Why are pure solids left out of Kc expressions even when they appear in the balanced chemical equation? This is one of the small conceptual questions that separates a memorised H2 Chemistry formula from a reliable understanding of equilibrium. A pure solid can participate fully in a reversible reaction without appearing as a variable concentration in its equilibrium expression. For the familiar thermal decomposition CaCO₃(s) ⇌ CaO(s) + CO₂(g), the school-level expression is Kc = [CO₂]—not [CaO][CO₂]/[CaCO₃].
For a student searching for an H2 Chemistry tutor in Hougang, JC1 or JC2 Kc calculations, heterogeneous equilibrium worked examples, solid versus gas concentration questions or A-Level Chemistry 9476 revision, my first diagnostic question is this: what changes when you add another piece of calcium oxide to a vessel that already contains both pure solids at equilibrium? The answer is not that the equilibrium constant gets bigger. If the temperature remains fixed and both solid phases continue to exist, adding more pure CaO does not alter the equilibrium partial pressure or equilibrium concentration of CO₂.
This guide follows the reasoning, not merely the rule. We will connect chemical activity, phase, reaction quotient Qc, equilibrium, the meaning of an excess solid and a worked mole calculation. It is suitable for JC students making the transition from writing equations correctly to using those equations as mathematical models.
Explore the Hougang Chemistry learning library · Check the official H2 Chemistry examination-year syllabus
The mistake begins when we copy every species into Kc
A student sees the reversible reaction CaCO₃(s) ⇌ CaO(s) + CO₂(g), then writes Kc = [CaO][CO₂]/[CaCO₃] by copying all stoichiometric coefficients into a products-over-reactants fraction. This makes the expression look neat but ignores the physical nature of the substances.
The symbols (s), (l), (g) and (aq) are not decorations. They describe phases. For pure solids and pure liquids, the thermodynamic activities are treated as constant and conventionally equal to one in an expression based on specified standard states. Their factors therefore disappear from the school-level equilibrium expression.
By contrast, the concentration or partial pressure of a gaseous species in the vessel can change as it reacts or as the vessel conditions change. That variable remains in Kc or Kp. Dissolved solutes similarly have concentration-dependent terms in the idealised textbook expressions.
One sentence solves many problems: include the variable activities represented by aqueous or gaseous species; omit pure solids and pure liquids in the familiar Kc treatment. This does not mean the omitted substances are chemically inactive. It means their pure-phase contributions are not separate variables in that expression.
Why the concentration of a pure solid is effectively constant
Think of a lump of calcium carbonate. Adding another lump increases the total amount and perhaps the exposed surface area. But it does not change the density of the pure solid. The number of formula units per unit volume within that pure solid remains essentially the same under fixed conditions.
We do not replace [CaCO₃] in Kc by the mass of solid divided by the whole vessel volume. That artificial ‘concentration’ would vary merely because we used a larger gas container around the same piece of rock. It would not represent the thermodynamic activity of the pure solid.
For an aqueous solution, by contrast, doubling the amount of dissolved solute in the same volume often does change its concentration, so the corresponding activity can change. Different phases behave differently in an equilibrium expression.
There are important advanced complications, such as non-ideal solutions, solid solutions, changes in composition and surface effects for very small particles. For the standard JC H2 Chemistry question about pure solids, you do not need to use these complications unless the problem explicitly introduces them. The school rule is based on a physically meaningful model, not a trick.
The main worked example: CaCO₃(s) ⇌ CaO(s) + CO₂(g)
Calcium carbonate can thermally decompose at suitable temperatures. The reversible overall chemical equation is:
CaCO₃(s) ⇌ CaO(s) + CO₂(g)
There is one mole of gas on the product side and no gaseous reactant. Under the standard idealised concentration expression, Kc = [CO₂].
To be clear, the rigorous thermodynamic equilibrium constant uses activities and is dimensionless. Writing Kc in terms of concentrations is the familiar A-Level working convention, with units or concentration-standard conventions determined by the form of the expression. For the calculations below, use the Kc value and units as supplied in the question.
The presence of both solid phases is essential to interpreting the simple fixed-temperature coexistence equilibrium. If the starting solid has completely reacted away, the system can have a different final composition because it has reached a material limit rather than the two-solid coexistence equilibrium.
