Heterogeneous equilibria — pure solids - liquids excluded: Question bank

Symbols used on this page — define once, use everywhere

Before the traps, here is every symbol you will meet, in plain words. Nothing below appears until it is named here.

Formula The vocabulary of

$K$

$a_{i}$ — the activity of species $i$ : its "effective concentration" measured as a pure number (a ratio to a reference amount), so it carries no units.
$P^{\circ}$ — the standard pressure, fixed at $1 bar$ . Every gas pressure is secretly divided by this.
$c^{\circ}$ — the standard concentration, fixed at $1 M$ . Every dissolved concentration is secretly divided by this.
$K_{p}$ — the equilibrium constant written with gas partial pressures: each gas contributes $P_{i} / P^{\circ}$ , so $K_{p}$ is dimensionless.
$K_{c}$ — the equilibrium constant written with molar concentrations: each species contributes $[i] / c^{\circ}$ , also dimensionless.
$K_{s p}$ — a special $K_{c}$ for a solid dissolving into ions (the solubility product), see Solubility Product (Ksp).
$R$ — the gas constant ( $8.314 J mol^{- 1} K^{- 1}$ ), the number linking pressure, volume, moles and temperature via $P V = n R T$ .
$T$ — absolute temperature in kelvin.
$Δ n_{g}$ — the change in the number of gas moles going from reactants to products: $Δ n_{g} = (gas moles of products) - (gas moles of reactants)$ .

Recall Why only gases live in

$Δ n_{g}$ (mini-derivation of $K_{p} = K_{c} (R T)^{Δ n_{g}}$ ) For an ideal gas, $P V = n R T$ , so its partial pressure is $P_{i} = \frac{n _{i}}{V} R T = [i] R T$ — pressure and concentration differ by exactly a factor $R T$ . Write $K_{p}$ as a product of $P_{i} / P^{\circ}$ terms and substitute $P_{i} = [i] R T$ : each gas species drags out one factor of $R T$ . Products give $+$ powers, reactants give $-$ powers, so the leftover exponent is $(gas product moles) - (gas reactant moles) = Δ n_{g}$ . Result: $K_{p} = K_{c} (R T)^{Δ n_{g}}$ . Solids and liquids never contributed a $P_{i} = [i] R T$ term (they were excluded, activity 1), so they cannot appear in $Δ n_{g}$ . Only gases can.

The figure below fixes the three intuitions the whole page rests on: constant solid activity, the saturation ceiling $K_{s p}$ , and how a Le Chatelier squeeze plays out. Refer back to it as you work.

Figure — Heterogeneous equilibria — pure solids - liquids excluded

True or false — justify

The word "pure" in "pure solid" is the whole game. Watch for it.

Adding more solid CaCO₃ to a sealed vessel at equilibrium increases the equilibrium

P_{CO_{2}}

False — the activity of a pure solid is fixed at 1 regardless of amount (see the flat red line in figure 1), so

K_{p} = P_{CO_{2}}

is untouched; you only added more of something whose "concentration" cannot change.

Water is always excluded from an equilibrium expression because it is a liquid.

False — water is excluded only when it is a pure liquid or a solvent in vast excess (

x_{water} ≳ 0.99

); when water is a reactant/product in a non-aqueous or concentrated setting (see Buffer Solutions (when water matters)), its activity varies and it stays in

K

A pure liquid and a pure solid are both dropped from

K

for essentially the same reason.

True — for both, the substance is its own standard state, so

a = [substance] / [substance]_{standard} = 1

; the ratio is trivially one.

If a solid catalyst appears in the reaction vessel, it must appear somewhere in

K_{p}

False — a catalyst is neither reactant nor product; it changes the rate of reaching equilibrium, not the position, so it never enters

K

(double-excluded here since it is also a pure solid).

For

AgCl (s) ⇌ Ag^{+} + Cl^{-}

, doubling the mass of undissolved AgCl doubles the solubility.

