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For the first order decomposition reaction :
$2 N_{2} O_{5} (g) ⟶ 4 {NO}_{2} (g) + O_{2} (g)$
Initially the total pressure is found to be 650 torr and after a very long time total pressure is found to be 1550 torr. If after 4 minutes from the start of the reaction partial pressure of $O_{2}$ = 100 torr, calculate half life of the decomposition in minutes. [Given: in 3 :: 1.1 and ln 2 = 0.7) [Assume initially $O_{2}$ is absent]

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answer is 7.

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Detailed Solution

$\underset{\underset{\underset{0}{P_{0} - x}}{P_{0}}}{2 N_{2} O_{5}} ⟶ \underset{\underset{\underset{2 P_{0}}{2 x}}{-}}{4 {NO}_{2}} + \underset{\underset{\underset{\frac{P_{0}}{2}}{\frac{x}{2}}}{-}}{O_{2}}$

$2 P_{0} + \frac{P_{0}}{2} + P_{inert} = 1550$

$\begin{array}{l} \Rightarrow P_{0} + P_{inert} = 650 \\ \Rightarrow P_{0} = 600, P_{inert} = 50 \end{array}$

Now, $\begin{array}{r} \frac{x}{2} = 100 \\ \Rightarrow x = 200 \end{array}$

$\begin{array}{l} \frac{\ln 2}{t_{1 / 2}} \times 4 = \ln (\frac{600}{400}) \\ \Rightarrow t_{1 / 2} = 7 \min \end{array}$

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