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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.)

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  • $\begingroup$ 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. $\endgroup$ – Ivan Neretin Aug 16 at 11:01
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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).

enter image description here

The graph looks like $y=x^2$ but only upto $T=\dfrac E{2R}$. But don't misinterpret the same.

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    $\begingroup$ A is also temperature dependent, that's why k goes up as well $\endgroup$ – user96208 Aug 16 at 18:37

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