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

This page rebuilds the parent result (topic 5.3.3) one picture at a time. We start from a single molecule getting shaken by heat, and we finish with the exact formula linking the fraction that fell apart ($\alpha$) to the equilibrium constant $K_p$ and the pressure $P$. Nothing is assumed — every symbol is earned on a figure before it is used, and the key relation $\Delta G^{\circ }=-RT\ln K_p$ is derived on this page, not imported.

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Step 1 — What "dissociation" even looks like

The reaction we will follow the whole page: $\underset{\text{1 particle}}{\underbrace{\mathrm{C}{\mathrm{O}}_2}}\rightleftharpoons \underset{\text{1 particle}}{\underbrace{\mathrm{CO}}}+\underset{\text{half a particle}}{\underbrace{\frac{1}{2}{\mathrm{O}}_2}}$

The symbol $\rightleftharpoons$ is a two-way arrow: pieces can break apart and re-join. At balance (equilibrium) both directions happen at the same rate.

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Step 2 — The tug-of-war that decides: enthalpy vs entropy

$\Delta G^{\circ }=\underset{\begin{array}{c}\text{cost to break}\\ (>0,\text{resists})\end{array}}{\underbrace{\Delta H^{\circ }}}-\underset{\begin{array}{c}\text{temperature}\\ (\text{lever})\end{array}}{\underbrace{T}}\underset{\begin{array}{c}\text{freedom gained}\\ (>0,\text{favours})\end{array}}{\underbrace{\Delta S^{\circ }}}$

Read it plainly: at low $T$ the lever is short, so the $\Delta H^{\circ }$ cost dominates → $\Delta G^{\circ }$ is large positive → bonds stay. At high $T$ the lever is long, $T\Delta S^{\circ }$ grows and can overpower $\Delta H^{\circ }$ → $\Delta G^{\circ }$ drops → bonds break.

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Step 3 — Deriving the bridge $\Delta G^{\circ }=-RT\ln K_p$ from scratch

The derivation, three honest lines. Start with the ideal-gas chemical potential of species $i$: ${\mu }_i=\underset{\begin{array}{c}\text{fixed part at}\\ p^{\circ }=1\text{bar}\end{array}}{\underbrace{{\mu }_i^{\circ }}}+\underset{\begin{array}{c}\text{pressure part}\\ (\text{integral of }1/p)\end{array}}{\underbrace{RT\ln \frac{p_i}{p^{\circ }}}}$

At equilibrium the total Gibbs energy cannot go down any further, so the weighted sum over all species is zero, using the stoichiometric coefficient ${\nu }_i$ ($+$ for products, $-$ for reactants; here ${\nu }_{\mathrm{CO}}=+1,{\nu }_{{\mathrm{O}}_2}=+\frac{1}{2},{\nu }_{\mathrm{C}{\mathrm{O}}_2}=-1$):

