therefore, Fall of concentration from \(1.28 \;\mathrm{mg}\, \mathrm{L}^{-1}\) to \(0.04\; \mathrm{mg} \,\mathrm{L}^{-1}\) involves five half-lives.
\(1.28 \stackrel{t_{1 / 2}}{\rightarrow} 0.64 \stackrel{t_{1 / 2}}{\rightarrow} 0.320 .16 \stackrel{t_{1 / 2}}{\rightarrow} 0.08 \stackrel{t_{1 / 2}}{\rightarrow} 0.04\)
therefore, Time required \(=5 \times t_{1 / 2}\)
\(=5 \times 138\; \mathrm{s}\)
\(=690\; \mathrm{s}\)
(આપેલું છે$: \ln 10=2.3, R =8.3 \,J\, K ^{-1} \,mol ^{-1}, \log 2=0.30$ )