Numerical example 1: how much carbonate must decompose?
Suppose a hypothetical rigid vessel has volume 2.00 dm³ at a fixed temperature where the equilibrium CO₂ concentration for CaCO₃(s)/CaO(s)/CO₂(g) coexistence is 0.125 mol dm⁻³. Initially the vessel contains enough CaCO₃ and CaO for both phases to remain present, and gaseous CO₂ has concentration 0.0500 mol dm⁻³.
The reaction quotient in the simplified model is Qc = [CO₂] = 0.0500, less than Kc = 0.125. So, if there is sufficient CaCO₃, net decomposition can occur until CO₂ reaches its required equilibrium concentration.
- Identify the required change: Δ[CO₂] = 0.125 − 0.0500 = 0.0750 mol dm⁻³.
- Convert concentration change to amount: Δn(CO₂) = 0.0750 × 2.00 = 0.150 mol.
- Apply stoichiometry: CaCO₃ → CaO + CO₂ is 1:1, so 0.150 mol of CaCO₃ must decompose.
- Check the inventory: at least 0.150 mol of CaCO₃ must be available to react, and some must still remain for the two-solid coexistence assumption to hold.
- Read the chemical answer: if enough CaCO₃ is present, equilibrium is reached at [CO₂] = 0.125 mol dm⁻³ regardless of the precise extra mass of CaCO₃ or CaO.
Using Mr(CaCO₃) ≈ 100.1, the carbonate consumed would be about 15.0 g. That is a large amount for a small teaching example and is simply chosen to make the mole accounting transparent. It is an illustration, not a laboratory procedure to attempt independently.
Numerical example 2: what if there is not enough solid?
Keep the same vessel and temperature but suppose there is initially only 0.100 mol of CaCO₃ available to decompose and no other reaction produces CO₂. The amount of CO₂ that can be generated is at most 0.100 mol, equivalent to a concentration increase of 0.100/2.00 = 0.0500 mol dm⁻³.
Starting from [CO₂] = 0.0500, complete decomposition of that available carbonate would give [CO₂] = 0.100 mol dm⁻³, which remains less than the two-solid coexistence value of 0.125 mol dm⁻³.
What happens? The CaCO₃ can be exhausted. You cannot continue the simple CaCO₃/CaO/CO₂ coexistence-equilibrium calculation when one required pure solid phase is absent. This is a material-balance limit, not a mysterious change of Kc.
The distinction matters in exams. If a problem states that solid CaCO₃ and CaO are both present at equilibrium, the pure-solid activities are fixed. If the information implies one solid has run out, check whether the requested coexistence equilibrium can actually be attained.
Will adding more CaO shift the equilibrium?
For the ideal closed system at constant temperature, if both CaCO₃ and CaO are already present, simply adding more pure CaO does not change the CO₂ equilibrium concentration. The amount of pure solid increases, but its activity does not.
Students frequently reason, ‘CaO is a product, so adding it makes equilibrium shift left.’ That wording takes Le Châtelier’s principle out of its physical context. Adding a pure solid product when that pure phase is already present is not equivalent to increasing the concentration of a dissolved or gaseous product.
If added solid changes the apparatus temperature, the surface area and reaction speed, or the available solid phases, then those effects must be considered separately. But the basic fixed-temperature equilibrium result does not move just because another chunk of the same pure solid is present.
What happens if CO₂ is added instead?
Adding CO₂ increases its gaseous concentration immediately, so Qc becomes greater than Kc at the same temperature. If both pure solids are present and can react, the system tends toward the reverse reaction, converting CaO and CO₂ into CaCO₃ until the equilibrium CO₂ value is restored or a phase runs out.
This is a genuine equilibrium disturbance because the activity of the gaseous species has changed. It is therefore different from adding more pure CaO. The important distinction is not ‘reactant versus product’ alone. It is whether you changed an activity that enters the equilibrium relationship.
Similarly, if CO₂ is removed at fixed temperature and both solids remain available, net decomposition may replace it. The system can act like a buffer for CO₂ pressure over a region where both pure solid phases coexist.
What if I add a catalyst or grind the solid into powder?
