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If the rate constant of a first order reaction is $4.606 \times 10^{-3} \mathrm{~s}^{-1}$, then find the time required for $400 \mathrm{~g}$ of the reactant to reduce to $50 \mathrm{~g}$.
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The correct answer is:
7.52 min
Given,
Rate constant, $k=4.606 \times 10^{-3} \mathrm{~s}^{-1}$
Concentration at, time $t=50 \mathrm{~g}$
Initial concentration $=400 \mathrm{~g}$
Now, formula,
$$
\begin{aligned}
k & =\frac{2.303}{t} \log \frac{A_0}{A_t} \\
t & =\frac{2.303}{4.606 \times 10^{-3}} \log \frac{400}{50} \\
& =\frac{2.303}{0.00460} \log 8 \\
& =\frac{2.303 \times 0.9030}{0.00460} \\
& =452.08 \mathrm{~s}=\frac{452.08}{60} \mathrm{~min} \\
t & =7.52 \mathrm{~min}
\end{aligned}
$$
Rate constant, $k=4.606 \times 10^{-3} \mathrm{~s}^{-1}$
Concentration at, time $t=50 \mathrm{~g}$
Initial concentration $=400 \mathrm{~g}$
Now, formula,
$$
\begin{aligned}
k & =\frac{2.303}{t} \log \frac{A_0}{A_t} \\
t & =\frac{2.303}{4.606 \times 10^{-3}} \log \frac{400}{50} \\
& =\frac{2.303}{0.00460} \log 8 \\
& =\frac{2.303 \times 0.9030}{0.00460} \\
& =452.08 \mathrm{~s}=\frac{452.08}{60} \mathrm{~min} \\
t & =7.52 \mathrm{~min}
\end{aligned}
$$
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