exchemist

There is no magic temperature at which the bonds suddenly break. At any given temperature, molecules have a statistical range of velocities. When you dissociate a substance by heating, what happens is some of the molecules get enough thermal kinetic energy for the bond holding the atoms together to break. If a dissociated atom later on encounters another dissociated atom then, unless the atoms have between them more energy than the bond energy, they will recombine. So what you have is a dynamic process, in which some molecules are splitting into their constituent atoms, and some atoms are recombining into molecules. This is a chemical equilibrium: N₂ <-> 2N. Whether most of the substance is in the form on the left hand side or the form on the right hand side depends on the bond energy, the entropy of the two states and the conditions (temperature and pressure). As you raise the temperature, the average velocity of the molecules increases. That means that the fraction of them with an energy greater than the bond energy increases. This results in a greater fraction of them being in the dissociated state at any given moment. A strong bond energy favours the left hand side, while a significant entropy of dissociation favours the right hand side, more so at higher temperatures. What I am pointing out is that - because the bond energy in this case is so high - it is not until you reach about 9000C that you will have equal amounts of molecular and atomic nitrogen. In fact, I've now got a more accurate value for the entropy of dissociation, 115J/K-mol, which leads to temperature of more like 8200C for a 50:50 mixture of atomic and molecular nitrogen. (This is in good agreement with @studiot's earlier post in this thread, in which he quoted a textbook saying that at 8000C, there would be 40% dissociation.) By the way, catalysts are irrelevant to this. A catalyst does not change the thermodynamics of a reaction, which is what determines the equilibrium state. It just accelerates the rate at which it gets to equilibrium from an initial set of reactants. In the Haber process, the catalyst accelerates the reaction by avoiding having to dissociate the nitrogen molecule into atoms before reacting with hydrogen. It does this by forming new bonds between nitrogen and the metal surface. The energy of the bound molecules and atoms stays lower, throughout the process, than if they had to be free atoms. As a result, a greater fraction of the molecules have enough energy to react - so it goes faster.

Well, let's first of all see if our poster is now satisfied with the answers we have given, or if he or she wants to go somewhere else with the topic. It is, after all, a chance to discuss some chemistry.

Quite right, so you did. I've just filled in a few more details for our poster, to help him or her understand why it is not what he or she thinks. 🙂

Not really. The way you describe it misrepresents the mechanism of this reaction. This is the Haber process. The N2 molecule is adsorbed on the surface of the catalyst, where it forms a kind of nitride with it, first end-on and then sideways-on, which then reacts, still on the surface, with adsorbed H to form an adsorbed NH, NH2 and finally NH3, before desorption of ammonia occurs. At no stage are free N atoms produced. This is not a route to formation of atomic nitrogen.

Where do you get 500C from, though? If we look at the equilibrium N2(g) <-> 2N(g), this will be in balance, with similar amounts of N and N2, when ΔG = 0. Since ΔG=ΔH-TΔS, that will happen when TΔS= ΔH, which will be at the temperature at which where T= ΔH/ΔS. ΔH =945kJ/mol, but according to what I have been able to dig up, ΔS seems to be around~100J/K-mol. That would imply an absolute temperature of 945 x 1000/100, 9450K, which is over 9,000C, well above the boiling point, repeat boiling point, of iron!

I think this will be condensation, from periods at which the temperature of the table and cover goes below the dew point of the ambient air. An uncovered table would also develop moisture, but the presence of the cover will hinder its re-evaporation when it warms up and/or the relative humidity drops. Typically, objects will get somewhat colder than the ambient air at night, as they radiate in the IR. This is especially pronounced when there is no cloud cover, so that there is no balancing IR radiation coming back down from the clouds. This explains for example why you can get a ground frost on clear winter nights, even though the air temperature can be 3-4C above freezing. So if the relative humidity is close to 100%, a table or other exposed surface may, for a while during the night, have a temperature below the dew point.

