Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: Maud on May 31, 2015, 10:35:38 AM
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I can not answer this question... what is it? Does it have to do with the leaving groups?
Thanks a lot in advance!
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Driving force for any spontaneous reaction is is ΔG factor
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Driving force for any spontaneous reaction is is ΔG factor
Yeap, thanks a lot Vidya. The thing is to explain why the ΔG factor is favourable; what is driving the reaction to happen? Why is it more energetically favorable the reaction to proceed to a polymer? I think the first explanation has to do with the length of the leaving groups parting from the tetrahedrical intermediate....
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Favorable and spontaneous do not have the same meaning. If a reaction must be heated to form the product, it can be favorable, yet will not occur spontaneously. What does this mean?
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Favorable and spontaneous do not have the same meaning. If a reaction must be heated to form the product, it can be favorable, yet will not occur spontaneously. What does this mean?
It means that energy needs to be provided for the reaction to occur
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Favorable and spontaneous do not have the same meaning. If a reaction must be heated to form the product, it can be favorable, yet will not occur spontaneously. What does this mean?
It means that energy needs to be provided for the reaction to occur
That's the answer to your question.
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Favorable and spontaneous do not have the same meaning. If a reaction must be heated to form the product, it can be favorable, yet will not occur spontaneously. What does this mean?
It means that energy needs to be provided for the reaction to occur
That's the answer to your question.
Buy, why? Why is it required to provide energy? what is the explanation for that?
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Wiki lists two processes for Pet production: transesterification and esterification
http://en.wikipedia.org/wiki/Polyethylene_terephthalate#Production
for which the driving force must differ, I read for instance:
"Methanol is removed by [permanent] distillation to drive the reaction forward" versus
"Water is eliminated in the reaction and is also continuously removed by distillation".
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Better expressed: "Water is eliminated in the reaction and is also continuously removed by [permanent azeoptropic] distillation".
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Thermodynamics don't work differently for polymerizations than other reactions. The driving force is jargon for the Gibbs energy (well, it does have a real meaning, but most people use it to mean the Gibbs energy).
Therefore, the driving force for polymerization of PET, or any polymer, is a function of the enthalpy change and the entropy change that occurs with each successive addition of a monomeric unit.
Let's take a look at PET.
Each step of a PET polymerization is basically a condensation between a carboylic acid group and an alcohol group to form an ester linkage. This essentially means you are breaking a C-O bond (of the carboxylic acid, say) and an O-H bond (of the alcohol) and forming a new C-O bond (ester) and new O-H bond (water). So you're breaking and forming a C-O bond and breaking and forming an O-H bond. At first glance, this seems like the enthalpy change should break even. But not so! We can call the C-O formation and breaking a wash if you want, but the O-H bond of an alcohol has a bond energy of ~436.0 kJ/mol, whereas the O-H bond of water is quite a bit stronger: about 498.7 kJ/mol. Meaning, based on these very rough estimates, that about 62 or so kJ/mol of heat energy is liberated per mole of polymerization steps.
The enthalpy change of polymerization of PET is exothermic, and negative.
If that's the case, why doesn't polymerization go on indefinitely?
The reason is because polymerization is entropically unfavorable. As the polymer grows, the system becomes more ordered: lots of monomers is more entropically favorable than one giant macromolecule. Eventually you add enough monomers and the rate of adding another monomer just about balances out the rate of removing a monomer from the existing polymer chain. At this point equilibrium is reached and the polymer no longer grows. Note that, because polymerization is entropically unfavorable, increasing the temperature will tend to favor shorter polymers. In fact, in the business there is what is called a ceiling temperature for each polymer type - at this temperature depolymerization is favored over polymerization, and it becomes difficult to grow polymers above this temperature. I'm not sure what this temperature is for PET.
http://en.wikipedia.org/wiki/Ceiling_temperature
So, the answer to the original question is quite simple: the driving force for PET polymerization, like any polymer, is heat liberation - that is, enthalpy. It would also be correct to say that PET polymerization is primarily driven by the fact that the OH bonds of water are more stable than the OH bonds of alcohols/carboxylic acids.
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Thermodynamics don't work differently for polymerizations than other reactions. The driving force is jargon for the Gibbs energy (well, it does have a real meaning, but most people use it to mean the Gibbs energy).
