show that time required for first order reaction to go 99.9%completion is 10 times the half life period.
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]