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The rate constant of a reaction at 500 K...

The rate constant of a reaction at `500 K` and `700 K` are `0.02s^(-1)`, respectively. Calculate the values of `E_(a)` and `A` at `500 K`.

Text Solution

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`logk_(2)/k_(1)=E_(a)/(2.323R)[(T_(2)-T_(1))/(T_(1)T_(2))]`
`"log"k_(2)/k_(1)=(E_(a)/(2.323xx8.314JK^(-1)"mol"^(-1)))[(700-500)/(700xx500)]`
`0.544= E_(a)xx 5.714xx10^(-4)//10.15`
`E_(a)=xx 0.544xx19.15//5.714xx10^(-4)=18230.8J`
Since ` k=Ae^(-Ea//RT`
`0.02=Ae^(-18230.8//8.314xx500`
`A= 0.02//0.02=1.61`
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