Heterogeneous equilibria — pure solids - liquids excluded

Core Concept

The Mathematical Why: Deriving the Exclusion Rule

Starting from First Principles

For any equilibrium, the thermodynamic equilibrium constant is:

$K = \prod \frac{(\text{activities of products})}{(\text{activities of reactants})}$

What is activity? Activity $a_i$ is the "effective concentration" that accounts for non-ideal behavior:

$a_i = \gamma_i \cdot \frac{[i]}{c^\circ}$

where $\gamma_i$ is the activity coefficient, $[i]$ is concentration/pressure, and $c^\circ$ is the standard state.

For Different Phases:

1. Gases: $a_i \approx \frac{P_i}{P^\circ}$ where $P^\circ = 1$ bar
2. Aqueous solutions: $a_i \approx \frac{[i]}{1 \text{ M}}$ (dilute approximation)
3. Pure solids/liquids: $a_i \approx 1$ at all times!

Why does $a = 1$ for pure substances?

For a pure solid or liquid, the standard state is chosen to be the pure substance itself. Activity is defined as a ratio to that standard state:

$a_{\text{pure}} = \frac{[\text{actual concentration of the pure phase}]}{[\text{concentration of the pure phase in its standard state}]}$

But the "actual" pure phase is the standard state. Its molar concentration is fixed:

$[\text{pure substance}] = \frac{\text{density}}{\text{molar mass}} = \text{constant (at given T, P)}$

Since the numerator and denominator are the same fixed quantity:

$a_{\text{pure solid/liquid}} = \frac{\text{(density/molar mass)}}{\text{(density/molar mass)}} = 1$

Adding more solid increases the total amount but not the density or molar mass, so the activity stays exactly 1.

The Exclusion in Action

Consider: $\text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g)$

Full expression: $K = \dfrac{a_{\text{CaO}} \cdot a_{\text{CO}_2}}{a_{\text{CaCO}_3}}$

Since $a_{\text{CaO}} = 1$ and $a_{\text{CaCO}_3} = 1$:

$K_p = \frac{1 \cdot P_{\text{CO}_2}}{1} = P_{\text{CO}_2}$

Include:

• Gases: use partial pressures $(P_i)$ for $K_p$ or concentrations for $K_c$
• Aqueous ions/molecules: use molar concentrations $[i]$

Exclude:

• Pure solids
• Pure liquids (including water in dilute aqueous solutions)
• Solvents in large excess

Relationship: $K_p = K_c(RT)^{\Delta n_g}$ where $\Delta n_g = \sum n_{\text{gas products}} - \sum n_{\text{gas reactants}}$

Worked Examples with Step Reasoning

Write $K_p$ for this equilibrium.

Step 1: Identify phases

• $\text{CaCO}_3$: solid → exclude
• $\text{CaO}$: solid → exclude
• $\text{CO}_2$: gas → include

Why this step? Only the CO₂ pressure can vary independently. The solids' "concentrations" are fixed by their densities.

Step 2: Write expression with included species only $K_p = P_{\text{CO}_2}$

Physical meaning: At a given temperature, limestone will decompose until CO₂ reaches a specific pressure. If you pump away CO₂, more limestone decomposes. If you add CO₂ above $K_p$, CaO and CO₂ react to form CaCO₃.

Numerical: At about 1100 K, $K_p \approx 1.0$ atm. (Calcium carbonate decomposes appreciably only at high temperatures, roughly 900 °C ≈ 1173 K, where $P_{\text{CO}_2}$ reaches 1 atm.) This means limestone spontaneously decomposes when the surrounding $P_{\text{CO}_2}$ is below the equilibrium value at that temperature.

Write $K_{sp}$ (solubility product).

Step 1: Identify phases

• $\text{AgCl}$: solid → exclude
• $\text{Ag}^+$: aqueous → include
• $\text{Cl}^-$: aqueous → include

Why this step? The undissolved AgCl is pure solid. Only dissolved ions have variable concentrations.

Step 2: Write expression $K_{sp} = [\text{Ag}^+][\text{Cl}^-]$

Physical interpretation: $K_{sp}$ tells us the maximum product of ion concentrations at saturation. No matter how much solid AgCl you add to water, the product $[\text{Ag}^+][\text{Cl}^-]$ cannot exceed $K_{sp}$.

Numerical: $K_{sp} = 1.8 \times 10^{-10}$ at 25°C. In pure water, if $[\text{Ag}^+] = s$, then $[\text{Cl}^-] = s$, so $s^2 = K_{sp}$, giving $s = 1.3 \times 10^{-5}$ M.

Write $K_p$ at high temperature.

Step 1: Identify phases

• $\text{C}$: solid (graphite) → exclude
• $\text{H}_2\text{O}$, $\text{CO}$, $\text{H}_2$: all gases → include

Why this step? At high T, water is steam (gas), not liquid. Solid carbon's activity is 1.

Step 2: Write expression $K_p = \frac{P_{\text{CO}} \cdot P_{\text{H}_2}}{P_{\text{H}_2\text{O}}}$

Key insight: Adding more graphite doesn't shift equilibrium—only changing gas partial pressures matters.

