Equilibrium products at high T — dissociation (H₂O ⇌ OH + H; CO₂ ⇌ CO + ½O₂)

Combustion does not stop at clean CO₂ and H₂O. At flame temperatures (2000–3500 K) the product molecules partly fall apart into radicals and smaller species. This costs energy and lowers the achievable flame temperature.

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1. Why does dissociation happen at all? (first principles)

Deriving the link between $\Delta G^\circ$ and $K_p$

Start from the chemical potential of an ideal gas species $i$: $\mu_i = \mu_i^\circ + RT\ln\!\left(\frac{p_i}{p^\circ}\right)$

Why this step? $\mu_i^\circ$ is the standard chemical potential (fixed $T$, $p^\circ=1\,\text{bar}$); the log term accounts for the actual partial pressure. This is the definition of an ideal-gas potential.

At equilibrium the total Gibbs energy is stationary, so $\sum_i \nu_i \mu_i = 0$ (with $\nu_i$ = stoichiometric coefficient, + for products, − for reactants):

$\sum_i \nu_i \mu_i^\circ + RT\sum_i \nu_i \ln\!\left(\frac{p_i}{p^\circ}\right)=0$

Why this step? We just substituted each $\mu_i$. The first sum is $\Delta G^\circ_{rxn}$ by definition.

$\boxed{\Delta G^\circ = -RT\ln K_p}\qquad K_p=\prod_i\left(\frac{p_i}{p^\circ}\right)^{\nu_i}$

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2. The two canonical dissociations

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3. Worked example A — CO₂ dissociation degree $\alpha$

Problem: 1 mol $CO_2$ at total pressure $P$, temperature $T$, dissociates as $CO_2 \rightleftharpoons CO + \tfrac12 O_2$. Fraction dissociated $=\alpha$. Find $K_p$ in terms of $\alpha$ and $P$.

Species	Initial	At equilibrium
$CO_2$	1	$1-\alpha$
$CO$	0	$\alpha$
$O_2$	0	$\alpha/2$

Total moles $n_{tot}=1-\alpha+\alpha+\alpha/2 = 1+\alpha/2$.

Why this step? Half a mole of $O_2$ appears per mole of $CO_2$ broken, so total gas count grows — a hallmark of dissociation (more particles).

Mole fractions → partial pressures $p_i = x_i P$: $p_{CO_2}=\frac{1-\alpha}{1+\alpha/2}P,\quad p_{CO}=\frac{\alpha}{1+\alpha/2}P,\quad p_{O_2}=\frac{\alpha/2}{1+\alpha/2}P$

