Equilibrium Constants

Core Concept

The equilibrium constant K is a number that quantifies the position of equilibrium. For the general reaction aA + bB ⇌ cC + dD, the law of mass action states that, at equilibrium, the ratio of product concentrations to reactant concentrations (each raised to its stoichiometric coefficient) is a constant at a given temperature:

K_c = [C]^c[D]^d[A]^a[B]^b

K is a measure of how far a reaction proceeds — its value fixes the balance between products and reactants.

Key Ideas

• K is written with products in the numerator, reactants in the denominator.
• Each concentration is raised to its stoichiometric coefficient.
• Pure solids and pure liquids are omitted (their "concentration" is constant and folded into K).
• K depends only on temperature; it is independent of starting amounts and of any catalyst.
• K is dimensionless (concentrations are understood as ratios to a standard state).

Equations and Variables

Concentration form (for reactions in solution or involving gases):

K_c = [C]^c[D]^d[A]^a[B]^b

Pressure form (for gas-phase reactions):

K_p = (P_C)^c(P_D)^d(P_A)^a(P_B)^b

• [X] = equilibrium molar concentration (M)
• P_X = equilibrium partial pressure (atm)
• a, b, c, d = stoichiometric coefficients

Omission rule: species written as (s) or pure (l) are left out of the expression entirely.

How It Works

The law of mass action follows from the equality of forward and reverse rates at equilibrium (see the Dynamic Equilibrium note). The exponents in K are the stoichiometric coefficients — unlike in rate laws, where exponents are experimental. Here the balanced equation does dictate the form of the expression.

For a homogeneous equilibrium, all species share one phase (all gas or all solution), and every species appears. For a heterogeneous equilibrium, multiple phases are present, and the pure solids and liquids are dropped. Example: CaCO₃(s) ⇌ CaO(s) + CO₂(g) gives simply K_c = [CO₂] (or K_p = P_CO₂). The solids' concentrations are constant, so they are absorbed into K.

Worked Example

(a) Homogeneous gas equilibrium. For 2 SO₂(g) + O₂(g) ⇌ 2 SO₃(g):

K_c = [SO₃]²[SO₂]²[O₂]   K_p = (P_SO3)²(P_SO2)²(P_O2)

(b) Heterogeneous equilibrium. For Fe₃O₄(s) + 4 H₂(g) ⇌ 3 Fe(s) + 4 H₂O(g), the solids Fe₃O₄ and Fe are omitted:

K_c = [H₂O]⁴[H₂]⁴   K_p = (P_H2O)⁴(P_H2)⁴

(c) Calculation. At 1000 K, K_c for COCl₂(g) ⇌ CO(g) + Cl₂(g) is 4.63 × 10⁻³. If at equilibrium [CO] = [Cl₂] = 0.0215 M, find [COCl₂].

K_c = [CO][Cl₂][COCl₂]   ⇒  [COCl₂] = (0.0215)(0.0215)4.63 × 10^-3 = 4.62 × 10^-44.63 × 10^-3 = 0.0998 M

How it works

The law of mass action follows from the equality of forward and reverse rates at equilibrium (see the Dynamic Equilibrium note). The exponents in K are the stoichiometric coefficients — unlike in rate laws, where exponents are experimental. Here the balanced equation does dictate the form of the expression.

For a homogeneous equilibrium, all species share one phase (all gas or all solution), and every species appears. For a heterogeneous equilibrium, multiple phases are present, and the pure solids and liquids are dropped. Example: CaCO₃(s) ⇌ CaO(s) + CO₂(g) gives simply K_c = [CO₂] (or K_p = P_CO₂). The solids' concentrations are constant, so they are absorbed into K.

Common confusions

• Including solids or liquids in K. Only aqueous and gaseous species belong.
• Putting reactants on top. Products always go in the numerator.
• Thinking K depends on starting amounts. It does not; only temperature changes K.
• Copying rate-law exponents into K. K's exponents are the stoichiometric coefficients.

Quick review

1. Write Kc for aA + bB ⇌ cC + dD.
2. Why are pure solids and liquids omitted from K?
3. What two variables, besides the balanced equation, does K depend on (and not depend on)?
4. Write Kp for 2 NO₂(g) ⇌ N₂O₄(g).
5. Why is K unitless?