12.5 Chemical Equilibrium: Keq & Le Chatelier's Principle

Most reactions relevant to the MCAT don't run to completion — they reach equilibrium, a dynamic state where the forward and reverse reaction rates become equal. At equilibrium, reactant and product concentrations stop changing, but the reaction itself hasn't stopped: molecules continue converting in both directions at exactly matched rates. This section builds on the thermodynamic favorability discussed in Sections 12.1–12.3 by supplying the quantitative tool for describing exactly where that balance point sits: the equilibrium constant. In physiology, the same ideas govern hemoglobin-oxygen binding, the CO2/bicarbonate buffer, and reversible metabolic steps near equilibrium in glycolysis.

Dynamic Equilibrium vs. Completion

A reaction that "goes to completion" has such a large Keq that leftover reactants are negligible; still, that is a limiting case of equilibrium, not a separate law. Dynamic equilibrium emphasizes that both directions continue: for A ⇌ B at equilibrium, rate_forward = rate_reverse, so d[A]/dt = 0 even though molecules still convert. Confusing static "stopped" with dynamic balance is a common wording trap.

The Law of Mass Action and the Equilibrium Constant

For a general reversible 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 own stoichiometric coefficient — is constant at a given temperature:

Keq = [C]^c[D]^d / ([A]^a[B]^b)

Two rules govern what belongs inside this expression. First, pure solids and pure liquids are omitted, because their effective concentration never changes and is instead folded into the constant itself. Second, only species that can actually vary in concentration — typically aqueous solutes or gases — appear in the expression. When every species is a gas, chemists often use Kp, written in terms of partial pressures instead of concentrations; Kp and Kc relate through Kp = Kc(RT)^Δn, where Δn is the change in moles of gas (moles of gaseous product minus moles of gaseous reactant).

Keq depends only on temperature. This is one of the most important traps on this topic: adding more reactant, removing product, or changing volume or pressure never changes the numerical value of Keq — those stresses shift the actual concentrations present at equilibrium, but the ratio the system settles back into is unchanged unless temperature itself changes.

The magnitude of Keq tells you where equilibrium lies without any arithmetic at all: Keq ≫ 1 means products dominate at equilibrium, Keq ≪ 1 means reactants dominate, and Keq ≈ 1 means comparable amounts of both are present. Remember from Section 12.2 that this magnitude maps onto ΔG° through ΔG° = −RT ln Keq.

Worked Example: Calculating Keq from Equilibrium Concentrations

For the Haber process, N2(g) + 3 H2(g) ⇌ 2 NH3(g), suppose equilibrium concentrations are measured as [N2] = 0.50 M, [H2] = 0.50 M, and [NH3] = 2.0 M. Write the expression and substitute directly:

Keq = [NH3]² / ([N2][H2]³) = (2.0)² / (0.50 × 0.50³) = 4.0 / (0.50 × 0.125) = 4.0 / 0.0625 = 64

Because Keq is much greater than 1, this equilibrium strongly favors ammonia — the product side — under these particular conditions.

Heterogeneous Equilibria

For CaCO3(s) ⇌ CaO(s) + CO2(g), the expression collapses to Keq = P_CO2 (or [CO2] if written as Kc), because pure solids drop out. Adding more solid CaCO3 does nothing to Keq or to equilibrium P_CO2 as long as some solid remains. This is a classic MCAT item pattern for solid-gas decompositions and for solubility equilibria written with solid pure phases omitted.

ICE Tables (Brief)

When only initial concentrations and Keq are known, an ICE table (Initial, Change, Equilibrium) tracks how concentrations move to equilibrium. For aA ⇌ bB with known Keq, define the change as x, write equilibrium concentrations as [A]0 − ax and [B]0 + bx, substitute into Keq, and solve for x. The MCAT rarely demands messy quadratic algebra; expect clean numbers, approximations when Keq is very small (x ≪ [A]0), or qualitative questions about the direction of shift rather than a full numerical solve.

