Combustion of hydrocarbons (RP-1 - kerosene, methane) and hydrogen: Worked examples

Child of Combustion of hydrocarbons and hydrogen. Here we do not learn new theory — we stress-test the theory against every kind of case a problem can throw at you: pure fuels, fuel-rich starvation, energy balances, ratios, and exam twists. Each example is labelled with the exact matrix cell it covers.

Before symbols appear, a promise: every letter is earned. $C_{x} H_{y}$ means a fuel built from == $x$ carbon atoms and $y$ hydrogen atoms== glued together — nothing else. $Δ H$ (say "delta H") means the heat energy change: a negative number means heat leaves (gets hot outside), a positive number means heat is soaked up. $n$ always means number of moles (a mole = a fixed huge count of molecules, $6.02 \times 1 0^{23}$ , the chemist's "dozen"). $M$ means molar mass — the mass in grams of one mole.

Two of these get a subscript later, and a subscript is just a tiny label that says "which kind of $Δ H$ ":

$Δ H_{f}$ — the little $f$ stands for ==formation. It is the heat change when one mole of a substance is built from its raw elements== (like building $H_{2} O$ from $H_{2}$ and $O_{2}$ ). Any pure element built from itself changes nothing, so $Δ H_{f} = 0$ for elements.
$Δ H_{c}$ — the little $c$ stands for ==combustion==. It is the special $Δ H$ for the reaction "one mole of fuel burns completely." So $Δ H_{c}$ is just a $Δ H$ wearing a label that says "this is a burning one."

We will re-anchor each symbol the first time it acts.

First, WHERE the $x + \frac{y}{4}$ formula comes from

Every example below leans on one master result, so we earn it once, in full, before using it. We start from the skeleton (products fixed, oxygen unknown): $C_{x} H_{y} + ? O_{2} ⟶ x C O_{2} + \frac{y}{2} H_{2} O$

Why are the product coefficients already $x$ and $\frac{y}{2}$ ? Because carbon has only one home ( $C O_{2}$ , one C each) so $x$ carbons force $x$ molecules of $C O_{2}$ ; and hydrogen has only one home ( $H_{2} O$ , two H each) so $y$ hydrogens force $\frac{y}{2}$ molecules of water. Those two are locked — no freedom.

Now count the oxygen atoms the right side demands: $2 O per C O_{2}, x of them 2 x + 1 O per H_{2} O, \frac{y}{2} of them \frac{y}{2} = 2 x + \frac{y}{2} O atoms .$ Each $O_{2}$ molecule delivers 2 oxygen atoms, so the number of $O_{2}$ molecules is that demand divided by 2: $# O_{2} = \frac{2 x + \frac{y}{2}}{2} = x + \frac{y}{4} .$

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

Read the figure as a two-sided budget. On the top-left (mint box) the $x$ molecules of $C O_{2}$ each carry 2 oxygens, so they demand $2 x$ O atoms. On the middle-left (lavender box) the $\frac{y}{2}$ waters each carry 1 oxygen, demanding $\frac{y}{2}$ O atoms. The two arrows funnel into the butter box: total demand $= 2 x + \frac{y}{2}$ . Now follow the arrow down to the coral box — the supplier: every $O_{2}$ molecule pays 2 oxygen atoms. The final mint box on the right does the only arithmetic: demand ÷ 2-per-molecule $= x + \frac{y}{4}$ . That is why the coefficient is a division: we are converting an atom-demand into a molecule-count, and the "divide by 2" is literally the " $O_{2}$ pays in twos" step drawn as an arrow.

The scenario matrix

Every combustion problem lives in one of these cells. Our job below is to hit all of them.

