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

What we do: balance one element at a time. Follow the cartoon below — each atom is tracked as it moves from fuel to products.

Step 1 — carbon (why first): every carbon atom must end up somewhere, and the only product holding carbon is $C{O}_2$. Ethane has $x=2$ carbons, so we need $2C{O}_2$. This decides the $C{O}_2$ coefficient — it is forced to equal $x$. Step 2 — hydrogen: the only product holding hydrogen is $H_{2}O$, and each water hides 2 H atoms. Ethane has $y=6$ hydrogens, so we need $\frac{y}{2}=3$ waters. The $H_{2}O$ coefficient is forced to $\frac{y}{2}$. Step 3 — count oxygen the products now demand: $C{O}_2$ carries 2 O each ($2x$ atoms) and $H_{2}O$ carries 1 O each ($\frac{y}{2}$ atoms). Total O atoms needed $=2x+\frac{y}{2}$. Step 4 — supply that oxygen: each $O_2$ delivers 2 atoms, so $n_{O_2}=\frac{2x+y/2}{2}=x+\frac{y}{4}.$ That is where the $x+\frac{y}{4}$ comes from — not a rule to memorise but a consequence of atom-counting. Here $x=2,y=6\Rightarrow n_{O_2}=2+\frac{6}{4}=3.5$. Products: $2C{O}_2$ and $3H_{2}O$. $C_2{H}_6+3.5O_2⟶2C{O}_2+3H_{2}O$

What we see: $CO$ (not $C{O}_2$), soot, and leftover $H_2$ — all signs the fuel could not fully oxidise. Why this means too little oxygen: full combustion pushes carbon all the way to $C{O}_2$. Stopping at $CO$/soot means oxygen ran out first. Answer: not enough oxidizer — this is incomplete (fuel-rich) combustion. See Incomplete Combustion and Soot Formation.

Step — carbon: 12 C → $12C{O}_2$. Step — hydrogen: 26 H → $13H_{2}O$ (each water holds 2 H). Step — oxygen needed: $2(12)+13=37$ O atoms $=18.5O_2$. $C_{12}{H}_{26}+18.5O_2\to 12C{O}_2+13H_{2}O$ Clear the fraction (×2): $2C_{12}{H}_{26}+37O_2\to 24C{O}_2+26H_{2}O$

What we do: read mole ratio off coefficients, then convert to mass with molar masses. Why mass, not moles: tanks hold mass of propellant, so engineers size tanks by mass ratio. $n_{O_2}=1,M_{O_2}=32$; $n_{H_2}=2,M_{H_2}=2$. $\text{O/F}=\frac{1\times 32}{2\times 2}=\frac{32}{4}=8.0$ So an $H_2/O_2$ engine needs 8 kg of oxygen per 1 kg of hydrogen — a big oxygen tank. See Specific Impulse $I_{sp}$.

Formula (Hess's law): $\Delta H_c=\sum \Delta H_f(\text{products})-\sum \Delta H_f(\text{reactants})$. Why this works: enthalpy is a state function — see Hess's Law and Enthalpy of Formation. The energy-level picture below shows fuel starting above the products; the drop is the heat released.

Gaseous water: $\Delta H_c=(-241.8)-[0+\frac{1}{2}(0)]=-241.8\text{ kJ/mol}$ Liquid water would give $-285.8$ kJ/mol (more heat, because condensing steam releases extra). Which to use: rocket exhaust is hot vapor, so use the gaseous value $-241.8$ kJ/mol.

Why per kg, not per mole: rockets are limited by mass, so specific energy (kJ per kg) is the fair comparison. Convert: energy per kg $=\frac{|\Delta H_c|}{M}\times 1000$ (since $M$ is g/mol, divide by grams then scale to kg). Methane: $\frac{802.3}{16}\times 1000=50,144$ kJ/kg $\approx 50.1$ MJ/kg. Hydrogen: $\frac{241.8}{2}\times 1000=120,900$ kJ/kg $\approx 120.9$ MJ/kg. Ratio: $120.9/50.1\approx 2.41$. Hydrogen releases about 2.4× more energy per kilogram — even though it releases less per mole.

The bar chart below makes the reversal visual: hydrogen's bar is shorter per mole in your head, but taller per kilogram because a mole of $H_2$ weighs only 2 g.

