Time required for 99.9% completion of a first order reaction is ________ times the time required for completion of 90% reaction.

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Solution

Q.
Time required for 99.9% completion of a first order reaction is ________ times the time required for completion of 90% reaction. (nearest integer) [2024]

Ans.
(3)

  Rate constant of a first order reaction is given as:   $k = \frac{2.303}{t} \log \frac{a_{o}}{a_{t}}$

  When reaction is 90% complete, then only 10% reactants remain i.e. at $t_{90 %}$ ,    $a_{t} = 0.1$ $a_{0}$

$k = \frac{2.303}{t_{90 %}} \log \frac{a_{0}}{0.1 a_{0}} = \frac{2.303}{t_{90 %}} \log 10 = \frac{2.303}{t_{90 %}} \times 1 = \frac{2.303}{t_{90 %}}$ ...(i)

When reaction is 99.9% complete, then only 0.1% reactants remain i.e. $t_{99.9 %},$ $a_{t} = 0.001$ $a_{0}$

   $k = \frac{2.303}{t_{99.9 %}} \log \frac{a_{0}}{0.001 a_{0}} = \frac{2.303}{t_{99.9 %}} \log 10^{3} = \frac{2.303}{t_{99.9 %}} \times 3 \log 10 = \frac{2.303}{t_{99.9 %}} \times 3$ ...(ii)

Equate I and II

$\frac{2.303}{t_{90 %}} = \frac{2.303}{t_{99.9 %}} \times 3$

   $t_{99.9 %} = t_{90 %} \times 3$

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