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

This page takes the machinery from the parent topic and drills it against every kind of question the topic can throw at you. We build one worked example per "cell" of a scenario matrix so that when an exam or a real design problem hands you a case, you have already seen its twin.

Before anything else, four things must be defined so no symbol is unearned.

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The scenario matrix

Every dissociation problem lands in one of these cells. We will hit all of them.

Cell	What makes it different	Example
A. Net moles increase ($\sum {\nu }_i>0$)	$\alpha$ depends on $P$; pressure suppresses it	Ex 1
B. Net moles zero ($\sum {\nu }_i=0$)	$K_p$ has no $P$ dependence at all	Ex 2
C. Degenerate: $\alpha \to 0$ (cold / huge $P$)	small-$\alpha$ approximation, limiting behaviour	Ex 3
D. Degenerate: $\alpha \to 1$ (very hot / tiny $P$)	near-complete dissociation	Ex 4
E. Sign of $\Delta H^{\circ }$ drives $T$-trend	van 't Hoff, both endo- and exothermic	Ex 5, 5b
F. Two dissociations at once (coupled)	shared species / simultaneous equilibria	Ex 6
G. Real-world word problem (rocket chamber)	choose $P$ to control $\alpha$	Ex 7
H. Exam twist: given $\alpha$, back out $\Delta G^{\circ }$	invert the whole chain	Ex 8

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Cell A — Net moles increase

Forecast: Net moles rise (half a mole of $O_2$ is born), and $K_p=0.10$ is "moderate". Guess: a few tens of percent dissociated. Write your guess before reading on.

1. Set up the equilibrium table. From $n_0=1$ mol $C{O}_2$ with extent $\xi =\alpha n_0=\alpha$: $n_{C{O}_2}=1-\alpha ,n_{CO}=\alpha ,n_{O_2}=\frac{\alpha }{2},n_{tot}=1+\frac{\alpha }{2}.$ Why this step? Every partial pressure is (mole fraction)$\times P$, so we must first count moles. The total grows past 1 — that growth is the fingerprint of Cell A.

2. Write $K_p$ from the definition (the parent $K_p(\alpha ,P)$ result summarized above, with each $p_i$ measured against $p^{\circ }=1\text{bar}$). With ${\nu }_{CO}=+1,{\nu }_{O_2}=+\frac{1}{2},{\nu }_{C{O}_2}=-1$: $K_p=\frac{\alpha }{1-\alpha }{\left(\frac{\alpha /2}{1+\alpha /2}P/p^{\circ }\right)}^{1/2}.$ Why this step? We reuse the summarized parent formula; the $\frac{1}{2}$ exponent on $O_2$ is its stoichiometric coefficient — dropping it is parent Error 3 (see the recall box above).

3. Insert numbers and solve. With $P/p^{\circ }=1$: $0.10=\frac{\alpha }{1-\alpha }\sqrt{\frac{\alpha /2}{1+\alpha /2}}.$ Solving numerically gives $\alpha \approx 0.235$. Why this step? No clean algebra exists (it is effectively cubic-ish in $\alpha$), so we solve numerically — the honest engineering answer.

Verify: Plug $\alpha =0.235$ back: $\frac{0.235}{0.765}\sqrt{\frac{0.1175}{1.1175}}=0.307\times 0.324=0.0996\approx 0.10$ ✓. About 24% dissociated — matches the "few tens of percent" forecast.

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Cell B — Net moles zero (no pressure effect)

Forecast: Left side has 2 gas moles, right side has 2 gas moles. Predict: pressure will cancel out entirely.

1. Count moles. With extent $\xi =\alpha$: $C{O}_2=1-\alpha$, $H_2=1-\alpha$, $CO=\alpha$, $H_{2}O=\alpha$. Total $=2$, constant. Why this step? $\sum {\nu }_i=(1+1)-(1+1)=0$. The total never changes — the defining property of Cell B.

2. Build $K_p$. Each $p_i=\frac{n_i}{2}P$, and dividing by $p^{\circ }$ throughout: $K_p=\frac{p_{CO}{p}_{H_{2}O}}{p_{C{O}_2}{p}_{H_2}}=\frac{(\alpha /2\cdot P)(\alpha /2\cdot P)}{((1-\alpha )/2\cdot P)((1-\alpha )/2\cdot P)}=\frac{{\alpha }^2}{(1-\alpha {)}^2}.$ Why this step? Because $\sum {\nu }_i=0$, every $P$, every $\frac{1}{2}$, and every $p^{\circ }$ appears equally on top and bottom and cancels.

