Longview, please read lenr-canr.org/acrobat/OrianiRAthephysica.pdf
Thanks for this interesting and informative reference! I'm traveling, so only a good time to read but not so much to reply.
My current favorite reference for this: Yuh Fukai's The Metal Hydrogen System, 2nd Edition, Springer, Berlin, 2005. See tables on page 16 and p. 38. The aggregated data presented portray surprises, for me at least.
My current favorite reference for this: Yu Fukai's The Metal Hydrogen System, 2nd Edition, Springer. See tables on page 16 and p. 38. The aggregated data presented portray surprises, for me at least.
I have the first edition, and there seems to be some significant differences, with more pages present in the second edition, so I don't know what tables you refer to specifically. I suspect they are Table 1.2 on pg. 15 and Table 1.6 on page 37 (1st ed. pgs.).
Please note that in section 1.4, Formation of Hydrides, p. 21 (1st ed.) Fukai writes "because they are practically the only ones that are occupied by H atoms." and on p. 22, "in order to locate the position of H atoms." - not molecular (diatomic) hydrogen...
Just to be clear, if Fukai writes VH2, he in not implying molecular hydrogen in vanadium, he is explicitly stating the chemical composition of 2 H atoms per unit cell of V. Since V is elemental, there is one metal atom per unit cell.
Have we cleared up the problem yet? Do you follow that molecular H2 will not be found inside bulk metal unless it is in a void of some sort?
I don't see that the energetics issues I raised are completely satisfied here. But, I can use an analogy / model to help me or any others "get over it". That model is the semiconductor "hole", that for many practical purposes stands as a positron, that is for charge and for mass. And indeed the "hole" is itself interstitial to metallic or semi-metallic matrices. The difference from a real positron is that this virtual positron has a much lower energy of formation and energy of annihilation. The most energetic annihilation of such an interstitial positron is currently a photon of around 200 nm (eg. the most energetic photon from a UV LED, several eV) .... that of course much less energy than a vacuum electron / positron mutual annihilation of 2 gamma photons at 511 keV each.
That model may be OK to intuit the enthalpy issues I raised, but it has the caveat that the "hole" is virtual, but the interstitial atomic hydrogen is presumably "real".
@'THHuxley
In metallic contexts effective masses for electrons are also measured and widely reported. These are apparently vectorial and are quantified by various means. It seems reasonable at least from your "bubble" and "absence" notion of holes to foresee that the effective mass in the vector coincident to a gravitational field (aka "weight") would be opposite in sign to that of an electron.
As you likely know, in a metallic or semi-metallic context, electrons have properties that are also "virtual".
Arranging for a metallic electron to appear fully formed at a surface.... no problem. Arranging to have a hole become an extra-metallic positron.... big problem.
For "atoms" of hydrogen in such metallic or semi-metallic contexts to appear at a surface as recombined H molecules.... big problem or not?
Appreciate the interest, Longview
Display Morehat is to say catalysis enables both the forward and the reverse reaction (no problem there, with a such a simple binary reaction). There is a large enthalpy involved, regardless of the catalysis. It takes about 486 kJ/mole to break molecular hydrogen to its atomic constituents. That energy is not an activation energy, it cannot be overcome by catalysis alone. That energy must be supplied as actual work. Either the Pd is not breaking the hydrogen-hydrogen bond, or something already much stranger than LENR / CF is happening.
kirkshanahan wrote:
Nope...H2 dissociation is very definitely exothermic. I gas load a variety of hydride materials and I always monitot the temperature rise that I get when exposing an empty material to H2. Sometimes the rise can be near to 100 degrees C (depends on a variety of parametes of course).
If this is so, then it must be an entropic function, that is the molecular hydrogen is yielding up degrees of freedom and hence the observed exothermy. No chance that atomic hydrogen formation in a gas / gas reaction is exothermic... look up in standard tables.
