$\mathrm{NO}_2$ required for a reaction is produced by decomposition of $\mathrm{N}_2 \mathrm{O}_5$ in $\mathrm{CCl}_4$ as by equation

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

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.

The rate of formation of $\mathrm{NO}_2$ is $\mathrm{x} \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}$, value of $\mathrm{x}$ is _________. (nearest integer)

<p>Rate of reaction (ROR)</p> <p>$$\begin{aligned} & =-\frac{1}{2} \frac{\Delta\left[\mathrm{N}_2 \mathrm{O}_5\right]}{\Delta \mathrm{t}}=\frac{1}{4} \frac{\left[\mathrm{NO}_2\right]}{\Delta \mathrm{t}}=\frac{\Delta\left[\mathrm{O}_2\right]}{\Delta \mathrm{t}} \\ & \mathrm{ROR}=-\frac{1}{2} \frac{\Delta\left[\mathrm{N}_2 \mathrm{O}_5\right]}{\Delta \mathrm{t}}=-\frac{1}{2} \frac{(2.75-3)}{30} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1} \\ & \mathrm{ROR}=-\frac{1}{2} \frac{(-0.25)}{30} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1} \\ & \text { ROR }=\frac{1}{240} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1} \end{aligned}$$</p> <p>Rate of formation of $$\mathrm{NO}_2=\frac{\Delta\left[\mathrm{NO}_2\right]}{\Delta \mathrm{t}}=4 \times \mathrm{ROR}$$</p> <p>$$=\frac{4}{240}=16.66 \times 10^{-3} \mathrm{molL}^{-1} \mathrm{~min}^{-1} \simeq 17 \times 10^{-3} \text {. }$$</p>