Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: Big-Daddy on April 27, 2013, 05:09:27 PM
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I was under the impression that only reaction equations where the reactant and product are the same had ΔHr°=0. But apparently this reaction also does:
[35Cl]2+[37Cl]2 ::equil:: 2 [35Cl-37Cl]
Under what conditions is ΔHr°=0 and why?
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I don't necessarily disbelieve you, but where did you find that the reaction enthalpy for this reaction is exactly zero?
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I don't necessarily disbelieve you, but where did you find that the reaction enthalpy for this reaction is exactly zero?
Where else but an IChO past question? ;)
There's an appealing question where they ask you to work out the K of this reaction with no other information - without the abundances of 35Cl and 37Cl I was clueless for a few moments, but then I tried it and something nice falls out: let f[35] be the fraction of Cl atoms which are 35Cl and f[37] the fraction which are 37Cl. Then the molar fractions are f[35]2 for [35Cl]2, f[37]2 for [37Cl]2 and 2*f[35]*f[37] for [35Cl-37Cl]. Now place this into the K expression and (because we square the numerator) we'll get f[35]2 and f[37]2 cancelling and K=22=4 regardless of the abundances.
Sorry for that digression - it's an introduction to the bit I'm stuck on, which was when they asked us to calculate the entropy change per mole of this reaction. Obviously I went for the K=e^(-ΔGr°/(RT)) approach, but without being given ΔHr° I was clueless on how to proceed. The mark scheme then said that ΔHr°=0. After this we just sub in ΔGr°=-ΔSr°*T so K=e^(-(-ΔSr°*T)/(RT)) and ΔSr° comes out as 11.53 JK-1mol-1.
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There's an appealing question where they ask you to work out the K of this reaction with no other information - without the abundances of 35Cl and 37Cl I was clueless for a few moments, but then I tried it and something nice falls out: let f[35] be the fraction of Cl atoms which are 35Cl and f[37] the fraction which are 37Cl. Then the molar fractions are f[35]2 for [35Cl]2, f[37]2 for [37Cl]2 and 2*f[35]*f[37] for [35Cl-37Cl]. Now place this into the K expression and (because we square the numerator) we'll get f[35]2 and f[37]2 cancelling and K=22=4 regardless of the abundances.
I didn't get this part. Can you elaborate? I don't see how you know K=4.
Maybe you are right but I don't see it yet.
What about other reactions of this same form. Or other isotopes. Your logic seems to say they will all have K=4?
My intuition says, your K will be some function of the translational, vibrational and rotational partition functions.
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Under what conditions is ΔHr°=0 and why?
To an approximation I can conjecture why ΔHr°=0: Isn't chemical bonding an electronic property. Therefore your left and right sides are essentially the same.
It's like saying a tank of Cl2 now and 24 hours later. Enthalphy hasn't changed even if some molecules exchange atoms mutually (assuming you indeed had some way to label atoms!)
OTOH, in a quantum sense I don't see why ΔHr°=0 in an exact measurement.
Vibrations of a 35-35 molecule are most assuredly different from a 37-37 molecule. So Zero Point Energies will be different. Unless there's a way to prove that everything compensates precisely. I don't know.
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Vibrations of a 35-35 molecule are most assuredly different from a 37-37 molecule. So Zero Point Energies will be different. Unless there's a way to prove that everything compensates precisely. I don't know.
No you are right, ZPE is different, which one of the reasons why the equilibrium constant isn't equal to exactly 1 (although it's usually close), at least for most isotope exchange reactions I'm familiar with. ΔS is, I believe, also nonzero, because scrambling of the isotopes is more "disordered". The change in ZPE is why isotopic substitution impacts reaction kinetics (kinetic isotope effect). So ΔG is definitely nonzero.
In addition, while adding a neutron doesn't affect the electrostatic properties of an atom, it CAN, rather ironically, affect the electrostatic properties of a molecule by causing asymmetries in a vibration. For instance, carbon dioxide has a dipole moment of zero when both oxygen atoms are the same isotope. However if one of them is heavy, the molecule has a slight dipole moment because of the asymmetry of the "symmetric" stretch. (See: http://jcp.aip.org/resource/1/jcpsa6/v39/i12/p3490_s1?isAuthorized=no). Differences in polarity would also obviously affect ΔH. Although this wouldn't apply to a diatomic molecule I don't think.
