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

For a first - order reaction ,
`t_(1//2) = (0.6932)/k " " …(i)`
Now , suppose a first - order reaction , represented as
`{:(a,0,"mitial concentration"),(A to ,"Product",),((a-0.999a),0.999a,):}`
` :. " " t = (2.303)/k log. a/(0.001a) `
` = (2.303)/k log 1000 = (2.303)/k log . 10^(3)`
` =(2.303xx3)/k = (6.909)/k" "....(1) `
From (1) and (2) , we see that t is about ten times of `t_(1//2)`