Show that in case of a first order reaction, the time required for `99.9%` of the reaction to take place is about `10` times that the required for half the reaction.

Explanation

`t_(1//2) = (0.69)/(k)`
`t_(99.9%) = (2.303)/(k)log.(a)/(a-x)`
`= (2.303)/(k)log.(100)/(100-99.9)`
`= (2.303)/(k) log10^(3)=(3 xx 2.303)/(k) = (6.9)/(k)`
`(t_(99.9))/(t_(1//2)) = (6.9)/(k) xx (k)/(0.69) = 10`
`:. t_(99.9%) = 10t_(1//2)`
Alternate methof
Using direct relation,
`(t_(99.9%))/(t_(1//2)) = (log((100)/(100-99.9)))/(0.3)`
`:. (t_(99.9%))/(t_(1//2)) = (log10^(3))/(0.3) = (3 log 10)/(0.3)`
`:. t_(99.9)% = 10t_(1//2)`