Anyway, this is why I asked. I couldn't find any real values for ΔH for this reaction. ΔH does seem to be nonzero for the similar H2 + D2 :rarrow: 2HD reaction (although I can't tell for sure - can't look at many actual papers from here at home because I don't have journal access). My intuition says it can't be exactly zero, but even if it's not, it's probably a decent not a bad approximation, provided that the gasses behave ideally.
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Vibrations of a 35-35 molecule are most assuredly different from a 37-37 molecule. So Zero Point Energies will be different. Unless there's a way to prove that everything compensates precisely. I don't know.
No you are right, ZPE is different, which one of the reasons why the equilibrium constant isn't equal to exactly 1 (although it's usually close), at least for most isotope exchange reactions I'm familiar with.
Is it close to 1? Or 4? I thought it was 4.
Perhaps I am confusing notations.
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There's an appealing question where they ask you to work out the K of this reaction with no other information - without the abundances of 35Cl and 37Cl I was clueless for a few moments, but then I tried it and something nice falls out: let f[35] be the fraction of Cl atoms which are 35Cl and f[37] the fraction which are 37Cl. Then the molar fractions are f[35]2 for [35Cl]2, f[37]2 for [37Cl]2 and 2*f[35]*f[37] for [35Cl-37Cl]. Now place this into the K expression and (because we square the numerator) we'll get f[35]2 and f[37]2 cancelling and K=22=4 regardless of the abundances.
I didn't get this part. Can you elaborate? I don't see how you know K=4.
Maybe you are right but I don't see it yet.
What about other reactions of this same form. Or other isotopes. Your logic seems to say they will all have K=4?
My intuition says, your K will be some function of the translational, vibrational and rotational partition functions.
Let f[35] be the fraction of all Cl atoms which are of the 35Cl form, and f[37] be the fraction of all Cl atoms which are of the 37Cl form. What are the relative abundancies, in the natural world around us, of [35Cl]2, [37Cl]2 and [35Cl-37Cl]? The answer is: f[35]2, f[37]2 and (because there are 2 possibilities) 2*f[35]*f[37] respectively.
Now these are the abundances of the different forms, in the natural world, which we can suppose is at equilibrium; thus:
K=(x[[35Cl-37Cl]])2/((x[[35Cl]2])*(x[[37Cl]2]))
K=(2*f[35]*f[37])2/((f[35]2)*(f[37]2))
K=22*f[35]2*f[37]2/(f[35]2*f[37]2)
K=22=4
Regardless of what f[35] and f[37] are.
Yes any isotope exchange of the form A2+B2 ::equil:: 2AB should follow this same logic (provided we assume nature - our source of the abundances - has reached equilibrium). At least I can't see a flaw in it ... and the mark scheme says it is correct ... but perhaps my explanation/understanding is wrong on a deeper level.
Edit: Maybe I should have defined K as Kx here? Though I don't understand what difference it would make really ...
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I feel something is amiss here ... the logic seems sound by surely it isn't true that every reaction of form A2+B2 ::equil:: 2AB has K=4? (indeed every reaction of form An+Bn ::equil:: nAB has K=nn).
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BD,
Again it's really hard to follow your train of thought with equations formatted this way. But from what I can tell you are trying to calculate K from abundances of the isotopes. However the equilibrium constant isn't affected by how much "stuff" is available to react. It's determined from the thermodynamics of the reaction. Thus it should be obvious by now that you can't generalize the equilibrium constant just by the stoichiometry or class of reaction.
For instance -
I couldn't find thermodynamic quantities for the title reaction of this thread, but data for this reaction are widely available
H2 + D2 --> 2HD
This is another isotopic substitution reaction. Standard heats for formation and standard entropies are available here:
http://cccbdb.nist.gov/
We can see the standard heats of formation for H2, D2 and HD are 0, 0 and 0.32 kJ mol-1, respectively, and the standard entropies are 130.600, 144.960 and 143.799 J K-1 mol-1, respectively. The entropies are at 295.15 K and I think the enthalpies are at the same.
Anyway from the data it is easy to calculate that ΔHf0 = 0.64 kJ mol-1 and ΔS = 12.038 J K-1 mol-1. Two comments here: the enthalpy change is obviously not zero. The reaction is endothermic, and probably this is mostly due to the difference in zero point energies between the reactants and products (more on that in a minute). Second, the enthalpy change is slightly positive, again as we'd expect because there are more possible combinations for HD than there are for the "pure" substances. Order increased, more entropy.
