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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

`log. (k_2)/(k_1)= E_a/(2.303R)([T_2-T_1])/(T_1T_2), ( :' R = 2 xx 10^(-3)Kcal)`
`:. 2.303log. (k_2)/(k_1)= 9/(2xx 10^(-3))[ 10/(298xx 308)]`
`log. (k_2)/(k_1)= 0.2129`
`:. K_2/k_1= 1.63, i.e., 63%` increase
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