Cell	Case class	What makes it tricky	Example
A	Balance a clean hydrocarbon (enough $O_{2}$ )	fractional $O_{2}$ coefficient	Ex 1
B	Balance a carbon-free fuel ( $H_{2}$ )	the "no carbon" degenerate case	Ex 2
C	Fuel-rich / insufficient $O_{2}$ (limiting reagent)	products change: $C O$ , soot	Ex 3
D	Energy from formation enthalpies	must ×coefficient, right water state	Ex 4
E	Compare two fuels' energy per kg (not per mole)	normalisation by mass	Ex 5
F	Mass O/F ratio (tank sizing)	moles → mass conversion	Ex 6
G	Real-world word problem (whole vehicle)	density vs $v_{e}$ trade-off	Ex 7
H	Exam twist: degenerate/zero input (pure carbon soot, $y = 0$ )	formula edge, no water at all	Ex 8

Prerequisite tools we lean on: Stoichiometry and Limiting Reagents, Hess's Law and Enthalpy of Formation, Specific Impulse $I_{sp}$, Incomplete Combustion and Soot Formation, Cryogenic Propellants, Rocket Equation (Tsiolkovsky).

Cell A — Balance a clean hydrocarbon

Worked example Ex 1 (Cell A): Balance the complete combustion of ethane

$C_{2} H_{6}$ . Forecast: guess the $O_{2}$ coefficient before reading. Will it be a whole number or a fraction?

Step 1 — Carbon. What: there are $x = 2$ carbons on the left, so we need $2 C O_{2}$ . Why this step? Every carbon atom must land somewhere on the right, and in complete combustion the only carbon-holder is $C O_{2}$ (one C each). So $2$ carbons force $2 C O_{2}$ .

Step 2 — Hydrogen. What: $y = 6$ hydrogens, and each water $H_{2} O$ holds 2 H, so we need $\frac{6}{2} = 3 H_{2} O$ . Why this step? Same logic: H has only one home, water, and water carries H in pairs.

Step 3 — Oxygen (the master formula in action). What: by the budget derived above, O demanded $= 2 x + \frac{y}{2} = 2 (2) + \frac{6}{2} = 4 + 3 = 7$ atoms, so $O_{2} = \frac{7}{2} = 3.5$ — exactly $x + \frac{y}{4} = 2 + 1.5$ . Why this step? Oxygen is fixed last because it is the only species we can freely tune — every other count is locked by C and H. We fill in exactly enough $O_{2}$ to supply the oxygen the products demand. $C_{2} H_{6} + \frac{7}{2} O_{2} ⟶ 2 C O_{2} + 3 H_{2} O$ Clear the fraction (×2): $2 C_{2} H_{6} + 7 O_{2} \to 4 C O_{2} + 6 H_{2} O$ .

Verify: Left atoms — C: $2 \times 2 = 4$ ; H: $2 \times 6 = 12$ ; O: $7 \times 2 = 14$ . Right atoms — C: $4$ ; H: $6 \times 2 = 12$ ; O: $4 \times 2 + 6 \times 1 = 8 + 6 = 14$ . All three match ✓. The master formula predicts $x + \frac{y}{4} = 2 + \frac{6}{4} = 3.5$ ✓.

Cell B — The carbon-free fuel

Worked example Ex 2 (Cell B): Balance hydrogen combustion, and explain why

$x = 0$ is not a "broken" case. Forecast: with no carbon, does the master formula $x + \frac{y}{4}$ still work?

Step 1 — Carbon. What: $x = 0$ , so we need $0 C O_{2}$ . No carbon term at all. Why this step? Hydrogen $H_{2}$ literally contains no carbon; the $C O_{2}$ box is simply empty. In the oxygen budget the $2 x$ term vanishes — this is the degenerate left end of the family, and the formula does not break, it just returns zero for that piece.

Step 2 — Hydrogen. What: one $H_{2}$ carries $y = 2$ H atoms → $\frac{2}{2} = 1 H_{2} O$ .

Step 3 — Oxygen. What: budget $= 2 x + \frac{y}{2} = 0 + 1 = 1$ O atom, so $O_{2} = \frac{1}{2}$ . $H_{2} + \frac{1}{2} O_{2} ⟶ H_{2} O ⟺ 2 H_{2} + O_{2} \to 2 H_{2} O$

Verify: Master formula $x + \frac{y}{4} = 0 + \frac{2}{4} = \frac{1}{2}$ ✓, matching what we balanced by hand. Atom count (×2 version): H — left $2 \times 2 = 4$ , right $2 \times 2 = 4$ ✓; O — left $2$ , right $2 \times 1 = 2$ ✓.

