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: `t=2.303/k log a/(a-x)`
Ist case: `a=100%n = 99%, (a-x) = (100-99) = 1%`
`t_(99%) = 2.303/k log 100/1 = 2.303/k log 10^(2)`
`=(2.303 xx 2)/k = 4.606/k`...........(i)
Iind case:`a=100%, x=90%(a-x) = (100-90) = 10%`
`t_(90%)= 2.303/k log 100/10 =2.303/klog 10 = 2.303/k`...........(ii)
Dividing eqn. (ii) by eqn(i) .
`t_(99%)/t(90%)= 4.606/k xx k/2.303 = 2`.