaction is produced by decomposition of N 2 O 5 in CCl 4 as by equation 2 N 2 O 5 ( g )…

The given reaction is,

$2 {\mathrm{N}}_2{\mathrm{O}}_{5( \mathrm{g})}\to 4{\mathrm{NO}}_{2( \mathrm{g})}+{\mathrm{O}}_{2( \mathrm{g})}$

The initial concentration of ${\mathrm{N}}_2{\mathrm{O}}_5$ is $3 \mathrm{mol} {\mathrm{L}}^{-1}$ and it is $2.75 \mathrm{mol} {\mathrm{L}}^{-1}$ after $30$ minutes.

Now, 

$$\begin{gathered} -\frac{\mathrm{\Delta }\left[{\mathrm{N}}_2{\mathrm{O}}_5\right]}{\mathrm{\Delta t}}=-\frac{(2.75-3)}{30-0} \mathrm{mol} {\mathrm{L}}^{-1} {\min }^{-1} \\ =\frac{0.25}{30} \end{gathered}$$

$$\begin{gathered} \frac{1}{2}\times \frac{\mathrm{\Delta }\left[{\mathrm{N}}_2{\mathrm{O}}_5\right]}{\mathrm{\Delta t}}=\frac{1}{4} \frac{-\mathrm{\Delta }\left[{\mathrm{NO}}_2\right]}{\mathrm{\Delta t}} \\ \frac{-\mathrm{\Delta }\left[{\mathrm{NO}}_2\right]}{\mathrm{\Delta t}}=\frac{0.25}{30}\times 2 \\ =1.66\times {10}^{-2} \mathrm{M}/\mathrm{Min} \end{gathered}$$

$\simeq 17\times {10}^{-3} \mathrm{M}/\mathrm{Min}$