We know, for a first order kinetics reaction,
where [a]° is the initial concentration of the reactant,
a is the concentration after time ‘t’,
k is the rate constant.
Let, initial concentration of the reactant be [a]°.
So, the concentration of the reactant after 90% completion of the reaction will be = ((100-90)/100)×[a]°
Thus, concentration of reactant at 90% = 0.1[a]°
So, time taken t = =
= [log 10 = 1]
Concentration of the reactant after 99% completion of the reaction will be= ((100-99)/100)×[a]°.
So, concentration of the reactant after 99% completion of the reaction will be = 0.01[a]°
Thus, time taken for 90% completion is
t’ = =
= = 2t [log 100 = 2 and t = ]
∴t’ = 2t.
Hence, the time taken to complete 99% of the first order reaction is twice the time required for the completion of 90% of the reaction.
Rate this question :