For a first order reaction,k show that the time required for `99%` completion is twice the time required for completion of `99%` reaction.

Explanation

Case I:
`If a =100, (a-x) =(100-99)=1`
`t_(99%)=(2.303)/(k )log ""(100)/(1)=(2.303)/(k )log10^(2)=(2.303 xx2)/(k)`
`k_(99%)=(4.602)/(k)-(i)`
Case II:
If `a =100, (a-x) =(100-90)=10`
For `99%` completion of the reaction
`t_(99%)=(2.303)/(k)log""(100)/(10)`
`=(2.303)/(k) log 10=(2.303)/(k)-(ii)`
on dividing eq (i) by eq (ii) we get `(t_(99%))/(t_(99%))=(4.603)/(k)xx (k)/(2.303)=2`
It means that time required for `99%` completion of the reaction is twice the time required to complete `90%` of the reaction