Now you know where we're going next. Standard Gibbs energy change for the reaction can be calculated from
ΔG0 = ΔHf0 - TΔS0
Easily it is shown that ΔG0 = -2.95 kJ mol-1 at 298.15 K, meaning the reaction is spontaneous (despite being endothermic - the reaction is entropically driven at 298.15K; you can see that the entropy term is much larger in scale than the enthalpy term). And finally we can calculate K from
ΔG0 = - RT ln K
K is found to be 3.29. Which means that if you start with 1 mole/L each of the reactants, at equilibrium you'll have approximately 0.35 mol/L of the starting materials and about 1.29 moles/L of the product. (And of course using the reaction quotient and calculating ΔG, you can determine the direction the reaction will go from any relative starting concentrations of products and reactants).
You see here that K quite obviously isn't 4.
I surmise that the K value will be a little lower for the reaction you inquired about that that calculated for this one with hydrogen isotopes (a little much closer to 1) because the difference in reduced mass between the starting materials and products will be smaller (mass difference between H2 and D2 is pretty large, relatively speaking). However as this is an entropically driven reaction, I think the K values may still be fairly close, because the reduced mass differences will probably mostly affect the enthalpy term.
We can also get an idea of the thermodynamics of this reaction based on the zero-point energies (ZPE). ZPE is the energy of the lowest lying vibrational state. To break a bond, you have to supply the energy difference between the ZPE and the transition state. (Actually, you have to supply less than this because not all molecules are in the lowest state, but we can still get an estimate of what's going on by pretending all molecules are in the lowest lying state.) Thus, to take a mole of H2 and a mole of D2 to form 2 moles of HD, you have to supply enough energy to bring 1 mole of H2 from its ZPE to the transition state and 1 mole of D2 from its ZPE energy to the transition state. Likewise, when we form 2 moles of HD, we gain an energy equivalent to the difference between the transition state energy and the ZPE of HD (times 2). The overall difference in energy is an approximation of the enthalpy change of the reaction, assuming the gas is dilute and behaves ideally.
Thus if the transition state energy is T, we have an approximation for DH:
ΔH = (T - ΣZPE{reactants}) - (T - ΣZPE{products})
Here, ΣZPE{reactants} = ZPE{H2} + ZPE{D2
and
ΣZPE{products} = 2*ZPE{HD}
Note here that the transition state energy cancels out so we get that ΔH is only dependent on the differences in the ZPEs for the reactants and products.
ΔH = - ΣZPE{reactants} + ΣZPE{products}
The sign on the ZPE for reactants is negative because it costs us energy to break bonds, and the sign on the ZPE for products is positive because we gain energy from forming them (remember, the sign convention here is opposite what you'd expect - a lower ZPE means MORE energy to break the bond because the bond energy is the difference between this number and the transition state energy, which is very large, relatively speaking): the net energy cost/gain (enthalpy) will be positive if the products have a higher combined ZPE than the reactants and vice-versa.
The ZPE for any species is equal (in the harmonic oscillator approximation) to be 0.5*hcν, where h is Planck's constant and ν is the vibrational frequency in wave numbers (inverse meter).
Again we turn to NIST for the vibrational frequencies and we find that ν for H2, D2 and HD is 4401 cm-1, 3116 cm-1 and 3813 cm-1 respectively. (Note, I found out later that the NIST pages actually supply ZPE values for you, so deriving them from vibrational frequencies isn't necessary - results are the same though.) You can see here that the frequency of HD is not exactly in between the pure substances, reflecting the fact that we deal with REDUCED mass, not total mass - this is the primary reason we will find that there is a nonzero ΔH in just a minute.) If we do a little math, we can find that the ZPEs for H2, D2 and HD are predicted to be 26.31, 18.63 and 22.79 kJ mol-1, respectively. Which leads to a ΔH of 0.656 kJ mol-1.
The ΔH value we calculate by this method is very close to that predicted from the heats of formation values. Any deviation would be because we have made a number of approximations (ideal gas behavior and especially at nonzero temperatures, we will have many states populated above the zero-point level; also we've approximated as harmonic oscillator) but you can see the approximations are really good. Point is that here again the ΔH value is predicted to be nonzero AND the equilibrium is not 4 (and it's not 1).
