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Topic: Endothermic and Exothermic Reaction Rates  (Read 21062 times)

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Offline ramboacid

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Endothermic and Exothermic Reaction Rates
« on: June 27, 2012, 01:19:21 AM »
For a forward exothermic reaction, the reverse reaction is always endothermic, and the activation energy of the forward exothermic reaction is always smaller than that of the reverse endothermic reaction. I would think that according to the Arrhenius equation this means that the exothermic reaction will always have a large rate constant k than its reverse endothermic counterpart, and therefore that the equilibrium constants of forward exothermic reactions would be greater than 1. It would also means that exothermic reaction rates are less affected by changes in temperature than their reverse endothermic reactions.

However, I feel that's not necessarily true, but I don't have any examples or counterexamples on hand. Looking at the equation I'd think that the Arrhenius constant would compensate for the higher activation energy of endothermic reactions. Since the constant is the product of cross-sectional area, a steric factor, and a number of other constants, I'd guess the steric factor and cross-sectional area would determine whether the endothermic or exothermic reaction rate would be higher, but then the reaction mechanisms of both the forward and reverse reactions would need to be known to make a decision of which is favored.

Does this make any sense, or are my assumptions incorrect? I feel I've thought myself into a fallacy, any suggestions are appreciated :)
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Offline Enthalpy

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Re: Endothermic and Exothermic Reaction Rates
« Reply #1 on: June 28, 2012, 03:27:52 PM »
A higher temperature favours the endothermal reaction (or makes it less defavoured).
But heat of reaction isn't the only player: entropy rules here, meaning that the number of molecules for instance is important, and some conditions can produce the endothermal compound.

Offline ramboacid

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Re: Endothermic and Exothermic Reaction Rates
« Reply #2 on: June 28, 2012, 04:12:55 PM »
I know entropy considerations play into the determination of spontaneity for a reaction, but does it play into calculating the rate the constant in any way? Unimolecularity and bimolecularity of elementary steps in a reaction mechanism is the most obvious place to consider the effect of entropy, as elementary steps in which the number of molecules decreases between the reactants and products would (at least in that step) result in a decrease in entropy (assuming no phase changes too).

But is there a place in the Arrhenius equation to reflect entropy concerns on the rate constant? I can only imagine it being in the parts of the Arrhenius constant that reflect cross-sectional area or the steric factor (in that smaller molecules have smaller cross-sectional areas maybe?). It would seem like a roundabout way of dealing with entropy if that's how it really worked...  ???
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Offline Bryan Sanctuary

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Re: Endothermic and Exothermic Reaction Rates
« Reply #3 on: June 28, 2012, 05:04:09 PM »
You do indeed make sense.  It is interesting to follow your thoughts.  First point is that when you have a reversible reaction,

     k1
A   ::equil:: Z
     k-1

that the reaction proceeds until equilibrium is reached.  In that case the equilibrium constant is related to the forward and reverse rate constants:
           
Kequil = [Z]equil/[A]equil = k1/k-1

If you use this, and plug in Arrhenius and any transition state theory quantities like entropy, internal energy, etc, most of what you are saying will make sense.

When you are discussing mechanisms, like collision cross sections, steric factors, then you are talking about the end goal of chemical kinetics: that is you what to know the steps that lead from reactants to products.  Now we can usually come up with a "reasonable" mechanism that is consistent with the rate law, but this consistency does not mean that the mechanism is correct.

But mechanisms are great to visualize steps.

It might help to look at chemical kinetics in the following way.

What is the real objective in understanding chemical changes?  You want to know how the particles react.  That information must be in the empirical rate equation obtained from experiments.

So let us consider that rate equation and take it apart.  First we find the way the data plots to give us the orders of the reaction. Usually pretty easily done.  So with that we can remove the concentration dependencies from our empirical rate law.  Less stuff to worry about.

Now we are left with the rate constant, k.  We study it and find that under different conditions, rates can change.  Hence the rate constant can be temperature, pressure and field (electric, magnetic) dependent.  Let's just use temperature.  Then Arrhenius found the exponential relation that allows us to separate out the dominant temperature dependence of the rate constant, leaving us with the Pre-exponential factor.  That is where the interest lies.

First you might ask why in Arrhenius you find the ratio of the activation energy to RT, 

k = A exp(-Ea/RT)

How RT is called the "thermal energy" which simply means it is the heat available around us.  For us it is RT= 8.314 X 300 = 25 kJ/mol .  Go out in space and it is 3 K or go to the sun and it is 6000 K.  This energy that surrounds us is available for use in processes.  So getting over a higher activation energy barrier is easier the more the thermal energy increases.

