A theoretical molecule that could match uranium mole-for-mole

XKCD #1162 piqued my curiosity. Obviously, there's nothing we generally use as a chemical fuel that can match the energy density of uranium fission. But I wondered what it would take, hypothetically, to match it mole-for-mole?

Looking at the bond enthalpy charts, I see that the carbon-oxygen triple bond has the highest enthalpy of formation (the energy required to break it and form something else, and by extension the energy produced when this bond is formed, minus the energy required to break the previous ones) at $$\pu{1072 kJ mol-1}$$. However, the carbon-oxygen single bond has an EoF of $$\pu{358 kJ mol-1}$$, and other bonds likely to be involved ($$\ce{H-C}$$: $$\pu{413 kJ mol-1}$$, $$\ce{H-O}$$: $$\pu{463 kJ mol-1}$$, etc.) have higher bond enthalpies, meaning the formation of carbon monoxide by combustion of an organic molecule is very likely a losing proposition.

Nitrogen, on the other hand, has an infamously high bond energy ratio; breaking a single nitrogen bond costs only $$\pu{163 kJ mol-1}$$, while forming a triple bond releases $$\pu{941 kJ mol-1}$$. Nitrogen single bonds, especially in high percentages of the total molecular mass, are notoriously explosive; the azide anion, covalently bonded, forms the backbone of the eminent Dr. Lowe's Things I Won't Work With. The nitro group ($$\ce{RNO2}$$), well-known in explosives, can be made even more potent by bonding it to another nitrogen to form a nitramine, and these are central to the well-known explosive cyclotrimethylenetrinitramine (better known in pure form as RDX, and with added plasticizers as C-4) and its big brother, HMX (cyclotetramethylenetetranitranime).

So, let's consider a theoretical molecule composed of only single-bonded nitrogen (eek). $$\ce{N4}$$, tetranitrogen, has been synthesized, but in a form theorized to be more like two pairs of double-bonded nitrogens that are joined by two single bonds (kind of hard to tell, since it only exists at very high temperature and for less than a microsecond at a time). Let's instead bond each nitrogen to each other one, forming an isohedron. When that structure reverts to diatomic nitrogen (and it will, oh, you betcha), we can expect a heat of detonation of

$$4\cdot\pu{163 kJ mol-1} - 2\cdot\pu{941 kJ mol-1} = \pu{-1230 kJ mol-1}$$

That is a pretty aggressive exotherm; RDX has a detonation energy of 1118 kJ/mol, and its molecular mass is four times that of our theoretical "isohedral tetranitrogen". RDX is as heavy as it is for stability; high molecular weights sacrifice pure energy density for a structure that won't detonate given the slightest excuse (or even no excuse at all).

However, bring in the champion:

$$\ce{^{235}_{92}U + ^1_0n -> ^{142}_{56}Ba + ^{91}_{36}Kr + 3 ^1_0n}$$

(in the atomic chaos of a nuclear reaction, a lot of other products are possible, but this is one of the most common). The masses of the reactants total $$\pu{236.0526 Da}$$, and the masses of the products, $$\pu{235.8658 Da}$$, for a delta of mass of $$\pu{0.1868 Da}$$ ($$\pu{3.102e-28 kg}$$). $$E = mc^2$$, so the energy produced by one uranium atom splitting this way is

$$E = \pu{3.102e-28 kg}\cdot(\pu{3.00e8 m s-1})^2 = \pu{2.79e-12 J}$$

which when multiplied by Avogadro's number and converted to kilojoules, gives us

$$\pu{2.79e-12 J}\cdot\pu{6.022e23 mol-1} = \pu{1.68e8 kJ mol-1}$$

That's 168 million kilojoules per mole available by fissioning uranium-235.

Assuming that double the nitrogen equals double the heat of reaction, in order to have the same heat of decomposition, mole-for-mole, our nitrogen molecule would instead have to be an isometric "sheet" of single-bonded nitrogen, or a geodesic dome (buckyazide?), consisting of $$546340$$ nitrogen atoms single-bonded to each other (a concept that, if ever seriously proposed, should have any student of chemistry falling out of their chair either laughing or horrified). One mole of this material would weigh $$\pu{14.0067 g mol-1}\cdot 546340 = \pu{7652.42 kg mol-1} \approx \pu{8.4 t}$$. One mole of U-235 weighs $$\pu{235 g}$$, just a little more than half a pound.

Long live the king.

The question

(thank you for reading this far)

Does the math look right? I've done it three times and come up with three different answers for the heat of detonation of my hypothetical isohetral tetranitrogen, and thus the size and weight of the resulting molecule, possibly based on the use of different enthalpy figures (I've seen the single bond energy of formation of nitrogen to be as little as $$\pu{158 kJ mol-1}$$, which produces a heat of detonation of $$\pu{1250 kJ mol-1}$$, meaning, all other things being equal, a required molecule of only $$537600$$ nitrogen atoms).

