Adiabatic flame temperature — calculation with enthalpies of formation

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1. The core idea — an energy bookkeeping problem

WHAT we conserve: enthalpy, at constant pressure. For a constant-pressure, adiabatic, no-work process the First Law gives:

$Q = \Delta H = 0 \quad\Rightarrow\quad H_{\text{products}}(T_{\text{ad}}) = H_{\text{reactants}}(T_{\text{in}})$

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2. Derivation from first principles

Split each species' enthalpy into two pieces (this is the key trick):

$H_i(T) = \underbrace{\Delta H^\circ_{f,i}}_{\text{chemical, at } 298\text{ K}} + \underbrace{\int_{298}^{T} C_{p,i}\,dT}_{\text{sensible heat}}$

Why this split? We only ever tabulate $\Delta H^\circ_f$ at a reference (298 K). Everything else is "how much hotter than 298 K is this molecule," which is sensible heat.

Apply $H_{\text{prod}} = H_{\text{react}}$. Take reactants entering at $T_{\text{in}} = 298$ K so their sensible term vanishes:

= \sum_{\text{react}} n_i\,\Delta H^\circ_{f,i}$$ Rearrange — group the formation enthalpies on one side: $$\boxed{\;\sum_{\text{prod}} n_j\!\int_{298}^{T_{\text{ad}}} C_{p,j}\,dT \;=\; -\,\Delta H^\circ_{\text{rxn}}\;}$$ where $\displaystyle \Delta H^\circ_{\text{rxn}} = \sum_{\text{prod}} n_j\Delta H^\circ_{f,j} - \sum_{\text{react}} n_i\Delta H^\circ_{f,i}$. > [!formula] Master equation > $$\sum_{\text{prod}} n_j \int_{298}^{T_{\text{ad}}} C_{p,j}\,dT = -\Delta H^\circ_{\text{rxn}}$$ > **Read it as:** "The sensible heat soaked up by the products (left) equals the heat the reaction would have released if cooled back to 298 K (right)." > With constant average $C_p$: $\;\sum_j n_j \bar C_{p,j}\,(T_{\text{ad}} - 298) = -\Delta H^\circ_{\text{rxn}}$. ![[5.3.02-Adiabatic-flame-temperature-—-calculation-with-enthalpies-of-formation.png]] --- ## 3. The HOW — a recipe 1. **Balance** the combustion equation (include any excess air/inert $N_2$ — they absorb heat!). 2. Compute $\Delta H^\circ_{\text{rxn}}$ from $\Delta H^\circ_f$ values. 3. Set sensible heat of products = $-\Delta H^\circ_{\text{rxn}}$. 4. **Solve for $T_{\text{ad}}$** (often iterate because $C_p$ rises with $T$). --- ## 4. Worked Example A — Hydrogen in stoichiometric O₂ $$\mathrm{H_2(g) + \tfrac12 O_2(g) \to H_2O(g)}, \quad \Delta H^\circ_f(\mathrm{H_2O,g}) = -241.8\ \text{kJ/mol}$$ - **Why gas-phase water?** At flame temperatures water is vapor; using the liquid value would wrongly add the latent heat. $\Delta H^\circ_{\text{rxn}} = -241.8 - 0 - 0 = -241.8$ kJ (elements have $\Delta H^\circ_f = 0$). Only product: 1 mol H₂O. Take $\bar C_p(\mathrm{H_2O,g}) \approx 0.045$ kJ/mol·K. **Why this step?** All released energy heats just that 1 mol of steam. $$1(0.045)(T_{\text{ad}} - 298) = 241.8 \Rightarrow T_{\text{ad}} \approx 298 + 5373 \approx 5670\ \text{K (unrealistically high)}$$ > [!mistake] The pure-O₂ number is *too high* > **Why it feels right:** the algebra is clean. **The fix:** at >2500 K, products **dissociate** ($\mathrm{H_2O \rightleftharpoons OH + H...