Heterogeneous equilibria — pure solids - liquids excluded: Exercises

This page is a self-test ladder. Each rung is harder than the last. Cover the answer, try the problem, then open the collapsible solution. Everything you need was built in the parent note Heterogeneous Equilibria and its prerequisites — we lean on Equilibrium Constant Expressions, Activity vs Concentration, Solubility Product (Ksp), and Le Chatelier's Principle.

Figure — Heterogeneous equilibria — pure solids - liquids excluded

Before we start, three reminders you will reuse constantly:

Level 1 — Recognition

You are just deciding what belongs in $K$ . No arithmetic yet.

Recall Solution 1.1

(a) Solids $CaCO_{3}$ and $CaO$ have activity $1$ → drop them. Only $CO_{2}$ (gas) survives: $K_{p} = P_{CO_{2}} .$ (b) $Fe_{3} O_{4}$ and $Fe$ are pure solids → drop both. $H_{2}$ and $H_{2} O$ are gases → keep, raised to their coefficients: $K_{p} = \frac{( P _{H_{2} O} ) ^{4}}{( P _{H_{2}} ) ^{4}} .$ (c) Liquid water is the solvent in huge excess → activity $1$ → drop it. Only the ions remain: $K_{w} = [H^{+}] [OH^{-}] .$

Recall Solution 1.2

Excluded (activity $= 1$ ): $AgCl (s)$ , $Hg (l)$ , $Cu (s)$ — all pure solids or a pure liquid.
Included: $Ag^{+} (a q)$ , $CO_{2} (g)$ , $Na^{+} (a q)$ — variable concentrations/pressures.

Level 2 — Application

Now we plug numbers into the expressions we just wrote.

Recall Solution 2.1

The reaction quotient uses the same rule as $K$ : $Q_{p} = P_{CO_{2}} = 0.40 atm$ . Compare to $K_{p} = 1.00$ . Since $Q_{p} < K_{p}$ , the system is "short" of $CO_{2}$ , so it moves forward (right): more $CaCO_{3}$ decomposes, raising $P_{CO_{2}}$ toward $1.00 atm$ . Answer: forward — limestone keeps decomposing until $P_{CO_{2}} = 1.00 atm$ .

Recall Solution 2.2

Carbon is solid → excluded. Gases: $1$ reactant gas, $2$ product gases, so $Δ n_{g} = 2 - 1 = 1.$ Then $K_{p} = K_{c} (R T)^{Δ n_{g}} = 6.00 \times (0.0821 \times 1000)^{1} = 6.00 \times 82.1 = 492.6.$ Answer: $K_{p} \approx 493$ atm.

Recall Solution 2.3

Each mole of $AgCl$ that dissolves gives one $Ag^{+}$ and one $Cl^{-}$ , so $[Ag^{+}] = [Cl^{-}] = s$ . $K_{s p} = s \cdot s = s^{2} ⟹ s = 1.8 \times 1 0^{- 10} = 1.34 \times 1 0^{- 5} M .$ Answer: $s \approx 1.3 \times 1 0^{- 5} M$ .

Level 3 — Analysis

Here you build ICE-style reasoning and interpret the result.

Recall Solution 3.1

Carbon excluded. Let $x$ = atm of $H_{2} O$ consumed. Since coefficients are all $1$ :

	$P_{H_{2} O}$	$P_{CO}$	$P_{H_{2}}$
start	$2.00$	$0$	$0$
change	$- x$	$+ x$	$+ x$
equil.	$2.00 - x$	$x$	$x$

$K_{p} = \frac{P _{CO} P _{H_{2}}}{P _{H_{2} O}} = \frac{x ^{2}}{2.00 - x} = 10.0.$ Rearranged: $x^{2} = 10.0 (2.00 - x) \Rightarrow x^{2} + 10 x - 20 = 0.$ $x = \frac{- 10 + 100 + 80}{2} = \frac{- 10 + 180}{2} = \frac{- 10 + 13.416}{2} = 1.708.$ We take the positive root (pressures can't be negative). Answer: $P_{H_{2} O} = 0.29 atm$ , $P_{CO} = P_{H_{2}} = 1.71 atm$ .

