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The rate of a reaction doubles when its ...

The rate of a reaction doubles when its temperature changes from 300K to 310K . Activation energy of such a reaction will be - `(R=8.314"J.K"^(-1)"mol"^(-1) and log2=0.310)`

A

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

B

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

C

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

D

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

Text Solution

Verified by Experts

The correct Answer is:
A
As `r=k[A]^(n)`
`(r_(2))/(r_(1))=(k_(2))/(k_(1))`
Since `(r_(2))/(r_(1))=2` (given)
`therefore (k_(2))/(k_(1))=2`
`log_(10).(k_(2))/(k_(1))=(E_(a))/(2.303R)[(T_(1)-T_(1))/(T_(1)-T_(2))]`
or, `log_(2)=(E_(a))/(2.303xx8.314xx10^(-3))[(310-300)/(310xx300)]`
or, `E_(a)=(0.301xx2.303xx8.314xx10^(-3)xx93xx10^(3))/(10)`
or, `E_(a)=53.6"kJ.mol"^(-1)`
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