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The time required for 10% completion of a first order reaction at 298 K is equal to that required for its 25% completion at 308 K.If the value ofA is 4xx10^(10)S^(-1).Calculate k at 318 K and E_(a) |
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Answer» Solution :`T_(1)`=298,time =`t_(1)` If initial concentration =`[R]_(0) mol L^(-1)` The reaction is COMPLETE 10% ,so 90% reactant will be reamain. `therefore[R]_(t)=0.9[R]_(0) mol L^(-1)` `therefore` log `k_(298)=(2.303)/(t_(1)) log [R_(0)]/(0.9 [R]_(0))=(2.303)/(t_(1))` log `(1)/(0.9)` `therefore t_(1)=(2.303)/(log k_(298))xx0.04575` `therefore t_(1)=(0.1054)/(log k_(298))` Inintial concentration =`[R]_(0) mol L^(-1),T_(2)`=308 K . The 25% reaction is complete in t time. `therefore` The concentration after completion of 75% reaction=75% of `[R]_(0)` `therefore [R]_(t)=0.75[R]_(0) mol L^(-1)` `k_(308)=(2.303)/(t_(2))xxlog ([R]_(0))/(0.75 [R]_(0))` `(2.303)/(t_(2))xxlog 1.3333=(2.303xx0.1249)/(t_(2))` `therefore t_(2)=(0.2878)/(log k_(308))`........(2) But `t_(1)=t_(2)` `therefore (0.1059)/(k_((298)))=(0.2878)/(k_((308)))` `therefore (K_((308)))/(k_((298)))=(0.2878)/(0.1059)=2.717` ..........(3) Calculation of `E_(a)` :`=4XX10^(10)s^(-1)` so, at 318 K temperature k=(?)`E_(a)` (?) `therefore (log K_(2))/(log k_(1))=(E_(a))/(2.303xx8.314)((308-298)/(308xx298))` `therefore` log 2.717=`(E_(a))/(2.303xx8.314)xx(10)/(308xx298)` `therefore E_(a)=75990.0 J mol^(-1)` logk=log A-`(E_(a))/(2.303 RT)` `=log 4xx10^(10)-(75990 mol^(-1))/(2.303xx8.314 JK^(-1) mol^(-1)xx318 K)` `therefore` k=0.013267 `s^(-1)~~0.013 s^(-1)` |
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