Your 486 kJ/mole *H* (make sure you always specify whether you are talking about 'per mole H', or 'per mole H2') is roughly correct. As you say, that energy must be supplied to break the diatomic hydrogen molecule into separate H atoms. That means it is endothermic and on an energy diagram, the state of "2H" lies well above that of "H2" (whose state is at 0 by definition). The heat of formation (enthalpy) of PdHis roughly 9.8 kcal/mole H2, by convention that value is negative, since H2 absorption is exothermic. Therefore PdH formation at the beta phase composition (~.67 H/Pd ratio) is spontaneous thermodynamically. As long as the reaction is not kinetically limited, it will proceed to form Pd-H. Entropy contributes of course, but it is given as ~22-23 cal/mole H2 and isn't too important to the overall exothermicity of the reaction. IOW, Pd + H2 spontaneously forms PdH at appropriate H/Pd ratios. The energy theoretically needed to break the H2 bond is returned plus some when the PdH is formed.
As noted previously there is no evidence for molecular H2 in the bulk Pd, just in voids or internal bubbles.
Entropy contributes of course, but it is given as ~22-23 cal/mole H2 and isn't too important to the overall exothermicity of the reaction
To the contrary: As you (Kirk) likely know, the relevant equation for coalescing enthalpy (delta H) with entropy (delta S), is "delta G = delta H minus T delta S" (delta G being Gibb's free energy, sometimes designated "delta F")
Your (Kirk's) given figure for entropy "delta S" is in calories (per degree K), so first let's metricate it. I'll take your figure as "22-23 cal/mole" times 4.184 to metricate to joules, getting 92 to 96 joules per degree K. So at 298 K (25°C, see below) giving a "minus T delta S" term of -27.4 to -28.6 kilo J/mol at room temperature.
Thus it is a considerable component of any negative (heat yielding) delta G, and that itself would rise in proportion to absolute temperature.
For comparison here is the earlier and appreciated quote given by you from Flanagan et al:
"Differential heats of absorption of hydrogen by palladium have been measured with an adiabatic calorimeter. For the first time calorimetric heats have been determined for bulk palladium in the single phase, α and β, regions....
The heat of absorption in the α-phase (25°C) was found to be 23.32 ± 4.45 kJ (mol H2)–1 and in the β-phase the isosteric heats decrease with hydrogen content from 46.5 (H-to-Pd = 0.61) to 27.3(H-to-Pd = 0.69).
Giving us every indication that the exothermicity you (Kirk) are citing is likely to be completely entropic in origin, as I had initially indicated:
If this is so, then it must be an entropic function, that is the molecular hydrogen is yielding up degrees of freedom and hence the observed exothermy. No chance that atomic hydrogen formation in a gas / gas reaction is exothermic... look up in standard tables.
And finally, a bit of a bone here:
Your 486 kJ/mole *H* (make sure you always specify whether you are talking about 'per mole H', or 'per mole H2') is roughly correct.
I clearly indicated "molecular hydrogen". I'm avoiding some warranted sarcasm here. Perhaps there are good reasons journal reviewers are rejecting some of your efforts. I'd be glad to preview, as some others here might also.
Giving us every indication that the exothermicity you (Kirk) are citing is likely to be completely entropic in origin, as I had initially indicated:
Longview: May be You sould once write down the complete equation: First splitting H2 then absorobing. After that the fog will dissapear...
Longview: May be You sould once write down the complete equation: First splitting H2 then absorobing. After that the fog will dissapear...
Apparently not so straightforward as a fog buster, that is essentially where my whole discussion began. That is, splitting H2 requires a plus (input) of 436 kJ/mol of real work, which far exceeds the minus (output) TdeltaS entropic Gibbs term (data now in hand from Flanagan, thanks to Kirk). Early on here I speculated that that real work of H2 dissociation might be supplied electrolytically under some circumstances... and that possibly then the return as heat of that minus 436 kJ/mol energy on re-association of the claimed atomic hydrogen would be impressive and might even explain some or many "heat after death" or HAD observations.
Now with Flanagan's credible Pd/H entropy numbers in hand, it appears that the delta H (enthalpic) portion of Gibbs is, in a passive Pd/H or Pd/D system, very modest or even non-existent. No surprises and most important it affirms for me that there is no need for highly unlikely "something-for-nothing" energy yield of say "burning" hydrogen in palladium.
I still have yet to see evidence, pro or con, as to whether it is possible that work energy might be stored as lattice embedded atomic hydrogen from say the electrolytic phase of a Pd / D F&P type experiment.
