InterviewSolution
Saved Bookmarks
| 1. |
What is the activation energy for a reaction if its rate doubles when the temperature is raised from `20^(@)C` to `35^(@)C`? `(R = 8.314 J "mol K"^(-))`A. 342`kj mol^(-1)`B. `269kjmol^(-1)`C. `34.7kj Mol^(-1)`D. `15.1 kj Mol^(-1)` |
|
Answer» Correct Answer - C given initial temperature `T_(1)=20+273=293K` final temperature `T_(2) = 35+273=308K` ` R= 8314mol^(-1) K^(-1)` since ,rate becomes double on raising temperature , `therefore I_(2) = 2r_(1) or (I_(2))/(I_(1))=2` As rate constant `k prop I` `therefore (K_(2))/(K_(1))=2` from arrherius equation , we know that `log (K_(2))/(K_(1))=-(E_(a))/(2.303R) [(T_(1)-T_(2))/(T_(1)T_(2))]` `log 2 =-(E_(a))/(2.303xx8.314)[(293-308)/(293xx308)]` ` 0.3010=-(E_(a))/(2.303xx8314)[(-15)/(293xx308)]` ` therefore E_(a) =(0.3010xx2.303xx8.314xx293xx308)/(15)` ` 34673.48 mol^(-1)=34.7 kj mol^(-1)` |
|