Question

Show that the time required for 99.9% completion in a first order reaction is 10 times of half-life (t_1/2) of the reaction [log 2 = 0.3010, log 10 = 1].

Answer 1

First-order reaction `k = 2.303/t log  a/(a - x)`

where a = original amount

a − x = amount left betund

t = time period

k = rate constant

We also know that

`k = 0.693/t_(1//2)`

t_1/2 = half-life period.

Now, x = 99.9% and x = 50%

We can say `t = 2.303/k log  a/(a - x)`

or `t_(99.9%)/t_(50%) = (2.303 //k log //100 - 99.9)/(2.303 //k log //100 - 50)`

or `t_(99.9%)/t_(50%) = (log  100/0.1)/(log  100/50)`

`t_(99.9%)/t_(50%) = (-log 1000)/(log 2)`

`t_(99.9%)/t_(50%) = (-log 1 xx 10^-3)/log 2`

`t_(99.9%)/t_(50%) = 3/0.3010`

`t_(99.9%)/t_(50%) = 3/0.3`

thus, t_99.9% = 10 × t_50%