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Derive the equation showing the relation between the concentration [R]_(1) and [R]_(2) at time t_(1) and t_(2) |
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Answer» SOLUTION :The intergrated rate equation for first order reaction is as, In [R]=-k`t_(1)+In [R]_(0)` ….V(A) So,time =`t_(1)` and that time concentration `[R]_(1)` then equation is In `[R]_(1)=-kt_(1)+IN[R]_(0)`and `t_(2)` the equation is……VII(A) In `[R]_(2)=-k t_(2)+In [R]_(0)` .......VII(B) Substracting VII (A) from VII (B), In `[R]_(1)-In [R]_(2)=-kt_(1)-(-kt_(2))` `THEREFORE In [R]_(1)-In [R]_(2)=-kt_(1)+kt_(2)` `therefore In ([R]_(1))/([R]_(2))=k(t_(2)-t_(1))` .......VIII(A) Taking log in this equation , 2.303 log `([R]_(1))/([R]_(2))=k(t_(2)-t_(1))` `therefore log ([R]_(1))/([R]_(2))=(k)/(2.303)(t_(2)-t_(1))` ....VIII(B) From the above equation the following equation for k `k=(1)/((t_(2)-t_(2))) In ([R]_(1))/([R]_(2))`.......IX(A) and `k=(2.303)/((t_(2)-t_(1)))` log `([R]_(1))/([R]_(2))` .......IX (B) In these equations .If `t_(1)` =zero then `[R]_(1)=[R]_(0)` and after t time ,`t_(2)` =t then `[R]_(2)=[R]` so,in this SITUATION above both equations are as under: `k=(1)/(t)` In `([R]_(0))/([R])` .......XA and `k=(2.303)/(t)` log `([R]_(0))/([R])` or .......XB  (The equation is like a form comparing with y=mx +c.The PLOT of log `([R]_(0))/([R])` against t is straight line with egative slope.) |
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