Combustion thermodynamics — stoichiometry, adiabatic flame temperature: Worked examples

Before we start, four symbols you will reuse (all defined in the parent, re-anchored here so nothing is assumed):

Definition The four quantities we juggle

$(\frac{O}{F})$ = oxidizer-to-fuel mass ratio = kilograms of oxidizer per kilogram of fuel. Bigger number = more oxidizer per unit fuel.
$ϕ$ = equivalence ratio = how the actual mixture compares to the perfect (stoichiometric) one. $ϕ = 1$ perfect, $ϕ > 1$ fuel-rich (extra fuel), $ϕ < 1$ fuel-lean (extra oxidizer).
$- Δ H_{r x n}$ = heat released per the balanced reaction (a positive number of kilojoules), which then heats the product gases up to $T_{a d}$ .
$M$ = average product molar mass = the mass of one "average mole" of exhaust gas, in grams per mole. If the exhaust were pure water vapour it would be $18$ ; leftover light hydrogen ( $2$ g/mol) drags this average down, heavy leftover oxygen ( $32$ ) pushes it up. We care about $M$ because exhaust speed scales as $T_{c} / M$ — light gas squirts out faster.

The scenario matrix

Every combustion problem lands in one of these cells. The examples below are labelled with the cell they hit, and together they cover all of them.

Cell	Case class	What makes it tricky	Example
A	Stoichiometric ( $ϕ = 1$ ) balance + O/F	the baseline; get atoms exact	Ex 1
B	Fuel-rich ( $ϕ > 1$ )	leftover fuel is dead mass, soaks heat	Ex 2
C	Fuel-lean ( $ϕ < 1$ )	leftover oxidizer is dead mass	Ex 3
D	Non-oxygen oxidizer (oxidizer carries its own O)	balance C, H, N and O across two compounds	Ex 4
E	Heat of reaction sign / product phase	liquid vs gas water changes $Δ H$	Ex 5
F	Constant- $c_{p}$ estimate of $T_{a d}$	naïve upper bound	Ex 6
G	Degenerate / limiting input ( $ϕ \to 0$ , $ϕ \to \infty$ )	what happens at the edges	Ex 7
H	Real-world word problem (tank sizing)	translate story → O/F → mass	Ex 8
I	Exam twist ("hotter ≠ better")	maximise $T / M$ , not $T$	Ex 9

Ex 1 — Cell A: the stoichiometric baseline

Forecast: guess — is the O/F ratio bigger or smaller than the value 8 we got for hydrogen? Jot a number.

Write the skeleton and conserve carbon. Why this step? Each atom on the left must reappear on the right — that is the whole rule. $C H_{4} + x O_{2} \to 1 C O_{2} + 2 H_{2} O$ One C on the left → one $C O_{2}$ . Four H on the left → two $H_{2} O$ (each holds 2 H).
Now conserve oxygen to find $x$ . Why this step? Oxygen is the only unknown coefficient left. Right side O: $2$ (from $C O_{2}$ ) $+ 2$ (from two $H_{2} O$ ) $= 4$ . So $2 x = 4 \Rightarrow x = 2$ . $C H_{4} + 2 O_{2} \to C O_{2} + 2 H_{2} O$
Convert mole ratio → mass ratio. Why this step? Engineers store mass in tanks, not molecules, so the useful ratio is by kilogram. $(\frac{O}{F})_{s t} = \frac{2 \cdot M _{O_{2}}}{1 \cdot M _{C H_{4}}} = \frac{2 \cdot 32}{1 \cdot 16} = 4$

Verify: left atoms = {1 C, 4 H, 4 O}; right atoms = {1 C (CO₂), 4 H (2×H₂O), 2+2=4 O}. Matched. Mass check: reactant mass $16 + 64 = 80$ g = product mass $44 + 2 \times 18 = 80$ g. Conserved. $(O / F)_{s t} = 4$ — smaller than hydrogen's 8, because methane already carries its own hydrogen cheaply.

