Combustion of hydrocarbons (RP-1 - kerosene, methane) and hydrogen: Visual walkthrough

A visual walkthrough of the parent's central result: Combustion of hydrocarbons and hydrogen. We start with nothing but "a rocket throws gas backward" and end with the rule that light exhaust means fast exhaust. Every symbol is earned before it is used.

Step 1 — What "combustion" gives us to work with

WHAT. Before we can talk about how fast the exhaust flies, we need to know what the exhaust is made of. From the parent note, burning a fuel $C_{x} H_{y}$ completely in oxygen gives:

fuel C_{x} H_{y} + how much oxidizer (x + \frac{y}{4}) oxygen gas O_{2} ⟶ one per carbon x C O_{2} + one per two H \frac{y}{2} H_{2} O

Read each symbol: $x$ is the number of carbon atoms in one fuel molecule, $y$ the number of hydrogen atoms. $C O_{2}$ is carbon dioxide (heavy), $H_{2} O$ is water vapour (light). The arrow $⟶$ means "turns into".

WHY. The products are the exhaust gas. Whether the exhaust is heavy or light depends entirely on which of these molecules come out. So the chemistry on the right of the arrow decides the physics of the jet.

PICTURE. The figure lines up the three parent fuels and shows what each one becomes.

Figure — Combustion of hydrocarbons (RP-1 - kerosene, methane) and hydrogen

Step 2 — A single hot gas particle bouncing in a box

WHAT. Forget rockets for a moment. Picture one gas molecule of mass $m$ trapped in a box, flying back and forth at speed $v$ . Its kinetic energy — the energy of motion — is

E_{kin} = constant \frac{1}{2} particle mass m speed squared v^{2}

The symbol $v^{2}$ means " $v$ times $v$ ". We square the speed because energy grows faster than speed — a molecule going twice as fast carries four times the punch.

WHY squared and not just $v$ ? Because that is what experiments and Newton's mechanics agree on: doubling speed quadruples the wallop a molecule delivers when it hits something. This is the tool we need because the exhaust jet's "kick" is exactly stored kinetic energy being released.

PICTURE. One molecule, its velocity arrow, and the $\frac{1}{2} m v^{2}$ tag.

Step 3 — Temperature IS average jiggle energy

WHAT. Now heat the box. Temperature $T_{c}$ (the "c" reminds us it is the combustion chamber temperature) is tied to the average energy of the molecules. For the simplest case — a gas whose molecules are single tiny balls that only fly (no spinning, no wobbling) — the translational (fly-through-space) energy obeys:

average fly-energy \frac{1}{2} m \overline{v^{2}} = 3 directions of motion \frac{3}{2} Boltzmann’s number k_{B} temperature T_{c}

The bar over $\overline{v^{2}}$ means "averaged over all the molecules". $k_{B}$ (Boltzmann's constant) is just a fixed conversion number that turns "degrees" into "energy". The $\frac{3}{2}$ appears because a molecule can fly in three independent directions (up-down, left-right, front-back), each carrying $\frac{1}{2} k_{B} T_{c}$ .

WHY this tool? We want to connect chamber temperature to molecule speed, and this equation is the bridge: it literally says "hotter chamber = faster average fly-speed". That is the first half of our target rule.

PICTURE. The same box, now hot, molecules jiggling harder; a thermometer reads $T_{c}$ and the arrows are longer.

Step 4 — Solve for speed: heavy particles are sluggish

WHAT. Take Step 3 and rearrange it to get the speed by itself. Multiply both sides by $2$ , divide by $m$ :

\overline{v^{2}} = \frac{3 k _{B} T _{c}}{m} ⟹ v_{typ} = \frac{3 k _{B} T _{c}}{m}

We took the square root () of both sides to undo the squaring in $\overline{v^{2}}$ — the square root asks "what number, times itself, gives this?". Now look at what sits where:

$T_{c}$ is on top → hotter chamber makes $v_{typ}$ bigger.
$m$ is on the bottom → heavier particle makes $v_{typ}$ smaller.