My understanding is that a router emits a stronger RF signal than a cellphone. Whether RF radiation at such levels can really pose a health hazard seems to be a matter of debate. Traditionally it was thought it had no effect on biological tissue, but I think there have been a few studies that challenge that assumption. Maybe someone else here knows more.

In what context?

There's a converter here that tells you how to get from a pressure to the corresponding altitude: https://www.mide.com/air-pressure-at-altitude-calculator

What evidence is there for these sound particles?

That's what I mean: fuel storage is energy storage. Fuel is a just source of chemical energy, after all.

You could, but using a highly compressed gas at over 500C might not be the most convenient or efficient form of energy storage.

I would think with DNA certainly cosmic rays will be one source of mutation, so they will have a role in evolution. With neurons I would think not. Neuronal processes do not seem to rely on individual molecules, in the way that genetic coding does. So damage to one molecule won’t affect their operation, I would have thought. The energy cosmic rays impart to organisms as a whole is tiny. But for an individual atom in an individual molecule, it is enough to break its bonds, tear off electrons, knock it out of position etc. Of all biological structures it seems to be only genetic material that relies on a single (very large) molecule. That’s why cosmic rays can make a difference there.

OK, I've looked this up and to be honest it is a bit confusing, between the collision rate for an individual molecule, total collisions for a given absolute number of molecules and a total collisions per unit volume. Also, a lot of the formulae are for reactions between 2 different species, so they are for molecules of type A colliding with molecules of type B, etc. But as far as I can see, and simplifying the formula as far as possible, it looks to me as if, for a gas consisting of a single type of molecule, the total number of collisions expected in unit time per unit volume is: Z = 2N²d²√(πkT/m), where: Z is the no. of collisions per unit volume per unit time N is the no of molecules per unit volume (i.e. number density) d is the effective diameter of the atom or molecule for interacting in a collision (atoms and molecules are not hard spheres of course) k is Boltzmann's constant m is the mass of the atom or molecule T is absolute temperature (The derivation of this formula is extremely hairy, by the way, as it has to allow for the fact that all the atoms of molecules are moving in random directions.) You can use 2 x atomic radius for d, which will be good enough for an approximate answer (order of magnitude). You can see the number of collisions goes up with the square of the number density, and with the square of the atomic or molecular diameter, and with the square root of temperature. Number density is fairly easily obtained from the ideal gas equation PV=nRT. I mole of any (ideal) gas occupies 22.4l at STP, viz. 273.15K (0C) and 1bar pressure. But I don't know what you are going to do with this information. You won't stop N atoms recombining at any practical temperature or pressure. In fact, if your idea is to use separated nitrogen atoms as an energy store, that can release energy when N2 is re-formed, I think you are better off to segregate the N atoms in molecules that when suitably brought together generate N2. For instance, ammonia is now talked about as a future fuel for ships. More about ammonia combustion here: https://www.sciencedirect.com/science/article/pii/S1540748918306345

Catalysts reduce activation energy, not increase it. Collision rate - that’s a kinetic theory question. You would also need to know the effective diameter of the atoms, but I’m sure that’s available. I’d need to think how to do that when I get out of bed and have moment later in the day. Someone else may offer an answer in the meantime.

Absolute zero and less than 1 mTorr.😁 Seriously,, forget it. There is almost no activation energy for this recombination. So you can only slow it down by lowering the rate of interatomic collisions.

Phew, so it was right! Yup, the Gibbs free energy is the energy available to do work, and electrical energy counts as "work" in this context. I must admit I was hesitant about thinking it through for the backwards reaction, but I think it makes sense. I suppose one can think of it as getting the reaction to go backwards using an 80% "push" from electrical input with the remaining 20% coming from the entropic "pull", due to 2 liquid phase molecules being replaced by 3 gas phase ones. This pull helps to force the bond formation to go in the "uphill direction, energetically speaking, so the balance of energy required will come from the environment. Of course all this is a bit theoretical, since what with electrode inefficiency losses and the desire for a rapid reaction rate, I'm sure a higher electrical input is required in practice than the mere Gibbs free energy theoretical amount.

