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What is the activation energy for a reac...

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"^(-1).K^(-1))`

A

`15.1 "kJ.mol"^(-1)`

B

`342 "kJ.mol"^(-1)`

C

`269 "kJ.mol"^(-1)`

D

`34.7 "kJ.mol"^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
D
`log.(k_(2))/(k_(1))=(E_(a))/(2.303R)((1)/(T_(1))-(1)/(T_(2)))`
`log.(2k_(2))/(k_(1))=log2=(E_(a))/(2.303xx8.314)((1)/(293)-(1)/(308))`
or, `0.3010=(E_(a))/(19.147)xx(15)/(293xx308)`
or, `E_(a)=34673"J.mol"^(-1)=34.67"kJ.mol"^(-1)`
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