Reaction Quotient (Q) versus Keq

The reaction quotient, Q, uses the exact same expression as Keq, but evaluated using concentrations at any arbitrary point in time — not necessarily at equilibrium. Comparing Q to Keq predicts which direction a reaction will shift to reach equilibrium:

Combined with ΔG = ΔG° + RT ln Q, when Q < Keq the reaction is spontaneous as written (ΔG < 0) until equilibrium is restored. When Q > Keq, the reverse is spontaneous. This is how cells keep pathways running: continuous product removal keeps Q small so ΔG stays negative even for steps with modest ΔG°.

Biological example: O2 binding to hemoglobin can be written in simplified form as Hb + n O2 ⇌ Hb(O2)n. In the lungs, high P_O2 makes Q large enough (or effectively drives binding) toward oxyhemoglobin; in tissues, low P_O2 and allosteric effectors shift the equilibrium toward deoxyhemoglobin. The "shift" language of Le Chatelier and the Q vs K comparison are the same idea applied to gas partial pressures.

Applying Le Chatelier's Principle

Le Chatelier's Principle states that when a system at equilibrium is disturbed by an outside stress, it shifts in whichever direction partially counteracts that stress.

Applying this to the Haber process — 4 total moles of gas on the reactant side, 2 moles on the product side — decreasing the container's volume raises the pressure and shifts equilibrium toward the side with fewer gas moles: toward ammonia. This is exactly why industrial ammonia synthesis is run under high pressure. If a reaction has equal moles of gas on both sides, pressure/volume changes do not shift the equilibrium position.

Temperature deserves special attention, because it is the one stress that actually changes the numerical value of Keq itself, rather than just shifting the position of equilibrium. Treating heat as though it were a reactant in an endothermic reaction (ΔH > 0), or a product in an exothermic reaction (ΔH < 0), lets the same shifting logic apply directly: raising the temperature of an endothermic reaction shifts it toward products and Keq increases; raising the temperature of an exothermic reaction shifts it toward reactants and Keq decreases.

A frequently tested trap belongs here too: a catalyst never appears as a stress that shifts position. Because a catalyst speeds up the forward and reverse reactions by an equal amount (Section 12.4), it changes only how quickly equilibrium is reached — never the equilibrium position and never Keq.

Physiological Le Chatelier Examples

• CO2 + H2O ⇌ H2CO3 ⇌ H+ + HCO3−: hypoventilation raises P_CO2, shifting the equilibrium toward more H+ (respiratory acidosis); hyperventilation lowers P_CO2, shifting toward less H+ (respiratory alkalosis).
• Hb(O2)n ⇌ Hb + n O2: high tissue CO2 and low pH (Bohr effect) effectively favor unloading by shifting oxygen-hemoglobin equilibrium toward free O2 and deoxyhemoglobin — a regulatory overlay on the same equilibrium framework.
• Removing product in a biosynthetic pathway (downstream enzyme consumption) continually shifts a reversible step forward without changing its Keq.

Keq and ΔG° — Closing the Thermodynamics Loop

As introduced in Section 12.2, the equilibrium constant connects directly to standard Gibbs free energy through ΔG° = −RT ln Keq. A large Keq (products favored) corresponds to a negative ΔG° — spontaneous as written under standard conditions — while a small Keq (reactants favored) corresponds to a positive ΔG°, and Keq = 1 gives ΔG° = 0. The practical takeaway is that everything covered here about writing and shifting Keq expressions is the same information already packaged into that thermodynamic relationship: a reaction driven strongly toward products by favorable energetics will also show a large equilibrium constant, and a reaction that barely favors products at all will show a Keq close to 1.

Kinetics still stands apart: a large Keq says nothing about how long it takes to get there. Catalysts and high temperature change rates; only temperature among common stresses changes Keq itself.

Common MCAT Traps

• Putting solids or pure liquids into Keq. They are omitted; only aqueous and gas species with variable concentration (or pressure) appear.
• Thinking concentration changes alter Keq. They alter Q and the position of equilibrium, not the constant's value at fixed T.
• Treating a catalyst as a Le Chatelier stress that shifts yield. Catalysts change rate, not equilibrium composition or Keq.
• Ignoring gas stoichiometry for pressure/volume stresses. Only unequal moles of gas on the two sides produce a shift when volume changes.
• Confusing Q with Keq. Q is the same expression at any moment; Keq is that expression only at equilibrium.
• Mixing kinetics with equilibrium favorability. Fast does not mean large Keq; large Keq does not mean fast.