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

5.3.3 · Chemistry › Combustion Chemistry (Propulsion Bridge)

Combustion sirf clean CO₂ aur H₂O pe nahi rukta. Flame temperatures (2000–3500 K) pe product molecules partly toot jaate hain radicals aur chhote species mein. Isse energy lagti hai aur achievable flame temperature kam ho jaati hai.

Intuition Badi picture

KYA: Jab tum fuel jalate ho, textbook products hote hain CO₂ aur H₂O. Lekin heat bonds ki dushman hai — high $T$ pe molecules itni violently vibrate karti hain ki ek fraction dissociate ho jaata hai — pieces jaise $O H$ , $H$ , $O$ , $C O$ , $O_{2}$ ban jaate hain. PROPULSION ke liye kyun zaroori hai: Har bond jo toot ta hai woh energy absorb karta hai (endothermic). Woh energy temperature badhane ke liye available nahi hoti → real adiabatic flame temperature "complete combustion" estimate se kam hoti hai. Rocket/engine designers ko yeh ZAROOR account karna padta hai warna woh thrust overpredict karte hain aur species (jaise atomic $H$ , $O H$ ) jo nozzle chemistry affect karti hain unhe underpredict karte hain. HOW hum ise quantify karte hain: Har dissociation ko ek reversible equilibrium maano jo equilibrium constant $K_{p} (T)$ se govern hota hai, jo $T$ ke saath steeply badhta hai.

1. Dissociation hoti kyun hai? (first principles)

$Δ G^{\circ}$ aur $K_{p}$ ke beech link derive karna

Ideal gas species $i$ ke chemical potential se shuru karo: $μ_{i} = μ_{i}^{\circ} + R T ln (\frac{p _{i}}{p ^{\circ}})$

Yeh step kyun? $μ_{i}^{\circ}$ standard chemical potential hai (fixed $T$ , $p^{\circ} = 1 bar$ ); log term actual partial pressure account karta hai. Yeh ideal-gas potential ki definition hai.

Equilibrium pe total Gibbs energy stationary hoti hai, isliye $\sum_{i} ν_{i} μ_{i} = 0$ (jahan $ν_{i}$ = stoichiometric coefficient, products ke liye +, reactants ke liye −):

$\sum_{i} ν_{i} μ_{i}^{\circ} + R T \sum_{i} ν_{i} ln (\frac{p _{i}}{p ^{\circ}}) = 0$

Yeh step kyun? Humne bas har $μ_{i}$ substitute kiya. Pehla sum hi $Δ G_{r x n}^{\circ}$ hai definition ke hisaab se.

$Δ G^{\circ} = - R T ln K_{p} K_{p} = \prod_{i} (\frac{p _{i}}{p ^{\circ}})^{ν_{i}}$

2. Do canonical dissociations

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

3. Worked example A — CO₂ dissociation degree $α$

Problem: 1 mol $C O_{2}$ total pressure $P$ aur temperature $T$ pe, $C O_{2} ⇌ C O + \frac{1}{2} O_{2}$ ke roop mein dissociate hota hai. Dissociated fraction $= α$ . $K_{p}$ ko $α$ aur $P$ ke terms mein nikalo.

Species	Initial	Equilibrium pe
$C O_{2}$	1	$1 - α$
$C O$	0	$α$
$O_{2}$	0	$α /2$

Total moles $n_{t o t} = 1 - α + α + α /2 = 1 + α /2$ .

Yeh step kyun? Har mole $C O_{2}$ tootne pe aadha mole $O_{2}$ banta hai, isliye total gas count badhta hai — yeh dissociation ki hallmark hai (zyada particles).

