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

WHAT we test: the sign of the temperature response, nothing numeric. Reasoning: the differential van 't Hoff form (derived in the intuition card above) says $\frac{d\ln K_p}{dT}=\frac{\Delta H^{\circ }}{R{T}^2}$. The denominator $R{T}^2$ is always positive. So the sign of the slope equals the sign of $\Delta H^{\circ }$. Here $\Delta H^{\circ }=+240>0$, so the slope is positive → $K_p$ increases with $T$. Plain words: heat is the enemy of bonds — more heat, more dissociation.

The stoichiometric coefficients ${\nu }_i$ are: $CO$ is $+1$, $O_2$ is $+\frac{1}{2}$, $C{O}_2$ is $-1$. ${\sum }_i{\nu }_i=(+1)+(+\frac{1}{2})+(-1)=+\frac{1}{2}>0.$ The product side has more gas particles than the reactant side. By Le Chatelier's Principle, raising $P$ pushes the system toward the side with fewer moles — that is back to $C{O}_2$. Conclusion: higher $P$ hurts (suppresses) dissociation.

WHY this tool: we want a number for $K_p$, and $\Delta G^{\circ }=-RT\ln K_p$ is the only bridge from a Gibbs energy to an equilibrium constant. Convert kJ → J so units match $R$: $\Delta G^{\circ }=118000\text{J/mol}$. $K_p=\exp \left(-\frac{\Delta G^{\circ }}{RT}\right)=\exp \left(-\frac{118000}{8.314\times 2500}\right).$ Denominator: $8.314\times 2500=20785$. Exponent: $-118000/20785=-5.677$. $K_p=e^{-5.677}\approx 3.4\times 10^{-3}.$ Interpretation: $K_p\sim 10^{-3}$ is small — only a few percent dissociate — but not zero. That "few percent radical pool" is exactly what matters for real flames.

WHY the two-point van 't Hoff, not the master formula again: we don't know $\Delta G^{\circ }$ at 3000 K, but we do know $\Delta H^{\circ }$. The finite-difference form (from integrating the differential form) needs only $\Delta H^{\circ }$ and the two temperatures to give a ratio. $\ln \frac{K_2}{K_1}=-\frac{\Delta H^{\circ }}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)=-\frac{240000}{8.314}\left(\frac{1}{3000}-\frac{1}{2500}\right).$ Inner bracket: $\frac{1}{3000}-\frac{1}{2500}=3.333\times 10^{-4}-4.000\times 10^{-4}=-6.667\times 10^{-5}$. $\ln \frac{K_2}{K_1}=-\frac{240000}{8.314}\times (-6.667\times 10^{-5})=+\mathrm{1.925.}$ $\frac{K_2}{K_1}=e^{1.925}\approx \mathrm{6.9.}$ Interpretation: just +500 K multiplies dissociation by ~7. This steepness is why flame temperature cannot be predicted from complete-combustion stoichiometry.

Step 1 — mole table (WHAT). Look at the figure: the bar splits into pieces. From 1 mol $C{O}_2$, a fraction $\alpha$ breaks, giving $\alpha$ mol $CO$ and $\frac{\alpha }{2}$ mol $O_2$; $(1-\alpha )$ mol $C{O}_2$ survives.

