Show that for a first order reaction, time required for completion of 99%of reaction is twice the time required for completion of 90% of reaction.

We know, for a first order kinetics reaction,

t =

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]

Similarly,

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.