Show that for a first order reaction the time required for 99.9% of the reaction to take place is ten times that required for half of the reaction

Explanation

For first order reaction
`k=(2.303)/(t) log (a)/(a-x)` or `t=(2.303)/(k) log (a)/(a-x)`
When `t=t_(1//2) x=a//2`
`t_(1//2)=(2.303)/(k) log (a)/(a-a//2)=(2.303)/(k) log 2`
`t_(1//2) =(0.693)/(k)`
dividing Eq (2) by Eq (1)
`(t_(99.9%))/(t_(1//2))=(6.909)/(0.693//k)=(6.909)/(0.693)=9.969`
or `(t_(99.9%))/(t_(1//2))=10` or `t_(99.9%)=10 t_(1//2)`
Hence for a first order reaction the time required for 99.9 % of the reaction to occur is ten times that required for half of the reaction