The time taken for $90 \%$ of a first order reaction to complete is approximately
- $1.1$ times that of half-life
- $2.2$ times that of half-life
- $3.3$ times that of half-life
- $4.4$ times that of half-life
Solution
$t_{50 \%}=\frac{2.303}{k} \log \frac{100}{100-50}$ (II)
Dividing $\frac{t_{90 \%}}{t_{50 \%}}=\frac{\log 10}{\log 2}$
$\therefore t_{90 \%}=3.3 t_{50 \%}$
Asked in: JEE-TOPICTESTS-CHEMISTRY