}$), absorbing energy and capping real AFT near ~3000 K. Constant-$C_p$ + no-dissociation always *overestimates*. --- ## 5. Worked Example B — Methane in air (more realistic) $$\mathrm{CH_4 + 2O_2 + 7.52\,N_2 \to CO_2 + 2H_2O(g) + 7.52\,N_2}$$ - **Why the $7.52\,N_2$?** Air is ~79% N₂ / 21% O₂, so per mol O₂ we drag in $79/21 = 3.76$ mol N₂; for 2 O₂ that's $7.52$. $\Delta H^\circ_f$: $\mathrm{CH_4}=-74.8$, $\mathrm{CO_2}=-393.5$, $\mathrm{H_2O(g)}=-241.8$ kJ/mol. $$\Delta H^\circ_{\text{rxn}} = [-393.5 + 2(-241.8)] - [-74.8] = -877.1 - (-74.8) = -802.3\ \text{kJ}$$ Product heat capacities (approx, kJ/mol·K): $C_p(\mathrm{CO_2})\approx0.055$, $C_p(\mathrm{H_2O})\approx0.045$, $C_p(\mathrm{N_2})\approx0.033$. $$\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}$$ **Why include N₂?** It's a huge thermal sponge — it's why **air flames are much cooler than O₂ flames**. $$T_{\text{ad}} = 298 + \frac{802.3}{0.393} \approx 298 + 2042 \approx 2340\ \text{K}$$ (Real value ~2230 K once $T$-dependent $C_p$ and minor dissociation are included — close!) > [!example] Quick sanity contrast > Remove the N₂ (burn CH₄ in pure O₂): denominator drops to $0.145$ → $T_{\text{ad}} \approx 298 + 5530 \approx 5800$ K. The N₂ alone slashed AFT by thousands of kelvin. **That's the dual-coding takeaway: the diagram's bar of "N₂ sensible heat" is the tallest.** --- ## 6. Forecast-then-Verify > [!recall] Predict before computing > Q: If I run methane with **excess air** (lean mixture), does AFT go up or down? > > Forecast → **down**: extra cold air/N₂ adds product moles to heat but releases no extra energy → bigger denominator, same numerator → lower $T_{\text{ad}}$. --- ## 7. Common mistakes (Steel-manned) > [!mistake] Using liquid-water $\Delta H^\circ_f$ > **Feels right:** $-285.8$ kJ/mol is the "famous" number (HHV). **Fix:** flame products are vapor → use $-241.8$ (LHV basis). Liquid value overstates released heat. > [!mistake] Forgetting nitrogen as a product > **Feels right:** N₂ "doesn't react." **Fix:** it doesn't react but it *heats up*, soaking energy. Omit it and you overestimate $T_{\text{ad}}$ massively. > [!mistake] Treating $C_p$ as constant > **Feels right:** simple algebra. **Fix:** $C_p$ rises ~30–50% from 300→2500 K. Constant-$C_p$ overestimates $T_{\text{ad}}$; for precision, iterate using $C_p(T)$ or enthalpy tables. --- ## 8. Flashcards #flashcards/chemistry What thermodynamic quantity is conserved for a constant-pressure adiabatic flame? ::: Total enthalpy: $H_{\text{prod}}(T_{ad}) = H_{\text{react}}(T_{in})$, i.e. $\Delta H = 0$. Into what two parts do we split each species' enthalpy? ::: Formation enthalpy at 298 K + sensible heat $\int_{298}^T C_p\,dT$. Why is the real AFT lower than the constant-$C_p$ estimate? ::: Dissociation of products absorbs energy and $C_p$ increases with temperature. How many mol N₂ accompany 1 mol O₂ in air? ::: 3.76 mol (79/21). Why use gas-phase (not liquid) water's $\Delta H^\circ_f$? ::: Products are vapor at flame temperature; liquid value would add unphysical latent heat. Does excess air raise or lower AFT? ::: Lowers it — inert N₂/air absorbs heat without adding chemical energy. Master equation for AFT in words? ::: Sensible heat soaked by products = $-\Delta H^\circ_{rxn}$. $\Delta H^\circ_f$ of any element in standard state? ::: Zero. --- > [!recall]- Feynman: