Show that for a first order reaction the time required for 99.9% completion is about 10 times its half life period.

Explanation

Given data : Time required for 99.9 % completion is about 10 times its half life period. Prove that for a first order reaction the time required for 99.9 % completion is about 10 times its half life period.
Formula : `k = (2.303)/(t)"log"(a)/(a-x)`
`t_(1//2)=(0.693)/(k)`
Solution : `t_(99.9)=(2.303)/(k)"log"(100)/(100-99.9)`
`=(2.303)/(k)"log"(100)/(0.1)`
`=(2.303)/(k)log 1000`
`t_(99.9)=(2.303xx3)/(k)`
`t_(99.9)=(6.909)/(k)`,
`t_(1//2)=(0.693)/(k) " " therefore k=(0.693)/(t_(1//2))`
`(t_(99.9%))/(t_(1//2))=(6.909)/(k)xx(k)/(0.693)` ,
`t_(99.9)=(6.909)/(0.693)xxt_(1//2)`
`t_(99.9)=100xxt_(1//2)`
Time required for 99.9% completion of the reaction `= 10 t_(1//2)`