# Arrhenius equation and rate constant

I am a high-school student and I am currently taking a course on chemical kinetics. Towards, what seems, end of the course at highschool level, the Arrhenius equation came in. Now, I'm not sure if what I'm presented is just a general form of Arrhenius equation and there's more to it, or this is it. Well, anyways, one of the things I am being told to learn is the plot between graph of rate constant $$k$$ vs temperature $$T$$. $$k=A\mathrm e^{-E_\mathrm a/(RT)}$$

The graph is sort of looks like with increasing $$T$$, the value of $$k$$ is skyrocketing (kinda like $$y=x^2$$ in first quadrant) but if I'm not mistaken, the rate constant $$k$$ should equal $$A$$ as temperature tends to infinity. Now I believe there must be some other factor in the equation that is also a function of temperature and is causing the graph to be the way it is. Can someone help me understand the same?

(In case you didn't get, because I am unable to attach an image, the looks like $$y=x^2$$ instead of $$x=y^2$$, which basically interrupts my intuition of Temperature on $$x$$ axis ever approaching infinity. $$k$$ on $$y$$ and $$T$$ on $$x$$ axis.)

• Imagine that you are small like an ant, only smaller, and looking at the initial portion of that graph. Now it will look convex all right. – Ivan Neretin Aug 16 at 11:01

Surely, as T tends to infinity K tends to A. The graph in this case will be exponential and not $$y= x^2$$ kind. You can plot the graph by differentiating the equation (twice to be particular).
The graph looks like $$y=x^2$$ but only upto $$T=\dfrac E{2R}$$. But don't misinterpret the same.