QJEEM25Chdd23: JEE Main 2025 Chemistry - Class 12 JEEM25Chdd23 Solution: For the thermal decomposition of at constant volume, the…

$\begin{aligned}& \mathrm{K}_{\mathrm{N}_2 \mathrm{O}_5}=2 \times 4.606 \times 10^{-2} \mathrm{~S}^{-1} \\& 2 \mathrm{~N}_2 \mathrm{O}_5(\mathrm{~g}) \longrightarrow 2 \mathrm{~N}_2 \mathrm{O}_4(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g})\end{aligned}$

$$\begin{array}{cccc}{P}_i& 0.6& 0& 0\\ P_f& 0.6-P& P& \frac{P}{2}\end{array}$$

$\begin{aligned}& 2 \times 4.606 \times 10^{-2}=\frac{2.303}{100} \log \frac{0.6}{0.6-\mathrm{P}} \\& \quad 4 \log _{10} \frac{0.6}{0.6-\mathrm{P}} \\& \quad 10^4=\frac{0.6}{0.6-\mathrm{P}} \\& \Rightarrow 0.6 \times 10^4-10^4 \mathrm{P}=0.6\end{aligned}$

$\begin{aligned}\begin{aligned}& \Rightarrow 10^4 \mathrm{P}=0.6\left(10^4-1\right) \\& \mathrm{P}=(6000-0.6) \times 10^{-4} \\&=5999 . \times 10^{-4} \\&=0.59994 \\& \mathrm{P}_{\text {Total }}=0.6+\frac{\mathrm{P}}{2} \\&= 0.6+0.29997 \\&= 0.89997 \\&=899.97 \times 10^{-3} \\& \text { Ans. } 900\end{aligned}\end{aligned}$

$\mathrm{P}_{\text {Total }}=0.6+\frac{\mathrm{x}}{2}$

As given in equation

$\mathrm{K}_{\mathrm{r}}=4.606 \times 10^{-2} \mathrm{sec}^{-1}$

(Here language conflict in question)

( $\mathrm{K}_{\mathrm{r}}=\frac{\mathrm{KA}}{2}$ not considered)

$\begin{aligned}\mathrm{K}_{\mathrm{r}} \mathrm{t} & =\ln \frac{0.6}{0.6-\mathrm{x}} \\4.606 & \times 10^{-2} \times 100=2.303 \log \frac{0.6}{0.6-\mathrm{x}} \\\mathrm{P}_{\text {Total }} & =0.6+\frac{0.594}{2}=0.897 \mathrm{~atm} \\\quad & =897 \times 10^{-3} \mathrm{~atm}\end{aligned}$