[arkantos]

The graphic of rate constant (k) x Temperature (T) from Arrhenius equation
is represented like these:


The same curve can be founded in reliable books like "Chemical Principles" (Peter Atkins/ Loretta Jones), as we can see:


But the graphic k(t)=Ae^-Ea/RT, has actually this aspect:

(just put A=Ea=R=1 to simplify the plot, which i gues there is no alteration in the "face" of the graphic, because in the reality A>>0 Ea>0 e R=8.31>0; just consider the asymptote "A" instead "1")

Even if we add the factor of proporcionality "sqrtT" in Arrhenius Equation: k(T)∝sqrtTAe^-Ea/RT, the graph would be still different (now there is no asymptote):


Why?

(obs.: sorry for my english)

[Corribus]

Well, why do you think?  Consider what the terms E_a and A mean, fundamentally.

EDIT: Also, don't arbitrarily add factors like sqrt(T) to make a function fit the way you think it should be.  This will lead you down a very wrong path. 

[arkantos]

First: thanks for the attention.
I guess I don't know the answer.
I guess the fact that "A" has a great magnitude order has something related. But in the other hand e^-Ea/R is the opposite.
Maybe the  x-axis (absolut temperature) in the first graphics is represented by a short interval and if we continued the graphic would assume the "expected face". Otherwise, i can't see how a graphic like e^-1/x, that has an asymptote can ""turn" into an exponential e^x

Also, I did not add the factor "sqrt(T)". It's from the equation

That simplified is:

[Corribus]

Yes but there are other constants there that modify it.  You set them all equal to one, you get an inflated importance of the square root dependency.

Anyway, what I meant was, think about the physical significance of E_a and A.  The first few plots you showed and the simulated plot showing the asymptotic limit are not in disagreement.  Most textbooks just cut off the plot without showing the limit as T goes toward infinity.  (I don't know why; it's instructive to see it and think about what it means.)

Consider: reaction rate is dependent on two things - probability of reactants colliding, and the amount of energy they have if a collision happens.  The former is embodied in the pre-exponential factor, A, and the latter is embodied in the activation energy.  The primary reason the reaction rate changes as a function of temperature is because as the temperature increases, so too does the average kinetic energy of reactant molecules, and likewise so does the probability that a given colliding molecule will have enough energy to surmount the reaction barrier and proceed toward reactants.  This effect is exponential and scaled by the Boltzmann factor, and thus as the temperature increases, so too does the rate, and this explains the trends shown in the first few figures you pasted in your post.

However, if the temperature increases to the point that the average kinetic energy of reactants have high probability of having enough energy to surmount the reaction barrier, then further increasing temperature would be expected to have little additional influence over the reaction rate.  If all molecules, in other words, have enough energy to pass the reaction barrier, giving them more energy isn't going to increase the rate further.  Thus we would expect at high temperatures (temperatures in which the average kinetic energy is higher than the activation energy), an assymptotic limit would be expected to be observed, and this is what the Arrhenius expression predicts - as T goes to infinity, the exponential term goes to 1.

And in that case, what are we left with?  Only A!  Which basically means, at the point that all molecules have enough energy to react, the rate is limited solely by the probability that reactants will collide with each other (in a geometry conducive to reaction).  rate = A, in other words.  Which is again what the Arrhenius expression predicts.

In general the temperature does impact the probability of collision (faster molecules have higher probability per unit time of colliding).  However under normal circumstances this weak temperature dependence is greatly overshadowed by the reaction barrier effect.  When the reaction becomes diffusion limited (when temperature is very high or when the reaction barrier is small), this weak temperature dependence can be observed, which is why the rate does continue to rise as temperature rises, even after there is unit probability that colliding molecules have enough energy to react.  This is what your last plot showed, although in your case the effect is "overly dramatic" because you've set all the other factors in collision expression arbitrarily to 1.  This is why you have to be careful when you are trying to see functional forms.  You can get the right form, but sometimes the effects are falsely magnified.

[curiouscat]

If I may ask a follow up question: Say two reactions have the same barrier does the shape of the barrier matter? i.e. I mean  the height of the Energy peak (the saddle point) being the same but the shape with respect to the reaction co-ordinate being different.

Not sure if this makes sense. But say, a diatomic reaction where the reaction coordinate simplifies to essentially bond length. Can two reactions have the same energy barrier height but a different shape, so to speak. Would rates be different?

By shapes I mean the functional form of Energy versus bond length between Initial and Transition states.

