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

$$\frac{dlnkº}{dT}=\frac{\Delta Hº}{RT^2}$$

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

$$\frac{dlnk}{dT}=\frac{E_a}{RT^2}\label{23}$$

Separating variables and integrating indefinitely

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

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

• 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

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

Who is  $E'_a$?

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

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

Differentiating $lnk$ with respect to at temperature

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

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

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

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