explain to a 12-year-old > Imagine a firework. The gunpowder has energy locked in its bonds. When it burns with the door shut (no heat sneaks out), all that energy has nowhere to go except making the smoke and gas *super hot*. The temperature it reaches is the "adiabatic flame temperature." If you mix in lots of cold air, that air also has to be warmed up — like adding cold water to hot soup — so everything ends up cooler. That's why rockets that breathe pure oxygen burn way hotter than ones using ordinary air. > [!mnemonic] Remember the steps > **"BRSS"** — **B**alance, **R**eaction-enthalpy, **S**ensible-heat-balance, **S**olve. > And: *"Nitrogen is a silent sponge."* ## Connections - [[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]] ## 🖼️ Concept Map ```mermaid flowchart TD A[Fuel burns, bonds release energy] B[Adiabatic Q=0, no work] C[First Law dH=0] D[H products = H reactants] E[Split Hi into parts] F[Formation enthalpy at 298K] G[Sensible heat integral Cp dT] H[Master equation] I[Delta H rxn from formation enthalpies] J[Adiabatic flame temperature Tad] K[Propulsion thrust ceiling] A -->|energy stays inside| B B -->|constant P, no work| C C -->|implies| D D -->|apply per species| E E -->|chemical part| F E -->|thermal part| G F -->|combine reactants minus products| I G -->|sum over products| H I -->|equals negative of| H H -->|solve for| J J -->|sets limit for| K ``` ## 🔊 Hinglish (regional understanding) > [!intuition]- Hinglish mein samjho > Dekho, adiabatic flame temperature ka funda bilkul simple hai: jab fuel jalta hai aur **koi heat bahar nahi nikalti** (adiabatic), toh saari released energy products ko *garam* karne mein lag jaati hai. Jo maximum temperature products reach karte hain, wahi adiabatic flame temperature (AFT) hai. Constant pressure pe First Law kehta hai $\Delta H = 0$, matlab products ka total enthalpy = reactants ka total enthalpy. Energy gayab nahi hui — bas "chemical" form se "thermal" (sensible heat) mein convert ho gayi. > > Calculation ka trick ye hai ki har molecule ka enthalpy do hisson mein todo: ek $\Delta H^\circ_f$ (298 K pe formation), aur doosra sensible heat $\int C_p\,dT$. Reactants ko 298 K pe lao toh unka sensible part zero. Phir master equation banti hai: products ka sensible heat $= -\Delta H^\circ_{rxn}$. Yaani jitni heat reaction release karti, utni hi products absorb kar lete. Isse $T_{ad}$ solve karo. > > Sabse important baat — **nitrogen ko mat bhoolna!** Air mein har 1 mol O₂ ke saath 3.76 mol N₂ aata hai. Ye N₂ react toh nahi karta, par heat zaroor absorb karta hai, isliye air flame ka temperature pure-O₂ flame se bahut kam hota hai (CH₄ in air ~2340 K, par pure O₂ mein ~5800 K). Yahi reason hai ki rockets jo pure oxygen use karte hain, woh zyada hot aur powerful exhaust dete hain. > > Ek aur cheez: constant $C_p$ wala answer hamesha thoda **zyada** aata hai, kyunki real life mein high temperature pe products **dissociate** ho jaate hain (jaise H₂O tutkar OH, H banta hai) aur energy soak karte hain. Toh exam mein constant-$C_p$ method theek hai, par yaad rakhna real AFT thoda kam hota hai. ![[audio/5.3.02-Adiabatic-flame-temperature-—-calculation-with-enthalpies-of-formation.mp3]]