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

$$\matrix{ {{P_i}} & {0.6} & 0 & 0 \cr {{P_f}} & {0.6 - P} & P & {{P \over 2}} \cr } $$

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

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

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

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)

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