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Bond energies are never equal unless the reactant and products are the same species. Remember that the bond energies that any textbook gives you are the average bond energies. Bond energies of a specific bond usually differ according the the chemical environment.

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No all reactions need thermal energy, or give out thermal energy. If you look at enthalpy, you will see how this works, whether its combustion c, reactant r, product p, or reversible t. some phenol reactions come close requiring very small amounts of energy. In the whole no matter how small that potential/fundamental energy is, it is still there.

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cool, i would like to two coppers too :D



Lets assume that you take some starting material run a reaction and the look at the products. The starting material must be different from the products otherwise we would be unable to tell wether or not there had been a reaction right? (for convienence, i will call the starting material "A" and the products "B" from now on).


Now we know that A and B must be different chemical species correct? But does this nessesarily mean that they are at different energies? NOt really. It could be (though extremely unlikely) that both A and B have exaclty the same energy. Thus, converting from A to B would seem as it this reaction should not liberate heat at all.


So can anyone prove that two chemicals must have different energies?


Just a question. I will have to think about it some too.

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You could still tell if there had been a reaction even if energies were the same, think bonds, think product, think property change. I think Pantano's Rule relate's to thermodynamics of chemical reactants, but take Ab [ab] / [bA]x[Ab] following equilibrium constant(s) in a reversible reaction "theoretically" the same energy would be re-created on its 2nd channel. (aB)--(Ba)---(cD)---(Dc)---(aB). or xy realting to yx.

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You could still tell if there had been a reaction even if energies were the same


That was kinda my point. The chemical species must be different in order for us to say that a reaction has taken place, becuase if the starting materials and products are the same, then we would not know if anything had happened.


But this says nothing (and i am not sure if anything does) about the relative energies of the starting material and products. Specifically, i am not sure that they are nessesarily of different energies. The reason why i bring this us is becuase of this quote...


Bond energies are never equal unless the reactant and products are the same species.


that is all, just a response to that quote. :D

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You cant have a reactant and a product the same they wouldnt react


Not true. Consider the following...


H3O+ + H2O --> H2O + H3O+


This is ment to be proton transer in a acidic solution.

here the reactants and the products are the same, however, we know that proton transfer occurs (or rather, we have strong evidence for this). Thus, a reaction is occuring. The problem is observing the reaction, that is there is no way to garuntee that a reaction has occured, even though it is quite likely that it has. So, though we suspect that there are reaction where the reactants and the products are identical, we have no way of testing this :/


And yes bond energies are never the same relating to principal quantum number(s)


would you care to explain this position more. I do not understand how this statement proves that energies of different molecules must be different.



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you can't lose/gain electrons without adding or subtracting energy, not in a reaction in any case



of course not. But that speaks only to the process of the reaction. While i was talking about the beginning vs ending products. the change in heat or free energy if you will for the reaction. if that is all we are concerned about, then we can ignore the energy that was required or put out durring the reaction, as both delta H and delta G are state variables. that is, the value of delta H and delta G are not dependent on the path you take to get there :D

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of course not.


I am going to take this back now. If we assume that the electron exists in an infintely high energy state within the atom/molecule, then it will take zero energy to strip it away from the molecule. Likewise, if an electron falls into an infinitely high state within the atom/molecule, then it will release no energy in this process.


Of course, the probablility for this happening is somewhat low (understatement of the year) but at least it is theoretically possible. cool, i think that is all for now :D

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Are you trying to tell me H20 and H30 are the same


not at all. the reaction was ment to be an example of a reaction in which the reactants and the products sides were identical. This applies if there is only one product and only one reactant as well.


For example...




in this case, the pairs of hydronuims can swap protons. Now you only have one species on the right and the same species on the left -- yet a reaction could still occur. I think it is fairly clear that delta G for this reaction should be zero. Perhaps i am wrong?


I apologize for any confusion my attempt at gereralization may have caused. But i still stand by my original acertion and that is reactions can in which the reagents and products are indistiuguishable. That is all :D

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just because you don't look at the energy transfer in the middle doesn't mean it isn't there


I think you are missing my point. My point is this; you can have a reaction in which there no heat of reaction. That is, the net delta H for the total reaction is zero. That is what I am saying.


You, however, seem to be addressing the process of the reaction. You are saying (mostly correctly) that stripping away an electron from a neutral atom or bonding orbital of a molecule requires energy, while adding an electron into a bonding orbital or a positive ion will give off energy. Since we move electrons about during the course of a reaction it is fair to assume that energy will be given off or required or both, during the course of the reaction. Like I said this is mostly correct.


However, this does not effect my assertion at all, nor does it affect the example I proposed. My assertion, that delta H for a reaction can be zero deals with the state of the reaction at the beginning and the end. It is not concerned with how you got there. I am not claiming that energy is not moved about during the process, I am merely claiming that when all is said and done the energy that you started with in the reactants is equal to the energy that you end with in the products (chemically speaking -- that is ignoring vibrational, rotational, and translational energy).


