For the first order reaction $2 N_{2} O_{5 (g)} \to 4 N O_{2 (g)} + O_{2 (g)}$ total pressure is 284.5 mm after 30min and 584.5 mm after complete reaction. Approximate rate constant of the reaction is. [log1.169=0.0678]

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$521 \times 10^{- 3} \min^{- 1}$

$52.1 \times 10^{- 3} \min^{- 1}$

$5.21 \times 10^{- 4} \min^{- 1}$

$5.21 \times 10^{- 3} \min^{- 1}$

answer is D.

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Detailed Solution

$2 N_{2} O_{5 (g)} \to 4 N O_{2 (g)} + O_{2 (g)}$
$\begin{array}{l} t = 0 P_{i} O O \\ t = 30 \min (P_{i} - x) 2 x \frac{x}{2} \\ a t t h e e n d (P_{i} - P_{i}) 2 P_{i} \frac{P_{i}}{2} \end{array}$
At the end total pressure is $= 2 P_{i} + \frac{P_{i}}{2} = 2.5 P_{i}$
$\begin{array}{l} = 584.5 m m \\ ∴ P_{i} = 233.8 m m \end{array}$
P total after 30 min Pi+1.5 $x$ =284.5 mm
$x$ =33.8 & (Pi- $x$ )=200 mm
$K = \frac{2.303}{30} \log \frac{233.8}{200} = 5.21 \times 10^{- 3} \min^{- 1}$