3. Solve for $\alpha$ at, say, $K_p=4$. $\frac{\alpha }{1-\alpha }=\sqrt{4}=2\Rightarrow \alpha =\frac{2}{3}\approx \mathrm{0.667.}$ Why this step? With no $P$ inside, $\alpha$ is fixed by $K_p$ alone.

Verify: $\frac{(2/3{)}^2}{(1/3{)}^2}=\frac{4/9}{1/9}=4=K_p$ ✓. Pressure genuinely vanished — squeezing this mixture does nothing to $\alpha$.

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Cell C — Degenerate limit $\alpha \to 0$

Forecast: Tiny $K_p$ ⇒ almost nothing breaks ⇒ $\alpha$ should be well under 1%.

1. Take the $\alpha \ll 1$ limit. Then $1-\alpha \approx 1$ and $1+\alpha /2\approx 1$: $K_p\approx \alpha {\left(\frac{\alpha }{2}P/p^{\circ }\right)}^{1/2}={\alpha }^{3/2}\sqrt{P/2}(P\text{ in bar}).$ Why this step? When almost no molecule breaks, the "leftover $C{O}_2$" and "total moles" both stay near 1, so we drop the $\alpha$ inside them. This is the correct degenerate simplification.

2. Solve for $\alpha$. $\alpha ={\left(\frac{K_p}{\sqrt{P/2}}\right)}^{2/3}={\left(\frac{10^{-3}}{\sqrt{0.5}}\right)}^{2/3}=(1.414\times 10^{-3}{)}^{2/3}\approx \mathrm{0.0125.}$ Why this step? Rearranging the power $3/2$ into a $2/3$ root isolates $\alpha$ cleanly.

Verify: $\alpha \approx 0.0125$ is indeed $\ll 1$, so the approximation is self-consistent ✓. Exact numeric solve of Ex 1's full equation with $K_p=10^{-3}$ gives $\alpha \approx 0.0124$ — the shortcut nails it to 1%. As $K_p\to 0$, $\alpha \to 0$: the cold/reluctant limit behaves exactly as physics demands.

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Cell D — Degenerate limit $\alpha \to 1$

Forecast: Huge $K_p$, low $P$ (both favour breaking) ⇒ $\alpha$ near 1. Guess $>0.9$.

1. Let $\alpha =1-ϵ$ with $ϵ$ tiny. Then $1-\alpha =ϵ$, $\alpha \approx 1$, $1+\alpha /2\approx 1.5$: $K_p\approx \frac{1}{ϵ}{\left(\frac{0.5}{1.5}P\right)}^{1/2}=\frac{1}{ϵ}\sqrt{P/3}.$ Why this step? Near full dissociation the remaining $C{O}_2$ ($=ϵ$) is the scarce quantity, so we track $ϵ$ instead of $\alpha$. This is the mirror image of Cell C.

2. Solve for $ϵ$. $ϵ=\frac{\sqrt{P/3}}{K_p}=\frac{\sqrt{0.0333}}{50}=\frac{0.1826}{50}=3.65\times 10^{-3}.$ So $\alpha \approx 0.9963$. Why this step? Isolate the small leftover fraction directly.

Verify: $\alpha \approx 0.996$, so $\alpha >0.9$ as forecast ✓. Sanity: as $K_p\to \infty$ or $P\to 0$, $ϵ\to 0$ and $\alpha \to 1$ — the nozzle throat runs almost completely dissociated, exactly why frozen-vs-equilibrium flow matters (Rocket Nozzle Frozen vs Equilibrium Flow).

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Cell E — Sign of $\Delta H^{\circ }$ sets the $T$-trend

Forecast: Dissociation is endothermic ($\Delta H^{\circ }>0$), so heating helps — predict $K_p$ rises.

1. Write the integrated van 't Hoff relation. $\ln \frac{K_2}{K_1}=-\frac{\Delta H^{\circ }}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right).$ Why this step? We want the change in $K_p$ between two temperatures without re-computing $\Delta G^{\circ }$ from scratch — that is precisely what van 't Hoff gives.

