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The rate constant for the 2nd order reaction shown below is $0.54 M^{–1}$ at 300$^{\circ}$ C. How many seconds would it take for the concentration of NO$_s$ to decrease from $0.62\; M$ to $0.28\; M$?

$\begin{array}{1 1} 3.6\;s \\ 6.2\;s \\ 1.8 \;s \\ 0.9\;s \end{array}$

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Answer: 3.6 s
From the integrated rate law for a second order reaction,$\large\frac{1}{[A]}$$ = \large\frac{1}{[A]_0}$$ + kt$
$\Rightarrow \large\frac{1}{0.28}$$ = \large\frac{1}{0.62}$$ + 0.54t$
$\Rightarrow 0.54 t = 3.57 - 1.61 = 1.96 \rightarrow t = 3.6\; s$
answered Jul 24, 2014 by balaji.thirumalai

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