Question
The time taken for $$90\% $$ of a first order reaction to complete is approximately
A.
1.1 times that of half-life
B.
2.2 times that of half-life
C.
3.3 times that of half-life
D.
4.4 times that of half-life
Answer :
3.3 times that of half-life
Solution :
$$\eqalign{
& {t_{90\% }} = \frac{{2.303}}{k}{\text{log}}\frac{{100}}{{100 - 90}}\,\,\,\left( {\text{I}} \right) \cr
& {t_{50\% }} = \frac{{2.303}}{k}{\text{log}}\frac{{100}}{{100 - 50}}\,\,\left( {{\text{II}}} \right) \cr
& {\text{Dividing}}\,\,\frac{{{t_{90\% }}}}{{{t_{50\% }}}} = \frac{{{\text{log}}\,10}}{{{\text{log}}2}} \cr
& \therefore \,\,{t_{90\% }} = 3.3{t_{50\% }} \cr} $$