Chemical Equilibrium

Equilibrium Constants

For a general reaction aA + bB ⇌ cC + dD, the equilibrium constant expression is Keq = [C]^c [D]^d / [A]^a [B]^b, where brackets denote equilibrium concentrations in mol/L (M). Kc uses concentrations; Kp uses partial pressures for gases. The two are related by Kp = Kc (RT)^Δn, where Δn = (moles gaseous products) − (moles gaseous reactants) and R = 0.08206 L·atm/mol·K. The magnitude of K tells you the equilibrium position: K >> 1 means the reaction is product-favored (equilibrium lies to the right); K << 1 means reactant-favored (equilibrium lies to the left); K ≈ 1 means significant amounts of both. Crucially, K is a constant at a given temperature—changing concentrations or pressures does NOT change K; only temperature changes K. For an exothermic reaction (ΔH < 0), increasing temperature decreases K (equilibrium shifts toward reactants). For an endothermic reaction (ΔH > 0), increasing temperature increases K (equilibrium shifts toward products). This is consistent with treating heat as a reactant or product: in an exothermic reaction, heat is a product (A + B ⇌ C + D + heat), so adding heat shifts equilibrium left, decreasing K. The MCAT frequently asks: 'Which change will alter the value of K?' The answer is ALWAYS temperature—and nothing else. K can be very large (10^30) or very small (10^−30); you must be comfortable with the exponential relationship between K and ΔG°: ΔG° = −RT ln K. At 298 K, every factor-of-10 change in K corresponds to roughly 5.7 kJ/mol change in ΔG°. If K = 10^10, ΔG° ≈ −57 kJ/mol—strongly product-favored. If K = 10^−5, ΔG° ≈ +28.5 kJ/mol—reactant-favored under standard conditions, but remember that the reaction may still proceed forward if Q is kept below K.

Reaction Quotient

The reaction quotient Q has exactly the same mathematical form as K—Q = [C]^c [D]^d / [A]^a [B]^b—but uses the current (non-equilibrium) concentrations, not equilibrium concentrations. Comparing Q to K tells you the direction the reaction must shift to reach equilibrium: if Q < K, the ratio of products to reactants is too low, so the reaction proceeds forward (→), consuming reactants and forming products until Q = K. If Q > K, there are too many products, so the reaction proceeds in reverse (←), consuming products and forming reactants until Q = K. If Q = K, the system is at equilibrium. This Q-versus-K comparison is the quantitative counterpart to Le Chatelier's principle. The MCAT often presents initial concentrations, asks you to calculate Q, and then determine the direction of shift. A classic trap: students forget that Q must be calculated from the actual balanced equation with correct exponents. For example, for 2 SO2 + O2 ⇌ 2 SO3, Q = [SO3]^2 / ([SO2]^2 [O2])—the exponent on SO2 is 2, not 1. Another application of Q: predicting precipitation. For a salt with Ksp, calculate the ion product Qsp = [cation][anion] (with exponents if needed). If Qsp > Ksp, precipitation occurs; if Qsp < Ksp, the solution is unsaturated and more solid can dissolve; if Qsp = Ksp, the solution is saturated at equilibrium. This is directly analogous to the general Q-vs-K comparison. The MCAT may also combine Q with acid-base equilibria: comparing Q (or the current [H+]) to Ka tells you whether an acid will dissociate further. In all cases, the logic is identical: calculate Q, compare to K, predict the shift.

