Chemistry 2e · Fundamental Equilibrium Concepts

Equilibrium Constants

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In 30 seconds
Why this matters
The college version
Eli explains
Key takeaway
Check yourself
Study tools
Sources & references

In 30 seconds

When a reversible reaction reaches equilibrium, the forward and reverse reactions are still occurring, but at equal rates, so the concentrations stop changing. The equilibrium constant (K) is a number that summarizes the composition of the mixture at that point. For a reaction written generally as

$a A + b B ⇌ c C + d D$

the equilibrium constant expression (the ) is

$K c = [C] c [D] d [A] a [B] b$

where the square brackets mean molar concentration (mol/L) and the exponents are the stoichiometric coefficients from the balanced equation. The subscript c indicates concentration-based; for gas reactions there is a pressure-based version, K_p, described below.

A key idea: K is a constant at a given temperature. It does not depend on how the equilibrium was reached — starting from pure reactants, pure products, or any mixture in between, the same value of K emerges. It does depend strongly on temperature, because temperature changes the relative energies of reactants and products.

Why this matters

The equilibrium constant turns a vague idea — "reactions can go both ways" — into a predictive tool. If you know K, you can predict:

Which side is favored. K > 1 means products dominate at equilibrium; K < 1 means reactants dominate.
What will happen when you start from a given mixture. Comparing the reaction quotient Q (same formula as K, but with current concentrations) to K tells you which direction the reaction must shift.
Whether a process is practical. The Haber process for ammonia, for example, is limited by an equilibrium that favors reactants at low pressure; engineers use K (and Le Châtelier's principle, the next topic) to design conditions that maximize yield.

Equilibrium constants also govern biological chemistry. The binding of oxygen to hemoglobin, the dissolving of kidney stones, and the carbonate equilibria that control ocean acidity are all described by equilibrium constants.

The college version

Core Concepts

The law of mass action: products over reactants, raised to coefficients

The equilibrium expression always places product concentrations in the numerator and reactant concentrations in the denominator, each raised to its balanced-equation coefficient. It is a ratio of concentrations at equilibrium, not at any other time. Write the expression only from the balanced equation — never from the stoichiometry of some other step or from the rate law. For example, for

$N 2 (g) + 3H 2 (g) ⇌ 2NH 3 (g)$

the expression is

$K c = [NH 3] 2 [N 2][H 2] 3$

Note the exponent 3 on [H₂], matching the coefficient 3 in the equation.

Homogeneous versus heterogeneous equilibria

In a , all reactants and products are in the same phase (all gas, or all in one solution). In a , more than one phase is present — for example, a solid decomposing to gases, or a gas dissolving in a liquid.

Pure solids and pure liquids do not appear in the equilibrium expression. Their concentrations are effectively constant (density does not change meaningfully), so they are built into the value of K. For the thermal decomposition of calcium carbonate,

$CaCO 3 (s) ⇌ CaO(s) + CO 2 (g)$

the equilibrium expression is simply

$K c = [CO 2]$

K_p: equilibrium constants for gases

For gas-phase reactions, it is often more convenient to measure partial pressures than concentrations. The pressure-based constant K_p has the same product-over-reactants structure, but with partial pressures (in atm) instead of molarities. The two constants are related by

$K p = K c (RT) Δ n$

where R = 0.08206 L·atm·mol^-1K^-1, T is the Kelvin temperature, and Δn is the change in moles of gas: moles of gaseous products minus moles of gaseous reactants. When the number of gas moles is unchanged by the reaction (Δn = 0), K_p = K_c.

What the magnitude of K tells you

K ≫ 1 (say > 10³): products strongly favored; equilibrium lies far to the right.
K ≪ 1 (say < 10^-3): reactants strongly favored; equilibrium lies far to the left.
K near 1: significant amounts of both reactants and products coexist.

This tells you about composition at equilibrium, not about speed. A reaction can have a huge K and still be impractically slow without a catalyst — equilibrium constants say nothing about kinetics.

The value of K depends on how the equation is written. Reversing the reaction inverts the constant: K_reverse = 1/K. Multiplying all coefficients by n raises the constant to the nth power: K' = Kⁿ. Adding reactions multiplies their constants. These rules matter because the same process can be written different ways in different sources; comparing K values requires comparing identically written equations.

How It Works / Step-by-Step Process

Worked example 1: writing an equilibrium expression (with dimensional analysis)

Problem. Write K_c for 2SO₂(g) + O₂(g) ⇌ 2SO₃(g).

Solution.

