Combustion thermodynamics — stoichiometry, adiabatic flame temperature: Exercises

Level 1 — Recognition

Can you name the tool and plug in the numbers? No analysis yet.

L1.1 — Balance an oxidizer-rich pair

Balance the reaction of methane with oxygen to $C O_{2}$ and water vapour: $C H_{4} +_O_{2} \to_C O_{2} +_H_{2} O$

Recall Solution

WHAT we do: conserve each atom (C, H, O) one at a time.

Carbon: 1 C on the left ⇒ 1 $C O_{2}$ .
Hydrogen: 4 H on the left ⇒ 2 $H_{2} O$ (gives 4 H).
Oxygen: right side now has $2$ (from $C O_{2}$ ) $+ 2$ (from $2 H_{2} O$ ) $= 4$ O atoms $= 2 O_{2}$ . $C H_{4} + 2 O_{2} \to C O_{2} + 2 H_{2} O$ Check: left $= 4$ H, $1$ C, $4$ O; right $= 4$ H, $1$ C, $4$ O. ✓

L1.2 — Stoichiometric O/F by mass

For the methane reaction above, compute the stoichiometric oxidizer-to-fuel mass ratio.

Recall Solution

WHY mass, not moles: tanks are filled by mass. The symbol == $O$ stands for the total mass of oxidizer and $F$ stands for the total mass of fuel==, so the ratio $O / F$ is literally "kilograms of oxidizer per kilogram of fuel". We convert the mole ratio to this mass ratio: $(\frac{O}{F})_{s t} = \frac{b M _{o x}}{a M _{f u e l}}$ where $a$ is the stoichiometric coefficient of the fuel (the number written in front of the fuel in the balanced equation) and $b$ is the stoichiometric coefficient of the oxidizer. From the balanced reaction $C H_{4} + 2 O_{2} \to \dots$ , the fuel $C H_{4}$ has coefficient $a = 1$ and the oxidizer $O_{2}$ has coefficient $b = 2$ . $M_{o x}$ and $M_{f u e l}$ are their molar masses. $(\frac{O}{F})_{s t} = \frac{2 \times 32}{1 \times 16} = \frac{64}{16} = 4$ So 4 kg O₂ per 1 kg CH₄.

L1.3 — Sign of an enthalpy

$Δ H_{f}^{\circ} (C O_{2}, g) = - 393.5 kJ/mol$ is negative. What does the negative sign physically mean?

Recall Solution

A negative $Δ H_{f}^{\circ}$ means forming that molecule from its elements ( $C + O_{2}$ ) releases heat — the product's bonds are lower in energy (more stable) than the loose elements. Positive would mean forming it costs heat. So $C O_{2}$ is a stable, heat-releasing "downhill" product — exactly what a fuel wants to end up as.

Level 2 — Application

Now chain two or three steps.

L2.1 — Heat of reaction for methane

Using Hess's law, find $Δ H_{r x n}$ for $C H_{4} + 2 O_{2} \to C O_{2} + 2 H_{2} O (g)$ .

Recall Solution

WHY Hess's law works: enthalpy is a state function — the total change depends only on start and end, so $Δ H_{r x n} = \sum_{p r o d} n_{i} Δ H_{f, i}^{\circ} - \sum_{r e a c t} n_{j} Δ H_{f, j}^{\circ}$ . Recall $Δ H_{f}^{\circ} (O_{2}) = 0$ (element in standard state). $Δ H_{r x n} = [1 (- 393.5) + 2 (- 241.8)] - [1 (- 74.9) + 2 (0)]$ $= (- 393.5 - 483.6) - (- 74.9) = - 877.1 + 74.9 = - 802.2 kJ$ $Δ H_{r x n} \approx - 802.2 kJ per mol CH_{4}$ Negative ⇒ exothermic, as a fuel must be.

L2.2 — Naïve adiabatic flame temperature

Estimate $T_{a d}$ for the methane burn treating $c_{p}$ as constant. Take average hot-gas $c_{p} (C O_{2}) \approx 57 J/mol\cdotK$ and $c_{p} (H_{2} O) \approx 45 J/mol\cdotK$ . The products are exactly $1 C O_{2} + 2 H_{2} O$ .

