The half life of a first order reaction is 2000 years. If the concentration after 8000 years is $0.02…
- $0.16 \mathrm{M}$
- $0.32 \mathrm{M}$
- $0.08 \mathrm{M}$
- $0.04 \mathrm{M}$
Solution
$\begin{aligned}
k & =\frac{0.693}{2000} \\
k & =\frac{2.303}{t} \log \frac{[\mathrm{A}]_o}{[\mathrm{~A}]_t} \\
\frac{0.693}{2000} & =\frac{2.303}{8000} \log \frac{[\mathrm{A}]_o}{0.02} \\
\therefore \quad[\mathrm{A}]_o & =0.32 \mathrm{M}
\end{aligned}$
Asked in: NEET 2022 (Phase 2)