Does it make sense to differentiate the Arrhenius equation with respect to temperature? I am trying to answer the question: &quot;between concentration and temperature, which has a greater significance on the rate of reaction&quot;
Hence, I'm trying to find a method to systematically compare the impact/significance of concentration and temperature on rate. This is where I ended up trying to differentiate the Arrhenius equation:
$$k = A\mathrm e^{-E_\mathrm a/(RT)}$$
I thought that deriving rate constant $k$ with respect to temperature is finding the rate of change of $k$ with respect to temperature. Is it conventionally and mathematically correct to do this?
If it is, then continuing the thought process:
I differentiated the equation assuming that the Arrhenius constant...is...constant... But is that true? Does $A$ change when temperature change (since my variable here is temperature)?
If everything was not non-sense until this point, I ended up with the derivative:
$$\frac{\mathrm dk}{\mathrm dT}=A\mathrm e^{-E_\mathrm a/(RT)}\times E_\mathrm a/(RT^2)$$
Is this correct chemistry/maths?
EDIT
Is there a better way to compare the significance of concentration with that of temperature on the rate of reaction? Maybe using percentage increase of each factor with the percentage increase of rate?
Note that $A$ is a kind of a collision frequency factor, which is temperature dependent via the mean molecule speed, even if far less then the Boltzmann exponential factor.
$$\mathrm{d}k/\mathrm{d}T = \frac {\mathrm{d}A}{\mathrm{d}T} \cdot \exp{(\frac {-E_\mathrm{a}}{RT})} + A \cdot \exp{(\frac{-E_\mathrm{a}}{RT})} \cdot \frac {E_\mathrm{a}}{RT^2} \tag{1}$$
$$A = A_0 \cdot {({\frac {T}{T_0})}}^{1/2} \tag{2}$$
$$\frac {\mathrm{d}A}{\mathrm{d}T} = A_0 \cdot \frac 12 \cdot {({\frac {T}{T_0})}}^{-1/2} \cdot \frac {1}{T_0}=\frac {A_0}2 \cdot {({\frac {T_0}{T})}}^{1/2} \cdot \frac {1}{T_0} =\frac {A_0}2 \cdot \frac 1{\sqrt{T \cdot T_0}}  \tag{3}$$
By subtitution in (1), using  (2) and (3):
$$\mathrm{d}k/\mathrm{d}T = \left( \frac 12 \cdot \frac 1{\sqrt{T \cdot T_0}}   +  {({\frac {T}{T_0})}}^{1/2}  \cdot \frac {E_\mathrm{a}}{RT^2} \right) \cdot A_0 \cdot \exp{(\frac{-E_\mathrm{a}}{RT})}\tag{4}$$
As AJKOER properly noted, temperature dependance  $A=A(T)$ can be usually neglected and we can use the simplified (1), as $\frac {\mathrm{d}A}{\mathrm{d}T} = 0$:
$$\mathrm{d}k/\mathrm{d}T = A \cdot \exp{(\frac{-E_\mathrm{a}}{RT})} \cdot \frac {E_\mathrm{a}}{RT^2} \tag{1a}$$
unless the activation energy is extremely low and reaction rate is driven by diffusion ( typical one is $\ce{H3O+(aq) + OH-(aq) -&gt; 2 H2O(l)}$  ).
In solvents, we cannot usually ignore intermolecular interactions, so temperature dependence of $A$ may get quite complicated.