Ex 2 — Cell B: fuel-rich mixture

Forecast: fuel-rich means extra fuel. Should the actual O/F be above or below 8? Decide before reading.

Recall exactly what $ϕ$ measures, then rearrange it. Why this step? The parent defined $ϕ = \frac{( O / F ) _{s t}}{( O / F ) _{a c t u a l}}$ — a fraction whose top is fixed (the perfect ratio, 8) and whose bottom is the number we actually run. So $ϕ$ is really the question "how many times more oxidizer-starved is my real mixture than the perfect one?" To get the real ratio, we invert that fraction: multiply both sides by $(O / F)_{a c t u a l}$ and divide by $ϕ$ . This isolates the one unknown. $ϕ = \frac{( O / F ) _{s t}}{( O / F ) _{a c t u a l}} ⟹ ϕ \cdot (O / F)_{a c t u a l} = (O / F)_{s t} ⟹ (O / F)_{a c t u a l} = \frac{( O / F ) _{s t}}{ϕ} = \frac{8}{1.5} = 5.33$ Sanity of the algebra: since $ϕ > 1$ we are dividing 8 by a number bigger than 1, so the answer must come out below 8 — exactly what "extra fuel, less oxidizer" should mean.
Interpret. Why this step? A number is useless without meaning. We now feed only $5.33$ kg O₂ per kg H₂, less oxygen than the 8 needed — so hydrogen is in surplus. Confirms fuel-rich.
Which leftover? Excess hydrogen survives the burn. Why it matters: leftover H₂ is very light ( $M = 2$ g/mol), which drags the average product molar mass $M$ down — good for exhaust speed (see Specific Impulse and Exhaust Velocity).

Verify: $ϕ > 1$ should give actual O/F below stoichiometric 8, and it does ( $5.33 < 8$ ). Cross-check via fuel/oxidizer form: $(F / O)_{a c t u a l} = 1/5.33 = 0.1875$ , $(F / O)_{s t} = 1/8 = 0.125$ , ratio $0.1875/0.125 = 1.5 = ϕ$ . ✓

Ex 3 — Cell C: fuel-lean mixture

Forecast: lean = extra oxidizer. Above or below 4?

Solve for actual O/F. Why this step? Same inversion as Ex 2 — $ϕ = (O / F)_{s t} / (O / F)_{a c t u a l}$ ⇒ $(O / F)_{a c t u a l} = (O / F)_{s t} / ϕ$ — but now $ϕ < 1$ , so we divide 4 by a number smaller than 1 and the answer must rise above 4. $(O / F)_{a c t u a l} = \frac{( O / F ) _{s t}}{ϕ} = \frac{4}{0.8} = 5$
Interpret. Why this step? We now supply 5 kg O₂ per kg fuel where only 4 are needed → excess oxygen survives. That is the meaning of lean.
Consequence. Leftover O₂ ( $M = 32$ g/mol) is heavy and unreacted — it raises the average $M$ and soaks heat without releasing any, lowering $T$ . Bad on both counts for a rocket, which is why lean is rare in launchers.

Verify: $ϕ < 1$ ⇒ actual O/F above 4 ⇒ we got 5. ✓ Consistency: $(F / O)_{a c t u a l} = 0.2$ , $(F / O)_{s t} = 0.25$ , ratio $0.2/0.25 = 0.8 = ϕ$ . ✓

Ex 4 — Cell D: a non-oxygen oxidizer (the oxidizer brings its own oxygen)

Forecast: with an oxidizer this heavy (92 vs O₂'s 32), do you expect the O/F ratio to be big or small compared with the H₂/O₂ value of 8?

