Worked examples — Heterogeneous equilibria — pure solids - liquids excluded

This page is the "no surprises" drill sheet for Heterogeneous equilibria — pure solids/liquids excluded. The parent note taught you the rule: pure solids and pure liquids have activity $=1$, so they vanish from $K$. Here we hit every kind of situation that rule can face — every phase mix, every degenerate case, every trap — one worked example per cell.

Before we start, one word we lean on constantly: activity. Think of it as the "effective concentration" a species actually contributes to the equilibrium tug-of-war. For a gas it is (roughly) its partial pressure in bar; for a dissolved ion it is (roughly) its molarity; for a pure solid or pure liquid it is exactly 1 — because a lump of pure chalk is "as concentrated as chalk can ever be," and adding a second lump doesn't make the chalk any chalkier. That last sentence is the whole chapter.

Prerequisites worth a click: Equilibrium Constant Expressions, Activity vs Concentration, Solubility Product (Ksp), Le Chatelier's Principle.

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The scenario matrix

Every problem this topic throws at you falls into one of these case classes. Each row is a distinct "shape" of situation; the last column points to the worked example that nails it.

#	Case class	What makes it distinct	Example
A	Solid ⇌ solid + gas	one gas only ⇒ $K_p=P_{\text{gas}}$	Ex 1
B	Solid ⇌ ions (dissolving)	$K_{sp}$, ions vary, solid excluded	Ex 2
C	Solid + gas ⇌ gases (mixed)	solid out, gases stay, $\Delta n_g\ne 0$	Ex 3
D	Degenerate: solid runs out	equilibrium law breaks — no solid phase	Ex 4
E	Zero / limiting input	pump gas away, or start at zero product	Ex 5
F	$K_p\leftrightarrow K_c$ conversion	which phases count in $\Delta n_g$	Ex 6
G	Common-ion / real-world word problem	solubility drops in shared-ion solution	Ex 7
H	Exam twist: "does X appear?"	catalyst, spectator solid, pure water	Ex 8

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Worked examples

Forecast: guess now — will more chalk mean more gas? Write down yes/no before reading on.

1. List phases, keep only the variable ones. ${\text{CaCO}}_3$ solid → out. $\text{CaO}$ solid → out. ${\text{CO}}_2$ gas → in. Why this step? Only the gas has an activity that can move; both solids sit pinned at activity $1$.
2. Write $K_p$. With the two solids gone: $K_p=P_{{\text{CO}}_2}$. Why this step? $K_p=\frac{a_{\text{CaO}}{P}_{{\text{CO}}_2}}{a_{{\text{CaCO}}_3}}=\frac{1\cdot P_{{\text{CO}}_2}}{1}$.
3. Solve. $P_{{\text{CO}}_2}=K_p=1.00\text{atm}$.
4. Double the chalk? No change: $K_p$ still equals $1.00\text{atm}$, so $P_{{\text{CO}}_2}$ is unchanged. See the figure — two flasks, different amounts of solid, same gas pressure.

Verify: $K_p$ carries units of pressure here because $\Delta n_g=+1$; a single-gas decomposition must give $K_p=P_{\text{gas}}$, and $P=1.00\text{atm}$ is a plausible chalk-decomposition pressure (chalk fizzes off ${\text{CO}}_2$ appreciably near $900{}^{\circ }\text{C}$). ✓

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Forecast: will $s$ be near $10^{-5}$ or near $10^{-10}$? Guess the order of magnitude.

1. Exclude the solid, include the ions. $K_{sp}=[{\text{Ag}}^{+}][{\text{Cl}}^{-}]$. Why this step? Undissolved $\text{AgCl}$ is a pure solid (activity $1$); only the dissolved ions vary.
2. Stoichiometry sets the ion concentrations. One $\text{AgCl}$ that dissolves makes one ${\text{Ag}}^{+}$ and one ${\text{Cl}}^{-}$, so $[{\text{Ag}}^{+}]=[{\text{Cl}}^{-}]=s$. Why this step? In pure water the ions arrive only from the dissolving solid, in a $1:1$ ratio.
3. Substitute and solve. $s\cdot s=s^2=K_{sp}\Rightarrow s=\sqrt{1.8\times 10^{-10}}=1.34\times 10^{-5}\text{M}$.

Verify: $(1.34\times 10^{-5}{)}^2=1.80\times 10^{-10}=K_{sp}$ ✓. Units: $\text{M}\times \text{M}={\text{M}}^2$, matching $K_{sp}$'s dimensionless-per-standard-state bookkeeping. ✓

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Forecast: with $K_p=10$ (products favoured), guess whether more or less than half the steam converts.

