Worked examples — Adiabatic flame temperature — calculation with enthalpies of formation

Before anything, one reminder of the single equation everything below plugs into (built in the parent from the First Law of Thermodynamics):

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The scenario matrix

Every AFT problem is really a choice on a few axes. This table lists every axis and the cell (a specific value) each worked example targets. Our job: leave no cell untouched.

Case class (the "axis")	Cell we must show	Hit by Example
Oxidizer identity	Pure O₂ (no diluent)	A
Oxidizer identity	Ordinary air (N₂ diluent)	B
Mixture ratio	Stoichiometric (exact)	A, B
Mixture ratio	Lean (excess air)	C
Mixture ratio	Rich (excess fuel)	D
Degenerate input	Zero heat release ($\Delta H_{\text{rxn}}^{\circ }=0$)	E
Sign of $\Delta H_{\text{rxn}}^{\circ }$	Endothermic ($\Delta H_{\text{rxn}}^{\circ }>0$) → $T_{\text{ad}}<T_{\text{in}}$	E2
Limiting behaviour	Infinite diluent → $T_{\text{ad}}\to 298$ K	F
Non-constant $C_p$	Two-step iteration with $C_p(T)$	G
Real-world word problem	Rocket chamber design read-off	H
Exam twist	Reactants pre-heated ($T_{\text{in}}\ne 298$)	I

We use one consistent data set so numbers are comparable:

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Example A — Pure O₂, stoichiometric (cell: pure oxidizer)

Forecast: No nitrogen sponge at all, so only 1 mol of steam absorbs everything → expect an absurdly high number (thousands of K above the air case). Guess before reading on.

1. Balance & find $\Delta H_{\text{rxn}}^{\circ }$. Why? The right side of our equation is $-\Delta H_{\text{rxn}}^{\circ }$; we can't proceed without it. $\Delta H_{\text{rxn}}^{\circ }=(-241.8)-0-0=-241.8\text{kJ}.$
2. List product heat budget. Why? The left side sums $n_j{\bar{C}}_{p,j}$ over products only. One product: 1 mol H₂O. $\sum n_j{\bar{C}}_{p,j}=1(0.045)=0.045\text{kJ/K}.$
3. Solve. Why? Set soak-up = hand-over and isolate $T_{\text{ad}}$. $T_{\text{ad}}=298+\frac{241.8}{0.045}=298+5373=5671\text{K}.$

Verify: $0.045\times (5671-298)=0.045\times 5373=241.8$ kJ ✓. Units: $(\text{kJ/K})\times \text{K}=\text{kJ}$ matches the right side. The number is physically unrealistic — see the parent's dissociation caveat via Chemical equilibrium and dissociation at high temperature. That's expected: constant-$C_p$, no-dissociation always overestimates.

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Example B — Air, stoichiometric (cell: air diluent)

Forecast: N₂ rides along, absorbs heat, releases none → far cooler than A. Guess "a bit over 2000 K."

1. $\Delta H_{\text{rxn}}^{\circ }$. Why? Same reason as A. $\Delta H_{\text{rxn}}^{\circ }=[-393.5+2(-241.8)]-(-74.8)=-877.1+74.8=-802.3\text{kJ}.$
2. Product heat budget — include N₂. Why? It doesn't react but it warms up, so it sits on the left. $\sum n_j{\bar{C}}_{p,j}=1(0.055)+2(0.045)+7.52(0.033)=0.055+0.090+0.248=0.393\text{kJ/K}.$
3. Solve. $T_{\text{ad}}=298+\frac{802.3}{0.393}=298+2042=2340\text{K}.$

Verify: $0.393\times 2042=802.5\approx 802.3$ kJ ✓.

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Example C — Lean mixture, 50 % excess air (cell: lean)

Forecast (predict first!): Same fuel → same energy released. But more cold gas to warm (extra O₂ + extra N₂). Bigger denominator, same numerator → lower than B.

1. $\Delta H_{\text{rxn}}^{\circ }$ unchanged. Why? Extra air are elements ($\Delta H_f^{\circ }=0$) that appear identically on both sides; they cancel. So still $-802.3$ kJ.
2. New product budget — now includes leftover O₂. Why? Unburnt O₂ is a product gas that heats up too. $\sum n_j{\bar{C}}_{p,j}=0.055+2(0.045)+1(0.035)+11.28(0.033).$ $=0.055+0.090+0.035+0.37224=0.55224\text{kJ/K}.$
3. Solve. $T_{\text{ad}}=298+\frac{802.3}{0.55224}=298+1452.8=1751\text{K}.$

Verify: $0.55224\times 1452.8=802.3$ kJ ✓. Lower than B's 2340 K — forecast confirmed. This is the lean-mixture penalty in numbers.

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Example D — Rich mixture, 20 % excess fuel (cell: rich)

Forecast: Only 1 mol actually burns (O₂-limited), so energy is the same as B. But the leftover 0.2 mol CH₄ is extra cold gas to heat. Slightly lower than B.