WHY take the square root at all? Because Step 3 gave us $v^{2}$ , but the rocket cares about actual speed $v$ . The root is the only operation that turns "speed squared" back into "speed".

PICTURE. Two boxes at the same temperature: one with light molecules (fast, long arrows), one with heavy molecules (slow, short arrows). Same heat, different speeds — purely because of mass.

Step 5 — From one molecule to a whole jet: swap $m$ for $M$

WHAT. A rocket exhausts moles of gas, not single molecules, so we must trade the tiny per-molecule mass $m$ for the per-mole mass $M$ . Two bookkeeping facts let us do this cleanly. Let $N_{A} = 6.02 \times 1 0^{23}$ be the number of molecules in one mole (Avogadro's number):

one molecule \times N_{A} = one mole m N_{A} = M and gas constant R = k_{B} N_{A}

The first says: take one molecule's mass $m$ , multiply by how many are in a mole, and you get the molar mass $M$ . The second defines the gas constant $R$ as Boltzmann's constant scaled up to a whole mole. Now multiply the top and bottom of Step 4's fraction by $N_{A}$ :

v_{typ} = \frac{3 k _{B} T _{c}}{m} = \frac{3 ( k _{B} N _{A} ) T _{c}}{m N _{A}} = \frac{3 R T _{c}}{M}

Each $N_{A}$ found its partner: $k_{B} N_{A} \to R$ on top, $m N_{A} \to M$ on the bottom. Dropping the fixed numbers $3$ and $R$ into a "proportional to" sign gives the exhaust rule:

v_{e} \propto \frac{T _{c}}{M}

The symbol $\propto$ means "proportional to" — "these rise and fall together, ignoring constant multipliers". $v_{e}$ is the exhaust velocity: how fast gas leaves the nozzle. $M$ is the average molar mass of the exhaust mixture from Step 1.

WHY drop the constants into " $\propto$ "? Because for comparing fuels the constants ( $3$ , $R$ ) are identical for everyone. Only $T_{c}$ and $M$ change from fuel to fuel, so those are all we keep.

PICTURE. The single bouncing molecule of Step 2 becomes a directed stream pouring out of a nozzle; the label reads $v_{e} \propto T_{c} / M$ .

Step 6 — Plug in the three fuels: hydrogen wins on $M$

WHAT. Now we apply the rule using the actual average molar masses of each exhaust, computed from Step 1's balanced products. The average molar mass of a product mixture is the total mass divided by the total moles of gas:

Methane $C H_{4} + 2 O_{2} \to C O_{2} + 2 H_{2} O$ makes $1$ mole $C O_{2}$ ( $44$ ) and $2$ moles $H_{2} O$ ( $18$ ):

M_{C H_{4} exh} = \frac{1 ( 44 ) + 2 ( 18 )}{1 + 2} = \frac{80}{3} \approx 26.7 g/mol

RP-1 $C_{12} H_{26}$ burns to $12 C O_{2}$ and $13 H_{2} O$ :

M_{RP-1 exh} = \frac{12 ( 44 ) + 13 ( 18 )}{12 + 13} = \frac{762}{25} \approx 30.5 g/mol

Hydrogen $2 H_{2} + O_{2} \to 2 H_{2} O$ makes only water:

M_{H_{2} exh} = 18 g/mol (lower still if run fuel-rich, as leftover H_{2} = 2)

Fuel	Exhaust products	Average $M$ (g/mol)
$H_{2}$	$H_{2} O$ ( $+$ leftover light $H_{2}$ )	$18$ (or lower)
$C H_{4}$	$C O_{2} + 2 H_{2} O$	$\approx 26.7$
RP-1 ( $C_{12} H_{26}$ )	$12 C O_{2} + 13 H_{2} O$	$\approx 30.5$

Because $M$ sits in the denominator under a square root, the fuel with the smallest $M$ gets the biggest $v_{e}$ — and hydrogen's water-only exhaust (molar mass $18$ , dragged even lower by leftover $H_{2}$ of molar mass $2$ ) is the lightest of all.

