No, pure solids are not included in the equilibrium constant expression. Whether you’re writing Kc (using concentrations) or Kp (using partial pressures), you leave solids out entirely. The same rule applies to pure liquids, including solvents. This isn’t just a shortcut or simplification. It comes from a core principle in thermodynamics: the activity of a pure solid is always equal to 1, so including it in the expression wouldn’t change the math.
Why Solids Are Left Out
Equilibrium constants are built on a concept called “activity,” which is essentially a measure of how available a substance is to participate in a reaction. For gases and dissolved species, activity changes with concentration or pressure. More gas in a container means higher pressure, which matters for how much reaction happens. But a pure solid doesn’t work that way. Its composition is fixed. A block of calcium carbonate has the same chemical character whether you have one gram or one kilogram. Its activity is always 1.
Since the activity is 1, including a solid in the equilibrium expression would just mean multiplying or dividing by 1, which has no effect on the value of K. So chemists leave solids out to keep the expression clean and meaningful. The same logic applies to pure liquids (like water as a solvent in aqueous reactions).
How This Looks in Practice
The decomposition of calcium carbonate is one of the classic examples. The balanced equation is:
CaCO3(s) → CaO(s) + CO2(g)
Two of the three species here are solids. The equilibrium expression for Kc includes only the carbon dioxide:
Kc = [CO2]
That’s it. No term for calcium carbonate, no term for calcium oxide. The entire equilibrium constant depends on just the concentration (or partial pressure) of the gas. If you were writing Kp instead, it would simply be Kp = PCO₂.
Another common example is the Boudouard reaction, where solid carbon reacts with carbon dioxide to form carbon monoxide:
C(s) + CO2(g) → 2CO(g)
You leave out the solid carbon entirely. The Kp expression is:
Kp = (PCO)2 / PCO₂
What About the Amount of Solid Present?
A reasonable follow-up question: if solids don’t appear in the equilibrium expression, does the amount of solid matter at all? The answer is nuanced. The amount of a solid does not change the equilibrium constant or the equilibrium position. Adding more calcium carbonate to the reaction above won’t shift the equilibrium or produce more CO2 at a given temperature. This is different from how adding more of a dissolved reactant or a gas would behave.
The surface area of a solid can affect how quickly equilibrium is reached. Grinding a solid into a fine powder exposes more surface for the reaction to occur, which speeds things up. But once the system reaches equilibrium, the concentrations of gases and dissolved species settle at the same values regardless of how much solid is present or how finely it’s divided. Speed of arrival changes; the destination doesn’t.
Common Mistakes to Avoid
The most frequent error in writing equilibrium expressions is accidentally including a solid or pure liquid in the K expression. If you see (s) or (l) next to a species in a balanced equation, that species does not appear in the Kc or Kp expression. Here are the key points to remember:
- Pure solids (s): Always excluded. Activity equals 1.
- Pure liquids (l): Always excluded for the same reason. This most commonly applies to water in aqueous reactions.
- Aqueous species (aq): Included in Kc using molar concentrations.
- Gases (g): Included in Kc using concentrations or in Kp using partial pressures.
Another subtle point: Kc is constant only at a specific temperature. Changing the temperature changes the value of K. But adding or removing a solid never changes K. The equilibrium expression simply doesn’t see solids.
Why This Matters Beyond Exams
Understanding why solids are excluded helps you predict how real chemical systems behave. Consider a kiln decomposing limestone (calcium carbonate) to make quicklime. The amount of limestone you load into the kiln doesn’t determine how much CO2 builds up at equilibrium. Temperature does. This is why industrial processes that involve solid-gas reactions focus so heavily on temperature control rather than on packing in more raw material. The equilibrium expression tells you exactly which variables matter, and solids simply aren’t among them.