Recall Solution

WHAT the governing equation says: all released chemical heat goes into raising the product gases, $- Δ H_{r x n} = \sum_{i} n_{i} c_{p, i} Δ T$ . Total heat capacity of the products (per mol of fuel burned): $\sum_{i} n_{i} c_{p, i} = 1 (57) + 2 (45) = 57 + 90 = 147 J/K$ $Δ T = \frac{- Δ H _{r x n}}{\sum _{i} n _{i} c _{p, i}} = \frac{802200 J}{147 J/K} = 5457 K$ $T_{a d} \approx 298 + 5457 \approx 5755 K$ Reality check: methane/oxygen really burns near $3500 K$ — the naïve number is a wild over-estimate for exactly the reasons in L5.1.

L2.3 — Equivalence ratio from a filled tank

An engine is loaded with $3 kg$ of $C H_{4}$ and $9 kg$ of $O_{2}$ . Is it fuel-rich, lean, or stoichiometric? Compute the equivalence ratio $ϕ$ .

Recall Solution

First, the definition. The ==equivalence ratio $ϕ$ == compares how much fuel you actually supplied to how much a perfect (stoichiometric) burn needs: $ϕ = \frac{( F / O ) _{a c t u a l}}{( F / O ) _{s t}} = \frac{( O / F ) _{s t}}{( O / F ) _{a c t u a l}}$ By construction $ϕ = 1$ is stoichiometric, $ϕ > 1$ is fuel-rich (more fuel than a perfect burn needs), and $ϕ < 1$ is fuel-lean (excess oxidizer).

Now the numbers. Actual mass ratio: $(\frac{O}{F})_{a c t} = \frac{9}{3} = 3$ . From L1.2, $(\frac{O}{F})_{s t} = 4$ . $ϕ = \frac{( O / F ) _{s t}}{( O / F ) _{a c t}} = \frac{4}{3} \approx 1.33$ $ϕ > 1$ ⇒ fuel-rich (there is less oxidizer than stoichiometry demands, so fuel is in excess). Intuition check: we only supplied $3$ of the $4$ "units" of oxidizer per fuel unit, so some methane cannot fully burn — consistent with rich.

Level 3 — Analysis

Now weigh competing effects and explain trends.

L3.1 — Why peak temperature sits at $ϕ = 1$

Here $T_{a d}$ is the adiabatic flame temperature — the temperature the product gases reach when all the released chemical heat stays inside them ( $Q = 0$ , no heat leaks out); it is the "how hot does the burn get" number built in the parent note. A student claims "adding more fuel always makes the flame hotter." Sketch/argue what really happens to $T_{a d}$ as $ϕ$ sweeps from lean ( $ϕ < 1$ ) through $ϕ = 1$ to rich ( $ϕ > 1$ ), and explain the shape. See the figure.

Figure — Combustion thermodynamics — stoichiometry, adiabatic flame temperature

Figure — $T_{a d}$ versus $ϕ$ (self-contained caption). The horizontal axis is the equivalence ratio $ϕ$ (dimensionless), sweeping from $0.4$ (lean, left) to $2.0$ (rich, right). The vertical axis is the adiabatic flame temperature $T$ in kelvin. The amber curve rises on the lean side, reaches a maximum near $ϕ = 1$ (dashed cyan line), and falls again on the rich side — an asymmetric hump. The white dot marks the peak; the LEAN block (left) notes "excess oxidizer soaks heat" and the RICH block (right) notes "excess fuel is dead mass". If the image does not render: picture a single rounded hump peaking just past $ϕ = 1$ , low at both ends.

Recall Solution

What the figure shows. The amber curve is $T_{a d}$ as a function of $ϕ$ : it rises from low values on the lean side, reaches a maximum near $ϕ = 1$ (the dashed cyan line marks stoichiometric), then falls again on the rich side — an asymmetric hump peaking just slightly rich. The white dot marks the peak; the left/right text blocks label the LEAN and RICH regions and why each is cooler.

