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

Options

A. `63%`
B. `50%`
C. `100%`
D. `10%`

Answer

A

Explanation

`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)/(308 xx 298)]`
`(k_(2))/(k_(1))=1.63, k_(2)=1.63 k_(1)`,
Increase in `k_(1)=(k_(2)-k_(1))/(k_(1)) xx 100`
`=(1.63 k_(1)-k_(1))/(k_(1)) xx 100=63.0%`