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The half-life of a substance in a certain enzyme-catalysed reaction is $138 \mathrm{~s}$. The time required for the concentration of the substance to fall from $1.28 \mathrm{mg} \mathrm{L}^{-1}$ to $0.04 \mathrm{mg} \mathrm{L}^{-1}$ is
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The correct answer is:
$690 \mathrm{~s}$
Enzyme-catalysed reactions follow first order kinetics.
$$
\begin{aligned}
& 1.28 \stackrel{t_{1 / 2}}{\longrightarrow} 0.64 \stackrel{t_{1 / 2}}{\longrightarrow} 0.32 \stackrel{t_{1 / 2}}{\longrightarrow} 0.16 \\
& \stackrel{t_{1 / 2}}{\longrightarrow} 0.08 \stackrel{t_{1 / 2}}{\longrightarrow} 0.04
\end{aligned}
$$
$$
\begin{aligned}
\therefore \text { Time required } & =5 \times t_{1 / 2} \\
& =5 \times 138 \mathrm{~s} \\
& =690 \mathrm{~s} .
\end{aligned}
$$
$$
\begin{aligned}
& 1.28 \stackrel{t_{1 / 2}}{\longrightarrow} 0.64 \stackrel{t_{1 / 2}}{\longrightarrow} 0.32 \stackrel{t_{1 / 2}}{\longrightarrow} 0.16 \\
& \stackrel{t_{1 / 2}}{\longrightarrow} 0.08 \stackrel{t_{1 / 2}}{\longrightarrow} 0.04
\end{aligned}
$$
$$
\begin{aligned}
\therefore \text { Time required } & =5 \times t_{1 / 2} \\
& =5 \times 138 \mathrm{~s} \\
& =690 \mathrm{~s} .
\end{aligned}
$$
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