Lean side ( $ϕ < 1$ ): excess oxidizer. That surplus $O_{2}$ does not react but still sits in the chamber absorbing heat (it has heat capacity). Energy released is fixed by the limiting fuel, but it is now shared among more moles ⇒ lower $Δ T$ .
Stoichiometric ( $ϕ = 1$ ): every fuel and oxidizer molecule reacts, no inert dead mass soaking heat ⇒ maximum energy per unit heat capacity ⇒ peak $T_{a d}$ .
Rich side ( $ϕ > 1$ ): excess fuel. Unburnt fuel is dead mass that heats up without releasing energy, and some carbon/hydrogen leaves partially oxidised (less energy released) ⇒ lower $T_{a d}$ . Conclusion: "more fuel = hotter" is false. Beyond $ϕ = 1$ , extra fuel is a heat sink, not a heat source. The real peak often lands slightly rich because dissociation (L5.1) is worst exactly at stoichiometric where $T_{a d}$ is highest.

L3.2 — Two fuels, same energy: which flies faster?

Here $T_{c}$ is the combustion chamber temperature — the temperature of the gas leaving combustion, essentially equal to the adiabatic flame temperature $T_{a d}$ built in the parent note. Fuel A produces exhaust at $T_{c} = 3600 K$ with product molar mass $M = 22 g/mol$ . Fuel B produces $T_{c} = 3000 K$ with $M = 12 g/mol$ . Using $v_{e} \propto T_{c} / M$ , which gives the faster exhaust? By what factor?

Recall Solution

WHY $T / M$ and not just $T$ : the nozzle converts thermal energy into directed kinetic energy; lighter molecules ( $M$ small) accelerate more for the same energy, so both hot and light matter. Compare the ratio: $\frac{v _{e, B}}{v _{e, A}} = \frac{T _{B} / M _{B}}{T _{A} / M _{A}} = \frac{3000/12}{3600/22} = \frac{250}{163.6} = 1.528 \approx 1.236$ $v_{e, B} \approx 1.24 v_{e, A}$ Fuel B wins despite being $600 K$ colder, because its exhaust is nearly half as heavy. This is exactly why hydrogen-rich exhaust ( $M$ low) beats hotter but heavier alternatives — the lesson behind Specific Impulse and Exhaust Velocity and Propellant Selection.

Level 4 — Synthesis

Combine stoichiometry + enthalpy + $T_{a d}$ + performance in one problem.

L4.1 — Full pipeline for H₂/O₂

Starting from raw elements, for $2 H_{2} + O_{2} \to 2 H_{2} O (g)$ : (a) confirm the stoichiometric O/F; (b) find $Δ H_{r x n}$ ; (c) estimate naïve $T_{a d}$ with $c_{p} (H_{2} O) \approx 45 J/mol\cdotK$ ; (d) find the average product molar mass; (e) form the performance figure $T_{a d} / M$ and compare with the methane case (use methane's naïve $T_{a d} = 5755 K$ from L2.2, product $M$ computed the same way).

Recall Solution

(a) O/F — WHY these numbers: from the balanced equation the fuel $H_{2}$ has coefficient $a = 2$ and the oxidizer $O_{2}$ has coefficient $b = 1$ ; converting mole ratio to mass ratio with $O / F = b M_{o x} / (a M_{f u e l})$ gives $(\frac{O}{F})_{s t} = \frac{1 \times 32}{2 \times 2} = 8 kg O_{2} per kg H_{2} . ✓$ This matches the parent note, confirming our bookkeeping.

(b) $Δ H_{r x n}$ — WHY Hess's law: enthalpy is a state function, so the reaction heat is (products − reactants) of formation enthalpies, and the elements $H_{2}, O_{2}$ have $Δ H_{f}^{\circ} = 0$ : $Δ H_{r x n} = 2 (- 241.8) - [2 (0) + 1 (0)] = - 483.6 kJ per 2 mol H_{2} .$ Negative ⇒ exothermic, releasing $483.6 kJ$ .

(c) Naïve $T_{a d}$ : products are $2$ mol H₂O, total heat capacity $2 \times 45 = 90 J/K$ . $Δ T = \frac{483600}{90} = 5373 K \Rightarrow T_{a d} \approx 298 + 5373 \approx 5671 K$

(d) Product molar mass: all product is water, so $M = 18 g/mol$ .

(e) Performance figure: $\frac{T _{a d}}{M}_{H_{2} / O_{2}} = \frac{5671}{18} = 315.1 \approx 17.75$ For methane products ${1 C O_{2}, 2 H_{2} O}$ , total mass $= 44 + 2 (18) = 80 g$ over $3$ mol ⇒ $M = 80/3 \approx 26.7 g/mol$ : $\frac{5755}{26.7} = 215.5 \approx 14.68$ Ratio: $17.75/14.68 \approx 1.21$ . Hydrogen/oxygen delivers about 21% higher exhaust speed — even at similar temperature — because its water-only exhaust is far lighter than methane's $C O_{2}$ -laden exhaust. This is the quantitative core of Propellant Selection and feeds straight into Nozzle Theory and Isentropic Expansion.

