`60%` of a first order reaction was completed in `60 min`. The time taken for reactants to decompose to half of their original amount will be

Options

A. 60 Minutes
B. 4 minutes
C. 50 minutes
D. 45 minutes

Answer

D

Explanation

If `t_(f)` is the time required to complete a fraction `'F'` of a first order reaction, then
`t_(f)=(2.303)/(k)log(1)/(1-f)`
For `60%` comp,etion, we have
`60` minutes `=(2.303)/(k)log(1)/(1-0.6)`
For `50%` completion, we have
`t_(1//2)=(2.303)/(k)log(1)/(1-0.5)`
Dividing Eq. `(1)` by Eq. `(2)` we get
`(60"minutes")/(t_(1//2))=(log1//0.4)/(log(1//0.5))=(log10-log4)/(log10-log5)`
`=(1-0.60)/(1-0.69)`
`=(0.40)/(0.31)`
Thus, `t_(1//2)=((60xx31)/(0.40))` minutes
`=46.5` minutes