Mole fractions → partial pressures $p_{i} = x_{i} P$ : $p_{C O_{2}} = \frac{1 - α}{1 + α /2} P, p_{C O} = \frac{α}{1 + α /2} P, p_{O_{2}} = \frac{α /2}{1 + α /2} P$

=\frac{\alpha}{1-\alpha}\left(\frac{\alpha/2}{1+\alpha/2}\,P\right)^{1/2}$$ **Yeh step kyun?** Har $p_i$ ko $K_p$ definition mein plug karo; $P^{1/2}$ bachta hai kyunki $\sum\nu_i = (1+\tfrac12)-1=+\tfrac12$ (net mole change). > [!formula] Pressure effect (Le Chatelier, quantitatively) > Kyunki $\sum\nu_i = +\tfrac12>0$ hai, **$P$ badhane se fixed $K_p$ pe $\alpha$ suppress hota hai**. High-pressure rocket chambers isliye low-pressure ones se *kam* dissociate karte hain — yeh ek genuine design lever hai. Chhote $\alpha\ll1$ ke liye: $K_p \approx \alpha^{3/2}\sqrt{P/2}$, isliye $\alpha\propto (K_p^2/P)^{1/3}$. --- ## 4. Worked example B — H₂O ke liye numbers **Problem:** $T=2500\ \text{K}$ pe, $\mathrm{H_2O\rightleftharpoons OH+\tfrac12 H_2}$ ke liye $\Delta G^\circ = +118\ \text{kJ/mol}$ hai. $K_p$ nikalo, comment karo. $$K_p=\exp\!\left(-\frac{118000}{8.314\times2500}\right)=\exp(-5.68)\approx 3.4\times10^{-3}$$ **Yeh step kyun?** Consistent SI units ke saath master formula ka direct use ($R$ J/mol·K mein). Result $K_p\sim10^{-3}$ matlab dissociation **chhoti lekin non-negligible hai** — exactly woh few-percent radical pool jo real flames carry karti hain. **Forecast-then-Verify:** *Calculate karne se pehle*, predict karo: $T$ ko 3000 K tak badhane se $K_p$ **bada** hona chahiye (endothermic, van 't Hoff). $\Delta H^\circ\approx 240\ \text{kJ/mol}$ ke saath: $$\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:** $K_p$ +500 K mein ~7× jump karta hai. Forecast sahi nikla: dissociation $T$ ke saath sharply badhti hai. --- > [!mistake] Common errors ko steel-man karna > **Error 1 — "Combustion complete hoti hai, products pure CO₂ + H₂O hote hain."** > *Kyun sahi lagta hai:* Stoichiometry equations yehi dikhate hain, aur room temperature pe re-combination ke liye $K_p$ astronomically large hota hai. *Fix:* $K_p$ **temperature dependent** hai; 2500–3500 K pe dissociation $K_p$ ab negligible nahi raha. Hamesha $K_p(T)$ ko *flame* temperature pe evaluate karo, 298 K pe nahi. > > **Error 2 — "Pressure matter nahi karta, $K_p$ sirf $T$ pe depend karta hai."** > *Kyun sahi lagta hai:* $K_p$ sach mein sirf $T$ ka function hai. *Fix:* lekin **dissociation degree $\alpha$** $P$ pe depend karta hai kyunki $\sum\nu_i\neq0$. Same $K_p$, different $\alpha$ — gas squeeze karo aur dissociation shrink hoti hai. > > **Error 3 — $O_2$ pe $\tfrac12$ exponent bhool jaana.** > *Kyun sahi lagta hai:* $p_{O_2}$ likhna tidier lagta hai. *Fix:* $K_p$ mein exponent **stoichiometric coefficient** hai, yahan $\tfrac12$, isliye $p_{O_2}^{1/2}$ hai. Yeh galat karne se $\alpha$ badly off ho jaata hai. --- > [!recall]- Feynman: 12-saal ke bacche ko samjhao > Socho LEGO castles (CO₂, H₂O). Agar box dheeray se hilao (warm), woh bane rehte hain. Agar *bahut zyada* shake karo (super hot flame), pieces snap off hote hain — loose bricks milte hain (OH, CO, O₂). Bricks snap karna **teri shaking energy use karta hai**, isliye box utna hot nahi ho sakta jitna tumne socha tha. Aur agar saare LEGO ek chhoti box mein tight pack karo (high pressure), woh itne pack hote hain ki kam toot te hain. Isliye hot engines kabhi 100% cleanly nahi jalate, aur isliye pressure cheezein ek saath rakhne mein help karta hai. > [!mnemonic] Direction yaad rakho > **"HEAT BREAKS, SQUEEZE MAKES (whole)."