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If energy of activation of the reaction ...

If energy of activation of the reaction is `53.6"kJ.mol"^(-1)` and the temperature changes from `27^(@)C " to " 37^(@)C` , then the value of `((k_(37^(@)C))/(k_(27^(@)C)))` is -

A

`2.5`

B

`1.0`

C

`2.0`

D

`1.5`

Text Solution

Verified by Experts

Given : `T_(1)=27^(@)C=27+273=300K`
`E_(a)=53.6 "kJ.mol"^(-1)`
`T_(2)=37^(@)C = 37 +273=310K`
`log ((k_(2))/(k_(1)))=(E_(a))/(2.303R)((T_(2)-T_(1))/(T_(1)T_(2)))`
`log ((k_(310K))/(k_(300K)))=(53.6xx10^(3))/(2.303xx8.314)((310-300)/(300xx310))=0.3010`
`therefore(k_(310K))/(k_(300K))=2`
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