The time required for completion of 93 . 75 % of a first order reaction is x minutes. The half life of it (in minutes) is

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Aicharya · Accepted Answer

$\begin{aligned} & t_{1 / 2}=\frac{0.0693}{k_1} \ & \text {Also, } t_{93.75}=\frac{2.303}{k_1} \log \cdot \frac{100}{100-93.75} \ & =\frac{2.303}{k_1} \log \cdot \frac{100}{6.25} \ & =\frac{2.303}{k_1} \log 2^4 \ & \frac{4 \times 2.303 \times \log 2}{k_1}=\frac{4 \times 0.693}{k_1}=4 t_{1 / 2}\end{aligned}$ $\begin{aligned} & 4 t_{1 / 2}=\mathrm{x} \ & t_{1 / 2}=\mathrm{x} / 4 \end{aligned}$ So correct option is (3) *

The time required for completion of 93 . 75 % of a first order reaction is x minutes. The half life of it…

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