Question

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

Answer 1

Case I If a = 100; (a-x) = (100 - 99) = 1
For $99\%$ completion of the reaction
$t_{99\%}=\frac{2.303}{k}log\frac{100}{1}=\frac{2.303}{l}log{10}^2=\frac{2.303\times 2}{k}{t}_{99\%}=\frac{4.602}{k}\dots \dots \left(i\right)$

Case II: If a = 100; (a-x) = (100-90) = 10
For $90\%$ completion of the reaction
$t_{90\%}=\frac{2.303}{k}log\frac{100}{10}=\frac{2.303}{k}log10=\frac{2.303}{k}\dots \dots \left(ii\right)$
On dividing Eq. (i) by Eq. (ii), we get
$\frac{t_{99\%}}{t_{90\%}}=\frac{4.603}{k}\times \frac{k}{2.303}=2$
It means that time required for $99\%$ completion of the reaction is twice the time required to complete $90\%$ of the reaction.