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The rate law for decomposition of $N_2O_5: N_2O_5\rightarrow 2NO_2+\large\frac{1}{2}$$O_2$ is consistent with observed rate law.If $k'=5\times 10^{-4}s^{-1}$ the time required for $[N_2O_5]$ be reduced to 10% of its original value.

$N_2O_5\;\;\;\overset{\overset{k}{\rightleftharpoons}}\;\;\;NO_2+NO_3$(fast)

$NO_2+NO_3\;\;\;\underrightarrow{k_1}\;\;\;NO_2+NO+O_2$(slow)

$NO+NO_3\;\;\;\underrightarrow{k_2}\;\;\;2NO_2$(fast)

'K' is rate constant of reaction: $N_2O_5\rightarrow 2NO_2+\large\frac{1}{2}$$O_2$

$\begin{array}{1 1}(a)\;5.2\times 10^3s&(b)\;4.9\times 10^3\\(c)\;4.6\times 10^3s&(d)\;6.1\times 10^3s\end{array}$

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1 Answer

Since slow step is rate determining step,hence
$r=k_1[NO_2][NO_3]$
$k=[NO_2][NO_3]/N_2O_5]$
$r=k_1k[N_2O_5]$
$\;\;=k'[N_2O_5]$
$r=k'[N_2O_5]$
This is the first order reaction.
So $t=\large\frac{1}{k}$$ln\large\frac{a}{a-x}$
$\Rightarrow \large\frac{1}{5\times 10^{-4}}$$ln\large\frac{100}{100-90}$
$\Rightarrow 4.6\times 10^3$s
Hence (c) is the correct answer.
answered Dec 19, 2013 by sreemathi.v
 

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