Limitations to replicator technology: energy and fuel

Replicators have become a staple of science fiction. However, many stories treat them as magical, or with inconsistent abilities.

This is not a magical device that makes anything for nothing. When fed a source of certain atoms (it can't transmute atoms), it will rearrange them into whatever it is programmed to create, such as food, microprocessors or rocket fuel from nearby asteroids. It is not a nuclear reactor.

Also, it can't transmute atoms (e.g: H > He), so it need all the types of atoms used by food (H, O, C, N...), microprocessors (Si, C, Au, H, O, Fe...) or rocket fuel (H, C, O) in order to assamble them.

Given a proper scientific basis, what realistic limitations might it have? Especially, what realistic limitations might it have in the energy consumption used to rearrange atoms or in the used fuel (resources to assemble things)?

• Wizzwizz4 posted this Q at my suggestion, so he’d have a place to hang his well-developed Answer. Such shill self-questions are indeed prone to being poor when read as normal posts — I’ve run into that myself. Considering that his real contribution is below, let’s help craft the Q to suit and be a solid question that could attract other answers. Nov 27 '17 at 1:03
• @JDługosz I don't think that shill questions as you call them should be given any leeway when compared to a normal question when considering whether to VTC. Nov 27 '17 at 1:54
• Nov 27 '17 at 3:59
• Why the downvote? It s a perfectly valid question
– Fred
Nov 27 '17 at 5:00
• @Fred Read the edit history. It's a tolerable question now but as it was originally written it was only there to allow for wizzwizz4's answer and was extremely low quality. Nov 27 '17 at 14:03

Your replicator is a programmable universal catalyst combined with a 3D printer. It can force any particular chemical reaction to occur over others and deposit the result of that chemical reaction in an arbitrary arrangement. However, it cannot violate conservation of energy.

So, how much energy is required to manufacture certain things? It depends on what you're making it from.

Enthalpy

Enthalpy is a word in chemistry that means the difference in amount of energy stored by the chemical bonds in two groups of molecules. The amount of energy stored in a chemical bond is always negative, meaning energy was released when the bonds were made. When you burn something, that heat energy gets put into breaking some chemical bonds, which then allows the atoms to bond to other atoms in a way that releases more energy. This energy then makes more surrounding molecules hot enough to do the same, resulting in the substance getting hot enough to make nearby air glow (fire).

Things in general tend to want to release energy more than they want to gain it. This is why hot things start to glow when they get really hot (like metal or the sun) and why things tend to fall down instead of floating in the air.

Calculating enthalpy

In order to work out the difference in enthalpy, you must first find some chemical bond diagrams, like this one for methane:

Methane has four C–H bonds. O₂ has one O=O bond, CO₂ has two C=O bonds and H₂O has two O–H bonds. Now that you have these, look them up in a bond energy table to work out the amount of energy stored in these bonds (remember that this will be negative!):

$$\begin{array}{|l|l|}\hline \text{Bond} & \text{Energy (kJ/mol)} \\ \hline \mathrm H- \mathrm O & -459 \\ \mathrm H- \mathrm C & -411 \\ \mathrm O- \mathrm O & -142 \\ \mathrm O= \mathrm O & -494 \\ \mathrm O- \mathrm C & -358 \\ \mathrm O= \mathrm C & -799 \\ \mathrm O≡ \mathrm C & -1072 \\ \hline\end{array}$$

This gives methane a total bond energy of 4 × −411 kJ/mol = −1644 kJ/mol. O₂ has −494 kJ/mol, CO₂ has −1598 kJ/mol and H₂O has −918 kJ/mol.

Balancing the equation

Now that you have the energy per mol, you must work out how much of each molecule you need to make one mol of methane. This reaction tends to go one way (methane is burned) so I'll write it that way as according to convention.

First, write the unbalanced equation:

$$\mathrm{CH_4} + \mathrm{O_2} → \mathrm{CO_2} + \mathrm{H_2O}$$

Now, count the number of each atom on both sides. On the left we have one C, four H and two O. On the right we have one C (check!), two O (oh dear), two H (this isn't good) and another O (making three in total).

