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The activation energy of a reaction is 9...

The activation energy of a reaction is `9.0 kcal//mol`.
The increase in the rate consatnt when its temperature is increased from `298 K` to `308 K` is

A

0.63

B

0.5

C

1

D

0.1

Text Solution

Verified by Experts

The correct Answer is:
a
2.303 log `(k_(2))/(k_(1)) = (E_(a))/(R) [(T_(2) - T_(1))/(T_(1)T_(2))]`
log `(k_(2))/(k_(1)) = (9.0 xx 10^(3))/(2.303 xx 2) [(308 - 298)/(308xx 298)]`
`(k_(2))/(k_(1)) = 1.63 , K_(2) = 1.63 K_(1) , (1.63 k_(1) - k_(1))/(k_(1)) xx 100 = 63.0` %
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