Question

show that for a first order reaction the time required 99.9% completion is 3 times the time required 90% completion?

Answer 1

let 100 mole be initially present.
and we know for first order reactio
$a = a |0| {e}^{ - kt}$
if 99.9 % completed that means 100-99.9=0.1 mol left
put in formula and get t.
$0.1 = 100 {e}^{ - kt} \\ log0.1 = log100 - kt \\ - log10 - 2log10 = - kt \\ \frac{3log}{10k} = t....(1)$
now if 90% completed that means 10 mol remaining ..again put in formula
$10 = 100 {e}^{ - kt |0| } \\ log10 = log100 - kt |0| \\ log10 - 2log10 = - kt |0| \\ t |0| = \frac{log10}{k} ...(2)$

compare (1)(2)...hope you get your answer....
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