Rate constant of the reaction at 27°C is 5×10^2 while at 127°C it is 5 ×10^4 calculate activation energy for the reaction?

Question

V Prakash · Answer

Use arrhenius equation K=Ae -x/RT where x is the activation energy. Use this equation two time for the two given rate constants and temperature.Note that activation energy and pre exponential factor are constant. Dividing those two equations will cancel out pre exponential factor. Then simplifying it and using logarithms will yield you x as 45.9kJ.

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Rate constant of the reaction at 27°C is 5×10^2 while at 127°C it is 5 ×10^4 calculate activation energy for the reaction?

Rate constant of the reaction at 27°C is 5×10^2 while at 127°C it is 5 ×10^4 calculate activation energy for the reaction?

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Rate constant of the reaction at 27°C is 5×10^2 while at 127°C it is 5 ×10^4 calculate activation energy for the reaction?

Rate constant of the reaction at 27°C is 5×10^2 while at 127°C it is 5 ×10^4 calculate activation energy for the reaction?

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