Step 2 — total moles (WHY it grows). $n_{tot}=(1-\alpha )+\alpha +\frac{\alpha }{2}=1+\frac{\alpha }{2}.$ Breaking bonds makes more particles — the total exceeds the starting 1 mol. Step 3 — partial pressures. Mole fraction $\times$ total pressure, $p_i=x_{i}P$: $p_{C{O}_2}=\frac{1-\alpha }{1+\alpha /2}P,p_{CO}=\frac{\alpha }{1+\alpha /2}P,p_{O_2}=\frac{\alpha /2}{1+\alpha /2}P.$ Step 4 — assemble $K_p$ (products over reactant, each to its ${\nu }_i$): $K_p=\frac{p_{CO}{p}_{O_2}^{1/2}}{p_{C{O}_2}}=\frac{\frac{\alpha }{1+\alpha /2}P\cdot {\left(\frac{\alpha /2}{1+\alpha /2}P\right)}^{1/2}}{\frac{1-\alpha }{1+\alpha /2}P}.$ Step 4a — show the cancellation explicitly (the missing WHY). Divide numerator by denominator by multiplying by the reciprocal of the denominator. The reciprocal of $\frac{1-\alpha }{1+\alpha /2}P$ is $\frac{1+\alpha /2}{(1-\alpha )P}$. So: $K_p=\frac{\alpha }{1+\alpha /2}P\cdot {\left(\frac{\alpha /2}{1+\alpha /2}P\right)}^{1/2}\cdot \frac{1+\alpha /2}{(1-\alpha )P}.$ Group the like factors. The $\frac{1}{1+\alpha /2}$ from the first factor meets the $(1+\alpha /2)$ from the reciprocal — they cancel to 1. The explicit $P$ in the first factor meets the $\frac{1}{P}$ in the reciprocal — they cancel to 1: $K_p=\underset{=1}{\underbrace{\frac{1+\alpha /2}{1+\alpha /2}}}\cdot \underset{=1}{\underbrace{\frac{P}{P}}}\cdot \frac{\alpha }{1-\alpha }\cdot {\left(\frac{\alpha /2}{1+\alpha /2}P\right)}^{1/2}.$ What survives (an untouched $P^{1/2}$ still sits inside the square-root factor, which we did not cancel): $\boxed{K_p=\frac{\alpha }{1-\alpha }{\left(\frac{\alpha /2}{1+\alpha /2}P\right)}^{1/2}}.$ The surviving $P^{1/2}$ comes from the net mole change $\sum {\nu }_i=+\frac{1}{2}$ — this is the algebraic fingerprint of the pressure effect: we cancelled one full power of $P$ but the extra half-power from $O_2^{1/2}$ has no partner to cancel with.

WHY approximate: the exact equation is messy in $\alpha$; when $\alpha \ll 1$ we can drop it wherever it is added to 1. Set $1-\alpha \approx 1$ and $1+\alpha /2\approx 1$. The boxed result collapses to: $K_p\approx \alpha {\left(\frac{\alpha }{2}P\right)}^{1/2}=\alpha \cdot \frac{{\alpha }^{1/2}}{\sqrt{2}}{P}^{1/2}=\frac{{\alpha }^{3/2}{P}^{1/2}}{\sqrt{2}}.$ Solve for $\alpha$ (with $P=1$): ${\alpha }^{3/2}=\sqrt{2}{K}_p=\sqrt{2}\times 3.4\times 10^{-3}=4.808\times 10^{-3}.$ $\alpha ={\left(4.808\times 10^{-3}\right)}^{2/3}\approx \mathrm{0.0285.}$ Check the approximation: $\alpha \approx 0.0285\ll 1$, so $1-\alpha =0.97$ and $1+\alpha /2=1.014$ are both close to 1 — dropping them cost under $3\%$. Approximation valid. About 2.8% dissociation.

WHY this works: $K_p$ depends only on $T$ (fixed here). So ${\alpha }^{3/2}{P}^{1/2}$ = constant, meaning ${\alpha }^{3/2}\propto P^{-1/2}$, i.e. $\alpha \propto P^{-1/3}$. $\frac{{\alpha }_2}{{\alpha }_1}={\left(\frac{P_1}{P_2}\right)}^{1/3}={\left(\frac{1}{50}\right)}^{1/3}=\frac{1}{50^{1/3}}.$ $50^{1/3}=3.684$, so $\frac{{\alpha }_2}{{\alpha }_1}=\frac{1}{3.684}\approx \mathrm{0.271.}$ Interpretation: squeezing to 50 bar cuts dissociation to ~27% of its 1-bar value — see the falling amber curve in the figure. High-pressure combustion chambers stay cleaner and hotter. This is the Le Chatelier pressure lever turned into a design rule.