Cell C — Fuel-rich starvation (limiting reagent)

The most important "scenario the textbook hides." When there is not enough oxygen, the clean equation lies. See Incomplete Combustion and Soot Formation.

Worked example Ex 3 (Cell C): 1 mol of methane

$C H_{4}$ is burned with only 1.5 mol of $O_{2}$ (not the 2 it needs). What comes out? Forecast: clean burning needs $2 O_{2}$ . We gave $1.5$ . Will we get $C O_{2}$ , or the lighter $C O$ , or soot? Guess the carbon product.

Step 1 — Fix the hydrogen first. What: hydrogen is "greedy" — it grabs oxygen before carbon fully oxidises. All 4 H become $2 H_{2} O$ , costing $2 H_{2} O \times 1$ O atom each $= 2$ O atoms $= 1 O_{2}$ . Why this step? In combustion, hydrogen's bond to oxygen is energetically favoured, so under starvation water forms preferentially. This is the physics behind the ordering.

Step 2 — Oxygen left for carbon. What: we started with $1.5 O_{2} = 3$ O atoms. Water took $2$ O atoms. Remaining $= 1$ O atom for the single carbon. Why this step? Oxygen is the limiting reagent — we spend it in priority order and see what the carbon gets.

Step 3 — What does 1 O atom make of the carbon? What: $C O_{2}$ needs 2 O per C; we only have 1. So the carbon can at most become carbon monoxide $C O$ (1 O per C). Why this step? With exactly 1 O atom and 1 C atom, the balanced product is $C O$ , not $C O_{2}$ . No leftover oxygen, no soot needed here. $C H_{4} + 1.5 O_{2} ⟶ C O + 2 H_{2} O$ This is the oxygen-budget figure run in reverse: instead of paying the full demand we now ration a fixed 3 O atoms — water first (2 atoms), carbon gets the leftover 1.

Verify: C — left $1$ , right $1$ ✓. H — left $4$ , right $2 \times 2 = 4$ ✓. O — left $1.5 \times 2 = 3$ , right (CO) $1$ + (2 H₂O) $2$ $= 3$ ✓. Note $C O$ has molar mass $28$ vs $C O_{2}$ 's $44$ — the lighter product is exactly why rocket engineers sometimes run fuel-rich to lower exhaust $M$ .

Cell D — Energy from formation enthalpies

Worked example Ex 4 (Cell D): Compute the combustion enthalpy

$Δ H_{c}$ for hydrogen combustion $H_{2} + \frac{1}{2} O_{2} \to H_{2} O (g)$ . Use formation enthalpies $Δ H_{f}$ : $H_{2} = 0$ , $O_{2} = 0$ , $H_{2} O (g) = - 241.8$ kJ/mol. Forecast: will it be near methane's $- 802$ , or much smaller in magnitude?

Step 1 — Recall the tool. What: $Δ H_{c} = \sum Δ H_{f} (products) - \sum Δ H_{f} (reactants)$ . (Reminder: the $c$ says "combustion", the $f$ says "formation from elements" — both are just labelled versions of the plain $Δ H$ from the top of the page.) Why this tool and not just "add bond energies"? Because enthalpy is a state function (Hess's Law and Enthalpy of Formation): the heat depends only on start and end, so we may pretend to tear reactants into elements ( $- Δ H_{f}$ of reactants) and rebuild products ( $+ Δ H_{f}$ of products). That is exactly the boxed formula.

Step 2 — Products. What: one water, coefficient 1: $1 \times (- 241.8) = - 241.8$ . Why this step? Multiply each $Δ H_{f}$ by its coefficient — here it's just 1, so no trap yet (the trap bites methane in the parent note, whose water is ×2).

Step 3 — Reactants. What: $H_{2}$ and $O_{2}$ are elements in standard form, so $Δ H_{f} = 0$ . Sum $= 0$ . Why this step? An element forming itself from itself changes nothing, so $Δ H_{f}^{\circ}$ of any pure element is defined as zero.

