# Does it make sense to differentiate the Arrhenius equation with respect to temperature?

I am trying to answer the question: "between concentration and temperature, which has a greater significance on the rate of reaction"

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?

• One problem is that dk/d(conc) and dk/dT will have different units, so you can't easily compare them quantitatively to each other. Oct 1, 2020 at 21:22
• @Andrew yes that makes total sense I did not think this through...I was also trying to ask how should I acutally go about comparing these two factors or if it's too complicated to compare them at all. Oct 1, 2020 at 21:26
• Does it mention anything about the order of the reaction?
– user98623
Oct 1, 2020 at 21:42
• @dval98 It's my vague research question so I can choose what order of reaction I want to look at. Oct 1, 2020 at 22:46

## 5 Answers

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.

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.

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.

• I remember van't Hoff ( I guess it was him ) empirical rule "Temperature increase of 10 degrees leads typically to chemical reaction rate increase 2-4 times". It would be 16-256 times per 40 degrees ( 340-298 = about 40 ) Oct 2, 2020 at 16:24
• @Poutnik Yes, that rule is nicely consistent with what I indentified as a typical Ea, since if Ea = 80 kJ/mol, and T_initial = 298 K, then increasing T by 10 K would increase k by 2.8 times, which is right in the middle of that rule's range. More broadly, if that rule assumes T_initial = 298 K, then its 2 - 4 x rate increase is tantamount to assuming an Ea range of 53 – 107 kJ/mol (298 K ->308 K with Ea = 53 kJ/mol => 2-fold rate increase; 298 K->308 K with Ea = 107 kJ/mol =>4-fold rate increase). Oct 2, 2020 at 19:26

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.

• Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown. Keep in mind \ce{…} macro is strictly for chemical equations and chemical formulae and should never be used for anything else. Symbols for math operators, mathematical constants and textual labels must be upright to avoid confusion with variables which are written in italics. Oct 2, 2020 at 6:27

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.

• You may want to double-check the indices in your chain rule.
– J.G.
Oct 4, 2020 at 18:46

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.