1. $\Delta H_{\text{rxn}}^{\circ }$. Why? The 0.2 mol CH₄ enters and exits unchanged — its formation enthalpy cancels. Only 1 mol CH₄ combusts. $\Delta H_{\text{rxn}}^{\circ }=-802.3\text{kJ}(\text{same as B}).$
2. Product budget with unburnt CH₄. Why include it? It's a warm product gas. We use ${\bar{C}}_p(\mathrm{C}{\mathrm{H}}_4)=0.070$ kJ/mol·K — the value already listed in Shared data above (largest on the list, because methane's many vibrational modes store extra heat). $\sum n_j{\bar{C}}_{p,j}=0.055+0.090+0.2(0.070)+7.52(0.033)=0.055+0.090+0.014+0.248=0.407\text{kJ/K}.$
3. Solve. $T_{\text{ad}}=298+\frac{802.3}{0.407}=298+1971.3=2269\text{K}.$

Verify: $0.407\times 1971.3=802.3$ kJ ✓. Slightly below B's 2340 K — forecast confirmed. Rich and lean both cool the flame; peak sits near stoichiometric.

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Example E — Degenerate: zero net heat release (cell: $\Delta H_{\text{rxn}}^{\circ }=0$)

Forecast: No energy released → nothing heats up → temperature shouldn't move.

1. Plug the zero. Why? To see the degenerate limit explicitly. $\sum n_j{\bar{C}}_{p,j}(T_{\text{ad}}-298)=-0=0.$
2. Interpret. Why? A product of two factors is zero. Since $\sum n_j{\bar{C}}_{p,j}>0$ (real gases have positive heat capacity), the other factor must vanish. $T_{\text{ad}}-298=0\Rightarrow T_{\text{ad}}=298\text{K}.$

Verify: $\text{(any positive number)}\times 0=0$ ✓. The flame temperature equals the inlet temperature — no combustion warmth. This is the sanity floor for exothermic or thermoneutral reactions: AFT never dips below inlet. But what if the reaction absorbs heat? That is the next case.

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Example E2 — Endothermic: $\Delta H_{\text{rxn}}^{\circ }>0$ (cell: positive sign → cooling)

Forecast: The bonds soak up energy rather than giving it out. With no external heat allowed (adiabatic), the only place that energy can come from is the products' own warmth → they must cool down → $T_{\text{ad}}<298$ K.

1. Plug into the same master equation. Why? Nothing about the equation assumed the sign of $\Delta H_{\text{rxn}}^{\circ }$ — it works for any sign. $0.050(T_{\text{ad}}-298)=-\Delta H_{\text{rxn}}^{\circ }=-(+200)=-200\text{kJ}.$
2. Solve — note the negative rise. Why? The right side is now negative, so $(T_{\text{ad}}-298)$ must be negative. $T_{\text{ad}}-298=\frac{-200}{0.050}=-4000\Rightarrow T_{\text{ad}}=298-4000=-3702\text{K}.$
3. Interpret the (absurd) number. Why? A negative kelvin is physically impossible, and that's the lesson: a strongly endothermic reaction cannot sustain itself adiabatically. In reality it simply stalls — it would need heat piped in from outside (violating $Q=0$). The model correctly signals infeasibility by producing a below-absolute-zero temperature.

Verify: $0.050\times (-4000)=-200$ kJ $=-\Delta H_{\text{rxn}}^{\circ }$ ✓ (algebra is self-consistent). Takeaway for the matrix: the sign of $\Delta H_{\text{rxn}}^{\circ }$ sets the direction of the temperature change — negative (exothermic) heats up, zero holds still (Example E), positive (endothermic) cools down and, if large, is adiabatically impossible.

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Example F — Limiting behaviour: flood with inert (cell: infinite diluent)

Forecast: Infinite cold sponge → temperature rise squeezed to zero → $T_{\text{ad}}\to 298$ K.

1. Write $T_{\text{ad}}$ as a function of $x$. Why? To take a limit we need the algebra, not a single number. $T_{\text{ad}}(x)=298+\frac{802.3}{0.393+0.033x}.$
2. Take the limit. Why? "Add infinite diluent" is precisely $x\to \infty$; the tool for "what value does this approach" is the limit. ${\lim }_{x\to \infty }\frac{802.3}{0.393+0.033x}=0\Rightarrow T_{\text{ad}}\to 298\text{K}.$
3. Spot-check a big finite $x$. At $x=1000$: $T_{\text{ad}}=298+\frac{802.3}{33.393}=298+24.0=322\text{K}$.

Verify: $802.3/(0.393+0.033\times 1000)=802.3/33.393=24.03$ K rise ✓ — already almost back to room temperature.

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Example G — Non-constant $C_p$: one iteration (cell: $C_p(T)$)

Forecast: Real $C_p$ rises with $T$, so the sponge is bigger than we assumed → the true $T_{\text{ad}}$ is lower than B's 2340 K.

