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Two I-order reactions have half-lives in the ratio 8 : 2. Calculate the ratio of time intervals $t_{1}$ , $t_{2}$ . Where $t_{1}$ is the time period for 25 % completion of the first reaction and $t_{2}$ , for 75 % completion of the second reaction.

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0.199 : 1

0.311 : 1

0.420 : 1

0.273 : 1

answer is A.

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Detailed Solution

The time intervals,

$\begin{array}{r} t_{1} = \frac{3}{\ln 2} \times \ln (\frac{100}{75}) \\ t_{2} = \frac{2}{\ln 2} \times \ln (\frac{100}{25}) \end{array}$

The ratio of time intervals,

$\frac{t_{1}}{t_{2}} = \frac{3}{2} [\frac{0 .6 - 0 .48}{0 .6}] \frac{t_{1}}{t_{2}} = 0.311$

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