Therefore, the driving force for polymerization of PET, or any polymer, is a function of the enthalpy change and the entropy change that occurs with each successive addition of a monomeric unit.
This problem will prove I am not an expert at thermochemistry. None the less, let me give my opinion, but don't count on it being correct.
Each step of a PET polymerization is basically a condensation between a carboylic acid group and an alcohol group to form an ester linkage. This essentially means you are breaking a C-O bond (of the carboxylic acid, say) and an O-H bond (of the alcohol) and forming a new C-O bond (ester) and new O-H bond (water). So you're breaking and forming a C-O bond and breaking and forming an O-H bond. At first glance, this seems like the enthalpy change should break even. But not so! We can call the C-O formation and breaking a wash if you want, but the O-H bond of an alcohol has a bond energy of ~436.0 kJ/mol, whereas the O-H bond of water is quite a bit stronger: about 498.7 kJ/mol. Meaning, based on these very rough estimates, that about 62 or so kJ/mol of heat energy is liberated per mole of polymerization steps.
The enthalpy change of polymerization of PET is exothermic, and negative.
I agree with the connection between polymerization and any reaction, but that is about all here. I don't want to discuss the bond energies, just the results. As described, it would appear that mixing acetic acid and ethanol will result in an exothermic reaction to form ethyl acetate and water. I don't think this happens. If you wait long enough, a small amount may form, but I don't think you will find anything like that predicted.
According to Wikipedia, polyethylene terephthalate is produced by heating terephthalic acid and ethylene glycol at 220-260°C. This result is more consistent with the expected reagents to produce an ester, namely, an acid chloride or anhydride. Clearly, it would take energy to prepare an acid chloride from an acid if the acid chloride were to be more reactive. Heat would also increase the energy level of the reactants.
If that's the case, why doesn't polymerization go on indefinitely?
The reason is because polymerization is entropically unfavorable. As the polymer grows, the system becomes more ordered: lots of monomers is more entropically favorable than one giant macromolecule. Eventually you add enough monomers and the rate of adding another monomer just about balances out the rate of removing a monomer from the existing polymer chain. At this point equilibrium is reached and the polymer no longer grows. Note that, because polymerization is entropically unfavorable, increasing the temperature will tend to favor shorter polymers. In fact, in the business there is what is called a ceiling temperature for each polymer type - at this temperature depolymerization is favored over polymerization, and it becomes difficult to grow polymers above this temperature. I'm not sure what this temperature is for PET.
http://en.wikipedia.org/wiki/Ceiling_temperature
So, the answer to the original question is quite simple: the driving force for PET polymerization, like any polymer, is heat liberation - that is, enthalpy. It would also be correct to say that PET polymerization is primarily driven by the fact that the OH bonds of water are more stable than the OH bonds of alcohols/carboxylic acids.
If one reacts acetic acid and ethanol with an acid catalyst, one will establish an equilibrium. If water were removed, the ester forms. If water is added, the ester is hydrolyzed. This is a simple le Chatelier's principle.
Without give a thermochemistry answer, but I'd say the driving force for the reaction is the heat being applied.
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Being in accordance with both Corribus and Orgopette, I would add:
Apart the thermodynamic reasons, reaction kinetics also plays an important role in PET polymerization.
Both esterification and transesterification are equilibrium reactions with given equilibrium constants. Thus and according Le Chatelier principle, by continuous removing of water and methanol respectively, the polymerization reaction is pushed to right side with high yields.
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@orgopete
First, it's important not to confuse kinetics with thermodynamics. A reaction can be thermodynamically favorable but take eons to reach equilibrium. Just because you mix two things and see little product conversion does not mean the reaction is (thermodynamically) unfavorable. Heat is certainly an important part of making any reaction go, and its relation to temperature means that it will affect all thermodynamic parameters involved in calculating the Gibbs energy. But I wouldn't call heat itself a driving force for the reaction - a spontaneous reaction will go forward (eventually) regardless of how high the reaction barrier is. So, I would say that while heat certainly affects the driving force, it isn't the driving force itself.
Second, most polymerization reactions employ catalysts to lower the activation barrier and make reactions kinetically favorable. If you just put the starting materials for PET in a flask and heated them, it's likely you wouldn't see a whole lot happen, either.
