Show that in a first order reaction, time required for completion of `99.9%` is 10 times of half-life `(t_1/2)` of the reaction

Explanation

For a first order reaction , rate constant,
`k=(2.303)/(t)log.(a)/(a-x)" " ...[1]and "half -life "t_(1//2)=(0.693)/(k) " "..[2]`
Let the time taken for 99.9% completion of the reaction = t
When 99.9 % of the reaction is completed , then
`x=axx(99.9)/(100)=0.999xxa`
`(a-x)=a-(0.999xxa)=0.001a`
From equation [1] we have,
`t=(2.303)/(k)log.(a)/(a-x)=(2.303)/(k)log.(a)/(0.001a)=(2.303)/(k)log1000`
`=(2.303)/(k)xx3=(6.909)/(k)" "...[3]`
Dividing equation [3] by equation [2], we get,
`(t)/(t_(1//2))=(6.909//k)/(6.909//k)=(6.909)/(0.693)~~10`
`thereforet=10xxt_(1//2)`