Show that in case of a first order reaction, the time taken for completion of 99.9% reaction is ten times the time required for half change of the reaction.

Explanation

To show that `t_(99.9%) = 10 xx t_(1//2)`
`k = (2.303)/(t) log ""([R]_(0))/([R])`
`(t_(99.9))/(t_(1//2)) = ((2.303)/(k)log""(100)/(0.1))/((2.303)/(k)xxlog""(100)/(50))rArr (t_(99.9%))/(t_(1//2)) = (3)/(0.3)`
`t_(99.9%) = 10 t_(1//2)`