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orthocresol
orthocresol
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It is easy to show an Arrhenius-like equation from the Eyring equation, but if you do this, you get that the activation energy is about equal to the activation enthalpy. However, the real approximation is that Ea is equal to delta H + RT$E_\mathrm{a} = \Delta H^\ddagger + RT$. Why is this?

It is easy to show an Arrhenius-like equation from the Eyring equation, but if you do this, you get that the activation energy is about equal to the activation enthalpy. However, the real approximation is that Ea is equal to delta H + RT. Why is this?

It is easy to show an Arrhenius-like equation from the Eyring equation, but if you do this, you get that the activation energy is about equal to the activation enthalpy. However, the real approximation is that $E_\mathrm{a} = \Delta H^\ddagger + RT$. Why is this?

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sweetandtangy
sweetandtangy
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Where does the RT term come from in the derivation for the activation enthalpy from the Eyring equation?

It is easy to show an Arrhenius-like equation from the Eyring equation, but if you do this, you get that the activation energy is about equal to the activation enthalpy. However, the real approximation is that Ea is equal to delta H + RT. Why is this?