Level 5 — Mastery

Now you must judge what is physically real, not just compute.

L5.1 — Why the naïve number is nonsense above 5000 K

Your L4 answer gave $T_{a d} \approx 5671 K$ for H₂/O₂, but real engines measure $\approx 3400 K$ . Explain two independent physical reasons the true value is far lower, and state what the naïve number actually represents.

Recall Solution

Reason 1 — Dissociation (endothermic sink). Above $\sim 2500 K$ product molecules tear apart: $H_{2} O ⇌ OH + H$ , $H_{2} ⇌ 2 H$ , $O_{2} ⇌ 2 O$ . Breaking these bonds absorbs energy that would otherwise raise $T$ . The hotter it gets, the more dissociates — a self-limiting brake (see Chemical Equilibrium and Dissociation). Reason 2 — Rising heat capacity. Real $c_{p} (T)$ grows with temperature as molecular vibration and (eventually) electronic modes activate. The same released energy now buys a smaller $Δ T$ because each degree costs more joules. Our constant $c_{p} \approx 45 J/mol\cdotK$ badly under-counts the true hot-gas capacity, so the real $Δ T$ is much smaller than we computed. What the naïve number actually represents. It is a strict upper bound on $T_{a d}$ — the temperature you would reach if no energy leaked into dissociation and $c_{p}$ never rose with temperature. Reality always sits below it. The correct route (temperature-dependent $c_{p} (T)$ plus Gibbs-energy minimisation over all possible product species, e.g. NASA CEA) rests on the First Law of Thermodynamics and Hess's Law and Enthalpy of Formation. So our $5671 K$ is not "wrong physics" — it is the ceiling the true $\approx 3400 K$ can never exceed.

L5.2 — Design judgement: pick the mixture ratio

For H₂/O₂ the stoichiometric O/F is 8, yet the RS-25 engine runs at O/F ≈ 6 (fuel-rich). Show that this is rich, and argue in one paragraph why a designer chooses to sit off the temperature peak.

Recall Solution

Is it rich? Actual $(O / F)_{a c t} = 6 < 8 = (O / F)_{s t}$ , so $ϕ = (O / F)_{s t} / (O / F)_{a c t} = 8/6 \approx 1.33 > 1$ ⇒ fuel-rich. There is leftover H₂. Why choose it? Three reasons stack:

Lower molar mass. Excess H₂ ( $M = 2$ ) mixes into the exhaust, dragging the average $M$ down. Since $v_{e} \propto T_{c} / M$ , a lower $M$ can raise $v_{e}$ even though $T$ drops slightly — the Specific Impulse and Exhaust Velocity trade.
Material survival. Peak $T_{a d}$ near $ϕ = 1$ can exceed what chamber walls and turbine blades tolerate. Backing off to $ϕ \approx 1.33$ cools the gas to a level the hardware survives, protecting the throat and injector face where heat flux is worst.
Less dissociation loss. Slightly cooler gas dissociates less (see L5.1), so more of the released chemical energy stays as usable directed flow rather than being locked up in torn-apart molecules. The verdict: the designer optimises $T_{c} / M$ subject to a wall-temperature limit, not raw $T$ — which is why the real operating point is deliberately, sensibly, off the temperature peak.

Active Recall

Recall Rapid self-check (hide answers)

Which atom do you balance first, and why? ::: The one appearing in only one product (e.g. C then H); oxygen last because it is spread across several products. $ϕ = 1.33$ means the mixture is…? ::: Fuel-rich (excess fuel, less oxidizer than stoichiometry needs). Why does hydrogen/oxygen beat methane/oxygen even at similar $T$ ? ::: Its water-only exhaust has lower molar mass $M$ , and $v_{e} \propto T_{c} / M$ . The naïve $T_{a d}$ is best described as…? ::: An upper bound; real $T$ is lower due to dissociation and rising $c_{p}$ . Why run an engine off the temperature peak? ::: To lower $M$ (faster exhaust) and stay under the wall-temperature limit.