For the thermal decomposition of N₂O₅(g) at constant volume, the following table can be formed, for the reaction mentioned below :

$2{\mathrm{N}}_2{\mathrm{O}}_5\left(\mathrm{g}\right)\to 2{\mathrm{N}}_2{\mathrm{O}}_4\left(\mathrm{g}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)$

S.No.	Time/s	Total pressure / (atm)
1	0	0.6
2	100	'x'

x = ________ × 10^–3 atm [nearest integer]
Given : Rate constant for the reaction is $4.606\times {10}^{-2}{\mathrm{s}}^{-1}.$

${\mathrm{K}}_{{\mathrm{N}}_2{\mathrm{S}}_5}2\times 4.606\times {10}^{-2}{\mathrm{S}}^{-1}$

$$\begin{gathered} 2{\mathrm{N}}_2{\mathrm{O}}_5\left(\mathrm{g}\right)⟶2{\mathrm{N}}_2{\mathrm{O}}_4\left(\mathrm{g}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right) \\ \begin{array}{l}{\mathrm{P}}_{\mathrm{i}}\hspace{0.25em} \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0.6\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em} 0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em} 0\end{array} \\ {\mathrm{P}}_{\mathrm{f}} 0.6-\mathrm{P} \mathrm{P} \frac{\mathrm{P}}{2} \end{gathered}$$

$\begin{array}{l}2\times 4.606\times {10}^{-2}=\frac{2.303}{100}\log \frac{0.6}{0.6-\mathrm{P}}\\ 4{\log }_{10}\frac{0.6}{0.6-\mathrm{P}}\\ {10}^4=\frac{0.6}{0.6-\mathrm{P}}\\ \Rightarrow 0.6\times {10}^4-{10}^4\mathrm{P}=0.6\\ \begin{array}{l}\Rightarrow {10}^4\mathrm{P}=0.6\left({10}^4-1\right)\\ \mathrm{P}=(6000-0.6)\times {10}^{-4}\\ =5999.\times {10}^{-4}\\ =0.59994\\ {\mathrm{P}}_{\text{Total }}=0.6+\frac{\mathrm{P}}{2}\\ =0.6+0.29997\\ =0.89997\\ =899.97\times {10}^{-3}\end{array}\end{array}$