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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

0.1

B

1

C

0.5

D

0.63

Text Solution

Verified by Experts

The correct Answer is:
D
`2.303 log K_(2)/K_(1) =E_(a)/R([T_(2)-T_(1)])/([T_(1)T_(2)])`
`therefore 2.303 log K_(2)/K_(1) = 9/(2 xx 10^(-3))[10/(298 xx 308)]`
`therefore K_(2)/K_(1) = 1.63 , i.e. 63%` increases
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