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]

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

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

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

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

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

answer is D.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

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
$K = \frac{2.303}{30} \log \frac{233.8}{200} = 5.21 \times 10^{- 3} \min^{- 1}$