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What is the activation energy for a reac...

What is the activation energy for a reaction if its rate constant doubles when the temperature is raised from `20^(@)C` to `35^(@)C`? (`R=8.314`Jmol/K)

A

34.7 kJ/mol

B

15.1 kJ/mol

C

342 kJ/mol

D

269 kJ/mol

Text Solution

Verified by Experts

The correct Answer is:
A
`"log"(k_(2))/(K_(1))=(E_(a))/(2.303R)(1/(T_(1))-1/(T_(2)))`
`log2=(E_(a))/(2.303xx8.314)(1/293-1/308)`
`0.3010=(E_(a))/(2.303xx8.314)(1/293-1/308)`
`0.3010=(E_(a))/19.147xx15/(293xx308)`
`E_(a)=34673` J/mol or 34.7kJ/mol
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