Any discussion of "Gibbs free energy" in a good, modern college chemistry text should shed a clear light on the principles I've been referring to.
Thus it is a considerable component of any negative (heat yielding) delta G,
OK, you are right, and I understated the entropic portion's importance...see...I do admit my mistakes...
Giving us every indication that the exothermicity you (Kirk) are citing is likely to be completely entropic in origin, as I had initially indicated:
What's funny is you chastize me for neglecting entropy and then go and make the exact opposite mistake. As you noted DeltaG = DeltaH -T*DeltaS. Numerically that is -9800 cal/moleH2 - T * 22 cal/moleH2/K. If we use 300 K , that neams t*DeltaS = 6600, which is 2/3rds of the DeltaH value, so it is not going to be a 'primary' reason as it is *roughly* equivalent (even at 1000K, the DeltaH term is not insignificant).
The original point (I thought) of your postings was to suggest molecular hydrogen absorption was 'energentically favored'. My point has been to show you it is not, in fact there is no data for absorbing molecular H2 into a metal, as it is not found to occur.
I clearly indicated "molecular hydrogen".
The 'you' was intended to be more generic. Sorry if I was unclear on that. I should have said 'one should always be clear about...' perhaps.
I'm avoiding some warranted sarcasm here. Perhaps there are good reasons journal reviewers are rejecting some of your efforts. I'd be glad to preview, as some others here might also.
You should avoid the sarcasm because everyone makes mistakes, including you. In fact sarcasm is of extremely limited value in a scientific discourse. And if you did make a sarcastic remark relating the fact I made a minor mistake here and trying to connect that to the two journal rejections I've talked about before, you would be mistaken. My first incident involved the J. Env. Monitor. case, and I was not even allowed to offer a rebuttal. No rebuttal written means no error made, right? The second case was Phys. Lett. A, where I received incorrect reviewer comments that I responded to (like 'author included no original data', when (a) I had a plot of data obtained by me in my lab in the paper, and (b) a Comment often doesn't include new, original data anyway, i.e. the reviewer was wrong to make the comment and was wrong in what the comment said), and where the editor then arbitrarily refused to further process my submission because he didn't want to publish more cold fusion papers. No 'error' on my part, just editorial choices, and nothing to be sarcastic about towards me.
From a layman's perspective, my understanding is that the following processes have been discussed in different experimental writeups:
Discussions of recombination have never to my knowledge pertained to a process occurring inside the palladium. How does the understanding above compare or contrast with what has been pointed out so far in this thread?
@Eric
The splitting of gaseous H2 is endothermic, which automatically implies recombination is exothermic. Adsorption into Pd, ie. Pd-Hx formation is exothermic, automatically implying that decomposition of Pd-Hx is endothermic. The only problem with that is that Pd-H is a ‘metallic bonding’ compound and it can have non-stoichiometric composition, i.e. PdH(sub 0.036) or PdH(sub .82) which in fact offers the possibility that the endo/exothermicity can vary with composition, as was noted in the Clewley, Flanagan reference I cited, so one has to be careful about that.
I’m not sure if any endo/exo decision has been stated re. migration of H into and out of Pd. That it occurs is an equilibrium thing, controlled by the chemical conditions. Low pressure H2 over highly loaded Pd-H is not an equilibrium condition, and if left to its own devices, the Pd-H will release some H2 to reach the equilibrium point, etc.
There have been computations of various types over the years for molecular H2 in the tetrahedral or octahedral holes of a metal lattice. In general the energetics are extremely unfavorable for this. But you also have various kinds of defects in a metal: missing atoms forming holes, mismatched lattices at grain boundaries, impurity atoms giving local distortions of the lattice, other interstitial adsorbates (like He, O or C), etc. As noted in particular, some of these ‘holes’ can sustain molecular H2 in them if they are big enough. The process of steam embrittlement is caused by H and O recombining in the metal to form H2O which either forms in bubbles or causes bubbles to appear. Subsequently those ‘protobubbles’ grow due to the (apparent, maybe the H and O diffuse instead) diffusion of more H2O to the bubble. Same thing happens with methane embrittlement except H recombines with C to form methane in bubbles.