Third, I did not mean to imply that this was the mechanism by which PET reactions occur. (Although, bear in mind that thermodynamic are state functions, and all that implies.) This was mostly a simplistic demonstration of how enthalpy and entropy play a role in polymerization. It is well accepted that most polymerizations are primarily driven by enthalpy changes, being at the same time entropically unfavorable. There is a delicate balance then in supplying enough heat to surmount a reaction barrier, and not too much to drive the reverse reaction. This is one of the reasons why catalysts are so important for polymerizations.
(Polymerizations are also challenged kinetically, because as the polymer grows, the relative number of available "ends" for reaction decreases.)
Ultimately, "driving force" is a term that most physical chemists formulate using thermodynamic considerations. I believe we are simply using the terms differently. You, and probably many experimental organic chemists, are using it in the context of a stimulus - something you are doing to bring about an experimental end. I heat the reaction flask up, and products are formed. Heat is therefore the force that brings success - like, pushing a ball up a hill. But physical chemists don't use the word in this active, deliberate way. Driving force for us is a term that refers to the factors the bring about a spontaneous change to a system without any deliberate intervention. It's a fundamental property of the system itself, not what you are doing to the system to hurry that change along. If that makes any sense.
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Re: thermochemistry
Corribus is correct in that I may view the driving force of a reaction differently as compared to a pchemist. None the less, I wish to defend my position. For the moment, let us assume esters are actually higher in energy and that their formation were endothermic. Heating the reaction to between 220-260°C could still cause the reaction to occur. I shouldn't even be surprised that a catalyst may not be needed, but a catalyst should only affect the rate of reaction and not the equilibrium. At 260°C, water would be driven from the ester shifting the equilibrium toward product. I would argue the formation of the product is driven by or caused by the high amount of heat being applied.
I acknowledge that heats of formation are commonly used to make predictions as Corribus has done here. You may sometimes see homolytic bond data used to explain heterolytic bond strengths. From some of my studies, I have been rather more skeptical of how this data may be used. For example, if rather than bond energy, I use heat of combustion.
EtOH, -1370.7 Kj/mol
HOAc, -876.1 Kj/mol
EtOAc, -2271.5 Kj/mol
I argue ethyl acetate should have roughly -2246.8 Kj/mol (-1370.7 + -876.1) heat released upon combustion. I don't know that the reported value of -2271.5 Kj/mol is significantly different, nor that it is actually less stable. I only present this data as a different value and one in which I can understand how it might have been collected.
A final note. I had been reluctant to add a comment as anyone may well understand with my thermochemical skills. However, "Driving force for PET polymerizing" just did not match how I thought the reaction actually occurred. Unlike the reaction of an acid chloride, I did not feel there was a strong enthalpic difference favoring the products, catalyzed or not.
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I don't really have much to add to my previous post, which I stand by, but:
You can in principle use heat of combustion values to calculate the same basic thermodynamic endpoints as you can from bond enthalpies or heats of formation. In a perfect world they should all yield the same answer. But these various reaction enthalpies are often measured under different conditions, so you have to pay close attention to where you got them from - and also what you're comparing them to. I'm not sure where you got your heat of combustion values from. NIST, which is just about the best source of information for this kind of stuff, does not list experimental combustion enthalpies for all three substances in the same phase (e.g., gas phase combustion values). Maybe you're getting them from somewhere else? Either way, I'd be wary of using combustion enthalpies for this kind of handwaving analysis if you're not sure how they are measured.
You can easily find gas phase heats of formation for the three organic molecules (plus water) at the NIST webbook - which I regard as far more reliable - and if you use these values, you get an enthalpy of reaction for ethanol and acetic acid condensation of about -20 kJ/mol. In liquid phase I found a value that was still negative (about -6.7 kJ/mol). Not quite as exothermic as the bond enthalpy calculation I did above, but I already admitted that one was incredibly rough. And this kind of estimation is rough to begin with.
And again, this is all aside from the issue of kinetics, which are only indirectly related to thermodynamical considerations.
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PETP is also produced from dimethyl terephthalate reacting with ethylene glycol, in which case any enthalpy change must be tiny, as the reaction replaces an alcohol plus ester by an other alcohol plus ester pair. At least then, removal of methanol looks like the driving force that pushes the first reaction step.
The second reaction step is also a transesterification, where glycol is distilled off.
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As a reminder, do not forget that entropy dramatically decreases during PET polymerization due to the formation of one highly long molecule with low flexibility by condensation of many small monomers and therefore, the free energy might finally be positive.
