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If the rate constant for a first order reaction is $k$, then find the time required for completion of $80 \%$ of the reaction.
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Verified Answer
The correct answer is:
$\frac{1.6}{k}$
First order rate constant is given as
$k=\frac{2.303}{t} \log \frac{\left[A_0\right]}{\left[A_t\right]}$
$80 \%$ completed, then
$\left[A_0\right]=100$
$A_t=100-80=20$
$\begin{aligned} & k=\frac{2.303}{t} \log \frac{100}{20} \\ & k=\frac{2.303}{t} \log \frac{10}{2} \\ & k=\frac{2.303}{t}(\log 10-\log 2) \\ & k=\frac{2.303}{t}(1-0.3) \\ & k=\frac{2.303 \times 0.7}{t} \Rightarrow t=\frac{1.6}{k} .\end{aligned}$
$k=\frac{2.303}{t} \log \frac{\left[A_0\right]}{\left[A_t\right]}$
$80 \%$ completed, then
$\left[A_0\right]=100$
$A_t=100-80=20$
$\begin{aligned} & k=\frac{2.303}{t} \log \frac{100}{20} \\ & k=\frac{2.303}{t} \log \frac{10}{2} \\ & k=\frac{2.303}{t}(\log 10-\log 2) \\ & k=\frac{2.303}{t}(1-0.3) \\ & k=\frac{2.303 \times 0.7}{t} \Rightarrow t=\frac{1.6}{k} .\end{aligned}$
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