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For a first order reaction, the time required for completion of 90% reaction is ‘x’ times the half life of the reaction. The value of ‘x’ is
(Given: in 10=2.303 and log2=0.3010)

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1.12

2.43

3.32

33.31

answer is C.

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Detailed Solution

$t_{0.90} = X t_{1 / 2}$

First order rate constant,

$K = \frac{\ln 2}{t_{1 / 2}} = \frac{1}{x t_{1 / 2}} \ln \frac{A_{0}}{A_{0} - A_{0} \times \frac{90}{100}}$

$\frac{\ln 2}{t_{1 / 2}} = \frac{\ln 10}{x t_{1 / 2}} \Rightarrow \frac{\ln 10}{\ln 2} = \frac{2.303}{2.303 \times 0.3010} = 3.32$

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