Is the Arrhenius Equation a solution to a differential equation?

Question

I just learned about the Arrhenius Equation,

$$k(T)=A\exp{\left(\frac{-E_\mathrm a}{RT}\right)}$$

and noticed that it appears to be in the same form as the solution to a linear homogeneous differential equation, so I attempted to find the associated differential equation.

Since $E_\mathrm a$ and $R$ are both constant parameters, you can treat it as a single constant. Defining $$\beta = -E_\mathrm a/R$$

we have $$\ln(k(T)) = \ln[\exp(\beta/T + \ln A)]$$

Take the derivative of each side and solve for $\mathrm dk/\mathrm dT$ ($A$ is treated as a constant of integration and goes away), and we get:

$$\frac{\mathrm dk}{\mathrm dT} = -\frac{\beta k}{T^2}$$

Here is the slope field for the differential equation when $β=-1$

and here is the slope field for $β=1$

However, since $k$ is a rate constant, I'm not sure if this method is actually applicable. Is this, or a similar differential equation actually used on this topic?

Answer 1

To answer the question of the title. In Adam-Gibbs theory the activation enegy is defined as [1]:

$$E_\mathrm a=-R\left[ \frac{\partial \ln k}{\partial (1/T)} \right]$$ Now clearly this is a first order linear differential equation, as you were looking for. This can easily be solved by seperation of variable: $$\int E_\mathrm a\, \mathrm d(1/T) = -R\int \mathrm d\ln k $$ $$\frac{E_\mathrm a}{T}=-R\ln k+A'$$ $$k(T)=A\exp\left( \frac{-E_\mathrm a}{RT} \right)$$ Here $A=\exp(-A'/R)$, now it can be seen that the Arrhenius equation has been regained.

[1] S. Petrucci; Ionic Interactions: From Dilute Solution to Fused Salts; equation 125