Also not mentioned so far is that the consideration of H2 dissociation as used in this thread so far is only of value to thermodynamic considerations. In actual fact, it is unlikely that free atomic hydrogen ever exists in the absorption process. Instead, most mechanisms consider the H2 to dissociate on the metal surface at specific sites and form _adsorbed_ H atoms (i.e. surface H atoms). These adsorbed atoms then transition to bulk absorbed H which is considered to be different from the surface H, and now become a participant in the metallic bonding and antibonding orbitals, where they become ‘screened’ protons (overall charge balance is retained).
The bottom line is that one has a selection of levels of sophistication that can be used in discussing the generic issue of H absorption. Choosing which level you work at is driven by what aspects you want to discuss and how accurate you need to be while doing that.
What's funny is you chastize me for neglecting entropy and then go and make the exact opposite mistake. As you noted DeltaG = DeltaH -T*DeltaS. Numerically that is -9800 cal/moleH2 - T * 22 cal/moleH2/K. If we use 300 K , that neams t*DeltaS = 6600, which is 2/3rds of the DeltaH value, so it is not going to be a 'primary' reason as it is *roughly* equivalent (even at 1000K, the DeltaH term is not insignificant).
The original point (I thought) of your postings was to suggest molecular hydrogen absorption was 'energentically favored'. My point has been to show you it is not, in fact there is no data for absorbing molecular H2 into a metal, as it is not found to occur.
Note: My, that is Longview's, first post to this thread is on September 1st, and another more detailed on September 2nd.
[Please use joules...no need to crash on Mars again!]
I don't follow your arithmetic above--- please look carefully at mine. My conclusion today is that there is no need at all for any delta H (work in or out) to understand passive Pd/hydrogen or Pd /deuterium association or "reaction". I am not bound to the idea that "molecular hydrogen absorption was energentically [sic] favored". But, certainly the work (delta H at +436 kJ/mol) of splitting molecular hydrogen is over ten times that supplied by the exothermic (negative) T delta S which is the only integral (internal) source to overcome that positive (i.e. work requiring) delta H seen in passive (e.g. non-electrolytic) Pd / H or D experiments-- I'm aware of. I suggested early on here, and now have written specifically, the entropic term (T delta S) appears to account neatly for those observed heats of lattice association (including Abd's Pd hydrogen "hunger") and dissociation.
The form that the hydrogenic isotopes take inside the lattice is something we can discuss much further. It is, I suspect, dependent on how one measures the hydrogen bonding energies. Wyttenbach suggested here that the crystallography is informative. I had suggested here that specific heat, electrical and thermal conductivities could be informative. I would add superconductivity transition T and/or Meissner effect behavior to those. But in the absence of those data we have good concurrence of the measured entropy function for formation and possibly its opposite.... giving a low, or no, enthalpy (delta H) scenario as the most plausible thermodynamics.
Re. my arithmetic, as you (Longview) wrote: delta G = delta H minus T delta S
Delta H = -9800 cal/mol H2, Delta S= 22 cal/molH2/K So at 300K, Delta G = (-9800 - 300* 22) cal/mole H2 = -16,400 cal/molH2 (neg delta G implies a favored thermodynamic reaction, i.e. likely to be spontaneous if not kinetically limited) So, 9800 = 60% of deltaG, 6600 = 40% (roughly a 2/3rds / 1/3rd split) –neither term is insignificant
At 1000K, deltaG = -9800 – 22000 = -31800 cal/moleH2, so deltaH term now = 31%, deltaS term = 69%, and neither term is insignificant, which is my point.
“But, certainly the work (delta H at +436 kJ/mol) of splitting molecular hydrogen” is not relevant to the mechanism of reaction. As I noted in my reply to Eric, the xH2 + Pd -> Pd-H(sub 2X) reaction is not thought to go through ‘free’ H. It’s fine to use that on a thermodynamic diagram of the reaction because of the assumption of path invariability, but that diagram has no relation to a mechanism. So to get H2 to absorb into Pd, you do not first have to supply the H2 dissociation energy and thereby dissociate H2. The mechanism would involve forming some kind of activated surface complex of one or two (or maybe 3) Pd atoms and the H2 molecule which then breaks down to surface H held by 1, 2, or 3 Pd atoms in some surface site such as on-top, bridge, or in-the-hole. That then further transforms to a subsurface H. That whole process is what you would need to map out theoretically to compare to some other proposed mechanism giving you molecular H2 in the Pd. It is unlikely that even the activation energy barriers approach the magnitude of the H2 dissociation energy, but I could always be wrong on that of course.