We have the right number of C, which suggests that we don’t need to change the number of anything with a C in it. H is only in one molecule that we don’t know the number of, so let’s add some more of that to get the right number.

$$\mathrm{CH_4} + \mathrm{O_2} → \mathrm{CO_2} + 2 \mathrm{H_2O}$$

Now we have the same number of H and C on both sides. On the right hand side we have 4 O, and on the left hand side we have...

$$\mathrm{CH_4} + 2 \mathrm{O_2} → \mathrm{CO_2} + 2 \mathrm{H_2O}$$

Yay! It’s balanced.

Calculating enthalpy change

Now that you know how much of each molecule there is, you can calculate the enthalpy change. The methane and O₂ have a combined enthalpy of:

−1644 kJ/mol + 2 × −494 kJ/mol = −2632 kJ/mol

The CO₂ and H₂O have a combined enthalpy of:

−1598 kJ/mol + 2 × −918 kJ/mol = −3434 kJ/mol

Performing a quick sanity check: Less energy is used making the bonds in CO₂ and H₂O than it took to break the bonds in methane and O₂, so this seems about right.

This means that you’ll have to put in −2632 kJ/mol − −3434 kJ/mol = 802 kJ/mol into the replicator order to produce methane (and oxygen) from water and carbon dioxide.

This value is probably not very useful at the moment. What this means is that, for every 802 kJ you put into the replicator you can produce 1 mol of methane and 2 mol of oxygen. In order to work out how much this weighs, you must turn to the Periodic Table!

Molecular Mass

The molar mass of an atom is written underneath the chemical symbol in this image. It is in the units g/mol. To calculate the molecular mass of a molecule, add together the molar masses of the atoms. For example, methane is:

15.999 g/mol + 4 × 1.0079 g/mol = 20.025 g/mol

Now we have energy per mole and mass per mole, and we want energy per mass. To work out what to do with these two values to get what we want, we can use algebra. Does multiplying them work?

$$\frac{\text{energy}}{\text{mole}} \times \frac{\text{mass}}{\text{mole}} = \frac{\text{energy} \times \text{mass}}{\text{mole}^2}$$

$$\frac{\text{energy}}{\text{mole}} \div \frac{\text{mass}}{\text{mole}} = \frac{\text{energy}}{\text{mole}} \times \frac{\text{mole}}{\text{mass}} = \frac{\text{energy}}{\text{mass}}$$

Yes! This means that we can produce the energy per mass:

802 kJ/mol ÷ 20.025 g/mol = 40.05 kJ/g

Getting energy

Here comes the main limitation — your replicator cannot violate conservation of energy, so it can't create compounds that require more than zero energy to synthesise. So, where can energy come from?

• An external energy source.
This one's obvious — any source of heat or electricity can be used as a source of energy. I'll ignore it because it's not specific to replicators.
• A battery.
The energy from exothermic (energy-releasing) reactions could be stored as electricity in a rechargeable battery. Or, better still, by performing an endothermic reaction to produce a stable, high-energy store that can later be used. It doesn't really matter what this is; a higher energy / mass ratio would be lighter but potentially more dangerous. (Then again, what isn't dangerous about a replicator?)
• A fuel source.
If you have some other material to spare, and don’t care what it’s made into, you could use that as a source of energy. One way of working this out is provided below.

Standard enthalpy of formation

There is a simple way of working out enthalpy using a value called the "standard enthalpy of formation" but it only works if you can find this value for every compound involved. This is a measure of the energy released when creating a compound from each constituent element's standard state. What this standard state is doesn't really matter, so long as every standard enthalpy of formation uses the same standard state (which it will).

If you take the standard enthalpy of formation of the fuel that you want to put into the machine (including any oxygen or other molecules) and subtract the standard enthalpy of formation of the products, you get the energy required to perform that chemical reaction. For octane:

$$\mathrm{C_8H_{18}} + 25 \mathrm{O_2} → 8 \mathrm{CO_2} + 9 \mathrm{H_2O}$$

$$(−250 kJ/mol + 25 \times 0 kJ/mol) - (8 \times 393.5 kJ/mol + 285.8 kJ/mol)$$

More coming soon.