Step 4 — Combine. $Δ H_{c} = (- 241.8) - (0) = - 241.8 kJ per mol H_{2}$ .

Verify: negative ⇒ exothermic ✓ (burning must release heat). Magnitude smaller than methane's $- 802$ — but remember this is per mole of a tiny 2 g molecule; Ex 5 shows why hydrogen still wins per kilogram.

Cell E — Energy per kilogram (normalisation twist)

Worked example Ex 5 (Cell E): Which releases more energy

per kilogram of fuel: methane ( $Δ H_{c} = - 802.3$ kJ/mol, $M = 16$ g/mol) or hydrogen ( $Δ H_{c} = - 241.8$ kJ/mol, $M = 2$ g/mol)? Forecast: per mole, methane released 3× more. Per kg, guess who flips ahead.

Step 1 — Convert per-mole to per-gram. What: divide energy by molar mass. Methane: $\frac{802.3 kJ/mol}{16 g/mol} = 50.14$ kJ/g. Hydrogen: $\frac{241.8}{2} = 120.9$ kJ/g. Why this step? The units force it: kJ/mol ÷ g/mol cancels the "mol" and leaves kJ/g — energy per unit mass, which is the honest way to compare two fuels of very different molecule weights.

Step 2 — Scale to per kg ( $\times 1000$ ): methane $\approx 50, 140$ kJ/kg; hydrogen $\approx 120, 900$ kJ/kg. Why this step? Rockets carry fuel by the kilogram, and $Δ v$ in the Rocket Equation (Tsiolkovsky) depends on mass ratios — so "per kg" is the number that actually predicts vehicle performance, not "per gram." The ×1000 is pure unit bookkeeping: 1 kg $=$ 1000 g, so kJ/g $\times \frac{1000 g}{1 kg} =$ kJ/kg (the grams cancel, leaving kilograms). No new chemistry — just re-labelling the same energy against a 1000× bigger mass unit.

Step 3 — Interpret. Hydrogen delivers roughly $120.9/50.14 \approx 2.4 \times$ the energy per kilogram. Why this step? This is the whole reason hydrogen gives the highest Specific Impulse $I_{sp}$ despite its small per-mole number: the molecule is so light that a kilogram of it is a huge pile of moles.

Verify: ratio check $120.9/50.14 = 2.41$ ✓. Sanity: hydrogen's real-world specific energy (~120 MJ/kg) matches textbook ✓.

Cell F — Mass O/F ratio (tank sizing)

Worked example Ex 6 (Cell F): Find the stoichiometric mass oxidizer-to-fuel ratio for hydrogen/oxygen.

Forecast: methane's O/F was $4.0$ . Hydrogen has no carbon — guess higher or lower.

Step 1 — Read mole ratio off the balanced equation. What: $2 H_{2} + O_{2} \to 2 H_{2} O$ means $n_{O_{2}} = 1$ per $n_{f u e l} = 2$ . Why this step? Coefficients ARE mole ratios — that is what a balanced equation encodes.

Step 2 — Convert moles to mass. What: $M_{O_{2}} = 32$ g/mol, $M_{H_{2}} = 2$ g/mol. $O/F = \frac{n _{O_{2}} M _{O_{2}}}{n _{f u e l} M _{f u e l}} = \frac{1 \times 32}{2 \times 2} = \frac{32}{4} = 8.0$ Why this step? Tanks are sized by mass, so we weight each mole by its molar mass.

Step 3 — Interpret. You need 8 kg of oxygen per 1 kg of hydrogen. Why this matters: combined with hydrogen's low density (Cryogenic Propellants), this large ratio means the oxygen tank dwarfs the fuel tank — a real design headache.

Verify: $32/4 = 8.0$ ✓. Cross-check methane from the parent note: $\frac{2 \times 32}{1 \times 16} = 4.0$ ✓, so hydrogen's O/F is exactly double — as expected.

