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 `1^(st)` order reaction
`k=(2.303)/(t)log""(a)/(a-x)" "[a to "initial conc."a -x to "remaining conc." x to "reacted"]`
`t99%=(2.303)/(k)log""(100)/(100-99)`
`t99%=(2.303)/(k)log10^(2)`
`t99%=(2.303)/(k)xx2xx1" "[log10=1]`
`t99%=(2.303)/(k)xx2" "...(i)`
`t99%=(2.303)/(k)xxlog 10`
`t99%=(2.303)/(k)" "...(ii)[log10=I]`
From equation (i) and (ii)
`t99%=2xxt920%`