Perhaps you are unfamiliar with the idea of a state variable. I will try to give a quick run down of what they are (I apologize if you are familiar with them). Basically, a state variable is a value whose value is path-independent. Elevation is an example of a state variable. That is, if I start at point A and move to point B, then there is a definite difference in elevation between my starting and ending positions. It does not matter how I arrived at point B, the difference in elevation between point A and point B is the same.


It also happens that enthalpy is a state variable. That means that it does not matter what path we take between reactants and products, the change in enthalpy for a certain chemical reaction will be the same.


Since change in enthalpy is path-independent, then we can ignore all processes that occur during the reaction. The way in which the reaction happens cannot change the value of enthalpy change.


Thus, in my reaction that I propose has a delta H of zero, it does not matter what we do during the reaction, if we have a zero change in enthalpy, then we will always have a change in enthalpy of zero. The path of the reaction cannot change this.


So I am not ignoring the energy change in the middle of the reaction. Rather, I am saying that the sum of all the energy changes that occur during the course of the reaction must add up to give us the total change in energy for the reaction as a whole. Thus, when I say that delta H is zero, we can ignore the course of the reaction (and any energy transfers that occur in the middle) as far as thermodynamics are concerned. This is because once we know what delta H is for the reaction, we already know what the sum of all energy changes during the reaction must add up to.


I hope that this helps somewhat in explaining why it is valid to ignore the middle of a reaction when we are talking about things such as change in enthalpy, which are state variables. Thus, I am not claiming that ‘just because you don't look at the energy transfer it isn’t there.’ Rather, I am saying that there is no need to concern yourself with these energy transfers and we are justified in ignoring them.



Now of course this only goes for thermodynamic considerations. If you want to talk about the kenetics of the reaction then that is a whole ‘nother ballgame. Then it is the energy “in the middle” of the reaction that is intimately important to the problem rather than the total change in energy.


Hope that helps :D

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Aside from the hydronium/water example, there's also:


H2O(l) + H2O(g) --> H2O(g) + H2O(l)


Where H2O(l) --> H2O(g) has a delta H of +40.66 kJ/mol


And H2O(g) --> H2O(l) has a delta H of -40.66 kJ/mol


Obviously the sum is 0 kJ/mol.

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liquid to Gas is NOT a chemical reaction, as mentioned above, it`s a mere change of state and need not apply.


the mercury oxide bit needs a little work though, a reaction does actualy take place there. but the 1`st part`s bang on :)

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Ok yeah, I just used it because I didn't know the delta H's of the water/hydronium, and it was just sposed to be an analogy of energy equilibrium. I.e. there's an energy transferance, but not change.

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Activation energy, Bond energy and Transferral energy (including reaction dynamics).


Hydronium ions (H30+), in water is by the process of Grotthuss mechanism, the protons transfer from one water molecule to the next. The transferral is directed to proton motion through H20. Transport of Hydronium ions in ice is very close say directionally proportional, to H30 transport in water. If Hydronium molecules are supposed to extinct movement at 190 K (theory), then this implies that H30 motion that is an activated process. Similar to the defect in the Hydrogen-bonded structure (the D defect).


Energy change, the result of breaking and forming bonds, the energy to break a bond is known as bond-dissociation energy. When this energy, needed for bond-dissociation is less than the energy produced by the bonds formed the reaction will give of energy. An exothermic reaction, is the process of releasing energy, so the P (product) bond-dissociation is greater than the R (reactant) bond-dissociation. Whereas an Endothermic reaction absorbs energy, the R bond-dissociation energy is greater than the energy released, by the P bond dissociation.


This energy change in a reaction, is known as the enthalpy of reaction, the symbol is H, the change of enthalpy is delta H, /_\H. Exothermic reactions, have negative enthalpy change, the molecules, atoms and bonds loose energy to the surroundings. On the other hand Endothermic reactions, have positive enthalpy change, the molecules, atoms and bonds absorb (gain), energy from their surroundings.


Law of conservation of energy IS the amount of energy in the universe id CONSTANT. Energy is ONLY transferred in chemical REACTIONS.


Entropy, S, is the (m) measure of chaos, large values for entropy, are equal to large amounts of disorder, this is not directed to the conservation law.


/_\G = /_\H - T/_\S


/_\G Free energy, Gibbs theorizing, the concept of free energy and entropy, it’s unit is calorie or kilocalorie, this energy is used for a specific process and is available from a process (a previous post of mine shows Gibbs equations).




All chemical REACTIONS, are accompanied by change in energy, EVEN in exergonic and endergonic, Reactant + reactant = Product + product + ENERGY RELEASED, the energy is still there.

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