1. Drop the solid, keep three gases. $K_p=\frac{P_{\text{CO}}{P}_{{\text{H}}_2}}{P_{{\text{H}}_2\text{O}}}$. Why this step? Graphite is a pure solid, activity $1$; the steam here is gas, so it stays.
2. Set up the change table with extent $x$. $P_{{\text{H}}_2\text{O}}=2-x$, and $P_{\text{CO}}=P_{{\text{H}}_2}=x$. Why this step? One unit of reaction consumes one steam and makes one $\text{CO}$ + one ${\text{H}}_2$.
3. Insert and clear the fraction. $10=\frac{x\cdot x}{2-x}=\frac{x^2}{2-x}\Rightarrow x^2+10x-20=0$. Why this step? Multiplying both sides by $(2-x)$ turns the equilibrium law into a quadratic we can solve.
4. Take the physical root. $x=\frac{-10+\sqrt{100+80}}{2}=\frac{-10+\sqrt{180}}{2}\approx 1.708\text{atm}$. (The negative root gives negative pressure — unphysical, discard.)

Verify: $P_{{\text{H}}_2\text{O}}=2-1.708=0.292$, $P_{\text{CO}}=P_{{\text{H}}_2}=1.708$; then $\frac{1.708^2}{0.292}=9.99\approx 10=K_p$ ✓. Adding more graphite would not change any of these numbers. ✓

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Forecast: trust the rule blindly, or check whether there's enough solid to sustain it? Decide first.

1. Ask how much ${\text{CO}}_2$ pressure $0.010\text{mol}$ could even make. If all the chalk decomposed, it releases $0.010\text{mol}$ ${\text{CO}}_2$. Why this step? The equilibrium law $P_{{\text{CO}}_2}=K_p$ only holds while some solid remains. If the solid is exhausted, there's no chalk left to buffer the pressure — the heterogeneous equilibrium no longer exists.
2. Compute the max achievable pressure via $PV=nRT$. With $R=0.0821{\text{L⋅atm⋅mol}}^{-1}{\text{K}}^{-1}$: $P_{\max }=\frac{nRT}{V}=\frac{0.010\times 0.0821\times 1100}{10.0}=0.0903\text{atm}.$ Why this step? This is the ceiling pressure if every last bit of chalk breaks down.
3. Compare to $K_p$. $0.0903\text{atm}<1.00\text{atm}=K_p$. The system can never reach equilibrium: all chalk decomposes and $P_{{\text{CO}}_2}$ stalls at $0.090\text{atm}$, below $K_p$.

Verify: $\frac{0.010\times 0.0821\times 1100}{10.0}=0.09031$ atm $<1.00$ atm ✓. This is the only way "how much solid you start with" can matter — when there's not enough to keep a solid phase present. ✓

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Forecast: which direction does the reaction run in each case?

1. Define the reaction quotient $Q_p=P_{{\text{CO}}_2}$. Same form as $K_p$, but for the current (not equilibrium) pressure. Why this step? Comparing $Q_p$ to $K_p$ tells us the direction: $Q_p<K_p$ → forward, $Q_p>K_p$ → reverse.
2. Pump away: $Q_p\to 0<K_p=1.00$. System runs forward — more chalk decomposes to refill ${\text{CO}}_2$ back toward $1.00\text{atm}$ (until chalk lasts). This is the Le Chatelier response: remove product, make more.
3. Inject to $2.00\text{atm}$: $Q_p=2.00>K_p=1.00$. System runs reverse — ${\text{CO}}_2$ + $\text{CaO}$ recombine into ${\text{CaCO}}_3$, dropping $P_{{\text{CO}}_2}$ back to $1.00\text{atm}$.

Verify: In both cases the target is $P_{{\text{CO}}_2}=K_p=1.00\text{atm}$: $Q_p<K_p\Rightarrow$ forward, $Q_p>K_p\Rightarrow$ reverse. The limiting input $Q_p=0$ drives maximal forward pull, exactly as an open kiln vents ${\text{CO}}_2$ and keeps decomposing limestone. ✓

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Forecast: does the solid count toward $\Delta n_g$? Guess yes or no.

