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A reaction has a half life of 1 min. The time required for 99.9% completion of the reaction is ___ min. (Round off to the nearest integer) [Use ln2 = 0.69, ln10 = 2.3]

Answer
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Hint: In radioactivity, the half-life defines the time interval required for the decaying of half of the nuclei. The representation of half-life is done by the symbol of \[{t_{1/2}}\]. Here, we have to use the integrated rate equation for the calculation of the time interval.

Formula used:
The integrated rate equation is,
\[t = \dfrac{1}{k}\ln \dfrac{{{A_0}}}{{{A_t}}}\]
Here, t stands for time, k stands for the rate constant, \[{A_0}\] stands for initial substrate concentration, \[{A_t}\] stands for the substrate concentration after t time.

Complete Step by Step Solution:
Let's solve the question. Given that, the half-life of the reaction is 1 min. So, we have to write the half-life equation using the above rate equation.
\[{t_{50\% }} = \dfrac{1}{k}\ln \dfrac{{100}}{{50}} = 1\,\min \] …… (1)

Now, we have to write the rate equation of the 99.9% completion of the reaction.
\[{t_{99.9\% }} = \dfrac{1}{k}\ln \dfrac{{100}}{{0.1}}\] ….. (2)

Now, we have to take the ratio of the two equations.
\[\dfrac{{{t_{99.9\% }}}}{{{t_{50\% }}}} = \dfrac{{\dfrac{1}{k}\ln \dfrac{{100}}{{0.1}}}}{{\dfrac{1}{k}\ln \dfrac{{100}}{{50}}}}\]

On rearranging the above equation, we get,
\[{t_{99.9\% }} = \dfrac{{\ln 1000}}{{\ln 2}} \times {t_{50\% }}\]
\[{t_{99.9\% }} = \dfrac{{\ln {{10}^3}}}{{\ln 2}} \times {t_{50\% }}\]
\[{t_{99.9\% }} = \dfrac{{3\ln 10}}{{\ln 2}} \times {t_{50\% }}\]

Given that, the values of ln 2=0.69 and ln 10=2.3 and the half-life is 1 min. So, we have to put these values in the above equation.
\[{t_{99.9\% }} = \dfrac{{3 \times 2.3}}{{0.69}} \times 1 = 10\,\min \]
Therefore, time required for the 99.9% completion of the reaction is 10 min.

Note: It is to be noted that an isotope's stability determines its decay rate. An isotope's half-life is not affected due to pressure, heat, or any other means. And the atoms that possess shorter half-lives are unstable in nature.