Write the skeleton with unknown coefficients on both fuel and oxidizer. Why this step? Unlike O₂ (a single element), here two compounds meet, so both need coefficients. We also have three elements to conserve — N, H, O — not two. $a N_{2} H_{4} + b N_{2} O_{4} \to c N_{2} + d H_{2} O$
Conserve hydrogen to tie $d$ to $a$ . Why this step? Only the fuel carries H, so hydrogen fixes the water coefficient immediately. Left H $= 4 a$ ; right H $= 2 d$ ⇒ $d = 2 a$ .
Conserve oxygen to tie $b$ to $d$ (hence to $a$ ). Why this step? Only $N_{2} O_{4}$ carries O, and only water carries it out — this is the step O₂-only problems never have. Left O $= 4 b$ ; right O $= d = 2 a$ ⇒ $4 b = 2 a \Rightarrow b = \frac{a}{2}$ .
Set $a = 1$ and read off the rest, then clear the fraction. Why this step? We pick the simplest fuel coefficient and scale. With $a = 1$ : $d = 2$ , $b = \frac{1}{2}$ . Conserve nitrogen for $c$ : left N $= 2 a + 2 b = 2 + 1 = 3$ , right N $= 2 c$ ⇒ $c = \frac{3}{2}$ . Multiply all by 2 to get whole numbers: $2 N_{2} H_{4} + N_{2} O_{4} \to 3 N_{2} + 4 H_{2} O$
Convert to mass ratio. Why this step? Tanks store mass; here $a = 2$ fuel, $b = 1$ oxidizer. $(\frac{O}{F})_{s t} = \frac{b M _{N_{2} O_{4}}}{a M _{N_{2} H_{4}}} = \frac{1 \cdot 92}{2 \cdot 32} = \frac{92}{64} = 1.44$

Verify: atoms — left {N: $2 \times 2 + 1 \times 2 = 6$ , H: $2 \times 4 = 8$ , O: $1 \times 4 = 4$ }; right {N: $3 \times 2 = 6$ , H: $4 \times 2 = 8$ , O: $4 \times 1 = 4$ }. All matched. ✓ Mass: reactants $2 \times 32 + 1 \times 92 = 156$ g $=$ products $3 \times 28 + 4 \times 18 = 84 + 72 = 156$ g. ✓ $(O / F)_{s t} \approx 1.44$ — far below H₂/O₂'s 8, because hydrazine is very oxygen-hungry-cheap and $N_{2} O_{4}$ delivers four O atoms per molecule.

Ex 5 — Cell E: heat of reaction, and the phase trap

Forecast: will $Δ H_{r x n}$ be negative (heat out) or positive? By roughly how much per mole of fuel?

Apply Hess's law: products minus reactants. Why this step? Enthalpy is a state function (path-independent), so the reaction's $Δ H$ is just the difference of the tabulated formation enthalpies — see First Law of Thermodynamics. $Δ H_{r x n} = [1 (- 393.5) + 2 (- 241.8)] - [1 (- 74.8) + 2 (0)]$
Crunch it. Why this step? Just arithmetic, but watch signs — a dropped minus flips exothermic↔endothermic. $= (- 393.5 - 483.6) - (- 74.8) = - 877.1 + 74.8 = - 802.3 kJ$
Read the sign. Why this step? Negative $Δ H_{r x n}$ = exothermic = heat leaves the chemistry and (in a chamber) heats the gases. So $- Δ H_{r x n} = + 802.3$ kJ released per mole of CH₄.

Verify: magnitude is sensible — methane's textbook lower heating value is ≈ 802 kJ/mol. ✓ The phase trap: if water condensed to liquid ( $Δ H_{f}^{\circ} (H_{2} O, l) = - 285.8$ ) you'd get more heat ( $- 890$ kJ, the higher heating value). In a rocket the water stays vapour, so use the gas value.

Ex 6 — Cell F: naïve constant- $c_{p}$ estimate of $T_{a d}$

Forecast: will this crude number be above or below the real ~3000 K? By how much?