Le Chatelier's Principle

Le Chatelier's principle states: if a system at equilibrium is subjected to a stress (change in concentration, pressure/volume, or temperature), the system shifts its equilibrium position to partially counteract the stress and re-establish equilibrium. The three types of stress and their effects: (1) Concentration change: adding a reactant shifts the reaction toward products (→); removing a product also shifts toward products. Adding a product shifts toward reactants (←); removing a reactant also shifts toward reactants. IMPORTANT: concentration changes shift the equilibrium position but do NOT change K. After the shift, the new equilibrium concentrations satisfy the SAME K value. The MCAT loves to test this with a numerical example: start at equilibrium, add more reactant, and ask for the new equilibrium concentrations—you must set up an ICE (Initial, Change, Equilibrium) table and solve using the original K. (2) Pressure/volume change (gases only): increasing pressure by decreasing volume shifts the equilibrium toward the side with FEWER moles of gas. Decreasing pressure (increasing volume) shifts toward the side with MORE moles of gas. If Δn = 0 (equal moles of gas on both sides), pressure changes have NO effect on equilibrium position. Adding an inert gas at constant volume does NOT change partial pressures of reacting gases and therefore does NOT shift equilibrium. Adding an inert gas at constant pressure does increase volume, which decreases partial pressures and shifts equilibrium toward the side with more moles. (3) Temperature change: this is the ONLY stress that changes K. For an exothermic reaction (ΔH < 0), heat is a product—increasing temperature is like adding a product, so the reaction shifts left and K decreases. For an endothermic reaction (ΔH > 0), heat is a reactant—increasing temperature shifts right and K increases. The magnitude of the K change is given by the van't Hoff equation: ln(K2/K1) = −(ΔH°/R)(1/T2 − 1/T1). A catalyst does NOT shift equilibrium—it speeds up both forward and reverse reactions equally, so equilibrium is reached faster but the position (and K) is unchanged.

Homogeneous and Heterogeneous Equilibria

A homogeneous equilibrium is one in which all reactants and products are in the same phase—all gases or all aqueous. Examples: N2(g) + 3 H2(g) ⇌ 2 NH3(g) (all gases); CH3COOH(aq) ⇌ CH3COO^−(aq) + H+(aq) (all aqueous). In homogeneous gas-phase equilibria, you can write K in terms of either concentrations (Kc) or partial pressures (Kp). A heterogeneous equilibrium involves species in more than one phase. The critical rule: the concentrations of pure solids and pure liquids are constant and are omitted from the equilibrium constant expression. For example, for CaCO3(s) ⇌ CaO(s) + CO2(g), K = P_CO2 (or Kc = [CO2])—only the gaseous species appears. The amounts of CaCO3 and CaO present do not affect the equilibrium CO2 pressure as long as both solids are present. For the dissolution of a sparingly soluble salt: AgCl(s) ⇌ Ag+(aq) + Cl^−(aq), Ksp = [Ag+][Cl^−]. The solid AgCl is omitted. Similarly, for the autoionization of water: 2 H2O(l) ⇌ H3O+(aq) + OH^−(aq), Kw = [H3O+][OH^−] = 1.0 × 10^−14 at 25°C. The concentration of liquid water (~55.5 M) is constant and incorporated into Kw. For reactions involving a liquid solvent that is also a reactant, like an esterification: CH3COOH(l) + C2H5OH(l) ⇌ CH3COOC2H5(l) + H2O(l), if water is not the solvent but a product, it IS included in K. The rule is: omit solids and liquids only when they are present as pure phases. A solute in aqueous solution IS included. The MCAT may test the subtlety that adding more solid CaCO3 to an equilibrium mixture at constant temperature does NOT shift the equilibrium—the amount of solid present does not affect the position as long as some solid remains.