Grinding the carbonate increases surface area and may affect how fast it decomposes or reacts. Catalysts, where suitable, can also change reaction pathways and kinetics. Neither action by itself changes the fixed-temperature thermodynamic Kc for the same overall reaction.
The distinction between equilibrium composition and rate of approach matters here. A fine powder may reach equilibrium faster than a large lump. The final CO₂ equilibrium concentration for a two-phase pure-solid coexistence system at the same temperature remains governed by the equilibrium condition.
Do not confuse ‘more surface’ with ‘more concentration in Kc’. Surface area enters many kinetic arguments; it is not inserted as a concentration power in the ordinary Kc expression.
A second example with gases on both sides
Consider the idealised high-temperature heterogeneous reaction:
Fe₃O₄(s) + 4H₂(g) ⇌ 3Fe(s) + 4H₂O(g)
Both Fe₃O₄ and Fe are pure solids. The concentration expression therefore includes only the gaseous species:
Kc = [H₂O]⁴ / [H₂]⁴
The powers come from the balanced coefficients of the gaseous species. A student who writes a factor of [Fe]³ or [Fe₃O₄] in the equilibrium fraction has mixed up the balanced equation with the activities of pure phases.
At school level, the bracket notation for H₂O here refers to water vapour, H₂O(g). Water existing as a pure liquid in another equilibrium would be treated differently. Reading the state symbol is not optional.
A third example: why liquid water may also be omitted
Take H₂O(l) ⇌ H₂O(g) in a closed system at a fixed temperature where pure liquid and vapour coexist. In a simple concentration-based equilibrium expression, the activity of pure liquid water is absorbed into the constant; the vapour contributes a variable term.
This resembles the familiar fact that, at a fixed temperature, a pure liquid’s equilibrium vapour pressure is characteristic of the substance while some liquid remains. Add more liquid to an already coexisting liquid-vapour system, and you have more liquid—but not necessarily higher equilibrium vapour pressure.
The analogy is useful: it shows that the ‘omit pure phase’ rule is a reflection of physical equilibrium, not just a notation convention invented for exams.
The difference between Kc and Qc when solids are present
Kc describes the equilibrium expression evaluated at equilibrium at a specified temperature. Qc is the corresponding reaction quotient calculated using current concentrations, whether or not the system has reached equilibrium.
For CaCO₃(s) ⇌ CaO(s) + CO₂(g), both quantities use the same form under the pure-phase assumptions: Kc is the equilibrium [CO₂], while Qc is the current [CO₂].
- Qc < Kc: the forward direction is thermodynamically favoured under the stated conditions, provided the reacting phases are available.
- Qc = Kc: the coexistence equilibrium condition is satisfied.
- Qc > Kc: the reverse direction is favoured, provided CaO and the other required species are available.
This is why the existence of each phase matters. The comparison tells us a driving direction, but it does not create missing material.
How the same principle appears in Ksp
For a sparingly soluble ionic solid such as AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq), the conventional solubility-product expression is Ksp = [Ag⁺][Cl⁻]. AgCl(s) is a pure solid and is not written as a concentration factor.
A precipitation question therefore asks you to calculate a dissolved-ion product and compare it with Ksp. It does not ask for the number of grams of solid to be placed in the denominator. Again, dissolved-ion activity can change; the activity of an existing pure AgCl solid is constant.
This is a bridge between chemical equilibria and JC2 solubility equilibria. The same physical idea appears under different chapter headings.
A diagnostic table for the first mistaken step
| Student answer | First conceptual error | Correction |
|---|---|---|
| Kc = [CaO][CO₂]/[CaCO₃] | Copied all substances regardless of phase | Omit the pure solids; Kc = [CO₂] |
| Adding CaO must shift equilibrium left | Treated pure solid as dissolved product concentration | Check whether its activity changed |
| Doubling CaCO₃ mass doubles Kc | Confused amount of solid with temperature-dependent constant | Kc is fixed at a specified temperature |
| A powdered solid gives a larger Kc | Confused kinetics and thermodynamics | Surface area may affect rate, not the equilibrium constant |
| Qc says decomposition continues even after CaCO₃ disappears | Ignored available material | Check phase coexistence and conservation of matter |
| Water always disappears from Kc | Ignored the phase symbol | Pure liquid water differs from H₂O(g) or aqueous solute terms |
Four changed questions to test real mastery
- Write Kc for C(s) + CO₂(g) ⇌ 2CO(g). Explain why C(s) is absent.