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Removal of one of the products (water or methanol) certainly helps drive the reaction forward, by lowering the statistical likelihood of a backward reaction event from occurring. You would not be able to form long polymers without this important aspect. You can, incidentally, drive even otherwise unfavorable reactions this way. But this is all kinetics and therefore is not, in the rigorous sense, related to a thermodynamic driving force.
It's hard to measure reaction enthalpies for polymerizations. I can't find a value for PET to settle the issue. Ultimately, it depends on what you mean for driving force. I think we can pretty much agree there is no thermodynamic driving force from entropy. A rough calculation shows there may be a driving force in the positive direction by a negative enthalpy change. I concede that this degree of exothermicity may be small enough to be inconsequential, and therefore polymerization in this case would be strictly kinetically controlled (e.g., via removal of a product during the course of the reaction). The kinetic aspects of polymerization may supercede thermodynamic considerations, no matter what the driving force is.
As a result of the confusion here, I submit that a better title for this thread would have been "Why does PET polymerize?" rather than one that uses a word like driving force, that implies a thermodynamic stimulus.
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[...] highly long molecule with low flexibility [...]
Longer molecules tend to have a higher heat capacity per mass unit than shorter ones. This is because they have vibration modes at lower frequencies, which are excited at a lower temperature hence store heat, for instance at room temperature.
The lower resonance frequency results from a bigger mass and compliance per molecule.
Compare for instance ethane and ethylene at room temperature:
Cp=53J/mol/K and Cp=43J/mol/K (from Air Liquide)
with polyethylene:
Cp=31J/K per mol of -CH2-
http://www.nist.gov/data/PDFfiles/jpcrd178.pdf
The extreme case are metals, where molecules are big enough to grant 3 degrees of vibration freedom to each atom, resulting in 3*RT/mol heat capacity.
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Apart the heat capacity, vibration freedom (rotation freedom) is also a question of polymer rigidity/flexibility. Thus, poly(ethylene terephthalate), [PET] is less flexible than poly(ethylene succinate), [PES] and poly(ethylene adipate), [PEA] but less rigid than poly(p-phenylene diamide terephthalate), [Kevlar].
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Vibration freedom that stores heat in a long chain does not need the freedom to rotate a full turn around a bond: in a molecule big enough, the rotation compliance and inertia sums over many bonds, which suffices to make the mode's energy accessible to room temperature.
The heat capacity goes even the other direction, since vibrations store RT while rotations store only RT/2.
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Question 1: Does the formation of one big molecule from condensation of a great number of small molecules, decrease entropy, Yes or No?
Question 2: Does polymer chain rigidity, decrease entropy, Yes or No?
Question 3: Are all above overcome by heat capacity increasing,Yes or No?
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Q1: you get confused with gases where PV determines many things.
Q2&3: apparently you don't understand that rigid molecules vibrate.
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You are probably right. So, I have to go back to my Physical Chemistry textbook.
Meanwhile, can you answer my questions by a Yes or by a No?
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Meanwhile, I have read in my old Physical Chemistry textbook that entropy is directly associated with the heat capacity, if volume is the only variable of the system that changes when the heat is supplied. But during polymerizations, a plethora of variables might change simultaneously, e.g. the number of moles n, during chain growth, disappearance of the initial H-bonds and creation of different new ones during polyesterification and polytransterification, creation of Van der Waals forces between polymer chains, even changes of the matter state from gas to solid during polyethylene formation, etc. As a conclusion, entropy cannot directly be associated with the heat capacity during polymerization, regardless if gaseous or liquid monomers are involved.
I have also read in my old Physical Chemistry textbook that indeed rigid molecules vibrate, too. But given that PET rigidity is due to the aromatic rings, I have also read in my old Organic Chemistry textbook that the aromatic rings are planar and that aromatic bonds are shorter than single bonds. As a consequence, benzene ring vibrates at a shorter distance and only in two directions. And as a conclusion, the thermal movements of the overall polymer molecule are quite restricted by the rigid benzene ring vibrations, when compared with the flexible cyclohexane ring vibrations.
However, the discussion went too far and dived into the deep waters of Theoretical Physical Chemistry. In terms of practice, the answer of the initial question is:
“The driving force for PET polymerization is the Le Chatelier principle.”