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How much faster would a reaction proceed...

How much faster would a reaction proceed at `25^(@)C` than at `0^(@)C` if the activation energy is `65 kJ` ?

A

`2` times

B

`5` times

C

`11` times

D

`16` times

Text Solution

Verified by Experts

The correct Answer is:
C
`"log"((K_(2))/(K_(1))) = (E_(a))/(2.303R)((T_(2) - T_(1)))/(T_(1)T_(2))`
`= (65 xx 10^(3)xx(298-273))/(2.303 xx 8.3 xx 298 xx 273)`
By calculation we fins `(K_(2))/(K_(1)) = 11`
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