Chemistry - Does it make sense to differentiate the Arrhenius equation with respect to temperature?
Solution 1:
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 $\ce{H3O+(aq) + OH-(aq) -> 2 H2O(l)}$ ).
In solvents, we cannot usually ignore intermolecular interactions, so temperature dependence of $A$ may get quite complicated.
Solution 2:
With respect to your question on the math, I start with the Arrhenius equation:
$$k = A\mathrm e^{-E_\mathrm{a}/(RT)}$$
and would recommend simplifying the math by first proceeding with the introduction of a natural log transformation (this can also be used to linearly estimate $E_\mathrm{a},$ see discussion here):
$$\ln k = \ln A - \frac{E_\mathrm{a}}{RT}$$
Then, take the derivative of $\ln k$ with respect to $T$ we have:
$$\frac{\mathrm d \ln k}{\mathrm dT} = \frac{E_\mathrm{a}}{RT^2}$$
Note the positive impact (on the natural $\log$ of $k)$ with respect to the activation energy scaled by the inverse square of the absolute temperature.
To answer your second question, "Does 'A' change when temperature change (since my variable here is temperature)?", here are some pertinent comments per Wikipedia, to quote:
Given the small temperature range of kinetic studies, it is reasonable to approximate the activation energy as being independent of the temperature. Similarly, under a wide range of practical conditions, the weak temperature dependence of the pre-exponential factor is negligible compared to the temperature dependence of the $\mathrm e^{-E_\mathrm{a}/(RT)}$ factor; except in the case of "barrierless" diffusion-limited reactions, in which case the pre-exponential factor is dominant and is directly observable.
So, 'A' apparently only displays a weak temperature dependence, to answer your question.
Solution 3:
You should start from an idea of the variation in T and concentration that might be encountered, call these $\delta T$ and $\delta c$. If you have a model of the rate $r(T,c)$ then the associated variation in $r$ with respect to the individual variables can be determined from the partial derivatives as
$$\delta r_i = \left(\frac{\partial r}{\partial x_i}\right)_j\delta x_i$$
The different $\delta r_i$ can be compared directly to see to changes in which variable $r$ is most sensitive.
Remember that whatever equation you end up using to describe $r$ is just a model, it may be regarded as correct insofar as it accurately predicts data. Since you do not discuss any particular data that you are attempting to model, presumably you may assume $A$ and $E_a$ are constant.
Solution 4:
The question lacks details, and is thus effectively asking for a coarse-grained answer.
So:
The answer depends strongly on how the activation energy compares with RT. Let's compare doubling the (absolute) temperature with doubling the concentration.
Concentration: If the reaction is first-order in the reactant in question, then we get a $2$-fold increase in rate; $2$nd-order => $4$-fold increase. If we have two reactants, and the reaction is first-order in one, and second-order in the other, and we double the concentrations of both, we get an $8$-fold increase. And so on.
Temperature: If we ignore the T-dependence of A and Ea, doubling the absolute temperature increases the rate by:
$$\frac{e^{\left(-\frac{\text{Ea}}{2 \text{RT}}\right)}}{e^{\left(-\frac{\text{Ea}}{\text{RT}}\right)}} = e^{\frac{\text{Ea}}{2 \text{RT}}}$$
If $\text{Ea} \approx \text{RT}$, then doubling T increases the rate by a factor of $e^{(1/2)} \approx 1.6 $. And if $\text{Ea} < \text{RT}$, doubling T increases the rate by a factor less than $1.6$. Thus, in this regime ($\text{Ea} \lessapprox \text{RT}$), one could say concentration is more important (except in the trivial case of a zeroth-order reaction, or in the unusual case where the order is between $0$ and $1$).
On the other hand, if Ea is much larger than RT (which is common for chemical reactions, if T is about room temperature*), then one could say temperature is more important. For instance, if $\text{Ea} \approx \text{10 RT}$, then doubling T would increase the rate by a factor of $e^{5} \approx 148$.
*At $298$ K, RT is only $\approx 2.5$ kJ/mol.
Indeed, if we consider a "typical" activation energy of $\approx 80$ kJ/mol**, then even a modest increase in T (from $298$ K to $340$ K) will increase k by about 50-fold.
**How to relate a reaction barrier to the time the reaction needs to proceed?
Another way to think of this is that, if RT is close to (or greater than) the activation energy, then there is abundant thermal energy relative to activation energy. Consequently, the availability of thermal energy isn't strongly limiting the rate of this reaction, so increasing the temperature doesn't have a strong effect on the rate.
However, if RT is much less than the activation energy, then the lack of available thermal energy is strongly limiting the rate of reaction. Consequently, increasing the temperature will have a large effect on the rate.
Solution 5:
Yes it does make sense to differentiate wrt temperature but in the Arrhenius model $A$ is a constant. However, if the pre-exponential $A$ is replaced with terms representing the bond that dissociates and other partition functions, i.e we replace $A$ with $A(T)$ a function that depends on temperature. The value of this term depends on the type or reaction, unimolecular, hard-sphere collision, atom + diatom etc. For two hard spheres $A(T) \sim T^{1/2}$ for example. This means, in effect, that the Arrhenius model is abandoned and transition-state theory is used instead.