Also, can anyone come up with something that would be more energetic to go up against the champ? There are molecules that have been successfully created in the lab, such as octanitrocubane, that are more powerful than RDX while being usably shock-insensitive, but are prohibitively expensive to manufacture.

• You could just take nitroylethylene and polymerise it to ultra-high molecular weight 250,000 monomers and same number of nitrooxy groups. – user2617804 Jul 18 '14 at 2:27
• The most violent "chemical" reactions have been proposed as rocket fuels (these guys don't think that chlorine trifluoride is beyond the pale). But the best in theory they have ever com up with is metallic hydrogen (and we don't even know it is possible). See this youtube video – matt_black Jan 28 '19 at 15:51

Getting a large exotherm per mole of molecules is a bit of a bad cheat if you're allowed to make molecules arbitrarily big! Energy per mass is the only honest way to do it (or energy per mole of atoms would work too I guess). However, there really is no chance to chemically match a nuclear exotherm; nuclear energies are simply in a class of their own.

Practically all chemical energy is merely a function of electric* repulsion and attraction on distance scales from around a picometre (being very generous, as that is quite small compared to the dimensions of any valence orbital) to perhaps a few micrometres (to include long-distance ion interactions). However, nuclear energies are based on electric attractions and repulsions on distances scales of a few femtometres or less. This makes quite a difference when you remember that electrostatic potentials go as $$1/r^n$$! Not only that, but the nucleus also plays a whole other ball game, displaying strong nuclear potentials. It is commonly mentioned that, apples to apples, the strong nuclear force is about three orders of magnitude stronger than the electrostatic force. Therefore, no chemical reaction involving structures stable at STP will ever reach the feet of nuclear energies. There is probably no "reasonable" suggestion like yours that could do better by a factor of $$10$$, much less a factor of $$10^5$$.

So what can we do about it? Well, if interaction distances being too large is part of the problem, one way to cheat is to simply add pressure, an unreasonable amount of pressure. This effectively brings atoms closer and closer together, so you climb the $$1/r^n$$ potential for each atom. Chemistry gets pretty weird at ultrahigh pressures. For one, lots of things (perhaps everything?) seem to like turning into a metal. With all those orbitals crammed up against each other, eventually bonds overlap extensively and you get bands.

Keep going, and the bands start to overlap, until there's no gap for electron excitation, and room temperature thermal energy can freely knock electrons all over the place. You talk of making huge polynitrogen molecules. Well, metallic nitrogen would be the natural limit of that. Apparently even at pressures of the order of a terapascal, metallic nitrogen has $$\ce{N-N}$$ single bonds around $$\pu{110 pm}$$, which is only a little smaller than STP $$\ce{N-N}$$ bonds, at around $$\pu{145 pm}$$.

I'm sure this would already create quite a bang if pressure were suddenly relieved. But how much pressure would it take to get down to N-N distances of $$\pu{10 pm}$$? And $$\pu{1 pm}$$? How much energy would be pent up per gram of whatever that material is? It's a crazy suggestion, but it takes beyond crazy to get electrons to carry energy comparable to nucleons. Keep going and at some point you'll arguably leave even the theoretical domain of chemistry and reach astrophysically relevant electron degenerate matter, the stuff in the core of relatively light dead stars.

Go even further, and eventually electronic potentials finally become comparable to nuclear ones, and so it actually becomes favourable for nuclei to gobble electrons up. This is the matter inside a neutron star. The fact that it takes the mass of a Sun in a sphere a few kilometres wide to press electron energies into the realm of nuclear ones should be a sobering thought.

* Magnetism is rarely, if ever, the dominant source of electromagnetic potential energy in chemical systems. Magnetic charges do not exist (except maybe for monopoles, but those aren't chemically relevant!), so all magnetism is induced by electric charges in motion (well, spin magnetism is a little more complicated). As far as I understand, which isn't much, magnetism induced from electric charges takes a $$1/c^2$$ hit in the equations of electromagnetism, and that's hard to recover from!

• Your first paragraph is basically my whole point and one of the reasons I used nitrogen; the idea of single-bonding half a million nitrogens together should itself be laughable, when a substance that has more than two of them is treated with the utmost care in the lab. Nice addition of what it would take to make electro-repulsion match the intranuclear forces. – KeithS May 6 '13 at 19:51
• @keithS: if you can go arbitrarily big with molecules, you don't need to go to extremes with specific energy. Taking one mole of uranium vs one mole of nitrocellulose, I wouldn't be quite sure where to put my money. If I can get nitrocellulose with ~10^6 mers per chain, I'm really exceeding uranium energies, mole per mole. Sure mole of uranium will be 238 grams, and mole of nitrocellulose will approach 100 tons, but guess which one will make a bigger boom? – SF. Nov 21 '16 at 0:17
• It's just hard because c is large. Waiting until it degrades to some more manageable magnitude is, getting into wild hypotheses, your best bet. – Oxy Mar 27 '19 at 14:59