Show that in a first order reaction time required for completion of 99.9%is 10 times of half time of the reaction

k  =  2.303 /t .  log [a] / [a-x]  --------------------------------------------A 

Where,

k ----rate of reaction

t  ----time taken 

a  ----initial concentration

x --  reactants those converted into products

We know that  

t1/2 =  half life 

k   =  0.693 / t50%  -------------------------------------------------------1 

For   99%  reaction to be completed we have  a = 100 

x =99.9

From equation (A)

k = 2.303/t  . log  [ 100] / [ 100-99.9]

k  =  2.303 /t 99% . log [ 100]/[ 0.1]

k =  2.303/t99% . log [ 103]

k = [2.303 . 3]  -----------------------------------------------------------------------2

Equate 1 and 2 

We have  k =  0.693 / t50%   =  2.303 *  3 / t99%

t99% /  t50% =   2.303  * 3 / 0.693  =  9.97 = 10[approx]