Show that for a first order reaction, time required for `99.99%` of the reaction to take place is 10 times the time required for the completion of half of the reaction.

Explanation

For first order reaction, `t=2.303/k log a/(a-x)`
1st part, `a=100, x=50,`
`(a-x) = 100-50 = 50`
`t_(1//2) = 2.303/k log 100/50 = 2.303/k log 2`
`=(2.303 xx 0.3010)/k = (0.693)/k`………(i)
2nd part:
`a=100, x=99.9 , (a-x) = (100-99.9) = 0.1`
`t_(99.9%) = 2.303/k log 100/0.1 =2.303/k xx 3`..(ii)
Dividing eqn. (ii) by eqn. (i)
`t_(99.9%)/t_(1//2) = (2.303 xx 3)/0.693 = 9.97` or 10