I am rambling, but we are here on Earth because our thermal energy is such that stable chemical bonds can occur.  Not many chemical bonds on the sun, and not enough energy in outer space to mount any activation energy. So we are lucky to have our Earth.

So finally the pre-exponential factor is what is left over after removing temperature, activation energy and concentration from the empirical rate equation.  That is where we learn about the mechanisms, find out what collisions might do, and learn about intermolecular forces.

BTW the entropy comes in at the pre-exponential level too, and really formalizes the more empirical steric factor.  Non-the-less calculations including collisions we can actually do, like hard spheres, give remarkable insight into what is going on.
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Offline ramboacid

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Re: Endothermic and Exothermic Reaction Rates
« Reply #4 on: June 28, 2012, 06:54:07 PM »
Thanks for the response. I found it rather enlightening than rambling :)

In a nutshell though, does this mean that, at constant temperature, the Arrhenius parameter adjusts for the higher activation energy of endothermic reactions when endothermic processes are favored over their reverse exothermic counterparts? Without the pre-exponential factor it seems that all exothermic processes would be favored over their reverse endothermic reactions.

Also, how does entropy play into the steric factor? Is there a term or equation I can search for more info?
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Offline Bryan Sanctuary

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Re: Endothermic and Exothermic Reaction Rates
« Reply #5 on: June 28, 2012, 09:50:15 PM »
Thanks for the response. I found it rather enlightening than rambling :)

In a nutshell though, does this mean that, at constant temperature, the Arrhenius parameter adjusts for the higher activation energy of endothermic reactions when endothermic processes are favored over their reverse exothermic counterparts? Without the pre-exponential factor it seems that all exothermic processes would be favored over their reverse endothermic reactions.

Also, how does entropy play into the steric factor? Is there a term or equation I can search for more info?

I would say the fundamental reason there ar both forward and reverse processes, is that measurement involves a huge number of particles and so all we can really do is describe chemical reactions statistically. 

From before the equilibrium constant is proportional to the ratio of the forward and reverse processes,

Kequil  = k1/k-1=exp[-(E1-E-1)/RT]

and depends on the difference of activation energies.  The Arrhenius equation says the higher the activation energy, the harder it is to get over, but this does not rule out reverse processes.  The reason is the statistical nature of measurement.  So the Arrhenius equation is really a probability that a particle will pass the barrier at a given temperature, and the same for the reverse process.

A small technical point: perhaps you are also asking why all the molecules don't end up as products?  The answer can be found if you think about the following equation:

 Kequil = [Z]equil/[A]equil = k1/k-1

Notice in the above that the pre-exponential factors cancel.  I have assumed they are equal.  That is the collisions are reversible, which is usually a good assumption. 

I mentioned before that the rate constant, k, is not really a constant because it varies with T, etc.  I forgot to say that it also varies with the use of catalysts.

Regarding entropy and steric effects.  Well it makes sense: steric effects mean that something sticks out and inhibits something from happening.  So the orientation of molecules is important in determining if an interaction is successful and forms products, or fails (called "reaction coordinates").  That is why enzymes are of such importance because they orient chemically reactive sites farourably.

Therefore steric effects mean that only specific orientations will work, and that means the entropy is low (more order needed).  If, in contrast, a collision from any direction would be successful, then any random direction will do, no steric effects, and high entropy.

That is the basic reason.  So entropy is a qualitative measure of order, or disorder,  It is a substance, as tangible as energy.
Chemistry Prof, McGill University, Canada. Co-Author of Physical Chemistry by Laidler, Meiser, Sanctuary. President, MCHmultimedia.com. Interactive e-learning advocate.

Offline ramboacid

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Re: Endothermic and Exothermic Reaction Rates
« Reply #6 on: June 29, 2012, 12:07:34 AM »
Quote
Notice in the above that the pre-exponential factors cancel.  I have assumed they are equal.  That is the collisions are reversible, which is usually a good assumption.

Thank you! That cleared up a lot. I was reasoning under the assumption that the Arrhenius parameters were different for the forward and reverse reactions, which obviously presented problems.

Quote
Therefore steric effects mean that only specific orientations will work, and that means the entropy is low (more order needed).  If, in contrast, a collision from any direction would be successful, then any random direction will do, no steric effects, and high entropy.

I see your point. To overcome steric effects, there must be a high degree of order in the system so the molecular orientations of the reactants work. However, shouldn't the reaction favorability be more dependent on ΔS than the actual entropy of the system? Or can we deduce something about the ΔS of reactions from the entropy of the system?
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Offline Bryan Sanctuary

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Re: Endothermic and Exothermic Reaction Rates
« Reply #7 on: June 29, 2012, 12:47:40 AM »
Earlier I said

"So entropy is a qualitative measure of order, or disorder"

I meant to say "a quantitative measure.."

Sorry about that.
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