Write the energy balance. Why this step? Adiabatic + constant pressure ⇒ all released heat goes into raising the product gases (parent §3). $- Δ H_{r x n} = (\sum_{i} n_{i} c_{p, i}) Δ T$
Total heat capacity of the products. Why this step? Every product mole soaks up heat; add their capacities. $\sum_{i} n_{i} c_{p, i} = 1 (57) + 2 (45) = 57 + 90 = 147 J/K$
Solve for $Δ T$ and add $T_{0}$ . Why this step? Divide energy by heat capacity — that's the definition of a temperature rise. $Δ T = \frac{802300}{147} = 5458 K, T_{a d} \approx 298 + 5458 = 5756 K$

Verify: units — J ÷ (J/K) = K. ✓ The number $\approx 5760$ K is roughly double the real ~3000 K, exactly the overshoot the parent warned about: this ignores dissociation and the rise of $c_{p}$ with temperature. It is an upper bound, not a prediction.

Figure 1 — the energy-into-one-bucket picture (read it now): In the figure below the two coloured bars are the heat capacities of the products: the violet bar is the single mole of CO₂ ( $57$ J/K), the orange bar is the two moles of H₂O ( $90$ J/K). Their heights add to the dashed magenta line at $147$ J/K — the combined bucket. The whole $802300$ J of released chemical energy (magenta arrow) is poured into that one bucket, and the temperature rise is simply energy ÷ bucket size. The pedagogical punch: a bigger bucket — more product moles, or new species created by dissociation — gives a smaller rise, which is exactly why the real flame comes out cooler than this crude number.

Figure — Combustion thermodynamics — stoichiometry, adiabatic flame temperature

Alt text: bar chart of product heat capacities (CO₂ 57 J/K, H₂O 90 J/K) summing to a dashed line at 147 J/K, with an arrow labelling the 802 300 J of released energy divided by that capacity to give a 5458 K rise.

Ex 7 — Cell G: the degenerate limits

Forecast: at which extreme is there no fire at all?

Limit $ϕ \to 0^{+}$ (infinitely lean). Why this step? Read the definition $(O / F)_{a c t u a l} = (O / F)_{s t} / ϕ$ . As $ϕ \to 0$ , $(O / F)_{a c t u a l} \to \infty$ : oceans of oxidizer, a trace of fuel. $lim_{ϕ \to 0^{+}} (O / F)_{a c t u a l} = + \infty$ The tiny fuel releases a tiny heat, spread across a huge cold mass of excess oxidizer ⇒ $T_{a d} \to T_{0}$ . Essentially no temperature rise.
Limit $ϕ \to \infty$ (infinitely rich). Why this step? Same formula: $(O / F)_{a c t u a l} \to 0$ , almost no oxidizer. $lim_{ϕ \to \infty} (O / F)_{a c t u a l} = 0$ With no oxidizer, almost nothing burns ⇒ again $T_{a d} \to T_{0}$ . No fire.
Where is the peak? Why this step? Both edges are cold, so the maximum $T_{a d}$ sits between them, near $ϕ = 1$ . This is the shape you must remember: a hump peaking around stoichiometric.

Verify: both limits give $T_{a d} \to T_{0}$ , consistent with the physical fact that you need both reactants present to release heat. The maximum-in-the-middle behaviour matches the mistake callout in the parent ("more fuel ≠ hotter"). ✓

Figure 2 — the temperature hump (read it now): In the figure below the magenta curve is $T_{a d}$ plotted against the equivalence ratio $ϕ$ (horizontal axis) in kelvin (vertical axis). Trace it from the far left: at $ϕ \to 0$ (left orange arrow, "lean") the curve hugs the dotted floor marked " $T_{0}$ , no fire". Now walk to the far right: at $ϕ \to \infty$ (right orange arrow, "rich") the curve sinks back to that same floor. The only high ground is the navy dot near the violet vertical line at $ϕ = 1$ . That single hump is the visual proof of "more fuel ≠ hotter": push $ϕ$ past the peak and you slide straight back downhill toward $T_{0}$ .

Alt text: a hump-shaped magenta curve of adiabatic flame temperature versus equivalence ratio, hugging a dotted "no fire" floor at both extremes and peaking with a navy dot near the violet phi=1 line, with orange arrows labelling the lean and rich cold limits.