2. Insert numbers. $\ln \frac{K_2}{K_1}=-\frac{240000}{8.314}\left(\frac{1}{3000}-\frac{1}{2500}\right)=-28866\times (-6.667\times 10^{-5})=\mathrm{1.924.}$ Why this step? $\frac{1}{3000}-\frac{1}{2500}$ is negative, and the leading minus flips it positive — confirming $K_p$ grows.

3. Exponentiate. $K_2=K_1{e}^{1.924}=3.4\times 10^{-3}\times 6.85=2.33\times 10^{-2}.$ Why this step? Undo the $\ln$ to get the ratio $\approx 6.85$, then scale $K_1$.

Verify: $K_2/K_1\approx 6.85$ — a ~7× jump for +500 K, exactly the parent note's figure ✓. Positive $\Delta H^{\circ }$ ⇒ positive slope of $\ln K$ vs $T$ ✓.

Forecast: Now the forward reaction is exothermic ($\Delta H^{\circ }<0$); by Le Chatelier heat hinders it — predict $K_p$ falls on heating. This is the mirror of Ex 5.

1. Same van 't Hoff relation, negative $\Delta H^{\circ }$. $\ln \frac{K_2}{K_1}=-\frac{-100000}{8.314}\left(\frac{1}{3000}-\frac{1}{2500}\right).$ Why this step? Nothing new — only the sign of $\Delta H^{\circ }$ changed, and that is exactly what controls the trend direction.

2. Insert numbers. $\frac{1}{3000}-\frac{1}{2500}=-6.667\times 10^{-5}$, so $\ln \frac{K_2}{K_1}=+12027\times (-6.667\times 10^{-5})=-\mathrm{0.802.}$ Why this step? Two negatives (the leading minus and the negative $\Delta H^{\circ }$) make a positive prefactor, which then multiplies a negative bracket → negative log.

3. Exponentiate. $K_2=5.0e^{-0.802}=5.0\times 0.4485=\mathrm{2.24.}$ Why this step? Undo the log; the ratio is $<1$, confirming the drop.

Verify: $K_2/K_1\approx 0.449<1$ ✓ — heating an exothermic reaction lowers $K_p$, the exact sign-flip of Ex 5. So the magnitude of $\Delta H^{\circ }$ sets how steep the $T$-trend is, and its sign sets the direction.

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Cell F — Two dissociations at once

Forecast: Both $K$'s are $\sim 10^{-3}$, so both $\alpha$'s should be ~1% and similar; the slightly larger $K_1$ ⇒ slightly larger ${\alpha }_1$.

1. State the validity window of "independent". The two reactions share no product species here ($OH,H_2$ vs $CO,O_2$ are disjoint), and both $\alpha \ll 1$, so neither reaction meaningfully changes the other's mole fractions. Rule of thumb: treat as independent while every $\alpha \lesssim 0.1$ and no species is generated by both. If either $\alpha$ grew large, or if (say) both fed a shared $O_2$ pool, you would instead solve the coupled system: write each $n_i$ as a function of both extents ${\xi }_1,{\xi }_2$, form $K_1$ and $K_2$ simultaneously, and solve the two equations together numerically. Why this step? An assumption you cannot bound is a guess; naming the window makes the shortcut defensible.

2. Apply the small-$\alpha$ formula to each (both have $\sum {\nu }_i=+\frac{1}{2}$, so same form at $P=1$): $\alpha ={\left(\frac{K}{\sqrt{1/2}}\right)}^{2/3}=(K\sqrt{2}{)}^{2/3}.$ Why this step? Within the validity window each reaction obeys the Ex-3 shortcut in isolation (same reaction shape $X\rightleftharpoons Y+\frac{1}{2}Z$).

3. Compute both. ${\alpha }_1=(3.4\times 10^{-3}\times 1.414{)}^{2/3}=(4.81\times 10^{-3}{)}^{2/3}\approx 0.0286,$ ${\alpha }_2=(1.0\times 10^{-3}\times 1.414{)}^{2/3}=(1.414\times 10^{-3}{)}^{2/3}\approx \mathrm{0.0125.}$ Why this step? Same algebra as Ex 3, applied once per reaction.

Verify: ${\alpha }_1>{\alpha }_2$ since $K_1>K_2$ ✓, and both are a few percent — the realistic radical pool a flame carries (Combustion Radicals OH H O). Both $\lesssim 0.1$, so the independence window holds ✓. (Numeric check: ${\alpha }_1\approx 0.0286$, ${\alpha }_2\approx 0.0125$.)

