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.

Explanation

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