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# A certain reaction is first order, and 540 s after initiation of the reaction, 32.5% of the reactant remains. At what time after initiation of the reaction of the reaction will 10% of the reactant remain?

$\begin{array}{1 1} 1110 \;s \\ 1100\;s \\ 1081\;s \\ 1006\;s \end{array}$

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## 1 Answer

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Answer: 1110 s
Before we calculate the time we need to calculate $k$ the rate constant first.
Given 32.5% of the reactant remains
$ln\;A = -kt + ln\;A_0 \rightarrow ln 0.325 = -k \times 540\;s + ln \; 1 \rightarrow k = 0.00208s^{-1}$
Going back to the integrated form of the rate law for first order reactions: $ln\;A = -kt + ln\;A_0$
$\Rightarrow ln\; 0.10 = - (0.0020813 s^{-1}) (t) + ln 1 \rightarrow t \approx 1110 \;s$
answered Jul 25, 2014
edited Jul 25, 2014

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