Time required for completion of 99.9% of a First order reaction is ______ times of half life of the reaction.

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Solution

Q.
Time required for completion of 99.9% of a First order reaction is ______ times of half life ( $t_{1 / 2}$ ) of the reaction. [2024]

Ans.
(10)

For first order reactions, rate constant $k$ is given by:

   $k = \frac{2.303}{t} \log \frac{a_{o}}{a_{t}}$

  When reaction is 99.9% complete, 0.1% reactants are left i.e.

   $a_{t} = \frac{0.1}{100} a_{o},$ $\frac{a_{o}}{a_{t}} = 1000$ , thus

$k = \frac{2.303}{t_{99.9 %}} \log 1000 = \frac{2.303}{t_{99.9 %}} \times 3$ (I)

When reaction is 50% complete, 50% reactants are left i.e.

$a_{t} = \frac{50}{100} a_{o},$ $\frac{a_{o}}{a_{t}} = 2$ , thus

$k = \frac{2.303}{t_{50 %}} \log 2 = \frac{2.303}{t_{50 %}} \times 0.301$ (II)

  Equate I and II

   $\frac{2.303}{t_{99.9 %}} \times 3 = \frac{2.303}{t_{50 %}} \times 0.301$

   $t_{99.9 %} \approx 10 \times t_{50 %}$

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