Show that in a first order reaction,the time required for completion of $99.9 \%$ is $10$ times the half-life $(t_{1/2})$ of the reaction.
For a first order reaction,the rate constant $k$ is given by $k = \frac{2.303}{t} \log \frac{[R]_0}{[R]}$.
When the reaction is $99.9 \%$ complete,the remaining concentration is $[R] = [R]_0 - 0.999[R]_0 = 0.001[R]_0 = 10^{-3}[R]_0$.
Substituting this into the rate equation:
$k = \frac{2.303}{t_{99.9}} \log \frac{[R]_0}{10^{-3}[R]_0} = \frac{2.303}{t_{99.9}} \log 10^3 = \frac{2.303 \times 3}{t_{99.9}} = \frac{6.909}{t_{99.9}}$.
Thus,$t_{99.9} = \frac{6.909}{k}$.
For the half-life of a first order reaction,$t_{1/2} = \frac{0.693}{k}$.
Taking the ratio:
$\frac{t_{99.9}}{t_{1/2}} = \frac{6.909 / k}{0.693 / k} \approx 10$.
Therefore,$t_{99.9} = 10 \times t_{1/2}$.
When the reaction is $99.9 \%$ complete,the remaining concentration is $[R] = [R]_0 - 0.999[R]_0 = 0.001[R]_0 = 10^{-3}[R]_0$.
Substituting this into the rate equation:
$k = \frac{2.303}{t_{99.9}} \log \frac{[R]_0}{10^{-3}[R]_0} = \frac{2.303}{t_{99.9}} \log 10^3 = \frac{2.303 \times 3}{t_{99.9}} = \frac{6.909}{t_{99.9}}$.
Thus,$t_{99.9} = \frac{6.909}{k}$.
For the half-life of a first order reaction,$t_{1/2} = \frac{0.693}{k}$.
Taking the ratio:
$\frac{t_{99.9}}{t_{1/2}} = \frac{6.909 / k}{0.693 / k} \approx 10$.
Therefore,$t_{99.9} = 10 \times t_{1/2}$.
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