I saw the following plot of the Arrhenius Equation on Wikipedia (cc-by Zivilverteidigung and Matthias M.).

NO2 Arrhenius k against T.svg (cc-by Zivilverteidigung and Matthias M.)

It is a concave upwards graph, but according to the equation $$k= A \mathrm e^\frac{-E_\mathrm a}{RT},$$ it is evident that the graph is of the form $$y= \mathrm e^\frac{-1}{x}.$$

graph of e^(-1/x)

I believe that this should have been the graph (please disregard the negative side of the $x$ axis). It is a concave downwards curve. Why is there a discrepancy? This maybe also suitable in Mathematics.se, but I have a gut feeling that the solution to this lies in Chemistry and not maths.

  • $\begingroup$ May not explain the whole thing but A isnt necessarily constant as it is temperature dependent; it has to do with reactant "collisions" that form the products, including steric factors as well as frequency (depends on T) $\endgroup$
    – rch
    Jul 3, 2014 at 8:00

2 Answers 2


I believe that this should have been the graph...(please omit the negative side of x axis), so concave downwards. Why is there a discrepancy?

Your analysis and your graph are correct. Let's examine why the discrepancy arises. I've drawn two graphs below. The top one is similar to your graph, only I'm examining a smaller range on the x-axis. My lower graph examines an even smaller range and it is a range close to x=0. You can see that this lower graph appears to be "concave-up" just like the Wikipedia graph you referenced.

enter image description here

Here's a third graph where I changed the numerator from -1 to -100, notice how it moves the curve further to the right along the x-axis.

enter image description here

Depending upon 1) what numbers we put in the numerator and denominator of our exponential equation and 2) depending upon how we scale our axes, we can dramatically change how our graph appears. There will always be a relatively flat part to this curve located close to x=0, then there will be a concave-up section, followed by a concave-down section, and finally another relatively flat section further out on the x-axis. You happened to choose a large range (up to x=60-70) for the x-axis of your graph. This compressed your graph, so it did not show the first two parts of the curve (the relatively flat part located close to x=0 and the concave-up section) clearly.

  • $\begingroup$ This may indeed be the experimental data, but acc to the maths, if I differentiate twice, why is it becoming always negative? $\endgroup$ Jul 3, 2014 at 15:50
  • $\begingroup$ @Saurabh Raje The slope is given by the first derivative. The first derivative of e^(-1/x) is (e^(-1/x)/(x^2) which is always positive. $\endgroup$
    – ron
    Jul 3, 2014 at 16:58
  • $\begingroup$ sir , i am sure that concavity is checked using the second derivative, which is negative $\endgroup$ Jul 3, 2014 at 17:28
  • $\begingroup$ the second derivative of e^(-1/x) is given by ((e^(-1/x))/(x^4))-((2*e^(-1/x))/(x^3)) which does change sign $\endgroup$
    – ron
    Jul 3, 2014 at 17:38
  • $\begingroup$ so for x<1, this is positive, thats why? $\endgroup$ Jul 4, 2014 at 8:50

The two problems are:

You must have an "power-intercept" with inverse-x as $$\ y(x)=e^{a-(b/x)}$$ Important but not necessary, as to provide a limiting value $e^a$ to $y$ at $\infty$.

Also what you are worried about the derivatives is simple: $$y'(x)=\frac{b}{x^2}e^{a-(b/x)}$$ $$y''(x)=\frac{b(b-2x)}{x^4}e^{a-(b/x)}$$with second derivate a root at $x=\frac b2$

An example graph with two resolutions: $$\text{legend: }[\color{red}{y(x)}\;,\color{purple}{y'(x)}\;,\color{black}{y''(x)}]\;\text{ with }a=2.5,b=2.2$$ So the derivative changes the sign at $1.1$,

enter image description here

enter image description here


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