Products: $CO$ ($-110.5$) and $2H_{2}O(g)$ ($2\times -241.8$). Reactants: $C{H}_4$ ($-74.8$) and $\frac{3}{2}{O}_2$ ($0$). $\Delta H_c=[(-110.5)+2(-241.8)]-[(-74.8)+0]$ $=(-594.1)-(-74.8)=-519.3\text{ kJ/mol}$ Energy left on the table: $|-802.3|-|-519.3|=283.0$ kJ/mol — this is the heat that would have been released by oxidising $CO\to C{O}_2$. Why engineers accept this loss: running fuel-rich lowers exhaust molar mass $M$ (leftover light $H_2$/$CO$) and cools the chamber walls, protecting the engine. See Incomplete Combustion and Soot Formation.

Where the formula comes from (why $\ln$ appears): the rocket throws exhaust backward at speed $v_e$. In a tiny instant it ejects mass $-dm$ (fuel burned), and momentum conservation says the momentum gained by the rocket equals the momentum carried off by that exhaust: $mdv=-v_{e}dm.$ What this line says: rocket mass $m$ times its speed gain $dv$ equals exhaust speed $v_e$ times the mass thrown out. Rearrange to isolate the speed gain: $dv=-v_e\frac{dm}{m}.$ Why the logarithm: the term $\frac{dm}{m}$ is the fractional mass change, and summing (integrating) fractional changes from the full mass $m_0$ down to the empty mass $m_f$ gives a natural log — that is exactly what $\ln$ measures: $\Delta v=-v_e{\int }_{m_0}^{m_f}\frac{dm}{m}=v_e\ln \frac{m_0}{m_f}.$ This is the Tsiolkovsky equation — see Rocket Equation (Tsiolkovsky). Now plug in with $\ln (4.0)=1.3863$:

• RP-1: $\Delta v=3050\times 1.3863=4228$ m/s.
• Hydrogen: $\Delta v=4400\times 1.3863=6100$ m/s. Comment: for the same mass ratio, hydrogen delivers ~1870 m/s more $\Delta v$ because its light exhaust ($M\approx 18$, water) flies out faster. But hydrogen's low density often forces a smaller achievable $m_0/m_f$, so in practice RP-1 can compete. See Cryogenic Propellants.

(a) O/F: $\frac{n_{O_2}{M}_{O_2}}{n_{fuel}{M}_{fuel}}=\frac{2\times 32}{1\times 16}=4.0$. ✓ (matches parent note)

(b) Exhaust moles & average M: products are $1C{O}_2+2H_{2}O=3$ moles of gas per mole $C{H}_4$. Total exhaust mass per mole fuel $=44+2(18)=80$ g. Why mass ÷ moles gives the average molar mass: molar mass is "grams per mole." The mixture holds $80$ g of gas spread over $3$ mol of molecules, so on average each mole weighs $\frac{80}{3}$ g — that ratio is, by definition, the mean molar mass. It is a mass-weighted blend of the heavy $C{O}_2$ (44) and lighter $H_{2}O$ (18), and it lands between them. $\bar{M}=\frac{80\text{ g}}{3\text{ mol}}=26.67\text{ g/mol}$ (Heavier than hydrogen's water-only exhaust of $M=18$ — why $H_2$ gives higher $v_e$.)

(c) Mass accounting for 100 kg $C{H}_4$:

• $O_2$ needed $=\text{O/F}\times 100=4.0\times 100=400$ kg.
• Moles of $C{H}_4$: $\frac{100,000\text{ g}}{16}=6250$ mol.
• Exhaust mass $=6250\times 80=500,000$ g $=500$ kg.
• Conservation check: in $=100+400=500$ kg fuel+oxidizer; out $=500$ kg exhaust. ✓ Mass is conserved — nothing vanished. See Stoichiometry and Limiting Reagents.

Reasoning chain:

1. Hydrogen wins on $v_e$ (light exhaust) and on energy/kg (L3.2, 2.4×).
2. But the constraint here is volume, not mass.
3. RP-1 is $810/71\approx 11.4\times$ denser — same volume tank holds ~11× more fuel mass.
4. More propellant mass in a fixed volume → larger achievable mass ratio $m_0/m_f$ → larger $\Delta v$ despite lower $v_e$. Decision: pick RP-1. The deciding number is the density ratio ≈ 11.4, which overrides hydrogen's $v_e$ advantage when volume is the binding constraint. See Cryogenic Propellants and Rocket Equation (Tsiolkovsky).

O/F for CH₄/O₂ ::: $4.0$ O/F for H₂/O₂ ::: $8.0$ $\Delta H_c$ of H₂ (gas water) ::: $-241.8$ kJ/mol Energy per kg: H₂ vs CH₄ ::: ~2.4× more for H₂ Incomplete CH₄ → CO releases ::: $-519.3$ kJ/mol (283 left on table) Why RP-1 despite lower $v_e$ ::: ~11.4× denser → fits tight tanks