Why don't pure solids or liquids appear in the Keq expression?

Keq is technically written in terms of activities, and the activity of a pure solid or pure liquid (including the solvent in dilute solution) is defined as 1 regardless of how much is present, since adding more doesn't change its concentration/density. So pure solids and liquids are simply left out of the expression entirely — only species whose effective concentration actually changes (gases and solutes) appear.

How does Le Chatelier's principle fit in with Q and K?

Every Le Chatelier's principle stress (adding reactant, removing product, changing volume/pressure for gases) works by instantly changing Q away from K. The system then reacts in whichever direction brings Q back toward K — exactly the same Q-vs-K comparison used here, just triggered by an external change instead of starting from pure reactant.

Chemical Equilibrium (Keq) Calculator

ICE table: A ⇌ B + C, starting from pure A

[A]₀ — initial concentration (M)

Keq

	[A]	[B]	[C]
Initial	0.500	0	0
Change	−x	+x	+x
Equilibrium	—	—	—

How the ICE table gets solved

Starting from pure A, every mole of A that reacts makes one mole of B and one mole of C, so all three concentrations move by the same x. Substituting the equilibrium row into Keq = [B][C]/[A] gives Keq = x²/([A]₀ − x), a quadratic in x — the exact same math this site's own weak-acid pH calculator uses for Ka = x²/(C − x), just relabeled. Solve with the quadratic formula, keeping only the physically possible (non-negative, less than [A]₀) root.

Reaction quotient: is this system at equilibrium?

Enter any current concentrations (not necessarily equilibrium ones) to find Q and see which direction the reaction still needs to shift.

[A] now (M)

[B] now (M)

[C] now (M)

Keq (same reaction)

Q vs Keq

—

Keq → ΔG°

Keq

Temperature (K)

ΔG°

—

Same relationship the Gibbs free energy calculator and Nernst equation calculator both build on — ΔG° = −RT ln(K), just with K in place of the reaction-quotient term.

Don't confuse ΔG° with ΔG. ΔG° is the standard-state value (all species at 1 M) and is fixed for a reaction at a given temperature. The actual free-energy change under real conditions is ΔG = ΔG° + RT ln Q — equivalently ΔG = RT ln(Q/K). This is the bridge between the two cards above: when Q < K, ln(Q/K) is negative, so ΔG < 0 and the forward reaction is spontaneous right now — the same forward shift the Q-vs-K comparison predicts. A reaction with a positive ΔG° can still run forward if Q is kept low enough, which is exactly how coupled reactions in metabolism work.

Practice problems

1. For A ⇌ B + C with [A]₀ = 1.00 M and Keq = 0.100, find the equilibrium concentrations.

Show answer

Keq = x²/([A]0 − x) x = (−Keq + √(Keq² + 4·Keq·[A]0)) / 2 = (−0.100 + √(0.0100 + 0.400)) / 2 = 0.270 M [A]eq = 1.00 − 0.270 = 0.730 M [B]eq = [C]eq = 0.270 M Check: (0.270)(0.270)/0.730 = 0.0999 ≈ 0.100 ✓

2. A reaction has Keq = 25. Right now, Q = 4. Which direction does the reaction shift, and what happens to Q as it does?

Show answer

Q (4) < Keq (25), so the reaction shifts forward (toward products) to catch Q up to K. As it shifts forward, [products] increase and [reactants] decrease, which makes Q itself increase — the reaction keeps shifting forward until Q rises all the way to meet Keq, at which point it's at equilibrium and stops net-shifting.

FAQ

What's the difference between Q and Keq?
Same expression, different moment: Keq uses equilibrium concentrations, Q uses whatever concentrations exist right now. Comparing them tells you which way the reaction still needs to shift.

Why don't pure solids or liquids appear in Keq?
Their activity is defined as 1 regardless of amount present, so they're left out of the expression entirely — only species whose effective concentration actually changes (gases, solutes) appear.

How exactly does Keq relate to ΔG°?
ΔG° = −RT ln(K). Large K gives a large negative ΔG°; small K gives a positive ΔG°; K=1 gives ΔG°=0 exactly.

Does a large Keq mean the reaction happens fast?
No — Keq is purely thermodynamic (where it ends up), not kinetic (how fast it gets there). A reaction can have a huge Keq and still be immeasurably slow without a catalyst.

How does Le Chatelier's principle fit in?
Every Le Chatelier stress works by instantly moving Q away from K. The system then shifts in whichever direction brings Q back toward K — the same Q-vs-K comparison used above, just triggered externally.