WHY hydrogen and not just "most energetic"? Notice the formula does not contain "energy per mole". It contains $M$ . Even though methane releases more heat per mole of fuel, the exhaust-speed contest is won on lightness, and that is why $H_{2}$ takes the crown. (See the parent's mistake box: energy per kg ≠ automatic win.)

PICTURE. Three nozzles side by side; arrow length ∝ $1/ M$ , hydrogen's the longest.

Step 7 — Edge and degenerate cases (never leave a gap)

WHAT. We check the corners of the formula so no scenario surprises the reader.

Cold limit, $T_{c} \to 0$ : $v_{e} \propto 0 = 0$ . No heat, no jet. Makes sense — an unlit rocket goes nowhere.
Very heavy exhaust, $M \to \infty$ : $v_{e} \to 0$ . If the products were enormously heavy, the jet would ooze out slowly. This is why sooty, $C O_{2}$ -rich exhaust (heavier) is slower.
Fuel-rich hydrogen (real engines): engines run slightly fuel-rich on purpose. Leftover unburnt $H_{2}$ (molar mass $2$ ) mixes in and lowers the average $M$ below pure water's $18$ → $v_{e}$ goes up. The formula predicts this correctly. See Incomplete Combustion and Soot Formation for the trade-off.
Density blind spot: the formula says nothing about tank size. A light gas may need a huge cryogenic tank. That is why Cryogenic Propellants and the whole-vehicle Rocket Equation (Tsiolkovsky) can still favour dense RP-1. The formula gives $v_{e}$ , not the final $Δ v$ .

WHY spell these out? Because a formula you only trust in the "normal" middle is a trap. Checking the ends ( $0$ and $\infty$ ) confirms it behaves sensibly everywhere.

PICTURE. A curve of $v_{e}$ vs $M$ (falling) and vs $T_{c}$ (rising), with the two limit points marked.

The one-picture summary

Everything above collapses into a single chain: chemistry decides the products → products decide $M$ → hot chamber decides $T_{c}$ → together they set $v_{e}$ . The final figure compresses all seven steps into one diagram — the two inputs ( $T_{c}$ up top, $M$ down bottom) flowing into the one output $v_{e}$ , with hydrogen's light exhaust marked as the winner.

Recall Feynman retelling — explain the whole walkthrough to a 12-year-old

Imagine a room full of tiny balls bouncing around. Heating the room makes them zip faster — that's temperature, it's really just fly-speed. Now, at the same heat, a room of light ping-pong balls has them flying much faster than a room of heavy bowling balls, because heavy things are lazy. A rocket makes hot gas by burning fuel and shoots those "balls" out the back to push itself forward. So two things make the jet fast: a hotter chamber (more zip) and lighter gas balls (less lazy). We wrote this as $v_{e} \propto T_{c} / M$ — hotness on top, heaviness on the bottom, square-rooted. Burning hydrogen makes the lightest exhaust (it turns into water, and leftover hydrogen is lighter still), so hydrogen's jet is the fastest. Kerosene makes heavy carbon-dioxide, so its jet is slower — but it's squished tighter into the tank, which is a whole different kind of win.

Recall Self-test

Where does $M$ sit in $v_{e} \propto T_{c} / M$ and what does that mean? ::: In the denominator — heavier exhaust means slower jet. If $T_{c} \to 0$ , what is $v_{e}$ ? ::: Zero — no heat, no exhaust speed. Why does slightly fuel-rich hydrogen raise $v_{e}$ ? ::: Leftover $H_{2}$ (molar mass 2) lowers the average exhaust $M$ . Why doesn't "most energy per mole" automatically win? ::: The speed formula depends on $M$ , not on per-mole energy. What is the average exhaust molar mass for methane combustion? ::: $(44 + 2 \cdot 18) /3 \approx 26.7$ g/mol.

Related: Specific Impulse $I_{sp}$ · Hess's Law and Enthalpy of Formation · Stoichiometry and Limiting Reagents