Equilibrium Shifts

The quantitative tool for predicting how equilibrium concentrations change in response to a stress is the ICE table (Initial, Change, Equilibrium). Set up the balanced equation, write the known initial concentrations, define the change in terms of x (using stoichiometric coefficients), express equilibrium concentrations as initial + change, substitute into the K expression, and solve for x. If K is very small (< 10^−3), you can often use the approximation that x is negligible compared to initial concentrations (the 5% rule: if x is less than 5% of the initial concentration, the approximation is valid). If K is large (> 10^3), the reaction goes essentially to completion and you work backward from the limiting reagent. The MCAT expects you to handle these calculations efficiently—usually the numbers are chosen so that the math is straightforward (perfect squares, negligible-x approximation valid). A common exam task: given initial concentrations and K, determine equilibrium concentrations. Another: given equilibrium concentrations, calculate K. Another: after a concentration or volume change at equilibrium, calculate the new equilibrium concentrations. In all these, the approach is the same—ICE table → K expression → solve. For acid-base equilibria, the same ICE logic applies: for a weak acid HA with Ka, the ICE table yields [H+] = √(Ka × [HA]_initial) when the approximation holds (x << [HA]_initial). The Henderson-Hasselbalch equation (pH = pKa + log([A^−]/[HA])) is a shortcut for buffer systems that skips the ICE table. The MCAT may also test equilibrium shifts through the lens of coupled equilibria: the overall K for a multi-step reaction is the product of the K values of individual steps (K_overall = K1 × K2 × ...). This follows from the rule that when you add reactions, you multiply their equilibrium constants.

Solubility Product

The solubility product constant Ksp is the equilibrium constant for the dissolution of a sparingly soluble ionic solid. For a salt with the general formula M_m X_n, dissolution is: M_m X_n(s) ⇌ m M^{n+}(aq) + n X^{m−}(aq), and Ksp = [M^{n+}]^m [X^{m−}]^n. Ksp is temperature-dependent. A smaller Ksp indicates lower solubility for salts of the same stoichiometry (e.g., comparing AgCl with Ksp = 1.8 × 10^−10 to AgBr with Ksp = 5.0 × 10^−13—AgBr is less soluble). But you CANNOT compare Ksp values across salts with different stoichiometries without calculating molar solubility first. For a 1:1 salt like AgCl, molar solubility s = √Ksp. For a 1:2 salt like PbCl2, Ksp = [Pb^2+][Cl^−]^2 = s × (2s)^2 = 4s^3, so s = (Ksp/4)^{1/3}. The common ion effect suppresses solubility: adding a salt that shares an ion with the sparingly soluble salt shifts the dissolution equilibrium left (Le Chatelier). For example, adding NaCl to a saturated AgCl solution adds Cl^−, shifting AgCl(s) ⇌ Ag+ + Cl^− left, decreasing [Ag+] and causing AgCl to precipitate. This is exploited in qualitative analysis: adding a common ion can selectively precipitate one salt while leaving another in solution. The opposite effect—increasing solubility by removing an ion—occurs when the ion participates in a secondary equilibrium. For example, AgCl is more soluble in aqueous NH3 because Ag+ reacts with NH3 to form the complex ion [Ag(NH3)2]+, removing free Ag+ and pulling the dissolution equilibrium right. Similarly, metal hydroxides are more soluble at low pH because H+ consumes OH^−, shifting M(OH)n(s) ⇌ M^{n+} + n OH^− right. The MCAT may ask you to calculate solubility in pure water versus in a solution containing a common ion, or predict whether a precipitate will form by calculating Qsp and comparing to Ksp.

How it works

Chemical equilibrium unifies kinetics and thermodynamics. At the molecular level, the forward and reverse reactions never stop—they just proceed at equal rates. The equilibrium constant K = k_forward / k_reverse is the ratio of the rate constants, which explains why a catalyst (which lowers activation energy for both directions equally) does not change K. The thermodynamic basis is ΔG° = −RT ln K: K is exponentially sensitive to ΔG°. A reaction with ΔG° = −40 kJ/mol has K ≈ 10^7 at 298 K—the equilibrium is far to the right. A reaction with ΔG° = +40 kJ/mol has K ≈ 10^−7—equilibrium is far to the left. But remember: ΔG, not ΔG°, determines actual direction. Q vs. K is the practical tool: calculate Q from current concentrations, compare to K, predict shift. Le Chatelier codifies this in qualitative rules. The MCAT expects you to be fluent in both approaches—qualitative (predicting shift direction from Le Chatelier) and quantitative (ICE tables, K calculations, Ksp determinations). The most important framing is that K is a CONSTANT at a given temperature; only T changes K. Everything else changes the position but not the value.