Ex 8 — Cell H: real-world tank sizing

Forecast: LOX is heavier per unit — will the oxidizer tank hold more or less mass than the fuel tank?

Translate the ratio into a mass. Why this step? $(O / F) = 8$ literally means 8 kg O₂ per 1 kg H₂ — multiply. $m_{O_{2}} = 8 \times m_{H_{2}} = 8 \times 12000 = 96000 kg$
Total propellant. Why this step? Both tanks contribute to lift-off mass. $m_{t o t a l} = m_{H_{2}} + m_{O_{2}} = 12000 + 96000 = 108000 kg$
Sanity on the split. Why this step? The fuel is only $12000/108000 = 11.1%$ of propellant mass, even though it does the heavy lifting energetically. This is why LOX tanks dwarf LH₂ tanks by mass (though LH₂ tanks are bigger by volume — hydrogen is fluffy). See Propellant Selection.

Verify: $96000/12000 = 8 = (O / F)_{s t}$ . ✓ Fuel fraction $12000/108000 = 0.1111$ . ✓ Units all kilograms.

Ex 9 — Cell I: the exam twist — hotter is not always better

Forecast: Mix P is hotter. Does hotter win?

Write the figure of merit. Why this step? Exhaust speed scales as $T_{c} / M$ , not as $T_{c}$ alone — so compare that quantity, not the temperatures. $score \propto \frac{T _{c}}{M}$
Score Mix P. Why this step? Plug the hotter, heavier pair in. $\frac{T _{c}}{M}_{P} = \frac{3600}{18} = 200 = 14.14$
Score Mix Q. Why this step? Same formula, the cooler but lighter fuel-rich pair. $\frac{T _{c}}{M}_{Q} = \frac{3300}{13} = 253.8 = 15.93$
Compare and conclude. Why this step? Higher score = higher exhaust speed. Since $15.93 > 14.14$ , the cooler, fuel-rich Mix Q wins, even though it is 300 K colder. The gain is $15.93/14.14 = 1.127$ , i.e. about 13% more exhaust speed, because dropping $M$ from 18 to 13 outweighs the temperature loss. This is exactly why real engines run slightly rich rather than at peak temperature.

Figure 3 — the Hot ÷ Heavy trade-off (read it now): In the figure below each bar is a mixture's exhaust-speed score $T_{c} / M$ (vertical axis). The violet bar (Mix P) is the hotter, heavier stoichiometric case; the magenta bar (Mix Q) is the cooler, lighter fuel-rich case. Despite being 300 K colder, the magenta bar is taller — the orange arrow marks that Q wins by about $12.7%$ . The lesson to burn in: you chase $T / M$ , not $T$ alone.

Alt text: two bars comparing exhaust-speed scores, violet Mix P at 14.14 and magenta Mix Q at 15.93, with an orange arrow noting the cooler-but-lighter mixture wins by 12.7 percent.

Verify: ratio of scores $15.93/14.14 = 1.127$ , a 12.7% gain for Q. ✓ Confirms the parent's "chase $T / M$ , not $T$ " mnemonic. See Nozzle Theory and Isentropic Expansion for the full $v_{e}$ formula.

Active Recall

Recall Which cell was which?

Fuel-rich means actual O/F is ::: below stoichiometric (less oxidizer per unit fuel), leaving excess light fuel.
Both $ϕ \to 0$ and $ϕ \to \infty$ give ::: $T_{a d} \to T_{0}$ — no complete reaction, so peak $T$ sits near $ϕ = 1$ .
The naïve constant- $c_{p}$ flame temperature is ::: an upper bound; dissociation and rising $c_{p}$ pull the real value far lower.
Fuel-rich H₂/O₂ can beat stoichiometric because ::: lower average molar mass $M$ raises $T_{c} / M$ more than the temperature drop lowers it.
A non-O₂ oxidizer like $N_{2} O_{4}$ forces you to ::: balance oxygen (and nitrogen) that lives inside a compound, not as free O₂.