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Calculate half-life for first-order reaction if 68% of a substance is reacted within 66 s.

$\begin{array}{1 1} 39.3\;s \\ 45.3\;s \\ 22.67\;s \\ 17.45\;s \end{array}$

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Answer: 39.3 s
Before we calculate the half-life, we need to calculate $k$ the rate constant in order to get the half-life.
Given 68% reacted $\rightarrow$ 32% remains:
$ln\;A = -kt + ln\;A_0 \rightarrow ln 0.32 = -k \times 66\;s + ln \; 1 \rightarrow k - 0.01726s^{-1}$
For a first order reaction, $t_{1/2} = \large\frac{ln\;2}{k}$$ =\large\frac{0.693}{0.01762}$$s = 39.3\;s$
answered Jul 25, 2014 by balaji.thirumalai

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