How it works

Chemical equilibrium unifies kinetics and thermodynamics. At the molecular level, the forward and reverse reactions never stop—they just proceed at equal rates. The equilibrium constant K = k_forward / k_reverse is the ratio of the rate constants, which explains why a catalyst (which lowers activation energy for both directions equally) does not change K. The thermodynamic basis is ΔG° = −RT ln K: K is exponentially sensitive to ΔG°. A reaction with ΔG° = −40 kJ/mol has K ≈ 10^7 at 298 K—the equilibrium is far to the right. A reaction with ΔG° = +40 kJ/mol has K ≈ 10^−7—equilibrium is far to the left. But remember: ΔG, not ΔG°, determines actual direction. Q vs. K is the practical tool: calculate Q from current concentrations, compare to K, predict shift. Le Chatelier codifies this in qualitative rules. The MCAT expects you to be fluent in both approaches—qualitative (predicting shift direction from Le Chatelier) and quantitative (ICE tables, K calculations, Ksp determinations). The most important framing is that K is a CONSTANT at a given temperature; only T changes K. Everything else changes the position but not the value.

Comparisons

• C/P (Thermodynamics): ΔG° = −RT ln K. At 298 K, ΔG° (kJ/mol) ≈ −5.7 log10 K. If K = 10^10, ΔG° ≈ −57 kJ/mol. This equation appears in both general chemistry and biochemistry passages. You must be able to solve for any variable given the others.
• C/P (Kinetics): At equilibrium, rate_forward = rate_reverse, so k_f [A]^a [B]^b = k_r [C]^c [D]^d, giving K = k_f / k_r. A catalyst increases both k_f and k_r equally, so K is unchanged—equilibrium is reached faster but the position is the same.
• C/P (Acids and Bases): Ka, Kb, and Kw are all equilibrium constants. Weak acid/base calculations use ICE tables with Ka or Kb. The Henderson-Hasselbalch equation (pH = pKa + log [A−]/[HA]) is derived from the Ka expression and the definition of pKa = −log Ka.
• C/P (Electrochemistry): The Nernst equation E = E° − (RT/nF) ln Q relates cell potential to concentrations. At equilibrium, E = 0 and Q = K, so E° = (RT/nF) ln K. This connects equilibrium constants to cell potentials—a common passage integration.
• C/P (Solutions): Ksp and the common ion effect explain selective precipitation in qualitative analysis. The relationship between Ksp and molar solubility depends on salt stoichiometry.
• B/B (Hemoglobin): O2 binding to hemoglobin is a classic Le Chatelier example: Hb + 4 O2 ⇌ Hb(O2)4. In the lungs (high pO2), equilibrium shifts right → O2 binds. In tissues (low pO2, high CO2/H+), the Bohr effect shifts equilibrium left → O2 is released.
• B/B (Enzyme kinetics): While Michaelis-Menten describes steady-state kinetics, the thermodynamics of enzyme-substrate binding is governed by equilibrium. The dissociation constant Kd = [E][S]/[ES] is an equilibrium constant; lower Kd means tighter binding.

