Prove that for a 1st order reaction, the time taken for 99% completion of the reaction is twice the time required for the completion of 90% of the reaction.

Explanation

For first order reaction , `t = (2.303)/K log. a/(a-x)`
Case I . If a = 100 , `(a-x) = (100 -99) = 1 `
For 99 % completion of the reaction ,
`t_(99)% =(2.303)/k log 100/l `
`= (2.303)/k log 10^(2) =(2.303 xx2)/k `
`t_(99%) = (4.606)/k " "...(i)`
Case II If a = 100 , `(a-x) = (100 - 90) = 10`
For 90 % completion of the reaction ,
`t_(90%) = (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_(90%)) = (4.606)/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 .