\;\;\Longrightarrow\;\; \underbrace{\sum_i \nu_i\,\mu_i^\circ}_{\;=\;\Delta G^\circ\ \text{(definition)}} + RT\sum_i \nu_i \ln\!\frac{p_i}{p^\circ} = 0$$ The first sum *is* $\Delta G^\circ$ by definition. Collect the logs (a sum of $\nu_i\ln$ is $\ln$ of a product of powers) and name that product $K_p$: $$\Delta G^\circ + RT\ln\underbrace{\prod_i\!\left(\frac{p_i}{p^\circ}\right)^{\nu_i}}_{\;\equiv\;K_p} = 0 \;\;\Longrightarrow\;\; \boxed{\;\Delta G^\circ = -RT\ln K_p\;} \;\;\Longleftrightarrow\;\; K_p=\exp\!\left(-\frac{\Delta G^\circ}{RT}\right)$$ - $R = 8.314\ \text{J/mol·K}$ — converts "energy per mole" into "per degree," so $\Delta G^\circ/RT$ is a **pure number** ready for $\exp$. - The **minus sign**: positive $\Delta G^\circ$ (hard to break) → negative exponent → *small* $K_p$. Exactly right. > [!formula] Why $K_p$ is dimensionless (and why textbooks look like it isn't) > Notice each pressure enters as the **ratio** $p_i/p^\circ$, divided by the reference $p^\circ = 1\ \text{bar}$. So $K_p$ is genuinely ==dimensionless==. If you *drop* the $p^\circ$ (as many books do) you get a $K_p$ that *appears* to carry units of $\text{bar}^{\sum\nu_i}$ — here $\text{bar}^{1/2}$, since $\sum\nu_i=+\tfrac12$. That is only because the $p^\circ=1\,\text{bar}$ was silently set to 1. **Keep pressures in bar and treat $K_p$ as a pure number**; the buried $(p^\circ)^{-\sum\nu_i}$ restores dimensional consistency. See [[Gibbs Free Energy and Spontaneity]] and [[Equilibrium Constant Kp and Kc]] for the same build at book length. --- ## Step 4 — Why the constant grows with temperature (van 't Hoff) > [!intuition] WHAT / WHY / PICTURE > **WHAT:** We claim $K_p$ rises steeply with $T$. Let us *see* the slope. > **WHY the derivative:** "How fast does $K_p$ change as $T$ changes?" is literally a question about a *rate of change*, and the tool that measures rate of change is the ==derivative== $\dfrac{d}{dT}$. Taking $\ln$ first (giving $\ln K_p$) is smart because it converts the messy exponential into a straight-ish line whose slope is easy. > **PICTURE:** Plot $\ln K_p$ against $T$. Because dissociation is endothermic ($\Delta H^\circ>0$), the curve **climbs** — the slope is positive everywhere. $$\frac{d\ln K_p}{dT} = \underbrace{\frac{\Delta H^\circ}{R\,T^2}}_{\substack{\text{positive because}\\ \Delta H^\circ>0}} \;>\; 0$$ Every symbol: $\Delta H^\circ>0$ is the bond-breaking cost (Step 2), $R$ the gas constant, $T^2$ always positive. Product of positives ⇒ **positive slope** ⇒ hotter flame, more dissociation. This is the [[Van 't Hoff Equation]], and it is [[Le Chatelier's Principle]] in equation form: add heat to an endothermic reaction, it shifts forward. --- ## Step 5 — Bookkeeping the pieces: the ICE table as a picture > [!intuition] WHAT / WHY / PICTURE > **WHAT:** Start with exactly **1 mole** of $\mathrm{CO_2}$. Let a fraction ==$\alpha$== (the ==degree of dissociation==, a number between 0 and 1) break apart. We track how many moles of each species remain. > **WHY:** $K_p$ is built from *amounts*. To get a formula in $\alpha$, we must first write every amount in terms of $\alpha$. This is pure counting — no chemistry hidden. > **PICTURE:** Three stacked bars — **I**nitial, **C**hange, **E**nd. Watch the total height (total moles) *grow* from 1 to $1+\alpha/2$, because half a particle of $\mathrm{O_2}$ is born for every $\mathrm{CO_2}$ lost. | Species | Initial | Change | End (equilibrium) | |---|---|---|---| | $\mathrm{CO_2}$ | $1$ | $-\alpha$ | $1-\alpha$ | | $\mathrm{CO}$ | $0$ | $+\alpha$ | $\alpha$ | | $\mathrm{O_2}$ | $0$ | $+\tfrac{\alpha}{2}$ | $\tfrac{\alpha}{2}$ | Total moles: $$n_{\text{tot}} = (1-\alpha) + \alpha + \tfrac{\alpha}{2} = \underbrace{1 + \tfrac{\alpha}{2}}_{\text{grows with }\alpha}$$ The $+\tfrac{\alpha}{2}$ is the fingerprint of dissociation: **more particles than we started with.