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. Oâ‚‚ has one O=O bond, COâ‚‚ has two C=O bonds and Hâ‚‚O 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. Oâ‚‚ has −494 kJ/mol, COâ‚‚ has −1598 kJ/mol and Hâ‚‚O 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 Oâ‚‚ have a combined enthalpy of:

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

The COâ‚‚ and Hâ‚‚O 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 COâ‚‚ and Hâ‚‚O than it took to break the bonds in methane and Oâ‚‚, 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?

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

No. What about dividing?

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

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.