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Chemical Kinetics

Question

For first order reaction show that time required for 99% completion is twice the time required for the completion of 90% of reaction.

Solution

For the first order reaction, time required for 99% completion.

t_{1 / 2} = \frac{2.303}{t} \log \frac{a}{a - x}

Ist case

a = 100 % x = 90 %, (a - x) = 100 - 90 = 10 %

t_{90 %} = \frac{2.303}{k} \log \frac{100}{10} = \frac{2.303}{k} \log 10 t_{90 %} = \frac{2.303}{k} . . . (i)

IInd case

a = 100 % x = 99 %, (a - x) = (100 - 99) = 1 % t_{12} = \frac{2.303}{k} \log \frac{100}{1} = \frac{2.303}{k} \times 2

t_{99 %} = \frac{2.303}{k} \times 2

...(ii)

Dividing eqn. (ii) by eqn. (i)

\frac{t_{99 %}}{t_{90 %}} = \frac{2.303}{k} \times 2 \times \frac{k}{2.303} = 2

Hence, time required for 99% completion is twice for the time required for the completion of 90% reaction.