`99%` at a first order reaction was completed in `32 min`. When will `99.9%` of the reaction complete.

Options

A. 48 min
B. 50 min
C. 35 min
D. 39 min

Answer

A

Explanation

Time required to complete any particular fraction of first order reaction is
`_(t//2)=(2.303)/(k)log(1)/(1-f)`
therefore `t_(99%)=(2.303)/(k)log(1)/(1-0.99)`
`=(2.303)/(k)log(1)/(0.01)`
`=(2.303)/(k)log10^(2)`
`t_(99.9%)=(2.303)/(k)log(1)/(1-0.999)`
`=(2.303)/(k)log(1)/(0.001)`
`=(2.303)/(k)log10^(3)`
Dividing Eq. `(1)` by Eq. `(2)` , we get
`(t_(99%))/(t_(99.9%))=(log10^(2))/(log10^(3))=(2log10)/(3log10)=(2)/(3)`
Thus, `t_(99.9%)=(3)/(2)t_(99%)`
`=(3)/(2)(32min)=48min`