An organic compound undergoes first order decomposition. If the time taken for the 60% decomposition is 540 s, then the time required for 90% decomposition will be is______s.
(Nearest integer).
Given: In 10 = 2.3; log 2=0.3

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answer is 1350.

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Detailed Solution

For first order reaction $Kt = 2.303 \log (\frac{A_{0}}{A})$
$\begin{array}{l} Case 1: A = 40 % A_{0} = 0.4 A_{0} K \times 540 = \ln (\frac{A_{0}}{0.4 A_{0}}) \\ \frac{540}{t} = \frac{\ln 0.4}{\ln 0.1} \\ \frac{540}{t} = \frac{\log 2^{2} - \log 10}{\log (10^{- 1})} = \frac{2 \log 2 - 1}{- 1} = \frac{2 \times 0.3 - 1}{- 1} = \frac{0.6 - 1}{- 1} = 0.4 \Rightarrow \frac{540}{t} = 0.4 t = 1350 \sec \end{array}$