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(i) Reactions with low activation energy are fast, while those of high energy of activation are slow. (ii) The reaction rate almost doubles for each 10^(@)C rise in temperature although frequency does not change appreciably. (iii) Calculate the rate constant for the reaction having the activation energy 39.3kcal mol^(-1) (iv) Calculate the factor by which the rate of reaction is increased for a temperature rise of 10^(@)C from 25^(@)C to 35^(@)C. The energy of activation is 35kcal mol^(-1). Given , E_(a)=35xx10^(3)calmol^(-1) R=1.987cal, T_(2)=308K, T_(1)=298K |
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Answer» SOLUTION :`(i)` Rate depends on rate constant and rate constant `=Ae^(-Ea//RT)`. Thus, if `E_(a)` is small, rate constant will have HIGHER value and thereby rate of reaction is more for low `E_(a)` value reactions. `(ii)` The reaction rate increase due to increase in energy of molecules which enables them to cross over the THRESHOLD energy barrier as well as an increase in collision rate. `(iii)` GIVEN , `A=1.11xx10^(11)sec^(-1)`, `R=1.987cal` `E_(a)=39.3xx10^(3)calmol^(-1)` `T=573K` `:. k=Ae^(-Ea//RT)` or `log_(10)k=log_(10)A-(E_(a)//2.303RT)` or `log_(10)k=log_(10)1.11xx10^(11)-(39.3xx10^(3))/(2.303xx1.987xx573)` `:.k=1.14xx10^(-4)sec^(-1)` `(iv) :.2.303log_(10).(k_(2))/(k_(1))=(35xx10^(3))/(1.987)[(308-298)/(308xx298)]` or `(k_(2)//k_(1))=6.812` or `k_(2)=6.812xxk_(1)` `:.r_(2)=6.812xxr_(1)` `(:. r_(2)//r_(1)=k_(2)//k_(1))` |
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