False — solubility is fixed by

K_{s p} = [Ag^{+}] [Cl^{-}]

; extra solid just sits at the bottom with activity 1, changing nothing (see Solubility Product (Ksp)).

The relation

K_{p} = K_{c} (R T)^{Δ n_{g}}

counts all species when finding

Δ n_{g}

False —

Δ n_{g}

counts gaseous moles only; only gases contribute a

P_{i} = [i] R T

factor, so solids and liquids (already excluded, activity 1) cannot appear in

Δ n_{g}

Excluding a solid means we are pretending it isn't chemically involved.

False — the solid is fully involved in the reaction; we exclude it because its activity is constant, and a constant factor is absorbed into

K

, not because it is inert.

Spot the error

Each statement below has one flaw. Find it, then reveal.

"For

CaCO_{3} (s) ⇌ CaO (s) + CO_{2} (g)

K_{p} = \frac{P _{CO_{2}}}{[ CaCO _{3} ]}

The denominator

[CaCO_{3}]

is wrong — a pure solid has activity 1 and is dropped, so correctly

K_{p} = P_{CO_{2}}

(strictly

P_{CO_{2}} / P^{\circ}

"Since

H_{2} O

is a liquid in

C (s) + H_{2} O (g) ⇌ CO (g) + H_{2} (g)

, we leave it out."

The water here is labelled

(g)

— it is steam, a gas, so it stays in the expression; the phase label, not the formula, decides inclusion.

K_{s p}

for

AgCl

[Ag^{+}] [Cl^{-}] [AgCl]

The

[AgCl]

term is illegitimate — the solid is excluded, so

K_{s p} = [Ag^{+}] [Cl^{-}]

with only the two dissolved ions.

"Activity of a pure solid is 1 because its concentration is zero."

The reasoning is backwards — its concentration is large and constant (

density / molar mass

), and activity is that value divided by the same standard-state value, giving 1, not because concentration vanishes.

"Adding graphite to the water-gas reaction shifts equilibrium toward more CO."

Graphite is a pure solid with activity 1; adding it changes nothing about the gas-phase ratio, so no shift occurs (a common Le Chatelier's Principle misfire).

"Because

K_{p} = P_{CO_{2}}

for limestone,

K_{p}

has units of atm."

The thermodynamic

K_{p}

is dimensionless — each

P

is divided by

P^{\circ} = 1 bar

, so quoting "1 atm" really means

P_{CO_{2}} / P^{\circ} \approx 1

, a pure number.

Why questions

Say the reasoning before revealing.

Why does the amount of solid not appear in

K

, even though amount clearly affects a reaction's total yield?

Because

K

is built from activities (intensive quantities), not amounts (extensive); the solid's activity stays 1 whether you have a grain or a mountain — total yield in moles can differ, but the equilibrium ratio cannot (link: Activity vs Concentration).

Why is the standard state for a pure solid chosen as the pure solid itself rather than 1 M?

Because a solid never dissolves to give a tunable molarity — its molar "concentration" is fixed by density and molar mass, so the natural, always-available reference is the pure substance, making

a = 1

automatically.

Why can pumping away CO₂ cause more limestone to decompose, if the solid amount doesn't matter?

Removing CO₂ drops

P_{CO_{2}}

below

K_{p}

, so the system responds by making more gas to restore

P_{CO_{2}} = K_{p}

; the gas is what re-equilibrates, and it happens to consume solid to do so.

Why do we say the reaction stops being a heterogeneous equilibrium if all the CaCO₃ decomposes?

With no solid phase remaining, there is no reactant of activity 1 to draw on; the "equilibrium"

P_{CO_{2}} = K_{p}

can no longer be maintained, so you have a single-phase gas that isn't at the heterogeneous equilibrium condition.

Why does temperature, but not initial solid mass, set the equilibrium CO₂ pressure?

K_{p}

depends on temperature through Gibbs Free Energy and Equilibrium (

Δ G^{\circ} = - R T ln K

); solid mass never enters

K_{p}

, so only

T

(and the value of

K_{p}

it produces) fixes

P_{CO_{2}}

Why is the "concert hall" analogy about constancy rather than bigness?