Cell G — Real-world word problem (whole vehicle)

Worked example Ex 7 (Cell G): A stage must burn 100 kg of hydrogen. How much liquid oxygen (LOX) mass is needed, and roughly what tank

volume does that hydrogen occupy? ( $ρ_{L H_{2}} \approx 71$ kg/m³.) Forecast: the mass of oxidizer is set by O/F; the volume problem is where hydrogen bites back.

Step 1 — Oxidizer mass from O/F (Ex 6). What: O/F $= 8.0$ , so LOX $= 8.0 \times 100 = 800$ kg. Why this step? We reuse the ratio directly — that's the point of computing it once.

Step 2 — Hydrogen tank volume. What: $V = \frac{m}{ρ} = \frac{100}{71} = 1.41$ m³. Why this step? Volume $=$ mass ÷ density; low density $\Rightarrow$ big volume, the puffy-balloon problem.

Step 3 — Contrast with RP-1. RP-1 ( $ρ \approx 810$ kg/m³): the same 100 kg would take $100/810 = 0.123$ m³ — about 11× smaller. Why this step? This is the density trade the parent note warns about: hydrogen wins on Specific Impulse $I_{sp}$ but loses on tank size/mass, so whole-vehicle $Δ v$ can favour dense RP-1.

Verify: $8 \times 100 = 800$ ✓; $100/71 = 1.408$ m³ ✓; $100/810 = 0.1235$ m³ ✓; ratio $1.408/0.1235 = 11.4 \times$ ✓.

Cell H — Degenerate exam twist: pure carbon (soot)

Worked example Ex 8 (Cell H): A fuel-rich chamber deposits pure carbon soot, C (

$x = 1, y = 0$ ). Balance its complete combustion, and check the master formula survives $y = 0$ . Forecast: with no hydrogen, will any water appear? What does $x + \frac{y}{4}$ predict?

Step 1 — Carbon. What: $x = 1$ carbon → $1 C O_{2}$ . Why this step? The single carbon has only one home, $C O_{2}$ , so it forces exactly one $C O_{2}$ .

Step 2 — Hydrogen. What: $y = 0$ , so $\frac{y}{2} = 0$ water. No water at all. Why this step? In the oxygen budget the $\frac{y}{2}$ term now vanishes — this is the opposite degenerate end from hydrogen (Ex 2): there the $2 x$ term died, here the water term dies. One formula covers both ends.

Step 3 — Oxygen. What: budget $= 2 x + \frac{y}{2} = 2 + 0 = 2$ O atoms, so $O_{2} = x + \frac{y}{4} = 1 + 0 = 1$ . Why this step? Two O atoms demanded, $O_{2}$ pays two at a time, so exactly one $O_{2}$ — the "divide by 2" arrow in the figure lands on a clean whole number here. $C + O_{2} ⟶ C O_{2}$

Verify: C — left $1$ , right $1$ ✓. O — left $2$ , right $2$ ✓. H — none on either side ✓. Formula $x + \frac{y}{4} = 1$ matches ✓. This closes the family: hydrogen ( $x = 0$ ), hydrocarbons ( $x, y > 0$ ), and pure carbon ( $y = 0$ ) all obey one equation.

Recall Which cell forces the product to change from

$C O_{2}$ to $C O$ ? Cell C — fuel-rich / insufficient $O_{2}$ . Hydrogen grabs oxygen first for water; leftover oxygen may only make $C O$ (or soot) instead of $C O_{2}$ .

Recall Why compare fuels per kilogram, not per mole?

Rockets carry propellant by mass and the rocket equation uses mass ratios; hydrogen's tiny molar mass makes it win per-kg even though it loses per-mole.

Recall What do the subscripts in

$Δ H_{f}$ and $Δ H_{c}$ mean? $f$ = formation (build one mole from raw elements); $c$ = combustion (burn one mole of fuel completely). Both are labelled kinds of the plain heat-change $Δ H$ .

Recall Where does the

$x + \frac{y}{4}$ oxygen coefficient come from? The products demand $2 x + \frac{y}{2}$ oxygen atoms; each $O_{2}$ supplies 2, so divide by 2 to get $x + \frac{y}{4}$ molecules.