• @EnderLook the OP is the same person as this A's author. The Q was a shill to have a place for this post, specifically! This is exactly what the question means to ask; feel free to edit the Q to improve it to this end. Nov 27 '17 at 1:40
• @JDługosz, impresive, this is the first time I see an "Auto-Answered" question, and it's really complete, great. Ok, I'll see if I can improve the question because I throught he wasn't asking this exactly. Nov 27 '17 at 2:03
• Interesting question, but I do not think that basic math from chemistry is the substitution to addressing the limitation such a device might have. Conservation of energy, ok we do know that for few hundred years, at least in chemistry(ok maybe 160 years). Sure it consumes energy, we kinda used to that with all our devices. The main question about the device how it manipulates all those atoms. And there are core limitations of it. Nov 27 '17 at 3:29
• @MolbOrg I'm getting to that! I was just trying to give this answer all of the information from the original three comments before I elaborated on that. Thanks for the reminder. Nov 27 '17 at 16:53
• Enthalpy isn't always negative- it's a measure of the heat required by the reaction. So exothermic reactions, those that produce heat, will have a negative enthalpy because they make their own heat; endothermic reactions require heat, so their enthalpy is positive. (You need x joules of heat to make this reaction happen) Nov 27 '17 at 19:40

One of the limitations often forgotten in the science fiction of replicators is the challenge that arises when the replicated object undergoes chemical reactions while being replicated. Many structure are very difficult to construct without them reacting with the environment, such as oxidizing. If you attempt to recreate a famous painting via this method, be prepared to have issues when the Linseed Oil in the paint burns in the air during replication. Indeed this could lead to patterns which are used to detect the original painting, distinguishing it from forgeries.

Proteins are notoriously difficult to develop. Even identifying their folded structures is difficult, but making them is nearly impossible unless you do it the old fashioned way: in a cell, surrounded by cytoplasm. As such, living things would be nearly impossible to replicate.

• it definitely helps to form proteins the way they are forming, because it works already, but I would say it is a stretch to say it requires cell machinery. yes, it is a complex structure and yes better to be done in particular fashion. Also do not forget, if we talk about food it notorious to be cooked, I'm just saying ))) But it is a not disclosed question which the thing is operating - atoms or blocks. Second seems to be more legit way, and it introduces a wide variety of limitations, but solves a lot of problems too. Nov 27 '17 at 23:44

I think you could save a lot of energy if you didn't try to do everything with atoms. A lot of stuff readily reacts with other stuff, requiring you to invest energy to keep your atoms pure. Just think about everything that tends to oxidize.

Often enough, you don't need pure atoms i the first place.
If you want to create food, you could work with the pure elements, but you could also start with hydrocarbbons, carbon oxides and so forth.
Look at the way chemistry is done. You use ingredients that are reasonably safe to store, and let them react with other ingredients to create your results. Adding catalysts or enzymes often takes a lot of complication out of those processes.

While there might be fancy stuff that would require pure atomic matter for it's creation, a lot of other things (like food, as mentionned), are easier to make from molecular ingredients, thus reducing your total power consumption both in your recyclers (where you obtain most of your raw matierals from) and in your replicators.

It might also make things a lot safer. Just imagine creating water from pure atoms: there are known processes for that, they are called rocket engines. I am sure there are other processes that are a bit better suited for indoor use, and that require a lot less energy for purifying the ingredients in the first place. After all, the energy that is freed in the process of oxidation has to be provided up front in the recycling process, and more energy needs to be invested in preventing unwanted reactions.

• definitely, upvote for identifying that atoms manipulation problem, but why talk about oxidizing agents. Where they come from if you do not introduce them and manipulate things at an atomic level. Same thing as in Cort answer. Access of outside atmosphere in reaction chamber probably is not required. Nov 27 '17 at 23:48
• @MolbOrg to avoid unintentional oxidizing, you need to evacuate the reaction chamber, which requires energy. you also need more complex storing and provisioning. Nov 28 '17 at 8:50
• ok, yes. and what? we talk about a device which potentially manipulates matter on atomic scale(or scale of blocks, does not matter). Compressors are known for a hundred years as for now, maybe more. Evacuate 1L of air requires about 100J at normal pressure to outside of the thing assuming normal pressure again. 100g of a breed and we definitely talk about more energy Nov 28 '17 at 12:45