1. Compute $\Delta n_g$ using only gas moles. Gas products $=1$ (${\text{CO}}_2$), gas reactants $=0$. So $\Delta n_g=1-0=+1$. Why this step? $\Delta n_g$ counts gases only — solids never enter it, mirroring their exclusion from $K$.
2. Apply $K_p=K_c(RT{)}^{\Delta n_g}$. So $K_c=\frac{K_p}{(RT{)}^{\Delta n_g}}=\frac{1.00}{(0.0821\times 1100{)}^1}$. Why this step? This is the standard bridge between pressure- and concentration-based constants; here the exponent is just $1$.
3. Evaluate. $RT=0.0821\times 1100=90.31$, so $K_c=\frac{1.00}{90.31}=0.01107{\text{mol⋅L}}^{-1}$.

Verify: $K_c(RT{)}^1=0.01107\times 90.31=1.00=K_p$ ✓. Units of $K_c$: concentration${}^{\Delta n_g}={\text{mol⋅L}}^{-1}$, consistent with one gas product. ✓

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Forecast: will the shared ${\text{Cl}}^{-}$ make $\text{AgCl}$ more or less soluble?

1. Write $K_{sp}$; identify the shared ion. $K_{sp}=[{\text{Ag}}^{+}][{\text{Cl}}^{-}]$. The $\text{NaCl}$ already floods the tank with ${\text{Cl}}^{-}$. Why this step? $\text{NaCl}$ is fully dissolved, so $[{\text{Cl}}^{-}]\approx 0.10\text{M}$ before any $\text{AgCl}$ dissolves — a common ion.
2. Let $s^{\prime }$ be the new solubility. $[{\text{Ag}}^{+}]=s^{\prime }$ and $[{\text{Cl}}^{-}]=0.10+s^{\prime }\approx 0.10$ (since $s^{\prime }$ will be tiny). Why this step? The approximation is safe if $s^{\prime }\ll 0.10$; we'll check it in Verify.
3. Solve. $s^{\prime }\times 0.10=1.8\times 10^{-10}\Rightarrow s^{\prime }=1.8\times 10^{-9}\text{M}$.
4. Compare. $\frac{1.34\times 10^{-5}}{1.8\times 10^{-9}}\approx 7.4\times 10^3$: solubility drops ~$7400\times$. Extra ${\text{Cl}}^{-}$ pushes the equilibrium back onto solid ($Q>K_{sp}$ momentarily), so less $\text{AgCl}$ dissolves — the common-ion effect.

Verify: $1.8\times 10^{-9}\times 0.10=1.8\times 10^{-10}=K_{sp}$ ✓, and $s^{\prime }=1.8\times 10^{-9}\ll 0.10$, so the approximation holds. ✓

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Forecast: three "extra" things (Fe, water, the catalyst role) — which, if any, enter $K_p$?

1. Fe is a pure solid AND a spectator. Excluded twice: pure solids leave $K$, and a catalyst is neither reactant nor product — it changes rate, never $K_p$. Why this step? A catalyst speeds forward and reverse equally, so the equilibrium position is untouched.
2. Liquid water is a pure liquid, not in the reaction. Activity $1$, and it's not a species in the balanced equation — excluded. Why this step? Only species written in the reaction whose activity can vary belong in $K$.
3. Write $K_p$ from the gases only. $K_p=\frac{(P_{{\text{NH}}_3}{)}^2}{P_{{\text{N}}_2}(P_{{\text{H}}_2}{)}^3}.$

Verify (dimension/exponent check): exponents match the stoichiometric coefficients $2,1,3$; $\Delta n_g=2-(1+3)=-2$, so $K_p$ has units ${\text{atm}}^{-2}$ — internally consistent. Neither Fe nor water appears. ✓

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Recall Rapid self-test

Case A gives $K_p$ equal to what? ::: The single gas's partial pressure, $P_{{\text{CO}}_2}$. Doubling the mass of a pure solid changes $K$ by how much? ::: Not at all — its activity stays $1$. When can starting amount of solid matter? ::: Only when the solid runs out entirely (Case D) — then no equilibrium exists. In $\Delta n_g$, do you count solids or liquids? ::: No — gases only. Common-ion effect does what to solubility? ::: Lowers it (extra shared ion pushes equilibrium back onto solid).

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Where this connects

• Dissolving-solid cases → Solubility Product (Ksp), Precipitation Reactions.
• Kiln/decomposition (Ex 1, 4, 5) → Cement Production, Limestone Caves and Stalactites, Metallurgy: Ore Reduction.
• $\Delta n_g$ conversions and $Q$ vs $K$ → Equilibrium Constant Expressions, Chemical Equilibrium Fundamentals.
• Why activity $=1$ underneath it all → Activity vs Concentration, Gibbs Free Energy and Equilibrium.