[Corribus]

Well, Arrhenius law remember is semiempirical and thus didn't originally have any first principles significance.  In complicated reaction mechanisms there are often multiple activation barriers, intermediate complexes, and so forth.  In addition, only energy funneled into certain vibrational degrees of freedom will usually lead to a successful reaction (this may be part of the "shape" of the potential energy surface you mention).  Most often these kinds of parameters are essentially bundled into the pre-exponential factor, but not always in an explicit manner.  I think for understanding those kinds of reaction dynamics, there are better kinetic models than the Arrhenius equation - which while it usually quantifies reaction rates pretty well, it doesn't always offer direct insight into mechanisms.

[arkantos]

[...]However, if the temperature increases to the point that the average kinetic energy of reactants have high probability of having enough energy to surmount the reaction barrier, then further increasing temperature would be expected to have little additional influence over the reaction rate.  If all molecules, in other words, have enough energy to pass the reaction barrier, giving them more energy isn't going to increase the rate further.  Thus we would expect at high temperatures (temperatures in which the average kinetic energy is higher than the activation energy), an assymptotic limit would be expected to be observed, and this is what the Arrhenius expression predicts - as T goes to infinity, the exponential term goes to 1.

Thank you man. This is what i tought. Rearing it from another people clarify the things, especially when this people seems to know more than me.

Let me ask you another question.
My original doubt came when my professor said that statement "The increase of temperature reduces the rate of an exothermic reaction" is true. I completely disagree. For what I understand, when we have an exothermic reaction and we provide heat, consequently increasing the temperature, the reaction becames faster. Even in reversible process, where the direct reaction is exothermic, and so the reverse is endothermic, when we heat we increase the rate of both reactions, but the increase in the endothermic reaction is bigger. This is why the equilibrium is deslocated in the reverse direction, consumig products and forming more reactants.
(Obs.: I know the existence of reactions that rate falls with the increase of the temperature, linke oxidation of NO, but it is not the case)



Am I wrong?

To explain this i go search for a mathematical explanation, and i think i the answer is in this phrase (from Atkins book):
"In Arrhenius plot of ln(k) against 1/_T is used to determine the Arrenius parameters of a reaction; a large activation energy (Ea) signifies a high sensitive of the rate constant (k) to changes in temperature"

In graphic therms:


Therefore, the bigger Ea for the reverse reaction (endothermic) makes k' (the reaction constant for the inverse reaction in a reversible process that is exothermic in forward direction) more sensible than k (the reaction constant for the direct reaction) to the variation of temperature.

This is kind of obvious to me, but i became insecure when my opinion diverged from my teacher.

Could you please confirm that i correct or show me why i could be wrong?

[Corribus]

I'm not sure of the context in which your professor made the statement - I'm always therefore loathe to tell someone that what their professor said is wrong - but in general this is not correct.  First of all, reaction rate is dependent really on the Gibbs energy change, not just the enthalpy change.  Therefore endothermicity/exothermicity is not directly related to rate.  The more appropriate term would be exergonicity/endergonicity.  However this little detail is often ignored for sake of simplicity, because why complicate things with entropy?

That aside, in most cases that obey Arrhenius kinetics, increasing the temperature increases the rates of both the forward AND backward reactions - both exergonic/exothermic and endergonic/endothermic reactions.  In an exergonic process, the forward reaction has a smaller reaction barrier than the backward reaction, and so the forward reaction rate is higher than the backward reaction rate - at least until equilibrium is reached.  Because rate does not only depend on the reaction barrier, but also on the amount of reactants present (collision frequency).

As usual, it's easy to confuse thermodynamics with kinetics.  They are related concepts, of course, but you can't generalize reaction rates based on thermodynamical starting or ending points.  There are non-thermodynamic considerations that impact the rate as well.

[curiouscat]

Well, Arrhenius law remember is semiempirical and thus didn't originally have any first principles significance.  In complicated reaction mechanisms there are often multiple activation barriers, intermediate complexes, and so forth.  In addition, only energy funneled into certain vibrational degrees of freedom will usually lead to a successful reaction (this may be part of the "shape" of the potential energy surface you mention).  Most often these kinds of parameters are essentially bundled into the pre-exponential factor, but not always in an explicit manner.  I think for understanding those kinds of reaction dynamics, there are better kinetic models than the Arrhenius equation - which while it usually quantifies reaction rates pretty well, it doesn't always offer direct insight into mechanisms.

All true what you say. What I meant was, let's say I gave you a  DFT calculated barrier. What's  an accurate way (not Arrhenius, as you say) to compute the exact rate expression for that step. For simplicity let's assume a simple diatomic step, say A + A  A₂

And in this calculation will shape enter the picture? Let's also assume we have DFT calculated vibrational frequencies for all states.

PS. I did some DFT and this was always a mystery to me.

[arkantos]

[...]but in general this is not correct. 
[...]in most cases that obey Arrhenius kinetics, increasing the temperature increases the rates of both the forward AND backward reactions[...]