Step 0 — count the gas honestly (WHY, before approximating). Dissociating a fraction $\alpha =0.03$ of 1 mol $H_{2}O$ leaves $1-\alpha =0.97$ mol $H_{2}O$ and makes $\alpha =0.03$ mol $OH$ plus $\frac{\alpha }{2}=0.015$ mol $H_2$. Total gas: $n_{tot}=0.97+0.03+0.015=1.015\text{mol}.$ So the "1 mol" is really $1.015$ mol — the extra half-mole-per-dissociation adds only $1.5\%$ more gas. This is why we may treat $n_{tot}\approx 1$ mol here: the correction is smaller than the other roughnesses in the estimate (constant ${\bar{C}}_p$, $\Delta H^{\circ }$ at flame $T$), so carrying it would be false precision. We keep it in the pocket and quote the answer to two significant figures. Step 1 — define ${\bar{C}}_p$ and the energy balance. Let ${\bar{C}}_p\approx 45{\text{J mol}}^{-1}{\text{K}}^{-1}$ be the average heat capacity per mole of product gas (energy needed to warm one mole of the mixture by 1 K). Energy soaked up by dissociation is energy not available to warm the gas: $q_{abs}=\alpha \Delta H^{\circ }=0.03\times 240000\text{J}=7200\text{J}.$ Step 2 — convert to a missing temperature rise. That energy would otherwise have raised the temperature of the $n_{tot}$ moles by $\Delta T=\frac{q_{abs}}{n_{tot}{\bar{C}}_p}\approx \frac{7200}{1.0\times 45}=160\text{K}.$ If we insist on the honest $n_{tot}=1.015$, we get $\Delta T=7200/(1.015\times 45)=157.6\text{K}$ — the same to two figures. ~160 K either way. Interpretation: a mere 3% dissociation robs the flame of ~160 K. This is precisely why the real Adiabatic Flame Temperature falls short of the naive complete-combustion value — the deficit is the energy locked into broken bonds.

(a) $K_p$ from master relation. $\Delta G^{\circ }=180000\text{J/mol}$. $K_p=\exp \left(-\frac{180000}{8.314\times 3000}\right)=\exp \left(-\frac{180000}{24942}\right)=e^{-7.216}\approx 7.35\times 10^{-4}.$ (b) Solve for $\alpha$. Small-$\alpha$ form with $P=20$: $K_p\approx \frac{{\alpha }^{3/2}{P}^{1/2}}{\sqrt{2}}\Rightarrow {\alpha }^{3/2}=\frac{\sqrt{2}{K}_p}{P^{1/2}}=\frac{1.4142\times 7.35\times 10^{-4}}{\sqrt{20}}.$ Numerator $=1.039\times 10^{-3}$; $\sqrt{20}=4.472$; so ${\alpha }^{3/2}=2.324\times 10^{-4}$. $\alpha =(2.324\times 10^{-4}{)}^{2/3}\approx 3.79\times 10^{-3}.$ Check: $\alpha \approx 0.0038\ll 1$ — small-$\alpha$ approximation is excellent. (c) Design consequence. At 20 bar dissociation is only ~0.38%. Had we run at 1 bar, $\alpha$ would be $20^{1/3}=2.71$ times larger, ~1.0%. The figure's two bars make this visible. High chamber pressure both suppresses $CO$ formation (cleaner exhaust) and keeps more energy in heat (higher $T$). This directly feeds the frozen-vs-equilibrium question downstream, since the radical/$CO$ pool set here is what the nozzle either freezes or re-equilibrates.

WHY exact, not approximate: $\alpha =0.10$ is borderline; let us not assume $\alpha \ll 1$. Exact formula with $\alpha =0.10$, $P=1$: $K_p=\frac{\alpha }{1-\alpha }{\left(\frac{\alpha /2}{1+\alpha /2}P\right)}^{1/2}=\frac{0.10}{0.90}{\left(\frac{0.05}{1.05}\right)}^{1/2}.$ First factor: $0.10/0.90=0.1111$. Inside root: $0.05/1.05=0.047619$; its square root $=0.21822$. $K_p=0.1111\times 0.21822=\mathrm{0.02425.}$ Back out $\Delta G^{\circ }$ from $\Delta G^{\circ }=-RT\ln K_p$: $\ln K_p=\ln (0.02425)=-\mathrm{3.719.}$ $\Delta G^{\circ }=-RT\ln K_p=-8.314\times 2800\times (-3.719)=+86580\text{J/mol}\approx +86.6\text{kJ/mol}.$ Interpretation: a measured dissociation degree, run backwards through the same two relations, yields the thermodynamic $\Delta G^{\circ }$. Forward and inverse use the identical toolkit — that is mastery.

The two-step engine used everywhere ::: (1) $K_p=\exp (-\Delta G^{\circ }/RT)$ turns thermodynamics into a number; (2) the mole-table gives $K_p=\frac{\alpha }{1-\alpha }{\left(\frac{\alpha /2}{1+\alpha /2}P\right)}^{1/2}$ turning that number into a physical degree of dissociation. Heat direction ::: endothermic $\Rightarrow$ $K_p$ and $\alpha$ rise with $T$ (van 't Hoff, both forms). Pressure direction ::: $\sum {\nu }_i=+\frac{1}{2}>0\Rightarrow$ higher $P$ lowers $\alpha$ ($\alpha \propto P^{-1/3}$ when small); $K_p$ unchanged.