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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%` |
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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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