So while this wasn't the reaction you asked about, I think it is instructive in many ways. I don't think the ΔH value for the chlorine isotopic scrambling reaction will be exactly zero. I think it will be close to zero, but close isn't exact! Equilibrium constant I predict will be a little smaller than the one for the hydrogen reaction, but it will depend a lot on the entropy term. I did all of the above pretty quickly, so I hope there aren't too many errors. :D
Anyway, I think that's a lot to digest, but if you have any questions, you know where to find me. :)
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the logic seems sound
Nope. I think your reasoning to deduce K=4 does not make sense.
It it fortuitous that you get that number 4.
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You explained that the logic isn't sound but not why it isn't sound. It seems to make sense that, in nature, there is an established equilibrium between the isotopes, and because this equilibrium exists in nature we can ascertain the abundances of each isotope (which means that, given the abundances of each isotope in nature's equilibrium, we should be able to find the equilibrium constant in the method I showed). Although I understand why and how you came up with the K values you did (using the enthalpies, entropies and Gibbs' energy), and this must be right, I cannot see the logical flaw in the path of thought I am following.
I have attached the original question (Problem 0, starting on Page 2) along with its solutions - perhaps you can shed some light on what I have misinterpreted?
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Think of it this way. Suppose through an experiment I create a new isotope of chlorine that's never been seen before. The natural abundance is obviously zero. Are you telling me this should affect the way it reacts in an isolated vessel in my lab when combined with an equal quantity of a naturally occuring isotope?
(Recall, natural abundance is affected by a lot of things - decay reactions, for instance - so it's not necessarily an isolated equilibrium, as you're making it out to be.)
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Think of it this way. Suppose through an experiment I create a new isotope of chlorine that's never been seen before. The natural abundance is obviously zero. Are you telling me this should affect the way it reacts in an isolated vessel in my lab when combined with an equal quantity of a naturally occuring isotope?
(Recall, natural abundance is affected by a lot of things - decay reactions, for instance - so it's not necessarily an isolated equilibrium, as you're making it out to be.)
Hmm I'm not completely convinced. By the logic of the second point, the natural abundances should be changing; the equilibrium is more a measure of where the abundances are (and using the abundances we might therefore get the constant) rather than something defining their position (since the latter in this case would be backward justification). As to the first, I suspect you cannot predict the isotope would instantly turn into its more commonly seen neighbour, but I lack lab experience to confirm this; and logically it doesn't seem unsound, given that we see none of the isotope in the world (because it has all been converted out to the others - and that is why the natural abundance is 0), as you say it is a "new isotope". But my view also seems lacking to me. :p
Perhaps if you look at the paper I attached and explain why the situation is different, or even where they might have made a mistake, it will make things clear for me. Because this was an unused problem from IChO 2008, it was not scrutinized in detail by hundreds of professors like IChO problems usually are and thus there may be a mistake in the solutions.
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As to the first, I suspect you cannot predict the isotope would instantly turn into its more commonly seen neighbour, but I lack lab experience to confirm this;
Every isotope has a half life. Suppose @Corribus has created a hypothetical Cl isotope with a t1/2 of 10,000 years.
How could it instantly turn into anything else? Radioactive decay (for most purposes) is a nuclear phenomenon and absolutely unconcerned with what else is in the mixture.
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I just read the IChO problem. Nothing wrong with that problem at all. But there's critical assumptions there that they explicitly mention that you left out.
I quote verbatim:
(1) The difference in the chemical behaviour of various isotopes is usually negligible, unless the relative change in the molecular mass is considerable.
(2) What are the mole fractions of these species in bromine at natural abundance
(3) What is the standard molar entropy change associated with this reaction, supposing that the chemical behaviour of the molecules involved is identical?
With those assumptions everything's fine! But those are assumptions. Givens. Not something that can be inferred.
All of @corribus's great exposition on ZPE's etc. was exactly what Point #1 was trying to tell you. So also the bit about natural abundance.
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I just read the IChO problem. Nothing wrong with that problem at all. But there's critical assumptions there that they explicitly mention that you left out.
I quote verbatim:
(1) The difference in the chemical behaviour of various isotopes is usually negligible, unless the relative change in the molecular mass is considerable.
(2) What are the mole fractions of these species in bromine at natural abundance
(3) What is the standard molar entropy change associated with this reaction, supposing that the chemical behaviour of the molecules involved is identical?
