exchemist

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.

Not in the least. The OP asked the simplest of questions: whether the energy released when a bond forms is the same as the energy required to break it. And the answer to that is an unambiguous "yes". All the rest of the stuff we have been discussing since is merely all the other things that can influence the course of a chemical reaction. But, interesting thoughtit all may be, that is not what was asked.

Electrostatic ion-solvent bonding. See solvation energy: https://en.wikipedia.org/wiki/Solvation The process of an ionic solid dissolving in a polar solvent requires the ionic bonds in the crystal lattice to be broken. This requires energy input. The energy to do it comes from the greater energy released, when polar solvent molecules are electrostatically attracted to the cations and anions and settle into a solvent cage around them. (We can argue semantically about whether these ion-solvent interactions are to be called a formal "chemical bond" or not, but they are in terms of physics no different from an ionic "bond" in a crystal, viz. a reduction in electrostatic potential energy by the approach of unlike electric charges.) None of this is to dismiss the role that entropy also plays in determining whether a solid dissolves or not. Famously, dissolving ammonium nitrate in water leads to a cooling of the solution: it is an endothermic process. In this example, the entropy gain when it dissolves outweighs the enthalpy gain, so the reaction still occurs, even though it is "uphill" from the point of view of internal energy change alone. In terms of the classic chemist's formula ΔG = ΔH - TΔS, even though ΔH is +ve, ΔG can still be -ve, if the effect of the TΔS term is big enough. As for catalysis etc, this is to do with kinetics of reaction rather than thermodynamics. Catalysts lower the activation energy for the reaction to proceed and thereby increate the reaction rate. They allow a lower energy path as the reactant bonds break, reducing the energy of the transition state between reactants and products. But energy input is still needed, to get to even this lower energy transition state.

You are right about the overall thermodynamics of course, but you are talking of the entropy term in the free energy of reaction. The OP question was about bond energy, which is the internal energy term, the major part of the enthalpy of reaction. G= U+PV-TS. You are talking about S while the OP is enquiring about U, isn’t it?

That has nothing to do with bond energy, however. Sure, dissolving solid NaOH in water is exothermic, like a great number of chemical reactions. That does not mean there is any bond that releases energy when it breaks, however. It simply means the new bonds that form release more energy than is absorbed by the bonds that are broken.

Yes, in any chemical bond the energy to break it is what you get when it forms. There is an intrinsic energy reduction due to the bonding electrons entering a state of lower energy than in the isolated atoms. Because energy is conserved, if you reverse the process that is the energy needed to break the bond again. But be aware that, in industrial processes in practice, there are always inefficiencies. So for example in electrolysis, you don't get anything like 100% conversion of the actual electricity input into splitting water molecules, as there are a lot of energy losses along the way.

Yes. When hydrogen is burnt with oxygen the reaction is 2 H2 + O2 -> 2H2O. You break 2 H-H bonds and an O=O double bond, which requires energy. But the energy you get back from forming 4 H-O bonds, i.e. H-O-H + H-O-H is greater, by 286kJ/mol of water. If you run the reaction backwards, as you do in electrolysis of water to make hydrogen and oxygen, this is the energy you have to input to do it.

The only rationalisation I can think of for @studiot's comment is that he is thinking of endothermic reactions, in which the internal energy of the products is greater than that of the reactants. That can happen if there is a sufficiently favourable entropy change, such that the free energy decreases even though the internal energy increases. ΔG= ΔU +PΔV -TΔS etc. But all that means is that there is less energy released by the new bonds that form than is absorbed by those that break.

Yes, it is identical. I don't understand your answer here. As far as I know there is no bond that releases energy when it breaks. If that were to happen the "bonded" state would have a higher energy than the unbonded state, which would be an antibond rather than a bond, wouldn't it? So the "bond" would never form in the first place. But maybe I'm misunderstanding what you are getting at. Can you give an example, to help clarify?

I think one difficulty would be that the salt would not drop back into the ocean. As each droplet progressively evaporated and shrank, the salt would become more and more concentrated, leading eventually to tiny particles of salt that would be carried inland by the breeze for long distances. I would propose instead to combine your idea with solar stills to generate fresh water for your pumps, returning brine to the ocean. In such hot deserts, one almost invariably gets a daily onshore breeze after about 11am, due to the rising of hot air over over the desert drawing in cooler air from above the sea. This normally persists until just before sunset. So it would be a fairly reliable process, I think. But I have no idea whether enough evaporation could be achieved to change the climate this way. I think I read some years ago that another way is simply to use the fresh water for irrigation and promote the growth of plants which, by their transpiration, start to alter the climate that way - and generate moisture-retentive soil, by the detritus on the ground that they create.

A controlled application of the strong nuclear strong interaction would be a nuclear power station or the nuclear bomb. I rather think the weak interaction is also involved, along the way. I can't think of a controlled application of the weak interaction on its own, though. A controlled application of gravity could be a clock pendulum or the weight mechanism of a grandfather clock.

While it is true that UV does eventually weaken fabrics, this is usually noticeable only for things such as curtains and sofas that may be exposed to a lot of direct sunlight for hours every day, for years on end. For clothes and bed linen, i.e. the items on your washing line, you can usually ignore this, as the exposure time is a few hours per week at most. Most clothes wear out for other reasons. Regarding dryers, it is noticeable that a significant amount of fluff collects on the filter of a dryer every time you run it. This is no doubt partly human skin cells, but a lot is fibres from the fabric. My guess would be this will have more of a weakening effect than sunlight on a clothes line. In my opinion, clothes dried outside smell fresher - and of course it is more environmentally friendly as well. So personally I always dry clothes outside, when the weather permits. I make an exception for towels, which are much softer if dried in a dryer.

But there is friction in the oceans, in the tidal currents. This is spinning the earth down. However, angular momentum is a conserved quantity, so if one part of a system spins down due to friction, something else has to gain the angular momentum lost. In this case it is the moon, which is moving to a higher orbit - getting farther away - as a result.

OK. What do you think will be the mass of 0.1 moles of it?