MCAT topics → Thermodynamics and Kinetics → Chemical Equilibrium and Keq

Chemical Equilibrium and Keq

MCAT trap: Includes pure solids and liquids in the Keq expression. Pure solids and pure liquids are excluded from the K expression because their activities are defined as 1 (constant concentration).

Chemical equilibrium is one of the highest-yield topics on the MCAT, appearing across general chemistry, biochemistry, and acid-base passages. It describes the state where forward and reverse reaction rates are equal — concentrations stop changing, but the reaction hasn't stopped. The equilibrium constant Keq (also written K) captures the ratio of products to reactants at that point, with each species raised to its stoichiometric coefficient., appearing in general chemistry passages, biochemistry contexts (think enzyme-substrate binding), and standalone questions. The exam hits it from multiple angles: pure definition recall, algebraic ICE table calculations, conversion between Kp and Kc, and reasoning about what belongs in the K expression in the first place.

What makes this topic genuinely tricky isn't the math — it's the conceptual traps. Students consistently mix up thermodynamics and kinetics here: a large K means the equilibrium position favors products, but it says absolutely nothing about how fast equilibrium is reached. That's a kinetics question. Similarly, students often treat Kp and Kc as interchangeable for gas-phase reactions, which is only valid when the moles of gas don't change across the reaction. The MCAT will hand you a reaction where Δn ≠ 0 and expect you to apply Kp = Kc(RT)^Δn correctly.

The other major trap is the heterogeneous equilibrium rule — pure solids and pure liquids get excluded from the K expression entirely. This isn't arbitrary: their 'concentration' is constant (activity = 1), so including them would just multiply both sides by a constant and absorb into K anyway. Students who memorize the rule without understanding why will misapply it under pressure. Build the right mental model here and the calculation steps follow naturally.

Common misconceptions

Common mistake
Wrong: Pure solids and pure liquids are included in the equilibrium expression because they are reactants or products.
Right: Pure solids and pure liquids are excluded from the K expression because their activities are defined as 1 (constant concentration).
Pure solids and pure liquids have fixed 'concentrations' that don't change as the reaction proceeds — their thermodynamic activity is defined as exactly 1. Because multiplying by 1 doesn't change anything, chemists absorb these terms into the equilibrium constant itself rather than writing them explicitly. If you include a solid in your K expression during an MCAT calculation, you're adding a variable that shouldn't be there and your answer will be wrong.
Common mistake
Wrong: Kp and Kc are always numerically equal for gas-phase reactions.
Right: Kp = Kc(RT)^Δn, so they are equal only when Δn = 0 (no change in moles of gas).
Kp uses partial pressures while Kc uses molar concentrations, and these are only numerically equal when the total moles of gas are the same on both sides of the reaction (Δn = 0). When Δn ≠ 0, the ideal gas law connects pressure and concentration — that's where the (RT)^Δn correction comes from. Whenever you see a gas-phase equilibrium problem, count the moles of gas on each side first before deciding whether to treat Kp and Kc as equal.
Common mistake
Wrong: A large equilibrium constant means the reaction reaches equilibrium quickly.
Right: Keq reflects the thermodynamic favorability (ratio of products to reactants at equilibrium) and says nothing about the rate at which equilibrium is reached.
Keq is a thermodynamic quantity — it tells you where the system ends up (the ratio of products to reactants at equilibrium), not how quickly it gets there. A reaction with K = 10^10 could still take centuries without a catalyst. Rate is governed by activation energy and kinetics, which are completely independent of the equilibrium position. The MCAT loves to test this distinction, so never use K to make any claim about reaction speed.
Common mistake
Gap: Sets up ICE table changes without applying stoichiometric coefficients to the change row
In an ICE table, the change row must reflect stoichiometric ratios (e.g., if [A] decreases by x, [B] decreases by 2x for A + 2B ⇌ C), and the equilibrium row is substituted into the K expression to solve for x.
In an ICE table, the change in concentration for each species must honor the stoichiometry of the balanced equation. If A decreases by x, then in the reaction A + 2B ⇌ C, B decreases by 2x and C increases by x. Using the same x for every species regardless of coefficients is the most common ICE table error. Once your equilibrium row is set up correctly with these scaled changes, substitute each expression into the K formula and solve — usually giving a quadratic or a simplifiable expression when K is very small.
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What the exam tests

  1. Write the correct Keq expression for a given reaction: products over reactants, each raised to its stoichiometric coefficient — and know why that structure comes from the law of mass action.
  2. Convert between Kp and Kc using Kp = Kc(RT)^Δn, where Δn is the change in moles of gas from reactants to products in a gas-phase reaction.
  3. Set up and solve an ICE table — correctly applying stoichiometric ratios to the change row — and substitute the equilibrium row into the K expression to find unknown concentrations.
  4. Identify which species belong in the K expression for a heterogeneous equilibrium: exclude pure solids and pure liquids, include dissolved species and gases.

Can you avoid these mistakes?

For the reaction N2(g) + 3H2(g) ⇌ 2NH3(g), write the correct Keq expression and determine whether Kp is greater than, less than, or equal to Kc at 298 K. Justify your answer using Δn.
Consider the heterogeneous reaction CaCO3(s) ⇌ CaO(s) + CO2(g). Write the correct K expression. A student writes K = [CO2] / ([CaCO3][CaO]) — what's wrong with this, and what is the correct form?
A reaction A ⇌ 2B has Keq = 4.0 × 10^-3. Starting with [A] = 0.500 M and [B] = 0, set up the ICE table (being careful about stoichiometric coefficients in the change row) and write the equation you would solve to find equilibrium concentrations.
A reaction has a very large Keq (≈ 10^8) but proceeds extremely slowly at room temperature. How do you explain this, and what would a catalyst do to Keq versus the reaction rate?

Related topics

Gibbs Free Energy and Spontaneity First Law of Thermodynamics and Internal Energy Enthalpy and Hess's Law Calorimetry and Heat Capacity Entropy and the Second Law

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