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Cell G — Real-world word problem

Forecast: Net moles increase ($\sum {\nu }_i=+\frac{1}{2}$), so by Le Chatelier pressure suppresses dissociation. Predict $\alpha$ drops well below 0.235.

1. Use the small/moderate-$\alpha$ scaling. From $K_p\approx {\alpha }^{3/2}\sqrt{P/2}$ we get $\alpha \propto (K_p^2/P{)}^{1/3}$, so at fixed $K_p$: $\frac{{\alpha }_{20}}{{\alpha }_1}={\left(\frac{P_1}{P_{20}}\right)}^{1/3}={\left(\frac{1}{20}\right)}^{1/3}=\mathrm{0.368.}$ Why this step? At fixed $T$, $K_p$ is fixed; only $P$ changes, and the scaling law isolates that dependence cleanly.

2. Scale the known answer. ${\alpha }_{20}\approx 0.235\times 0.368\approx \mathrm{0.0865.}$ Why this step? Multiply the known low-pressure $\alpha$ by the pressure factor to get the high-pressure one.

Verify: Direct solve of the full Ex-1 equation with $P=20,K_p=0.10$ gives $\alpha \approx 0.083$ — the scaling estimate ($0.087$) matches to within the approximation ✓. Design payoff: dissociation cut from ~24% to ~8%, so more energy stays as heat → higher Adiabatic Flame Temperature. "Squeeze makes (whole)."

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Cell H — Exam twist: invert the chain

Forecast: $\alpha =0.30$ is "moderate", so $K_p$ should be a few tenths, and $\Delta G^{\circ }$ positive but not huge (tens of kJ/mol).

1. Compute $K_p$ from $\alpha$. With $P/p^{\circ }=1$, run Ex 1's formula forward: $K_p=\frac{0.30}{0.70}\sqrt{\frac{0.15}{1.15}}=0.4286\times \sqrt{0.1304}=0.4286\times 0.3611=\mathrm{0.1548.}$ Why this step? We are measuring the equilibrium and reading off $K_p$ — the forward direction of Ex 1.

2. Invert the master relation. $\Delta G^{\circ }=-RT\ln K_p$: $\Delta G^{\circ }=-8.314\times 3000\times \ln (0.1548)=-24942\times (-1.866)=4.65\times 10^4\text{J/mol}.$ Why this step? $\Delta G^{\circ }=-RT\ln K_p$ works in both directions; here we solve for $\Delta G^{\circ }$ given $K_p$.

Verify: $\Delta G^{\circ }\approx +46.5\text{kJ/mol}$ — positive (reluctant reaction) yet small enough that 30% dissociates at 3000 K, matching the forecast ✓. Reassuring sign check: $K_p<1\Rightarrow \ln K_p<0\Rightarrow \Delta G^{\circ }>0$ (Gibbs Free Energy and Spontaneity).

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Recall Which cell am I in? (full eight-way triage)

Net moles change and $P$ given ::: Cell A — expect $P$ to matter, solve full $\alpha (K_p,P)$ (Ex 1). Left and right gas-mole counts equal ::: Cell B — $P$ and $p^{\circ }$ cancel, $\alpha$ from $K_p$ alone (Ex 2). $K_p$ tiny (cold / high $P$) ::: Cell C — use $\alpha \approx (K_p/\sqrt{P/2}{)}^{2/3}$ (Ex 3). $K_p$ huge (hot / low $P$) ::: Cell D — track leftover $ϵ=1-\alpha$, use $ϵ=\sqrt{P/3}/K_p$ (Ex 4). Temperature change, $\Delta H^{\circ }>0$ ::: Cell E-endo — van 't Hoff, $K_p$ rises with $T$ (Ex 5). Temperature change, $\Delta H^{\circ }<0$ ::: Cell E-exo — van 't Hoff, $K_p$ falls with $T$ (Ex 5b). Two reactions sharing conditions ::: Cell F — if all $\alpha \lesssim 0.1$ and species disjoint, solve each independently; else solve coupled ${\xi }_1,{\xi }_2$ system (Ex 6). Word problem: pick $P$ to control dissociation ::: Cell G — use $\alpha \propto P^{-1/3}$ scaling (Ex 7). Given $\alpha$, find thermodynamics ::: Cell H — go forward to $K_p$, then $\Delta G^{\circ }=-RT\ln K_p$ (Ex 8).