Interesting that Liz Cheney is the one wielding the lance. From her concession speech it looks as if this will be her big project for the next year or so. She's using all the language that has been deafening by its absence. At last someone is prepared to say it. The Republican party has indeed been reduced to "a cult of personality". I do hope she gets some traction with the more sensible segment of Republican voters.

True. But the boil must be lanced.

Careful. We'll start thinking you are a 'bot. 😁

What do you mean you "read that too"? It's your own post! From 2 years ago.

To comment, you have to describe the theory to us. You have not done that. Referring us off-site is not allowed by forum rules. You would have to explain it here. But I must say that from the little you have posted so far, it looks like nonsense. What can possibly be meant by a "periodic table of humanity in space?

Yes, 941 (or is it 945?) kJ/mol will be released if 2 nitrogen atoms are allowed to form N2. Regarding the electrolysis question yes I agree it is confusing. I think it works like this, but I'm open to correction on it as it is a bit tricky to get right. The lower number, 237.24kJ/mol is ΔG, the increase in Gibbs free energy involved in the dissociation and formation of hydrogen and oxygen from one mole of water, i.e. for the reaction H20 -> H2 + 1/2 O2. The higher number, 285.83 kJ/mol is ΔH, the change in enthalpy. The change in entropy ΔS is 163J/K mol, which at 25C, i.e. 298K gives 163 x 298 = 48.57kJ/mol. You will see this is the difference between the two figures (apart from a minor rounding error) It is ΔH that is closer to the change in the bond energies. (But it is not exactly equal to the change in bond energies, as there is PV work done, due to one molecule generating 1 and 1/2 molecules, i.e. 2 molecules of water generate 3 molecules, 2 of hydrogen plus one of oxygen. This occupies a greater volume so, under atmospheric pressure, work is done against the atmosphere to make room for the extra gas molecules created. Remember H = U + PV.) The actual energy input required is ΔH, but the theoretical amount of electrical input needed is given by ΔG. The remainder of the enthalpy is taken in by the system from the environment, i.e. the system will get colder and will suck heat in from around it, like my example of ammonium nitrate. At least, I think that is how it works out.

Haha, yes. In my case we had a related formula dinned into us at school in the 1st yr 6th form: Log "how far" = -ΔH/RT +ΔS/R, (R being the gas constant) which we used to predict what reactions would go effectively to completion and which would not, based on looking up enthalpies and entropies of formation in something called the Rubber Book. This formula came (as we later learnt) from the relation between Gibbs Free Energy change and the equilibrium constant for the reaction: ΔG =-RT lnK. That would have been in 1970.....

No it doesn't affect the energy released or needed but it does affect whether or how far the reaction proceeds. What determines whether or not a chemical reaction should proceed is whether the free energy of the products is lower than that of the reactants. We normally uses the Gibbs Free Energy, G. The relevant relationship is one I have mentioned in this thread, namely ΔG = ΔH - TΔS, in which S is Entropy, T is absolute temperature and H is something called the Enthalpy, which is the internal energy in the chemical bonds, U, plus any PV work done during the reaction against the atmosphere (this can be important if there is a big volume change during the reaction for instance if 2 molecules of gas produce 3 molecules.) You can see from this that whether the change in G is +ve or -ve can depend on the entropy change as well as the enthalpy change. In an earlier post I gave the example of a reaction (dissolving ammonium nitrate) that proceeds even though the products have a higher internal energy than the reactants. This is fun to do - the solution actually gets cold as the salt dissolves! Comparing the enthalpies of products and reactants tells you that if the reaction proceeds energy input is required. However the reaction occurs spontaneously in spite of this, and it sucks the energy it needs from the environment - because of the effect of the entropy change. The system moves to a state of lower free energy but, in spite of this, an energy input occurs.