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The half life of a first order reaction is $6.0$ hour. How long will it take for the concentration of reactant to decrease from $0.4 \mathrm{M}$ to $0 \cdot 12 \mathrm{M} ?$
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The correct answer is:
$10.42 \mathrm{~h}$
For first order reaction,
$\begin{array}{l}
k=\frac{0.693}{t_{1 / 2}}=\frac{0.693}{6.0 h}=0.1155 h^{-1} \\
\text { Here, }[A]_{0}=0.4 M,[A]_{t}=0.12 M \\
\therefore t=\frac{2.303}{k} \log _{10} \frac{[A]_{0}}{[A]_{1}}=\frac{2.303}{0.1155 h^{-1}} \times \log _{10} \frac{0.4}{0.12}=\frac{2.303}{0.1155} \times 0.5224 \\
\therefore t=10.42 h
\end{array}$
$\begin{array}{l}
k=\frac{0.693}{t_{1 / 2}}=\frac{0.693}{6.0 h}=0.1155 h^{-1} \\
\text { Here, }[A]_{0}=0.4 M,[A]_{t}=0.12 M \\
\therefore t=\frac{2.303}{k} \log _{10} \frac{[A]_{0}}{[A]_{1}}=\frac{2.303}{0.1155 h^{-1}} \times \log _{10} \frac{0.4}{0.12}=\frac{2.303}{0.1155} \times 0.5224 \\
\therefore t=10.42 h
\end{array}$
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