Equilibrium Expression

6 min read Updated Mar 26, 2026

Now that you understand what equilibrium looks like, let’s put a number on it. The equilibrium constant tells you exactly where the balance point lies - does this reaction barely make any products, or does it essentially go to completion?

The answer comes from the law of mass action. For a generic reversible reaction:

aA + bB ⇌ cC + dD

This ratio is constant at a given temperature. It does not matter how much of each substance you start with - once the system reaches equilibrium, this ratio always equals K.

Where Does Keq Come From?

The law of mass action connects equilibrium to kinetics. Consider a one-step reversible reaction:

2A ⇌ B + C

The forward rate is $\text{rate}_f = k_f[A]^2$ and the reverse rate is $\text{rate}_r = k_r[B][C]$ . At equilibrium, these rates are equal:

$\displaystyle k_f[A]^2 = k_r[B][C]$

Rearranging gives:

$\displaystyle K_{eq} = \dfrac{k_f}{k_r} = \dfrac{[B][C]}{[A]^2}$

The equilibrium constant is simply the ratio of the forward and reverse rate constants. A large K_eq means $k_{f}$ >> $k_{r}$ , so the forward reaction is much faster and products are heavily favored.

What Goes in the Expression?

Not every species appears in the equilibrium expression:

Species	Include in K?	Why
Aqueous solutes (aq)	Yes	Concentration can vary
Gases (g)	Yes	Partial pressure or concentration can vary
Pure solids (s)	No	Activity = 1 by definition
Pure liquids (l)	No	Activity = 1 by definition

Interpreting the Size of K

The value of K tells you which side the equilibrium favors:

K value	Meaning	Equilibrium position
K >> 1 (e.g., $10^6$ )	Products strongly favored	Far to the right
K ≈ 1	Neither side strongly favored	Roughly balanced
K << 1 (e.g., $10^{-8}$ )	Reactants strongly favored	Far to the left

Properties of K to Memorize

Reverse the reaction - the new K is 1/K_original
Multiply the reaction by a factor n - the new K is (K_original)^n
Add two reactions - the new K is K₁ × K₂
K is temperature-dependent - changing T is the ONLY way to change K
K has no units on the MCAT - it is technically based on activities, which are dimensionless

Write the equilibrium expression for: CaCO₃(s) ⇌ CaO(s) + CO₂(g)

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K_eq = [CO₂] (or K_p =

P_{\text{CO}}

₂). CaCO₃ and CaO are pure solids, so they do not appear in the expression. Only the gaseous CO₂ remains.

If the equilibrium constant for A ⇌ B is K = 100, what is the equilibrium constant for 2B ⇌ 2A?

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$K = 1/100^2 = 1/10{,}000 = 10^{-4}$ . Reversing the reaction gives 1/K =

\frac{1}{100}

. Multiplying by 2 raises it to the second power:

(1/100)^2 = 10^{-4}