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

For a order reaction,show that time required for 99% completion is twi

1.	For a order reaction,show that time required for 99% completion is twice the time required for the completion of 90% of reaction.
Answer» Solution :90% reaction is complete so,CALCULATION of t (90%): Here ,90 % reaction is complete `therefore` REMAINING concentration is 10% If the initial concentration =`[R]_(0)` So, concentration after completion of 90% Reaction=`([R]_(0)xx10)/(100)=0.1 [R]_(0)=[R]_(90)` So,`t_(90)=(2.303)/(K)` log `([R]_(0))/([R]_(t))` `=(2.303)/(k)` log `(1)/(0.1)`{But `(1)/(0.1)`=10.0} `therefore t_(90)=(2.303)/(k)(1)=(2.303)/(k)` 99% reaction is complete,so , calculation of t(99%) : 99% reaction is complete so,1% reactant is remaining so,If the initial concentration =`[R]_(t)(99)` `[R]_(t)=1% of [R]_(0)=(1)/(100)[R]_(0)=0.01[R]_(0)` `t_((99%))=(2.303)/(k)` log `([R]_(0))/(0.01[R]_(0))=(2.303)/(k)xx2` `therefore (t_(99%))/(t_(90%))=((2.303)/(k))xx2xx((k)/(2.303))=(2)/(1)` So,if the first order reaction then time required for 99% completion is twice then time required for the completion of 90% of reaction.