Arrhenius' Law is obtained from Van't Hoff's Law.

\begin{equation} \frac{dlnkº}{dT}=\frac{\Delta Hº}{RT^2} \end{equation}

Changing the enthalpy increase to the activation energy and the equilibrium constant to the kinetics, we obtain:

\begin{equation} \frac{dlnk}{dT}=\frac{E_a}{RT^2}\label{23} \end{equation}

Separating variables and integrating indefinitely

\begin{equation} \int dlnk=\int\frac{E_a}{RT^2}dT \;\rightarrow \; \; lnk=-\frac{Ea}{RT}+C \end{equation}

Calling the constant C lnA and solving for k yields: \begin{equation} k=Ae^{-E_a /RT} \end{equation}

  • The pre-exponential factor, A, represents the frequency of collisions.
  • $e^{-E_a /RT}$ represents the fraction of collisions that are reactive.

Another form of Arrhenius' Law

\begin{equation}\label{25} k=aT^me^{-E'_a/RT} \end{equation}

Who is  $E'_a$?

Taking neperians in the equation $[\ref{25}]$

\begin{equation}\label{26} lnk=lna+mlnT-\frac{E'_a}{RT} \end{equation}

Differentiating $lnk$ with respect to at temperature

\begin{equation}\label{27} \frac{dlnk}{dT}=\frac{m}{T}+\frac{E'_a}{RT^2} \end{equation}

Solving for $ E_a$ from the equation $[\ref{23}]$ and substituting the equation $[\ref{27}]$

\begin{equation} E_a=RT^2\frac{dlnk}{dT}=RT^2 \left(\frac{m}{T}+\frac{E'_a}{RT^2}\right) \end{equation}

Adding the fractions gives:

\begin{equation} E_a=E'_a+mRT \end{equation}

The relationship between the pre-exponential factor and the constants a and m is obtained by substituting k for $[\ref{25}]$ in Arrhenius. \begin{equation} A=ke^{E_a/RT}=aT^me^{-E'_a/RT}e^{E_a/RT}=aT^m \cancel{e^{-E'_a/RT }}\cancel{e^{E'_a/RT}}e^m \end{equation}


\begin{equation} A=aT^me^m \end{equation}

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