Thank you man.  I know reaction rate is dependent really on the Gibbs energy change, not just the enthalpy change, but I guess that for simplificated analysis we can simplify ΔG=ΔH-TΔS to ΔG=ΔH and work just with enthalpy without bigger problems.
I unterstand you concern in correcting another people, but I think my professor really made a mistake.
As you said, for reactions that obey Arrhenius behavior (the most, I guess), an increase in temperature increases both exothermic and endothermic reaction rates.

Let me ask you one question about the time needed to reach the equilibrium with the temperature increasing or decreasing:
Let's suppose a generic ENDOthermic reaction where reactants (R) turn into products (P) in a closed system (the exothermic just would be the opposite). Let's take two ideal gases to simplify:


Let's analyse it qualitatively. The numbers are just to illustrate.

In the black curve, we have the inicial situation:
We put the reactants together in a Temperature T=T1. After 14 hours they reach the equilibrium at the indicated concentrations

In the blue curve (a new situation):
We put the same amount of reactants, but we keep the system at a temperature T2>T1.
We know this will favor the direct reaction (endothermic), consuming more reactants and forming more product.
My question is: the equilibrium will be reacheted faster, only after 8 hours (in the proposal sittuation)?

And so, in the red curve, the opposite:
we put the same amount of reactants, but the system in a temperature T3<T1. We will have more reactant and less Product. But the equilibrium will take more time to be reached (18 hours). Am I right?

Thank you for the attention. Hope I am not abusyng your good will.

[Corribus]

@curious

While I've built some potential energy surfaces and estimated reaction barriers by DFT, I've never tried to calculate actual reaction rates from this information.  Because there are other factors involved beyond what the DFT calculation will predict.  Although, you can get a qualitative idea from the shape of the potential energy surface what vibrational modes need to be activated to lead to a reaction.  Still, once the number of degrees of freedom increases beyond two, it becomes hard to picture a meaningful true reaction coordinate (the limitation being, of course, that the most dimensions you can plot on a piece of paper in a single graph is three). 

@arkantos

Ok -

When you increase the temperature, two things happen.  First, equilibrium is reached faster.  Second, the equilibrium constant changes.

Your reaction is endothermic*, and thus at T1, you can see that equilibrium tends to favor reactants to products, as expected.  When you increase the temperature to T2 (blue), not only does the overall reaction speed up (reach equilibrium faster), but it also begins to favor the products a little more than what was observed at T1 (still more R than P, but the ratio of R to P is less).  What this essentially means is that when you increase the temperature from T1 to T2, both the forward and backward reaction rates increase, but the backward reaction rate increases LESS. 

Why? 

The transition state energy for both reactions is the same, which means the activation barrier for the reverse reaction is smaller than that of the forward direction.  So generally speaking the reverse reaction (exothermic) happens faster than the forward (endothermic) reaction because at a given temperature, the probability of colliding product molecules having enough energy to go back to reactants exceeds the probability of the colliding reactants having enough energy to go forward to products. 

But recall that as the temperature increases, the effect of temperature on any reaction rate approaches an assymptotic limit for reasons explained above.  In our equilibrium where we have two reactions going on at one time, as the temperature increases the reaction with the smaller reaction barrier (the reverse, exothermic reaction) is closer to this "diffusion limited case" than tht reaction with the larger reaction barrier (the forward, endothermic reaction).  Therefore increasing the temperature speeds up the endothermic reaction to a greater extent than the backward, exothermic reaction, and thus increasing the temperature will favor the formation of products in an overall endothermic equilibrium process.  (The reverse is true for an overall exothermic process - increasing temperature will favor reactants in this case, although it's just a matter of bookkeeping at this point.  Since we're talking about an equilibrium, there's always both an exothermic and endothermic process going on simultaneously.  All that changes is which one is "forward" and which is "backward", an arbitrary distinction governed typically by what materials are initially put in the reaction flask.)

As you can guess, the exact opposite occurs when you lower the temperature.  This will favor an exothermic reaction, and in your endothermic reaction will tend to favor formation of reactants, which is what your red plot shows.

My guess is that your professor was alluding to this effect when he/she said that increasing temperature reduces the rate of an exothermic reaction.  Probably the professor just misspoke.  Increasing temperature doesn't reduce the rate of an exothermic reaction.  The rate still increases: it just increases less than the (simultaneous) endothermic counterpart, so that the endothermic products are - relative to the lower temperature reaction - favored.

Note that this reasoning really only works for small temperature changes.  If the temperature changes a lot, then the actual value of ΔH begins to change by significant degree, and the situation is more complicated.

*Here we run into the annoying endothermic/endergonic issue again.  And again we'll ignore it!

[arkantos]

I have no more doubts (at least now, hehe) in this subject and I was correct in my interpretations.
Thank you "Corribus" for your attention and explanations.