The point isn't that solids are large but that their activity doesn't respond to adding or removing material — a huge, effectively unchanging quantity contributes a constant factor that folds into

K

Why does a catalyst leave

K

untouched but speed things up?

It lowers the activation barrier for both forward and reverse directions equally, so it changes only how fast equilibrium is reached, not the free-energy difference that fixes the ratio

K

Why do only gases show up in

Δ n_{g}

K_{p} = K_{c} (R T)^{Δ n_{g}}

Because the substitution

P_{i} = [i] R T

(from

P V = n R T

) is what pulls out each factor of

R T

; only gases have a partial pressure to substitute, so only they contribute to the exponent.

Edge cases

The boundaries where the rule bends or breaks.

What happens to the exclusion rule if the "solid" is actually a solid solution (an alloy or mixed crystal)?

It no longer applies cleanly — in a mixture the component's activity is less than 1 and varies with composition, so it must be kept (with an activity coefficient) rather than dropped.

In a very concentrated aqueous solution, can water still be treated as activity 1?

Not reliably — once solutes are abundant (

x_{water}

falls below

\sim 0.99

), water is no longer in vast excess and its activity dips below 1 (this is where Activity Coefficients in Concentrated Solutions and Buffer Solutions (when water matters) matter).

For

AgCl

dissolving, what breaks if you keep adding NaCl to the water?

The solid stays activity 1, but added

Cl^{-}

raises the reactant side, driving precipitation (common-ion effect) — the ions respond while the solid activity does not, exactly as the rule predicts (see Precipitation Reactions).

If you compress the gas in

CaCO_{3} (s) ⇌ CaO (s) + CO_{2} (g)

at fixed

T

, what re-equilibrates?

Only the CO₂: compression raises

P_{CO_{2}}

above

K_{p}

, so gas reacts back with CaO to reform solid until

P_{CO_{2}} = K_{p}

again; the solids act only as an activity-1 reservoir.

What if the temperature is so low that the "gas" water in the water-gas reaction condenses to liquid?

Then that water becomes a pure liquid (activity 1) and drops out of the expression — the same species changes its treatment purely because its phase changed.

Is a pure liquid reactant (not solvent) always excluded?

Yes, as long as it is genuinely a pure liquid phase (activity 1); the moment it becomes a component of a mixture whose composition can shift, its activity varies and it re-enters

K

Does

Δ n_{g}

change if you add a solid product to a reaction?

No —

Δ n_{g}

tallies only gaseous moles, so adding solids or liquids to either side leaves

Δ n_{g}

, and therefore the

K_{p}

–

K_{c}

conversion, unchanged.

Does the "activity = 1" argument still hold for a supercritical fluid or a molten salt?

Not automatically — a supercritical fluid has a tunable, non-constant density (so its activity varies with pressure and is not fixed at 1), and a molten salt is a liquid mixture of ions whose activities depend on composition; both need explicit activity treatment, not blind exclusion.

What about a polymeric or amorphous solid whose "molar mass" is ill-defined — is its activity still exactly 1?

Only approximately — if it is a single pure phase of essentially constant composition its activity is still

\approx 1

, but blends, swollen gels, or composition-varying polymers break the "fixed density/molar mass" premise and can require correction factors.

Recall Quick self-test before you leave

Pure solid/liquid → activity ::: exactly $1$ , so it drops out of $K$ . Water stays in $K$ when ::: it is a reactant/product not in vast excess ( $x_{water} < 0.99$ ), or the solution is concentrated. $Δ n_{g}$ counts ::: gaseous moles only, because only gases carry a $P_{i} = [i] R T$ factor. A thermodynamic $K_{p}$ has units of ::: none — it is a pure ratio $P_{i} / P^{\circ}$ . Adding more solid shifts equilibrium ::: never — as long as some solid remains present.