Q. 75.0( 2 Votes )

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Answer :


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.


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