=\frac{\alpha}{1-\alpha}\left(\frac{\alpha/2}{1+\alpha/2}\,P\right)^{1/2}$$ **Why this step?** Plug each $p_i$ into the $K_p$ definition; the $P^{1/2}$ survives because $\sum\nu_i = (1+\tfrac12)-1=+\tfrac12$ (net mole change). > [!formula] Pressure effect (Le Chatelier, quantitatively) > Because $\sum\nu_i = +\tfrac12>0$, **raising $P$ suppresses $\alpha$** at fixed $K_p$. High-pressure rocket chambers therefore dissociate *less* than low-pressure ones — a genuine design lever. For small $\alpha\ll1$: $K_p \approx \alpha^{3/2}\sqrt{P/2}$, so $\alpha\propto (K_p^2/P)^{1/3}$. --- ## 4. Worked example B — numbers for H₂O **Problem:** At $T=2500\ \text{K}$, $\Delta G^\circ = +118\ \text{kJ/mol}$ for $\mathrm{H_2O\rightleftharpoons OH+\tfrac12 H_2}$. Find $K_p$, comment. $$K_p=\exp\!\left(-\frac{118000}{8.314\times2500}\right)=\exp(-5.68)\approx 3.4\times10^{-3}$$ **Why this step?** Direct use of the master formula with consistent SI units ($R$ in J/mol·K). The result $K_p\sim10^{-3}$ means dissociation is **small but non-negligible** — exactly the few-percent radical pool real flames carry. **Forecast-then-Verify:** *Before computing*, predict: raising $T$ to 3000 K should make $K_p$ **bigger** (endothermic, van 't Hoff). With $\Delta H^\circ\approx 240\ \text{kJ/mol}$: $$\ln\frac{K_2}{K_1}=-\frac{\Delta H^\circ}{R}\!\left(\frac1{T_2}-\frac1{T_1}\right)=-\frac{240000}{8.314}\!\left(\frac1{3000}-\frac1{2500}\right)\approx +1.92$$ $\Rightarrow K_2/K_1\approx e^{1.92}\approx 6.8$. **Verified:** $K_p$ jumps ~7× for +500 K. Forecast correct: dissociation rises sharply with $T$. --- > [!mistake] Steel-manning the common errors > **Error 1 — "Combustion goes to completion, products are pure CO₂ + H₂O."** > *Why it feels right:* That's what stoichiometry equations show, and at room temperature $K_p$ for re-combination is astronomically large. *The fix:* $K_p$ is **temperature dependent**; at 2500–3500 K the dissociation $K_p$ is no longer negligible. Always evaluate $K_p(T)$ at the *flame* temperature, not 298 K. > > **Error 2 — "Pressure doesn't matter, $K_p$ depends only on $T$."** > *Why it feels right:* $K_p$ truly is a function of $T$ alone. *The fix:* but the **degree of dissociation $\alpha$** depends on $P$ because $\sum\nu_i\neq0$. Same $K_p$, different $\alpha$ — squeeze the gas and dissociation shrinks. > > **Error 3 — Forgetting the $\tfrac12$ exponent on $O_2$.** > *Why it feels right:* Looks tidier to write $p_{O_2}$. *The fix:* the exponent in $K_p$ is the **stoichiometric coefficient**, here $\tfrac12$, so it's $p_{O_2}^{1/2}$. Getting this wrong throws $\alpha$ off badly. --- > [!recall]- Feynman: explain to a 12-year-old > Imagine LEGO castles (CO₂, H₂O). If you shake the box gently (warm), they stay built. If you shake it *really hard* (super hot flame), pieces snap off — you get loose bricks (OH, CO, O₂). Snapping bricks apart **uses up** some of your shaking energy, so the box can't get as hot as you hoped. And if you press all the LEGO tightly into a small box (high pressure), they're packed so they break apart less. That's why hot engines never burn 100% cleanly, and why pressure helps keep things together. > [!mnemonic] Remember the direction > **"HEAT BREAKS, SQUEEZE MAKES (whole)."** > Heat ↑ → dissociation ↑ (endothermic). Pressure ↑ → dissociation ↓ (more moles on product side). > And $\Delta G^\circ=-RT\ln K$: "**G**ets **R**eally **T**iny when bonds break easily" — big positive $\Delta G$ ⇒ tiny $K$. --- ## Active recall #flashcards/chemistry Why does dissociation increase with temperature (thermodynamic reason)? ::: Dissociation is endothermic ($\Delta H^\circ>0$); the $-T\Delta S^\circ$ term in $\Delta G^\circ$ grows with $T$, lowering $\Delta G^\circ$ and raising $K_p$ (van 't Hoff: $d\ln K_p/dT=\Delta H^\circ/RT^2>0$). State the master relation between $\Delta G^\circ$ and $K_p$. ::: $\Delta G^\circ=-RT\ln K_p$, i.e. $K_p=\exp(-\Delta G^\circ/RT)$. For $CO_2\rightleftharpoons CO+\tfrac12 O_2$ with degree $\alpha$ from 1 mol, give $K_p(\alpha,P)$. ::: $K_p=\dfrac{\alpha}{1-\alpha}\left(\dfrac{\alpha/2}{1+\alpha/2}P\right)^{1/2}$. Does