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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:
`(K_2)/(K_1) = "Anti log "[(Delta E)/(2.303 RT)]" " (i)`
Given, `Delta E = 25 xx 10^3 J , R = 8. 314 J K^(-1) "mol"^(-1)`
`T = 25^@ C = 298 K`
Substituting all the values in Eq. (i) , we get `(K_2)/(K_1) = "Anti log" [ (25 xx 10^3)/(2.303 xx 8.314 xx 298)] = 24069`
`therefore K_2 = K_1 xx 24069 ,` Therefore the rate increases by 24069 times.
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