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

Use direct relation,
`(t_(99%))/(t_(90%))=(log((100)/(100-90)))/(log((100)/(100-90)))=(log100)/(log10)=2`
`:.t_(99%)=2t_(90%)`