1. Iteration 0 — the constant-$C_p$ guess. Why start here? We need a temperature to evaluate $C_p(T)$; B gives $T^{(0)}=2340$ K.
2. Evaluate $C_p$ at the guessed $T$. Why? $C_p$ depends on $T$; use the current best $T$. The blow-up factor is common to all species: $1+\beta (T^{(0)}-298)=1+1.5\times 10^{-4}(2042)=1+0.3063=\mathrm{1.3063.}$ So $\sum n_j{\bar{C}}_{p,j}(T^{(0)})=0.393\times 1.3063=0.51338\text{kJ/K}.$
3. Re-solve for a better $T$. Why? Same balance, updated denominator. $T^{(1)}=298+\frac{802.3}{0.51338}=298+1562.8=1861\text{K}.$

Verify: $0.393\times 1.3063=0.51338$ ✓; $0.51338\times 1562.8=802.3$ kJ ✓. The estimate dropped from 2340 K to ~1861 K after one iteration — confirming a hotter-$C_p$ correction always lowers $T_{\text{ad}}$. Real flames need a few more iterations to converge.

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Example H — Real-world word problem: rocket chamber (cell: application)

Forecast: We already found A ≈ 5671 K. The real cap ≈ 3200 K, so dissociation removes ~2400 K.

1. Ideal ceiling. Why? It's the theoretical maximum the material could face if nothing dissipated the energy — worst-case for thrust and material limits. $T_{\text{ad,ideal}}=5671\text{K}(\text{Example A}).$
2. Dissociation buy-back. Why subtract? Bond-breaking at high $T$ re-absorbs chemical energy, lowering the real temperature. $\Delta T=5671-3200=2471\text{K}.$
3. Design read-off. Why? Chamber walls need cooling rated for the real ~3200 K, not the ideal 5671 K, but the ideal number bounds error if cooling fails.

Verify: $5671-3200=2471$ K ✓. Interpretation: over 40 % of the ideal temperature rise ($2471/(5671-298)=0.46$) is "eaten" by dissociation — a huge propulsion-relevant correction, exactly why the parent flags it.

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Example I — Exam twist: pre-heated reactants (cell: $T_{\text{in}}\ne 298$)

Forecast: Reactants arrive already warm → they start with a head-start of thermal energy → $T_{\text{ad}}$ rises above B's 2340 K.

1. Re-open the enthalpy balance with a non-zero reactant sensible term. Why? The parent set $T_{\text{in}}=298$ so that term vanished; now it doesn't. From $H_{\text{prod}}=H_{\text{react}}$: ${\sum }_{\text{prod}}{n}_j{\bar{C}}_{p,j}(T_{\text{ad}}-298)=-\Delta H_{\text{rxn}}^{\circ }+\underset{\text{extra heat carried in}}{\underbrace{\sum _{\text{react}}{n}_i{\bar{C}}_{p,i}(T_{\text{in}}-298)}}.$
2. Build the reactant heat-capacity sum from Shared data. Why? It must come from the actual species entering, not a mystery number. The reactants are 1 mol CH₄, 2 mol O₂, 7.52 mol N₂: ${\sum }_{\text{react}}{n}_i{\bar{C}}_{p,i}=1(0.070)+2(0.035)+7.52(0.033)=0.070+0.070+0.248=0.388\text{kJ/K}.$
3. Compute the head-start term. Why? It adds to the right-hand "hand-over." $\text{extra}=0.388\times (500-298)=0.388\times 202=78.4\text{kJ}.$
4. Solve. Why? Same product denominator as B (0.393), bigger numerator. $T_{\text{ad}}=298+\frac{802.3+78.4}{0.393}=298+\frac{880.7}{0.393}=298+2241.0=2539\text{K}.$

Verify: $0.393\times 2241.0=880.7$ kJ ✓; and $2539>2340$ — pre-heating raised the ceiling, forecast confirmed.

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Recall Self-test: match the cell to the effect

Adding inert N₂ moves $T_{\text{ad}}$ which way? ::: Down (bigger denominator, same numerator). Rich vs lean vs stoichiometric — which is hottest? ::: Stoichiometric (both excesses add cold, non-reacting gas). Pre-heating reactants moves $T_{\text{ad}}$ which way? ::: Up (extra sensible heat enters on the reactant side). What is $T_{\text{ad}}$ when $\Delta H_{\text{rxn}}^{\circ }=0$? ::: The inlet temperature (298 K), the sanity floor. What does $\Delta H_{\text{rxn}}^{\circ }>0$ (endothermic) predict for $T_{\text{ad}}$? ::: A temperature below inlet; if large, an impossible sub-zero kelvin — signalling the reaction cannot run adiabatically. As inert diluent $\to \infty$, $T_{\text{ad}}\to$? ::: 298 K (inlet temperature).

Connections

• 5.3.02 Adiabatic flame temperature — calculation with enthalpies of formation (Hinglish)
• Enthalpy of formation
• Hess's Law
• Heat capacity Cp and its temperature dependence
• Chemical equilibrium and dissociation at high temperature
• Stoichiometric vs lean vs rich mixtures
• Rocket propulsion — specific impulse
• First Law of Thermodynamics