Question

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

Answer 1

Correct Answer - d
`2.303 "log" K_(2)/K_(1)=E_(a)/R[(T_(2)-T_(1))/(T_(1)T_(2))]`
`:. 2.303 "log" K_(2)/K_(1)=9/(2xx10^(-3))[(10)/(298xx308)]`
`:. K_(2)/K_(1)=1.63, i.e., 63%` increse.