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Gaseous $N_{2} O_{5}$ decomposes according to the following equation:
$N_{2} O_{5} (g) \to 2 {NO}_{2} (g) + \frac{1}{2} O_{2} (g)$
The experimental rate law is $- Δ [N_{2} O_{5}] / Δt = k [N_{2} O_{5}]$ . At a certain temperature the rate constant is $k = 5 .0 \times 10^{- 4} \sec^{- 1}$ . In how many seconds will the concentration of $N_{2} O_{5}$ decrease to one-tenth of its initial value?

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$2 .1 \times 10^{2} s$

$4 .6 \times 10^{3} s$

$1 .4 \times 10^{3} s$

$2 .0 \times 10^{3} s$

answer is B.

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Detailed Solution

$\begin{array}{l} t = \frac{2 .303}{5 \times 10^{- 4}} \log \frac{A_{0}}{A_{0} / 10} \\ = 4 .6 \times 10^{3} \sec . \end{array}$

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