** > Heat ↑ → dissociation ↑ (endothermic). Pressure ↑ → dissociation ↓ (product side pe zyada moles). > Aur $\Delta G^\circ=-RT\ln K$: "**G**ets **R**eally **T**iny when bonds break easily" — bada positive $\Delta G$ ⇒ tiny $K$. --- ## Active recall #flashcards/chemistry Dissociation temperature ke saath kyun badhti hai (thermodynamic reason)? ::: Dissociation endothermic hai ($\Delta H^\circ>0$); $\Delta G^\circ$ mein $-T\Delta S^\circ$ term $T$ ke saath badhta hai, $\Delta G^\circ$ ghata ta hai aur $K_p$ badhata hai (van 't Hoff: $d\ln K_p/dT=\Delta H^\circ/RT^2>0$). $\Delta G^\circ$ aur $K_p$ ke beech master relation batao. ::: $\Delta G^\circ=-RT\ln K_p$, yani $K_p=\exp(-\Delta G^\circ/RT)$. 1 mol se degree $\alpha$ ke saath $CO_2\rightleftharpoons CO+\tfrac12 O_2$ ke liye $K_p(\alpha,P)$ do. ::: $K_p=\dfrac{\alpha}{1-\alpha}\left(\dfrac{\alpha/2}{1+\alpha/2}P\right)^{1/2}$. Total pressure badhane se yahan $\alpha$ badhega ya ghategga, aur kyun? ::: Ghategga — net mole change $\sum\nu_i=+\tfrac12>0$ hai, isliye high $P$ equilibrium ko kam moles ki taraf shift karta hai ($CO_2$ ki taraf). $K_p$ fixed rehta hai. Real adiabatic flame temperature "complete combustion" value se kam kyun hoti hai? ::: Endothermic dissociation woh energy absorb karti hai jo otherwise $T$ badhati, isliye achievable temperature drop hoti hai. H₂O dissociation kaun sa radical supply karta hai, aur yeh important kyun hai? ::: $OH$ — woh key chain-carrier radical jo combustion kinetics drive karta hai. $K_p$ nikalo agar 2500 K pe $\Delta G^\circ=+118$ kJ/mol ho. ::: $K_p=\exp(-118000/(8.314\times2500))\approx 3.4\times10^{-3}$. Flame products ke liye 298 K $K_p$ values use kyun NAHI karni chahiye? ::: $K_p$ strongly $T$-dependent hai; flame $T$ (2000–3500 K) pe dissociation $K_p$ 298 K ke comparison mein many orders of magnitude larger hota hai. --- ## Connections - [[Adiabatic Flame Temperature]] — dissociation main reason hai ki measured $T_{ad}$ < ideal hota hai. - [[Equilibrium Constant Kp and Kc]] — $K_p=e^{-\Delta G^\circ/RT}$ engine jo yahan use hua. - [[Le Chatelier's Principle]] — $\alpha$ ke pressure aur temperature shifts explain karta hai. - [[Van 't Hoff Equation]] — quantify karta hai ki $K_p$ $T$ ke saath kaise badhta hai. - [[Combustion Radicals OH H O]] — woh species jo dissociation produce karti hai. - [[Rocket Nozzle Frozen vs Equilibrium Flow]] — kya dissociated species nozzle mein recombine karti hain. - [[Gibbs Free Energy and Spontaneity]] — $\Delta G=\Delta H-T\Delta S$ ka origin. ## 🖼️ 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] ```

1. Dissociation hoti kyun hai? (first principles)

ΔG∘ aur Kp​ ke beech link derive karna

2. Do canonical dissociations

3. Worked example A — CO₂ dissociation degree α

$Δ G^{\circ}$ aur $K_{p}$ ke beech link derive karna

3. Worked example A — CO₂ dissociation degree $α$