Numerical reasoning: If $K_p = 10$ at 1000 K and initially $P_{\text{H}_2\text{O}} = 2$ atm, $P_{\text{CO}} = P_{\text{H}_2} = 0$:

• Let $x$ = extent of reaction
• At equilibrium: $P_{\text{H}_2\text{O}} = 2-x$, $P_{\text{CO}} = P_{\text{H}_2} = x$
• $10 = \dfrac{x^2}{2-x}$ → $x^2 + 10x - 20 = 0$ → $x = \dfrac{-10 + \sqrt{180}}{2} \approx 1.71$ atm

Why this step? We take the positive root since pressure must be positive; $\sqrt{180} \approx 13.42$, so $x \approx (-10 + 13.42)/2 \approx 1.71$ atm.

Does the iron catalyst appear in $K_p$?

Answer: No. Catalysts speed up both forward and reverse reactions equally—they don't shift equilibrium position. Even though Fe is solid, it's not even a reactant or product, so it's doubly excluded.

$K_p = \frac{(P_{\text{NH}_3})^2}{P_{\text{N}_2} \cdot (P_{\text{H}_2})^3}$

Why this matters: You can add tons of catalyst to speed up reaching equilibrium, but the final pressures depend only on $K_p$, T, and initial amounts.

Common Mistakes and Steel-Manning

Why it feels right: In everyday experience, more starting material → more product. The confusion stems from thinking solids are "used up" like gases.

The truth: Once any solid is present, adding more doesn't change the equilibrium position. The solid's activity is already at its maximum (1). Think of it like this: the solid surface provides sites for reaction, but all available sites are already being used. More solid = more sites, but same rate per site, so same equilibrium.

Correct: The amount of CO₂ at equilibrium depends only on temperature (which determines $K_p$), not on how much CaCO₃ you started with—as long as some CaCO₃ remains unreacted.

Caveat: If you start with so little CaCO₃ that it all decomposes, then you don't have equilibrium (no solid phase left). This is a different scenario—the system is no longer heterogeneous.

Why it feels right: In most aqueous chemistry (acids, bases, salts dissolving), water is the solvent in huge excess, so its activity stays ≈ 1.

The truth: Water is excluded only when it's a pure liquid or solvent in large excess. If water is:

• A gas (steam): include it! Example: $\text{H}_2(g) + \text{CO}_2(g) \rightleftharpoons \text{H}_2\text{O}(g) + \text{CO}(g)$ → $K_p = \frac{P_{\text{H}_2\text{O}} \cdot P_{\text{CO}}}{P_{\text{H}_2} \cdot P_{\text{CO}_2}}$
• In non-aqueous solution where its concentration varies significantly: include it!
• Consumed/produced in significant amounts relative to its initial quantity: include it!

Decision rule: If the substance's amount changes enough to meaningfully affect its concentration/activity, include it. If it's an ocean that barely ripples, exclude it.

Why it feels right: We're used to ratios—if I double the numerator, the ratio changes.

The truth: $K$ depends only on temperature. It's a fundamental thermodynamic property. Adding solid changes neither the forward nor reverse rate constants ($k_f, k_r$), and since $K = k_f/k_r$, K stays the same.

What actually changes: The amount of time to reach equilibrium might change (more surface area → faster), but the equilibrium position (final pressures/concentrations) is identical.

Analogy: $K$ is like a "finish line" set by temperature. You can take different paths (different initial amounts), but you always end up at the same place.

Active Recall Practice

In that saturated water, the concentration of dissolved sugar is the same whether you have 1 extra cube sitting at the bottom or 100 extra cubes. The water is "full" either way.

For chemical reactions, pure solids and liquids are like those extra undissolved cubes. They're there, they're part of the reaction, but adding more doesn't change what's dissolved in the solution or floating around as gas. The equilibrium "balance" only cares about the parts that can move around freely—the gases and dissolved stuff—not the big chunks of solid sitting at the bottom.

So when we write the equilibrium math formula (called $K$), we only include the things whose amounts actually change and matter: gases and dissolved chemicals. The solids and pure liquids are always "full strength," so we leave them out of the formula to keep it simple and accurate.

Connections and Extensions

Prerequisites:

• Chemical Equilibrium Fundamentals
• Equilibrium Constant Expressions
• Le Chatelier's Principle
• Activity vs Concentration

Related Concepts:

• Solubility Product (Ksp)
• Precipitation Reactions
• Phase Diagrams
• Gibbs Free Energy and Equilibrium

Applications:

• Metallurgy: Ore Reduction
• Cement Production
• Limestone Caves and Stalactites
• Buffer Solutions (when water matters)

Advanced:

• Ellingham Diagrams for Metal Oxides
• Activity Coefficients in Concentrated Solutions
• Multi-phase Equilibria

#flashcards/chemistry

Why are pure solids and liquids excluded from equilibrium expressions? :: Their activity is constant (≈1) regardless of amount present, since concentration = density/molar mass doesn't change for a pure substance. Including a constant in K is redundant—it's absorbed into the equilibrium constant value itself.

Write the equilibrium expression for: CaCO₃(s) ⇌ CaO(s) + CO₂(g) :: $K_p = P_{\text{CO}_2}$ (solids excluded; only the gas partial pressure is included)

For C(s) + H₂O(g) ⇌ CO(g) + H₂(g), write Kp :: $K_p = \frac{P_{\text{CO}} \cdot P_{\text{H}_2}}{P_{\text{H}_2\text{O}}}$ (solid carbon excluded; water is a gas here, so included)

Relate Kp and Kc for a heterogeneous equilibrium :: $K_p = K_c(RT)^{\Delta n_g}$ where $\Delta n_g$ = (moles of gas products) − (moles of gas reactants). R = 0.0821 L·atm/(mol·K) or 8.314 J/(mol·K).

Concept Map

Hinglish (regional understanding)

Hinglish (regional understanding)