Question

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 1

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)`