Show that the time required for 99% completion of a first order reaction In twice the time required for the completion of 90%.

Explanation

For a first order initial concentration
`t = 2.303 // t log [A]_(0) //[A]`
Let a M be the initial concentration .
When reaction is 99% complete .
`[A] = a - 99 a//100 = 0.01a`
`t_(99) = 2.303// k log a //0.01 a `
` t_(99) = 2.303 // k xx 2 " "...(1)`
When reaction is 90% complete .
`[A] = a - 90a//100 = 0.1 a`
`t_(90) = k2.303 xx 1" "...(2)`
Divide equaiton (1) by (2) we get.
`t_(99)//t_(90) = 2 `
` t_(99) = 2 t_(90)`