Identify products and reactants from the balanced equation: products = SO₃, reactants = SO₂ and O₂.
Place products over reactants and raise each concentration to its coefficient:

$K c = [SO 3] 2 [SO 2] 2 [O 2]$

Check units: each molarity is mol/L, so K_c has units of (mol/L)²(mol/L)²(mol/L) = L/mol. In practice, chemists report K as a pure number using standard concentrations, but tracking the units confirms the exponent structure is correct.

Worked example 2: converting K_c to K_p

Problem. For H₂(g) + I₂(g) ⇌ 2HI(g), K_c = 50.2 at 445 °C. Find K_p.

Solution.

Convert temperature to kelvin: T = 445 + 273 = 718 K.
Compute Δn = moles gaseous products − moles gaseous reactants = 2 - (1+1) = 0.
Substitute into K_p = K_c(RT)^Δn:

$K p = 50.2 \times (0.08206 \times 718) 0 = 50.2 \times 1 = 50.2$

Because the reaction uses the same number of gas moles on both sides, pressure cancels and K_p = K_c. For a reaction with Δn ≠ 0, the (RT)^Δn factor adjusts the units from molarity to pressure: (mol/L × K)¹ = atm when Δn = 1, matching partial-pressure units.

Common Confusions

Do Not Confuse	With	Difference
Equilibrium constant	Rate constant (k)	K describes final composition; k describes how fast equilibrium is reached. They are unrelated numbers.
Concentrations in K	Initial concentrations	Only equilibrium concentrations go into K; initial values are used to compute Q or to set up ICE tables (next topic).
Coefficients in the expression	Experimental rate-law orders	Coefficients in K come from the balanced equation; orders in rate laws come from experiments.
Excluding solids from K	Solids being irrelevant to the reaction	Solids are still reactants/products; they just have constant concentration and drop out of the expression.
K changing with concentration	K changing with temperature	Adding reactant shifts the mixture but the value of K stays fixed; only temperature changes K.

Eli explains

The same idea, in plain words

Explain it like I’m 10

Imagine a seesaw where kids keep jumping on and off. At equilibrium, the number of kids on each side isn't changing — but kids are still moving. The equilibrium constant is a scoreboard that tells you which side usually "wins" at a given temperature. If the scoreboard shows a big number, the right side (products) is winning; a tiny number means the left side (reactants) stays in charge.

Key takeaways

K_c = product concentrations (each raised to its coefficient) divided by reactant concentrations (each raised to its coefficient).
K depends on temperature only — not on initial amounts, pressure, or the presence of a catalyst.
Pure solids and pure liquids are omitted from equilibrium expressions.
K_p = K_c(RT)^Δn for gas reactions, with Δn = (moles gaseous products) − (moles gaseous reactants).
K > 1 favors products; K < 1 favors reactants; magnitude says nothing about rate.
Reversing a reaction gives K' = 1/K; multiplying coefficients by n gives K' = Kⁿ.
The reaction quotient Q uses the same formula as K but with current (non-equilibrium) concentrations; Q < K means the reaction proceeds forward.

Check yourself

5 review questions from the chapter. Try each one, then open the answer.

Write the equilibrium expression for 2NO₂(g) ⇌ N₂O₄(g).
Show answer
K_c = [N₂O₄][NO₂]².
Why do pure solids and pure liquids not appear in an equilibrium expression?
Show answer
Pure solids and liquids have effectively constant concentration (fixed density), so they are incorporated into the value of K rather than appearing in the expression.
For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), what is Δn? Will K_p be larger or smaller than K_c?
Show answer
Δn = 2 - (1 + 3) = -2. Since (RT)^Δn = (RT)^-2 < 1, K_p < K_c.
If K = 10^-5 for a reaction, which side is favored at equilibrium?
Show answer
K < 1 means reactants are favored; equilibrium lies far to the left.
A reaction is written forward with K = 4. What is K for the reverse reaction?
Show answer
K_reverse = 1/K = 1/4 = 0.25.

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Study tools & related lessonsKey vocabulary · Related

Key vocabulary

equilibrium constant (K): A temperature-dependent ratio of product-to-reactant concentrations (or pressures) at equilibrium
law of mass action: The rule that K = products over reactants, each raised to its balanced coefficient
homogeneous equilibrium: Equilibrium where all species share one phase
heterogeneous equilibrium: Equilibrium involving more than one phase
Kp: Pressure-based equilibrium constant using partial pressures in atm
reaction quotient (Q): The same ratio as K, but with current, possibly non-equilibrium concentrations

Sources & references

openstax.org — Chemistry 2e

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