Recall Solution 3.2

$NaCl$ pre-loads chloride: $[Cl^{-}] \approx 0.10 M$ (the tiny bit from $AgCl$ is negligible). Let $s = [Ag^{+}]$ . Then $[Cl^{-}] = 0.10 + s \approx 0.10$ . $K_{s p} = [Ag^{+}] [Cl^{-}] = s (0.10) = 1.8 \times 1 0^{- 10} ⟹ s = \frac{1.8 \times 1 0 ^{- 10}}{0.10} = 1.8 \times 1 0^{- 9} M .$ Answer: $s \approx 1.8 \times 1 0^{- 9} M$ — about $7000 \times$ less soluble than in pure water. The extra $Cl^{-}$ pushes the equilibrium back onto the solid.

Level 4 — Synthesis

Combine multiple ideas: phase judgement + Le Chatelier + quantitative pushes.

Recall Solution 4.1

Mixing doubles the volume, so each ion is halved: $[Ag^{+}] = \frac{0.0020}{2} = 1.0 \times 1 0^{- 3} M, [Cl^{-}] = 1.0 \times 1 0^{- 3} M .$ Ionic product $Q = [Ag^{+}] [Cl^{-}] = (1.0 \times 1 0^{- 3})^{2} = 1.0 \times 1 0^{- 6} .$ Compare: $Q = 1.0 \times 1 0^{- 6} ≫ K_{s p} = 1.8 \times 1 0^{- 10}$ . Since $Q > K_{s p}$ , the solution is supersaturated → precipitate forms until $Q$ falls back to $K_{s p}$ . See Precipitation Reactions. Answer: Yes, $AgCl (s)$ precipitates.

Recall Solution 4.2

(a) No effect on equilibrium position. $CaCO_{3}$ is a pure solid; its activity stays $1$ . More solid ≠ more product. (b) More $CaO$ . Removing $CO_{2}$ drops $Q_{p} = P_{CO_{2}}$ below $K_{p}$ ; the system replaces the lost gas by decomposing more $CaCO_{3}$ , producing more $CaO$ . (c) More $CaO$ . For an endothermic reaction, heat behaves like a reactant; raising $T$ raises $K_{p}$ , shifting right. See Le Chatelier's Principle.

Level 5 — Mastery

Full derivations and the subtle "activity = 1" foundation.

Recall Solution 5.1

Dissolving $s$ mol/L of $PbCl_{2}$ gives $[Pb^{2 +}] = s$ and $[Cl^{-}] = 2 s$ (two chlorides per formula unit). $K_{s p} = [Pb^{2 +}] [Cl^{-}]^{2} = (s) (2 s)^{2} = 4 s^{3} .$ $K_{s p} = 4 (1.6 \times 1 0^{- 2})^{3} = 4 (4.096 \times 1 0^{- 6}) = 1.638 \times 1 0^{- 5} .$ Answer: $K_{s p} \approx 1.6 \times 1 0^{- 5}$ . Note the exponent $2$ on chloride — it comes from the coefficient, exactly like the $Δ n_{g}$ rule but for concentrations.

Recall Solution 5.2

Activity of a pure condensed phase is $a = \frac{c _{actual}}{c ^{\circ}},$ measured against its own standard state $c^{\circ}$ . For a pure solid the "concentration" is its molar density: $c = \frac{density}{molar mass} = \frac{ρ}{M} .$ Because $ρ$ and $M$ are fixed properties of the substance at given $T, P$ , and the chosen standard state $c^{\circ}$ is that same $ρ / M$ : $a = \frac{ρ / M}{ρ / M} = 1.$ Doubling the amount doubles the total moles and total volume equally, so $ρ / M$ is unchanged — $a$ stays exactly $1$ . Hence the solid multiplies into $K$ as $\times 1$ and disappears. This is the Activity vs Concentration idea in one line. Contrast: activity $= 1$ also means $ln a = 0$ , so a pure solid contributes $0$ to the Gibbs-energy expression $Δ G = Δ G^{\circ} + R T ln Q$ — see Gibbs Free Energy and Equilibrium.

Recall Solution 5.3

Solids $FeO$ , $Fe$ excluded. $Δ n_{g} = 1 - 1 = 0$ , so total gas pressure is conserved: $P_{CO} + P_{CO_{2}} = 1.00$ throughout. Let $x$ = atm of $CO$ converted. $K_{p} = \frac{P _{CO_{2}}}{P _{CO}} = \frac{x}{1.00 - x} = 0.40.$ $x = 0.40 (1.00 - x) ⟹ x + 0.40 x = 0.40 ⟹ 1.40 x = 0.40 ⟹ x = 0.2857.$ Answer: $P_{CO_{2}} = 0.286 atm$ , $P_{CO} = 0.714 atm$ . Because $Δ n_{g} = 0$ , here $K_{p} = K_{c}$ — no $(R T)$ factor needed.