Show that for a first order reaction, `t_(99.9%)=10t_(1//2)`. 

Explanation

For first order reaction
`t=(2.303)/(K)log""([R]_(0))/([R])`
The time required for `99.9%` completion is
`t_(99.9%)=(2.303)/(K)log""(100)/(100-99.9)`
`=(2.303)/(K)log""(100)/(0.1)`
`=(2.303)/(K)log 1000`
`=(2.303)/(K)log10^(3)"               "[logm^(n)=n log m]`
`t_(99.9%)=(3xx2.303)/(K)rArr (6.909)/(K)"       ….(1)"`
For the first order reaction line required line required to complete `50%` is
`t_(1//2)=(0.693)/(K)"         …(2)"`
`(eq(1))/(eq(2))`
`(t_(99.9%))/(t_(1//2))=(((9.909)K))/((0.909)/(K))`
`rArr (6.909)/(K)xx(K)/(0.693)`
`(t_(99.9%))/(t_(1//2))=10"       "rArr"       "t_(199.9%=10t_(1//2)`