With those assumptions everything's fine! But those are assumptions. Givens. Not something that can be inferred.
All of @corribus's great exposition on ZPE's etc. was exactly what Point #1 was trying to tell you. So also the bit about natural abundance.
I see, sorry. I should have dropped in the problem verbatim (reason I avoided this is because I only needed help with a later part, and didn't want you to be discouraged from helping).
By the same logic, though, my point about K for the reaction I listed in the OP being 4 also works, at natural abundance. Indeed, for any isotopic exchange reaction, it is 4 at natural abundance. Problems with the logic in there? I suppose that was something which I failed to state, but I was aware of.
1) and 3) raises my issue. "the chemical behaviour of the molecules involved is identical" can you explain how this necessarily means that ΔHr°=0? Because it's that which I don't understand.
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1) and 3) raises my issue. "the chemical behaviour of the molecules involved is identical" can you explain how this necessarily means that ΔHr°=0? Because it's that which I don't understand.
Your reaction's of the form:
A2 + B2 ::equil:: 2 AB
If A=B (=X , say) (follows from " chemical behaviour of the molecules involved is identical" ) then the reaction becomes:
X2 + X2 ::equil:: 2 XX
i.e.
2 X2 ::equil:: 2 X2
This, almost tautologically, has to have ΔHr°=0. (or for that matter ΔY =0 where Y is essentially any "strictly chemical" property. I'm not so sure about the last one though. Let's see if someone objects.. )
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1) and 3) raises my issue. "the chemical behaviour of the molecules involved is identical" can you explain how this necessarily means that ΔHr°=0? Because it's that which I don't understand.
Your reaction's of the form:
A2 + B2 ::equil:: 2 AB
If A=B (=X , say) (follows from " chemical behaviour of the molecules involved is identical" ) then the reaction becomes:
X2 + X2 ::equil:: 2 XX
i.e.
2 X2 ::equil:: 2 X2
This, almost tautologically, has to have ΔHr°=0. (or for that matter ΔY =0 where Y is essentially any "strictly chemical" property. I'm not so sure about the last one though. Let's see if someone objects.. )
Ah ok ... so if the chemical behaviour is said to be equal then all the molecules can be called the same thing. In that case of course there is no change happening. However if chemical behaviour being equal means we can write down all the molecules as the same to reach 2 X2 ::equil:: 2 X2, then our entropy change should also be 0. I thought the physical properties might not be included when it says "chemical behaviour is equal" - so then do the physical properties have no relationship (negligible) to enthalpy, meaning that only chemical properties affect enthalpy (to within a good approximation)?
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And the equilibrium constant MUST be 1 in that case. (Because ΔG0 = 0)
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And the equilibrium constant MUST be 1 in that case. (Because ΔG0 = 0)
In the case where chemical properties are the same across all molecules so ΔH°=0, and ΔS°=0 (i.e. physical properties are the same), you mean? Whereas the calculation will go as in the IChO problem solution if only ΔH°=0 as only chemical properties are the same (differences are negligible).
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As I said in my earlier, longer post while ΔH isn't exactly zero, it's not a terrible approximation. However it IS a terrible approximation for the entropic term. But I don't think I'd make a distinction that one represents chemical properties and one physical properties. At this point, what is the difference between chemical and physical?
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As I said in my earlier, longer post while ΔH isn't exactly zero, it's not a terrible approximation. However it IS a terrible approximation for the entropic term. But I don't think I'd make a distinction that one represents chemical properties and one physical properties. At this point, what is the difference between chemical and physical?
I think there is a distinction and that's what makes ΔH=0 and ΔS≠0.
Let's assume perfect chemical identity. That's not a bad assumption for something like Cl or Br with a large electronic shell. Let's also assume ZPE changes are negligible. That's mean no differences in any sort of bonding, attraction etc. Ergo ΔH=0.
But K=1 is not a good assumption. (I think) In fact, look at experimental values of K for diatomic isotopic exchange. Except H where ZPE etc. deviations are relatively large, K is indeed always very close to 4. Not 1. Empirically.
Hence I think it's perfectly reasonable to use ΔH=0 and ΔS≠0 both at the same time. Hence ΔG≠0.