Common confusions

• Thinking concentration or pressure changes alter K: K changes ONLY with temperature. Adding reactant shifts the position but the new equilibrium concentrations still satisfy the same K. If a problem asks 'What changes the value of K?' the answer is temperature—period.
• Including solids and liquids in K expressions: For CaCO3(s) ⇌ CaO(s) + CO2(g), K = P_CO2 only. Adding more CaCO3(s) does not shift equilibrium. Students routinely write K = [CaO][CO2]/[CaCO3]—this is wrong.
• Misidentifying the temperature effect: If ΔH < 0 (exothermic), T ↑ → K ↓ (shift left). If ΔH > 0 (endothermic), T ↑ → K ↑ (shift right). Memorize: increasing temperature favors the endothermic direction—the reaction absorbs some of the added heat.
• Forgetting to square/cube concentrations: For 2 NO2 ⇌ N2O4, K = [N2O4]/[NO2]^2. The exponent is the stoichiometric coefficient. A common error: writing K = [N2O4]/[NO2] or [N2O4]/2[NO2].
• Confusing Q and K: Q uses current concentrations; K uses equilibrium concentrations. Q vs. K tells you direction of shift. If Q < K, forward; if Q > K, reverse. Students sometimes reverse this.
• Comparing Ksp values across different stoichiometries: AgCl (Ksp = 1.8 × 10^−10) is a 1:1 salt; Ag2CrO4 (Ksp = 1.1 × 10^−12) is a 2:1 salt. You MUST calculate molar solubility to compare. For 1:1 salts, s = √Ksp; for 1:2, s = (Ksp/4)^{1/3}.
• Assuming a catalyst shifts equilibrium: A catalyst speeds up both forward and reverse reactions equally. Equilibrium is reached faster, but K and the equilibrium concentrations are unchanged. The MCAT loves this trap.
• Adding an inert gas at constant volume: This does NOT change partial pressures of reacting gases—no shift. Adding inert gas at constant pressure does increase volume, decreasing all partial pressures equally; then the shift depends on Δn (toward side with more moles of gas).
• Ignoring the stoichiometry in ICE tables: The 'Change' row must reflect stoichiometric ratios. For 2A ⇌ B, if A decreases by 2x, B increases by x. Getting the coefficients wrong in the Change row cascades into a wrong answer.
• Applying Le Chatelier to temperature without thermodynamic context: The van't Hoff equation ln(K2/K1) = −(ΔH°/R)(1/T2 − 1/T1) quantifies how K changes with T. For an endothermic reaction (ΔH° > 0), increasing T increases K.

Quick review

• K = [C]^c[D]^d / [A]^a[B]^b — products over reactants, each raised to its stoichiometric coefficient. Omit pure solids and liquids.
• Kp = Kc(RT)^Δn where Δn = mol gaseous products − mol gaseous reactants. R = 0.08206 L·atm/mol·K.
• Q has the same form as K but uses current (non-equilibrium) concentrations. Q < K → forward; Q > K → reverse; Q = K → at equilibrium.
• K changes ONLY with temperature. Concentration and pressure/volume changes shift the position but NOT the value of K.
• ΔG° = −RT ln K. At 298 K: ΔG° (kJ/mol) ≈ −5.7 log10 K. K > 1 ⇔ ΔG° < 0 (product-favored). K = 10^(−ΔG°/5.7) at 298 K.
• Le Chatelier: add reactant → shift right (→). Add product → shift left (←). Remove reactant → shift left. Remove product → shift right.
• Pressure/volume (gases only): ↑ pressure (↓ volume) → shift toward side with FEWER moles of gas. ↓ pressure (↑ volume) → shift toward side with MORE moles. Δn = 0 → no shift.
• Temperature: exothermic (ΔH < 0) → T ↑ shifts left, K ↓. Endothermic (ΔH > 0) → T ↑ shifts right, K ↑. Treat heat as product (exothermic) or reactant (endothermic).
• Catalyst: speeds up both forward and reverse equally. NO shift in equilibrium position. K unchanged. Equilibrium is reached faster.
• Inert gas at constant V: no shift (partial pressures of reactants/products unchanged). Inert gas at constant P: volume ↑, similar to pressure decrease.
• Ksp = [M^{n+}]^m [X^{m−}]^n for M_mX_n(s). Molar solubility s: for 1:1 salt, s = √Ksp; for 1:2 salt, s = ³√(Ksp/4).
• Qsp > Ksp → precipitation. Common ion effect: adding a shared ion decreases solubility (shifts dissolution equilibrium left).
• ICE table: Initial, Change (in stoichiometric ratio with x), Equilibrium = Initial + Change. Substitute into K expression, solve for x.
• Van't Hoff: ln(K2/K1) = −(ΔH°/R)(1/T2 − 1/T1). Used to calculate K at a new temperature given ΔH°.
• Overall K for multi-step reaction = product of individual K values: K_overall = K1 × K2 × K3 × ...