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

Explanation

For a first order reaction, `t=(2.303)/(k)log. (a)/(a-x)`
99% completion means , x = 99% of a =0.99a
`therefore" "t_(99%)=(2.303)/(k)log.(a)/(a-0.99a)=(2.303)/(k)log.10^(2)=2xx(2.303)/(k)`
99% completion means , x = 99% of a =0.99a
`therefore" "t_(99%)=(2.303)/(k)log.(a)/(a-0.90a)=(2.303)/(k)log10=(2.303)/(k)`
`therefore (t_(99%))/(t_(90%))=((2xx2.303)/(k))//((2.303)/(k))=2 " or, "t_(99%)=2xxt_(99%)`