Question

For a first order reaction, show that the time required for `99%` completion is twice the time required for the completion of `90%` of reaction.

Answer 1

For a first order reaction , the time required for `99%` completion is
`t_(1) = (2.303)/(k) "log" (100)/(100 - 99)`
= `(2.303)/(k) "log" 100`
= `2 xx (2.303)/(k)`
For a first order reaction , the time required for `90%` completion is
`t_(2) = (2.303)/(k) "log" (100)/(100 - 90)`
= `(2.303)/(k) "log" 10`
= `(2.303)/(k)`
Therefore , `t_(1) = 2t_(2)`
Hence , the time required for `99%` completion of a first order reaction is twice the time required for the completion of `90%` of the reaction .