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Prove that for a 1st order reaction, the...

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.

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Case I if `a= 100, (a-x) = (100 - 99) = 1`
For `99%` completion of the reaction,
`t_(99%) = (2.303)/( k) log"" (100)/(1) = (2.303)/( k ) log 10^(2)`
`= (2.303 xx 2)/( 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.
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