Note that the whole dissociation process described above happens on any clean metal surface (to my knowledge) and is why platinum is a good hydrogenation catalyst. There, instead of forming subsurface metallic bonded H, the surface H reacts with impinging chemicals to form useful and desired reaction products.(So getting back to Eric’s questions…H2 dissociation on clean metal surfaces at room temp and above is generally considered to be spontaneous, so I guess that process must be exothermic.)
But detailed discussions like these are something of a waste of time, as they postulate a lot of things that can't be tested experimentally (in particular, transition state complexes) in many cases. We stick with the observations...Pd + H2 is exothermic, and we see no molecular H2 in Pd-H.
And yes, folding in all those other observations is an activity of chemists...
Re. my arithmetic, as you (Longview) wrote: delta G = delta H minus T delta S
Delta H = -9800 cal/mol H2,
Where is this "-9800" coming from? That is 41 kJ/mol by the way--- looking much like an integrated or mean of Flanagan et al's isosteric heat for the range of atom ratios of H to Pd of 0.61 up to 0.69.
Where is this "-9800" coming from? That is 41 kJ/mol by the way--- looking much like an integrated or mean of Flanagan et al's isoteric heat for the range of atom ratios of H to Pd of 0.61 up to 0.69.
Alefeld and Volkl, "Hydrogen in Metals II", Springer-Verlag, 1978, Chapter 3, "Hydrogen in Palladium and Palladium Alloys", p. 81 (that's in section 3.2.2 "Phase Diagram, Methods, and Results.", authors of Chap. 3 are E Wicke and H. Brodowsky) They are comparing and contrasting different results from different authors. The numbers can vary depending on the temperature ranges used and the specific materials and methods. Generally people in the field don't attribute too much precision to the numbers.
These numbers were obtained via the van't Hoff equation, which plots the natutal log of pressure vs. the reciprocal temperature. The slope is used to derive the enthalpy and the intercept the entropy.
Where is this "-9800" coming from? That is 41 kJ/mol by the way--- looking much like an integrated or mean of Flanagan et al's isoteric heat for the range of atom ratios of H to Pd of 0.61 up to 0.69.
@Logview: Now, when we all see that the PdDxy (they mostly use D not H which has different values from - in Pd - for dH dS) is very complex and contains many free variables, which only can be measured empirically, we should stop any discussion which lead to polarization.
PdD has two different crystal structures, which may coexist to some extent and there is a highly Temperature dependent T * dS Term which might or might not dominate the overall reaction Enthalpy.
May be somebody can find some load diagrams (%D in Pd) with a 'fixed' Temperature (the load most go on slowly.. mostly exothermic) which shows the corresponding dG values for each % step.
Here discussions are of no help. Just read a table and get the answer!
May be somebody can find some load diagrams (%D in Pd) with a 'fixed' Temperature (the load most go on slowly.. mostly exothermic) which shows the corresponding dG values for each % step.
Here discussions are of no help. Just read a table and get the answer!
For equilibrium conditions, phase diagrams are creaed from isotherms. See here for Pd-H, Pd-D phase diagram:
http://www.technology.matthey.com/article/38/3/112-118/
The dotted line is supposed to be the phase boundary. In Storms' Student Guide (lenr.com) he had a Pd-D phase diagram but I don't like the dotted line on it. Seems to not go high enough...
PdD has two different crystal structures, which may coexist to some extent and there is a highly Temperature dependent T * dS Term which might or might not dominate the overall reaction Enthalpy.
Lasser shows a plot of the van't Hoff relationships for H, D, and T in Pd in Fig. 5.14 on pg. 65 of his book. He doesn't seem to give numerics, but he references where to get them.
Tritium and Helium-3 in Metals, R. Lasser, Springer-Verlag, 1989