The entropic question is an interesting one. I think that is a variant of the Gibbs paradox. Normally Cl2 molecules are indistinguishable. That is key to the derivation that tells us that mixing two containers of Cl2 will lead to no entropy change (unlike mixing two different gases where there indeed is an entropy of mixing).
Opening or closing a shutter between two partitions of Cl2 leaves the system unchanged. (since there is no way to label Chlorine molecules)
OTOH isotopic labeling allows us a way to make Cl2 molecules distinguishable! That minor change by itself is sufficient to lead to an entropy of mixing. Distinguishibility matters. In a way it's a bit weird and counter intuitive. It's like just being able to color some molecules differently will change entropy even if mass, forces and all other interactions remain the same.
Again, I'm a tad speculative here. I might be wrong.
PS. On second thoughts the physical / chemical distinction I made earlier is arbitrary. I just found the fact that Entropy can change merely be labeling very interesting.
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What I meant was - calling enthalpy "chemical property" and entropy "physical property" is a strange distinction to make, because entropy obviously affects the outcomes of chemical change (i.e., reaction). I agree with pretty much everything else you wrote.
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What I meant was - calling enthalpy "chemical property" and entropy "physical property" is a strange distinction to make, because entropy obviously affects the outcomes of chemical change (i.e., reaction).
Agree.
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As an aside, it'd be interesting to try and get the same answer for ΔS using a more direct approach. i.e. counting possible configurations etc.
The equilibrium approach is an interesting shortcut.
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What I meant was - calling enthalpy "chemical property" and entropy "physical property" is a strange distinction to make, because entropy obviously affects the outcomes of chemical change (i.e., reaction). I agree with pretty much everything else you wrote.
How about enthalpy being the "chemical properties" and entropy "both physical and chemical properties"? In other words, only if there really is no reaction going on is the entropy change 0 (i.e. A ::equil:: A), whereas the enthalpy change is 0 to within a good approximation if the constituents are merely chemically equivalent?
I'm not sure what else to take away from this problem ...
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Also, if we were looking at the molar fraction of ozone at natural abundance. Let's say we specify that 85% of O is 16O and 15% of O is 18O isotopes. Then is it reasonable to say that the abundance of [18O-16O-18O], the molecule with 2 18O atoms and 1 16O atom, will be (3!/2!)*0.152*0.85, and that of [16O-18O-18O] will be (3!/2!)*0.852*0.15? It seems to be a stats problem of how many arrangements there are of each molecule, multiplied by each atom in the molecule which has been raised to the power of how many times it shows up in the molecule (if we're looking at the bare maths).
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Also, if we were looking at the molar fraction of ozone at natural abundance. Let's say we specify that 85% of O is 16O and 15% of O is 18O isotopes. Then is it reasonable to say that the abundance of [18O-16O-18O], the molecule with 2 18O atoms and 1 16O atom, will be (3!/2!)*0.152*0.85, and that of [16O-18O-18O] will be (3!/2!)*0.852*0.15? It seems to be a stats problem of how many arrangements there are of each molecule, multiplied by each atom in the molecule which has been raised to the power of how many times it shows up in the molecule (if we're looking at the bare maths).
Yes, I think.
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How about enthalpy being the "chemical properties" and entropy "both physical and chemical properties"? In other words, only if there really is no reaction going on is the entropy change 0 (i.e. A ::equil:: A), whereas the enthalpy change is 0 to within a good approximation if the constituents are merely chemically equivalent?
Semantics. It's confusing, probably wrong and very unlikely to be useful.
I'm not sure what else to take away from this problem ...
I say you've pretty much flogged it for all it's worth.
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Sorry to bring this up again, but I was reading over it and it's thrown up a few more queries.
1) I don't think I understand the point about isotopic exchange reactions having K=4 when we take the assumption that ΔHr°=0. You pointed out that this is because it was "at natural abundance", but natural abundance only shows that the system has reached equilibrium. So then, for any isotopic exchange reaction, the constant is independent of isotopic abundances after all (so long as these abundances are at equilibrium, they just cancel out, so we do not actually need the abundances to derive the constant). I showed this math a few posts ago (sorry it wasn't in LaTeX, it was a multi-step thing).
2) curiouscat wrote a few posts above:
it'd be interesting to try and get the same answer for ΔS using a more direct approach. i.e. counting possible configurations etc.
I assume by "configurations" you mean isotopic combinations (as opposed to microstates or something quantum, which is probably better avoided for me). How would we go about doing this?