1.

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

Answer» 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.


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