** --- ## Step 6 — From moles to pressures > [!intuition] WHAT / WHY / PICTURE > **WHAT:** $K_p$ uses ==partial pressures== $p_i$ — the share of the total pressure $P$ that each gas contributes. The rule is $p_i = x_i\,P$, where $x_i$ = (moles of $i$)/(total moles) is the ==mole fraction==. > **WHY:** In a gas mixture, each species pushes on the walls in proportion to how many of it there are. That "share of the push" is its partial pressure — and $K_p$ is written in pressures because pressure is what a real chamber measures. > **PICTURE:** A pie chart of the mixture at some $\alpha$: three slices ($\mathrm{CO_2}$, $\mathrm{CO}$, $\mathrm{O_2}$). Each slice's *fraction of the pie* is $x_i$; multiply by total pressure $P$ to get its $p_i$. $$p_{\mathrm{CO_2}}=\frac{1-\alpha}{1+\alpha/2}P,\qquad p_{\mathrm{CO}}=\frac{\alpha}{1+\alpha/2}P,\qquad p_{\mathrm{O_2}}=\frac{\alpha/2}{1+\alpha/2}P$$ Each numerator is the mole count from Step 5; each denominator is the total $n_{\text{tot}}$; the $P$ converts a fraction into a pressure. --- ## Step 7 — Assembling $K_p$ and reading off the pressure lever > [!intuition] WHAT / WHY / PICTURE > **WHAT:** Plug the three partial pressures into the definition of $K_p$ for this reaction. The exponents on each pressure are the **stoichiometric coefficients** ($+1$ for $\mathrm{CO}$, $+\tfrac12$ for $\mathrm{O_2}$, $-1$ for $\mathrm{CO_2}$). > **WHY the $\tfrac12$ exponent:** $\mathrm{O_2}$ enters with coefficient $\tfrac12$, so it appears as $p_{\mathrm{O_2}}^{1/2}$. This is not decoration — it is the source of the crucial $P^{1/2}$ that gives pressure its grip on $\alpha$. > **PICTURE:** Two curves of $\alpha$ versus $P$ at fixed $K_p$ — the curve **falls** as $P$ rises. Squeeze the gas, and less of it dissociates: the quantitative face of [[Le Chatelier's Principle]]. Definition, then substitution (writing pressures in bar so $p^\circ=1$ stays hidden, per Step 3): $$K_p=\frac{p_{\mathrm{CO}}\;p_{\mathrm{O_2}}^{1/2}}{p_{\mathrm{CO_2}}} =\frac{\alpha}{1-\alpha}\left(\frac{\alpha/2}{1+\alpha/2}\,P\right)^{1/2}$$ Term by term: - $\dfrac{\alpha}{1-\alpha}$ — the ratio of $\mathrm{CO}$ made to $\mathrm{CO_2}$ left (the two totals $1+\alpha/2$ cancelled here). - $\left(\dots P\right)^{1/2}$ — the leftover half-power that carries $\mathrm{O_2}$ **and** the total pressure $P$. - Net exponent of $P$: $\sum \nu_i = (1+\tfrac12) - 1 = +\tfrac12 > 0$. **Positive** ⇒ raising $P$ (with $K_p$ fixed) forces $\alpha$ down. > [!formula] The parent's central result, now fully pictured > $$\boxed{\,K_p=\dfrac{\alpha}{1-\alpha}\left(\dfrac{\alpha/2}{1+\alpha/2}\,P\right)^{1/2}\,}$$ --- ## Step 8 — Every case: $\alpha\to 0$, $\alpha\to 1$, and small-$\alpha$ shortcut > [!intuition] WHAT / WHY / PICTURE > We must never leave the reader at a scenario we didn't show. Check the boundaries. > **PICTURE:** The full $K_p(\alpha)$ curve on $0\le\alpha<1$, with the three regimes flagged. - **$\alpha \to 0$ (nothing breaks, cold flame):** numerator $\alpha \to 0$ and the bracket $\to 0$, so $K_p \to 0$. Matches Step 3: tiny $K_p$ ⇒ almost no dissociation. ✔ - **$\alpha \to 1$ (everything breaks, extreme heat):** $\dfrac{\alpha}{1-\alpha}\to\infty$, so $K_p\to\infty$. A huge $K_p$ is needed to fully dissociate — consistent. ✔ - **Small $\alpha\ll 1$ (real flames, the useful regime):** set $1-\alpha\approx 1$ and $1+\alpha/2\approx 1$: $$K_p \approx \alpha \left(\tfrac{\alpha}{2}P\right)^{1/2} = \alpha^{3/2}\sqrt{\tfrac{P}{2}}$$ Now solve for $\alpha$. Squaring gives $K_p^2 = \alpha^3\,\tfrac{P}{2}$, so $\alpha^3 = \dfrac{2K_p^2}{P}$ and therefore $$\alpha \approx \left(\frac{2\,K_p^{2}}{P}\right)^{1/3}$$ **Keep the factor of 