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The ratio of rate constants of a reactio...

The ratio of rate constants of a reaction at 300K and 291K is 2. Calculate the energy of activation. `("Given "R = "8.314JK"^(-1)" mol"^(-1))`. 

Text Solution

Verified by Experts

To find activation energy for the reaction
Given `(K_(2))/(K_(1))=2, R=8.314, T_(1)=291K, T_(2)=300K`
`log"" (K_(2))/(K_(1))=(E_(a))/(2.303R)[(T_(2)-T_(1))/(T_(1)T_(2))]`
`log2=(E_(a))/(2.303xx8.314)[(300-291K)/(300xx291K)]`
`E_(a)=(2.303xx8.314xxlog2xx300xx291)/(9)=55.9kJ`
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