Calculate the activation energy of a rea...

Calculate the activation energy of a reaction whose reaction rate at `300 K` double for `10 K` rise in temperature.

40 K cal/mole

10 K cal/mole

20 Kcal/mole

`2^((-20/(RT)K."cal"//"mole")`

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The correct Answer is:

it is first order reaction

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Calculate the activation energy of a reaction whose reaction rate at 310 K gets doubled for 320 K rise in temperature.

Arrhenius studies the effect of temperature on the rate of a reaction and postulted that rate constant varies with temperature exponentially as k=Ae^(E_(a)//RT) . Thuis method is generally used for finding the activation energy of a reaction. Keeping temperature constant, the effect of catalyst on the activation energy has also been studied. If the rate of reaction doubles for 10^(@)C rise of temperature form 290K to 300K, the activation energy of the reaction will be approximately :

The rate of a reaction quadruples when the temperature changes from 293K to 313K . Calculate the energy of activation of the reaction assuming that it does not change with temperature.

The rate of reaction triples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction assuming that it does not change with temperature.

Arrhenius studied the effect of temperature on the rate of a reaction and postulated that rate constant varies with temperture exponentially as K=A.e^(-Ea//RT) . This method is generally used for finding the activation energy of reaction . Keeping temperature constant , effect of catalyst on activation energy has also been studied. If the rate of reaction double for rise of temperature from 500K to 1000K, the activation energy of the reaction will be approximately [ln2=0.7]

What is the activation energy (KJ/mol) for a reaction if its rate constant doubles when the temperature is raised from 300 K to 400 K ? (R="8.314 J mol"^(-1)"K"^(-1))

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