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A catalyst lowered the activation energy...

A catalyst lowered the activation energy by `25 kJ mol^(-1)` at `25^(@)C`. By how many times will the rate grow?

A

14069

B

24069

C

16049

D

19049

Text Solution

Verified by Experts

The correct Answer is:
B
The rate of reaction is related to the activation energy by the following relation:
Given `triangleE=25 xx10^(3) j ,R =8.314 j k^(-1) mol^(-1)`
substituting all the values in Eq (i) get `(k_(2))/(k_(1))` = Antilog `[(25xx10E^(3))/(2.303xx8.314 xx298)]` =24069
`therefore k_(2) =k_(1) xx24069` Therefore the rate increased by 24069 times
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