Show that for a first order reaction the time required for 99.9% of the reaction to take place is ten times that required for half of the reaction

Explanation

For first order reaction,
`k = (2.303)/(t)log""(a)/(a-x)` or `t = (2.303)/(k)log""(a)/(a-x)`
When `t = t_(1//2), x = a//2`
`t_(1//2) = (2.303)/(k) log""(a)/(a-a//2) = (2.303)/(k) log 2`.
`t_(1//2) = (0.693)/(k)` …..(1)
When `t = t_(99.9%) x = 99.9%` of a = 0.999a
And `t_(99.9%) = (2.303)/(k)log""(a)/(a-0.999a) = (2.303)/(k)log""(1)/(0.001)`
`= (2.303)/(k)log""(1)/(0.001) = (2.303)/(k) log 1000 = (2.303)/(k) xx 3`
`t_(99.9%) = (6.909)/(k)` ......(2)
Dividing Eq. (2) by Eq. (1)
`(t_(99.9%))/(t_(1//2) = (6.909 //k)/(0.693)//k) = (6.909)/(0.693) = 9.969`
or , `(t_(99.9%))/(t_(1//2)) ~~ 10` or, `t_(99.9%) = 10 t_(1//2)`
Hence, for a fist order reaction, the time required for 99.9% of the reaction to occur is ten times that required for half of the reaction.