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for the first order reaction, \mathrm{A \rightarrow P, t_{1 / 2}} (half life is 20 days. The time required for \mathrm{\frac{1}{6}} conversion of A (in days) is

Option: 1

8


Option: 2

9


Option: 3

11


Option: 4

5


Answers (1)

best_answer

The half life \mathrm{t_{1 / 2}=20} days

The decay constant

\mathrm{ k=\frac{0.693}{t_{1 / 2}}=\frac{0.693}{20 \text { days }}=\frac{0.03465}{\text { day }^{-1}} }
The time required for one sixth Comersion

\mathrm{ \begin{aligned} & t=\frac{2.303}{k} \log _{10} \frac{a}{a-x} \\\\ & t=\frac{2.303}{0.03465 \text { day }^{-1}} \log _{10} \frac{1}{1-(1 / 6)} \\ & t=5 \\ & \end{aligned} }

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Divya Prakash Singh

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