raising total pressure increase or decrease $\alpha$ here, and why? ::: Decrease — net mole change $\sum\nu_i=+\tfrac12>0$, so higher $P$ shifts equilibrium toward fewer moles (back to $CO_2$). $K_p$ stays fixed. Why is the real adiabatic flame temperature lower than the "complete combustion" value? ::: Endothermic dissociation absorbs energy that would otherwise raise $T$, so the achievable temperature drops. Which radical does H₂O dissociation supply, and why is it important? ::: $OH$ — the key chain-carrier radical that drives combustion kinetics. Compute $K_p$ if $\Delta G^\circ=+118$ kJ/mol at 2500 K. ::: $K_p=\exp(-118000/(8.314\times2500))\approx 3.4\times10^{-3}$. Why must you NOT use 298 K $K_p$ values for flame products? ::: $K_p$ is strongly $T$-dependent; at flame $T$ (2000–3500 K) dissociation $K_p$ is many orders of magnitude larger than at 298 K. --- ## Connections - [[Adiabatic Flame Temperature]] — dissociation is the main reason measured $T_{ad}$ < ideal. - [[Equilibrium Constant Kp and Kc]] — the $K_p=e^{-\Delta G^\circ/RT}$ engine used here. - [[Le Chatelier's Principle]] — explains pressure & temperature shifts of $\alpha$. - [[Van 't Hoff Equation]] — quantifies how $K_p$ climbs with $T$. - [[Combustion Radicals OH H O]] — the species dissociation produces. - [[Rocket Nozzle Frozen vs Equilibrium Flow]] — whether dissociated species recombine in the nozzle. - [[Gibbs Free Energy and Spontaneity]] — origin of $\Delta G=\Delta H-T\Delta S$. ## 🖼️ Concept Map ```mermaid flowchart TD HighT[High T flame 2000-3500 K] -->|drives| DISS[Dissociation of products] DISS -->|breaks| H2O[H2O to OH plus H] DISS -->|breaks| CO2[CO2 to CO plus half O2] H2O -->|yields| OH[OH radical] CO2 -->|yields| CO[CO species] DISS -->|absorbs energy endothermic| ENDO[Energy absorbed] ENDO -->|lowers| FLAME[Adiabatic flame temp lower] ENDO -->|comes from| ENT[Entropy gain breaks bonds] ENT -->|via| GIBBS[Delta G = Delta H minus T Delta S] GIBBS -->|links to| KP[Kp = exp of minus Delta G over RT] KP -->|rises with T| VANT[van 't Hoff dlnKp/dT positive] FLAME -->|matters for| PROP[Propulsion thrust prediction] ``` ## 🔊 Hinglish (regional understanding) > [!intuition]- Hinglish mein samjho > Dekho, jab hum fuel jalate hain to textbook bolti hai ki products sirf CO₂ aur H₂O hote hain. Lekin flame ka temperature bahut high hota hai (2000–3500 K), aur itni garmi me molecules itne zor se vibrate karte hain ki kuch bonds toot jaate hain. Isse OH, H, O, CO, O₂ jaise tukde ban jaate hain. Ise hi kehte hain **dissociation**. Yaad rakho: bond todna **endothermic** hai — energy chahiye — to ye energy temperature badhane me kaam nahi aati. Isiliye real flame temperature, ideal "complete combustion" wale temperature se hamesha kam hota hai. > > Kyun hota hai? Gibbs energy dekho: $\Delta G^\circ=\Delta H^\circ-T\Delta S^\circ$. Bond todne me $\Delta H$ positive (energy lagti hai) lekin tukde zyada bante hain to $\Delta S$ bhi positive (zyada entropy). Low $T$ pe $\Delta H$ wala term jeetta hai, bonds bachte hain. High $T$ pe $-T\Delta S$ wala term jeet jaata hai, $\Delta G$ chhota ho jaata hai, aur $K_p=\exp(-\Delta G^\circ/RT)$ bada ho jaata hai. Matlab garmi badhao to dissociation badhega — yeh van't Hoff bhi confirm karta hai. > > Propulsion ke liye ek aur cheez important hai: **pressure**. $CO_2\rightleftharpoons CO+\tfrac12 O_2$ me product side pe moles zyada hain (net $+\tfrac12$). To Le Chatelier ke hisaab se pressure badhao to equilibrium wapas $CO_2$ ki taraf push hota hai — dissociation **kam** ho jaata hai. Isiliye high-pressure rocket chambers me thoda kam dissociation hota hai, jo designers ke liye ek useful lever hai. > > Exam tip: degree of dissociation $\alpha$ wala table banao (initial, equilibrium), total moles $1+\alpha/2$ nikaalo, partial pressures $x_iP$ likho, aur $K_p$ formula me daal do. $O_2$ pe $\tfrac12$ power lagana mat bhulna — yeh sabse common galti hai! ![[audio/5.3.03-Equilibrium-products-at-high-T-—-dissociation-(H2O-⇌-OH-+-H;-CO2-⇌-CO-+-1⁄2O2).mp3]]