Homogeneous & Heterogeneous Equilibria

Phase: the dividing idea

A phase is a region of matter that is uniform in both chemical composition and physical state. A mixture of gases is a single phase, however many gases it contains, because the molecules are intermingled at the molecular level. A clear aqueous solution is likewise one phase. A lump of solid sitting in a gas, by contrast, presents two phases with a sharp boundary between them. The number of phases present at equilibrium is the single fact that sorts every equilibrium system into one of two classes.

When all reactants and products occupy the same phase, the equilibrium is homogeneous. When the system contains more than one phase, the equilibrium is heterogeneous. This is not a cosmetic label: it changes which species are permitted to appear in the equilibrium-constant expression, and therefore changes the numerical value and even the units of $K$.

Homogeneous equilibria

In a homogeneous system every reactant and product is in the same phase, so every concentration term is genuinely variable and every species appears in $K$. NCERT gives the all-gas synthesis of ammonia as the canonical example:

$$\ce{N2(g) + 3H2(g) <=> 2NH3(g)}$$

All three species are gases, so all three enter the constant. In terms of molar concentrations and partial pressures respectively,

$$K_c = \dfrac{[\ce{NH3}]^2}{[\ce{N2}][\ce{H2}]^3}, \qquad K_p = \dfrac{p_{\ce{NH3}}^2}{p_{\ce{N2}}\,p_{\ce{H2}}^3}$$

Homogeneous equilibria need not be gaseous. NCERT also cites reactions that are entirely in aqueous solution, such as the esterification equilibrium and the formation of the blood-red thiocyanate complex, where every species is dissolved in the same solution phase:

$$\ce{CH3COOC2H5(aq) + H2O(l) <=> CH3COOH(aq) + C2H5OH(aq)}$$ $$\ce{Fe^3+(aq) + SCN^-(aq) <=> [Fe(SCN)]^2+(aq)}$$

The first of these is worth a closer look, because it contains liquid water. When water is the bulk solvent of a dilute solution it behaves as a near-pure liquid and is treated as a constant, so in practice the working $K_c$ for hydrolysis or esterification omits the water term. The principle that does this omission is exactly the one that governs heterogeneous systems, taken up below.

Heterogeneous equilibria

Equilibrium in a system having more than one phase is called heterogeneous equilibrium. NCERT §6.5 opens with the simplest possible case — the equilibrium between liquid water and its own vapour in a closed container:

$$\ce{H2O(l) <=> H2O(g)}$$

Here a liquid phase and a gas phase coexist. A second standard example is a sparingly soluble solid in contact with its saturated solution, which spans a solid and an aqueous phase:

$$\ce{Ca(OH)2(s) <=> Ca^2+(aq) + 2OH^-(aq)}$$

Heterogeneous equilibria very often involve a pure solid or a pure liquid, and that is precisely what allows their equilibrium expressions to be simplified. The reasoning is given its own section because it is the most heavily tested idea in this subtopic.

Why pure solids and liquids drop out

The molar concentration of a pure solid or a pure liquid is fixed — it equals density divided by molar mass, both of which are constants for a given substance at a given temperature. Crucially, this value is independent of the amount present: a gram of calcium carbonate and a kilogram of it have the same molar concentration. NCERT states the rule directly — if a substance "X" is involved, then $[\ce{X(s)}]$ and $[\ce{X(l)}]$ are constant whatever amount of X is taken, whereas $[\ce{X(g)}]$ and $[\ce{X(aq)}]$ vary as the amount in a given volume varies.

Because the concentration of a pure condensed phase is a constant, it is absorbed into the equilibrium constant rather than written as a variable. In the modern language of activities, the activity of any pure solid or pure liquid is taken as exactly 1, so it contributes a factor of 1 and disappears from both $K_c$ and $K_p$.

Apply this to NCERT's headline example, the thermal dissociation of calcium carbonate:

$$\ce{CaCO3(s) <=> CaO(s) + CO2(g)}$$

Writing the full expression and then striking out the two constant solid terms gives a strikingly compact result:

$$K_c = \dfrac{[\ce{CaO(s)}]\,[\ce{CO2(g)}]}{[\ce{CaCO3(s)}]} \;\longrightarrow\; K_c' = [\ce{CO2}], \qquad K_p = p_{\ce{CO2}}$$

The same logic strips constant terms from other multi-phase systems. In the purification of nickel via its volatile carbonyl, the nickel metal is a pure solid and is omitted:

$$\ce{Ni(s) + 4CO(g) <=> Ni(CO)4(g)}, \qquad K_c = \dfrac{[\ce{Ni(CO)4}]}{[\ce{CO}]^4}$$

And in a reaction that mixes solid, aqueous and liquid species, only the aqueous terms survive — the solid oxide and the liquid water both drop out:

$$\ce{Ag2O(s) + 2HNO3(aq) <=> 2AgNO3(aq) + H2O(l)}, \qquad K_c = \dfrac{[\ce{AgNO3}]^2}{[\ce{HNO3}]^2}$$

Worked example: CaCO₃ decomposition

Homogeneous vs heterogeneous: K expressions

The table below collects the standard NCERT examples and shows, for each, the full expression before omission and the simplified expression after pure condensed phases are removed. Comparing the two columns is the fastest way to internalise the rule.

Δn and Kp = Kc(RT)^Δn

The general bridge between the two constants, derived in NCERT §6.4.1 from the ideal-gas relation $p = [\text{gas}]RT$, is

$$K_p = K_c\,(RT)^{\Delta n}, \qquad \Delta n = \big(\text{moles of gaseous products}\big) - \big(\text{moles of gaseous reactants}\big)$$

The phrase "gaseous" is doing real work here. Because solids and liquids never enter either $K_p$ or $K_c$, they also play no part in $\Delta n$. For $\ce{CaCO3(s) <=> CaO(s) + CO2(g)}$ the only gas is the single mole of $\ce{CO2}$ on the product side, so $\Delta n = 1$ and $K_p = K_c(RT)$. For $\ce{H2(g) + I2(g) <=> 2HI(g)}$ the gas moles balance ($2 - 2 = 0$), so $\Delta n = 0$ and $K_p = K_c$, both dimensionless. This is why the NEET 2024 item below singles out $\ce{PCl5 <=> PCl3 + Cl2}$, the one reaction in its list with a nonzero $\Delta n$.