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

`2.303log.(k_(2))/(k_(1))=(E_(a))/(R)[(T_(2)-T_(1))/(T_(1)T_(2))]`
`log.(k_(2))/(k_(1))=(9.0xx10^(3))/(2.303xx2)[(308-298)/(308xx298)]`
`(k_(2))/(k_(1))=1.63`, `K_(2)=1.63K_(1)` , `(1.63k_(1)-k_(1))/(k_(1))xx100=63.0%`