- A vessel at constant temperature contains both CaCO₃(s) and CaO(s) at equilibrium. More CaCO₃ is added. Does Kc change?
- In CaCO₃(s) ⇌ CaO(s) + CO₂(g), Qc = 0.150 and Kc = 0.125 in the same concentration convention. Which net direction is favoured if both solid phases exist?
- A question asks for Kc of H₂O(l) ⇌ H₂O(g). Which water species enters the usual expression and why?
Answers
1. Kc = [CO]²/[CO₂]. Solid carbon has constant pure-phase activity. 2. No, not merely from adding more pure solid at unchanged temperature, provided the required solid phases continue to exist. 3. The reverse direction consumes gaseous CO₂ and CaO to form CaCO₃ until equilibrium is restored, if sufficient CaO is available. 4. H₂O(g) contributes the variable vapour term; pure H₂O(l) has constant activity.
If any answer is uncertain, return to the physical question: what exactly changed in the chemical system? That is more useful than trying to remember a long list of exceptions.
A study routine from JC1 understanding to JC2 independence
Start with eight balanced equations that mix solids, liquids, gases and dissolved ions. Before calculating anything, circle state symbols and identify which species belong in Kc or Kp. Next, calculate a Kc from equilibrium data in a gas-only problem. Then move to a heterogeneous reaction and explain why its solids are omitted.
Two days later, complete a Qc versus Kc direction question and a limiting-solid question. In JC2, connect the same reasoning to Ksp and precipitation. Each practice session should include a changed reaction that cannot be solved by copying the previous line.
I would rather see five correctly explained equilibria than twenty expressions obtained by pattern-matching the position of substances in an equation.
What parents in Hougang should ask of an H2 Chemistry tutor
Bring one incorrect equilibrium answer from a school tutorial and ask the tutor to identify the first broken assumption. Does the student misunderstand the meaning of a phase symbol, the origin of powers in Kc, the distinction between Qc and Kc, or the limits of the amount available?
A useful three-student tutorial or private lesson should end with each learner doing a new equation independently. The tutor should be able to tell you exactly which misconception changed, rather than merely saying that the student ‘covered equilibrium’.
Check current H2 syllabus familiarity, teaching arrangement, venue, travel time and class availability with the provider. A Hougang-search-focused article does not by itself confirm a physical classroom in Hougang.
Why this is more than an exam rule
Geochemical carbonate decomposition, industrial reduction of metal oxides, materials processing and precipitation in water treatment all involve multiple phases. The amount of a phase and its thermodynamic activity are different concepts. Chemists and engineers must know when more raw material changes the possible production total, when it changes the speed, and when it does not change an equilibrium ratio.
The university or career path may lead into chemistry, materials science, engineering or environmental science. The lasting skill is the same: read the physical system before writing a mathematical expression.
Frequently asked questions
Why are solids omitted from Kc?
Pure solids have constant thermodynamic activities under the model used, so their factors are absorbed into the equilibrium constant rather than appearing as variable concentration terms.
Do all solids get omitted?
The standard rule applies to pure solids in these textbook equilibria. More complex materials such as solid solutions may need different treatments at advanced levels.
Does adding more CaCO₃ shift its decomposition equilibrium?
Not if both relevant pure solids are already present and temperature and gas conditions remain the same. Adding material can alter what is available to react but not the pure-solid activity.
Is Kc independent of the amount of solid?
At a fixed temperature in the pure-phase coexistence model, yes. The quantity of solid may determine whether that equilibrium state is attainable, which is a separate mass-balance question.
Does this matter for Ksp?
Yes. The pure solid is omitted; dissolved-ion terms determine the ion-product comparison.
Continue learning
Why Qc is not Kc until equilibrium · H2 Chemistry equilibrium and Kc tutorials · JC2 solubility product and common-ion effect · Hougang Chemistry library · Contact eduKate Singapore for current class information.
This article uses idealised high-temperature equilibrium examples to teach reasoning. It does not instruct readers to heat carbonate or metal oxide mixtures outside a supervised chemistry laboratory.