2** — it comes from the $\tfrac{P}{2}$ inside the square root and is easy to drop by accident. This shows cleanly: $\alpha$ **grows** with $K_p$ (hence with $T$, Step 4) and **shrinks** with $P$ (the $P^{-1/3}$). See [[Rocket Nozzle Frozen vs Equilibrium Flow]] and [[Adiabatic Flame Temperature]] for why this few-percent $\alpha$ steals real thrust. --- ## The one-picture summary Everything above, compressed into a single storyboard: heat lengthens the entropy lever (Step 2) → $\Delta G^\circ$ drops → the derived bridge $\Delta G^\circ=-RT\ln K_p$ makes $K_p$ climb (Steps 3–4) → the ICE count (Step 5) and pressure shares (Step 6) turn that into $\alpha$ → the master box (Step 7), tamed at the edges (Step 8). > [!recall]- Feynman retelling — the whole walkthrough in plain words > Picture a box of LEGO castles. Every castle is a $\mathrm{CO_2}$ molecule. Two forces fight over each castle: gluing it together costs effort ($\Delta H$, orange), but a pile of loose bricks has way more ways to be arranged, which nature likes ($\Delta S$, green). Temperature is the *hand* shaking the box — the harder you shake (bigger $T$), the more the "loose is better" side wins. > > We invented one scorecard, $\Delta G = \Delta H - T\,\Delta S$, that tells us who's winning. To turn that score into something countable, we wrote down each gas's Gibbs energy per mole (its chemical potential), set the weighted sum to zero at balance, and out fell $\Delta G^\circ = -RT\ln K_p$ — no magic, just the integral of $1/p$ giving a logarithm. Flip it and you get $K_p=\exp(-\Delta G^\circ/RT)$: big positive score ⇒ tiny $K_p$ ⇒ castles stay whole; the score dropping ⇒ $K_p$ balloons ⇒ lots break. > > Then we just *count bricks*. Start with 1 castle; let a fraction $\alpha$ crumble. Because crumbling makes extra pieces (a spare half-$\mathrm{O_2}$), the total pile grows to $1+\alpha/2$. Convert counts to pressures (each divided by the 1-bar reference, so $K_p$ stays a pure number), slot them into $K_p$, and out pops the master formula. Its shape tells two stories at once: shake harder (hotter) and $\alpha$ rises; **squeeze the box smaller (higher pressure)** and the bricks are packed too tight to fly apart, so $\alpha$ falls. Heat breaks, squeeze makes whole — and now you've watched every step of why. --- ## Active recall Which mathematical tool answers "how fast does $K_p$ change with $T$," and what is its sign for dissociation? ::: The derivative $d\ln K_p/dT = \Delta H^\circ/(RT^2)$; positive, because $\Delta H^\circ>0$ (endothermic). Where does the logarithm in $\Delta G^\circ=-RT\ln K_p$ actually come from? ::: From the ideal-gas chemical potential $\mu_i=\mu_i^\circ+RT\ln(p_i/p^\circ)$; the $\ln$ is the integral of $RT/p$, and setting $\sum\nu_i\mu_i=0$ at equilibrium collects the logs into $\ln K_p$. Is $K_p$ dimensionless, and why might it look otherwise? ::: Yes — each pressure enters as the ratio $p_i/p^\circ$ with $p^\circ=1\,\text{bar}$. Dropping $p^\circ$ makes it *appear* to carry units $\text{bar}^{\sum\nu_i}=\text{bar}^{1/2}$ here, but that reference is just silently set to 1. In the ICE count, why does total moles become $1+\alpha/2$ instead of $1$? ::: Each dissociated $\mathrm{CO_2}$ yields $\mathrm{CO}$ plus half an $\mathrm{O_2}$, adding $\alpha/2$ extra particles. Where does the $P^{1/2}$ in the master formula come from? ::: From the net stoichiometric change $\sum\nu_i=+\tfrac12$ (the $\tfrac12$ exponent on $p_{\mathrm{O_2}}$), so $\alpha$ depends on pressure. In the small-$\alpha$ limit, how does $\alpha$ scale with $K_p$ and $P$? ::: $\alpha\approx(2K_p^2/P)^{1/3}$ — rises with $K_p$, falls with $P$; keep the factor of 2. What happens to $K_p$ as $\alpha\to 1$? ::: It diverges to $+\infty$, since $\alpha/(1-\alpha)\to\infty$.