For a first order reaction, $\mathrm{A} \rightarrow \mathrm{P}, \mathrm{t}_{1 / 2}$ (half-life) is 10 days.…
For a first order reaction, $\mathrm{A} \rightarrow \mathrm{P}, \mathrm{t}_{1 / 2}$ (half-life) is 10 days. The time required for $\frac{1}{4}^{\text {th }}$ conversion of $\mathrm{A}$ (in days) is: $(\ln 2=0.693, \ln 3=1.1)$.
$3.2$
$2.5$
$4.1$
5
Solution
The half life $\mathrm{t}_{1 / 2}=10$ days The decay constant
$
\mathrm{k}=\frac{0.693}{\mathrm{t}_{1 / 2}}=\frac{0.693}{10 \text { days }}=0.0693 \text { days }^{-1}
$
The time required for one fourth conversion
$
\begin{aligned}
&\mathrm{t}=\frac{2.303}{\mathrm{k}} \log _{10} \frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}} \\
&=\frac{2.303}{0.0693 \text { day }^{-1}} \log _{10} \frac{1}{1-(1 / 4)}=4.1 \text { days }
\end{aligned}
$