Exam Weight

7% – 9% of AP Exam

Topics Covered

14 Standard Topics (7.1 – 7.14)

Core Concepts

Q vs K, ICE Tables, Le Châtelier, Ksp

Launch Free Lab

Pillar 1: The Equilibrium State & Constants

Dynamic Reversibility, Q vs K, and Equilibrium Constants (7.1–7.6)

Equilibrium represents a dynamic macroscopic steady state where forward and reverse reaction rates are precisely balanced.

Topics 7.1 & 7.2 Foundation & Definitions

Introduction to Dynamic Equilibrium

A reversible reaction reaches chemical equilibrium when the forward rate equals the reverse rate (\text{Rate}_{fwd} = \text{Rate}_{rev}). At this point, reactant and product concentrations remain constant over time, but molecular exchange never ceases.

Equal Rates, NOT Equal Amounts: The forward and reverse rates are strictly equal, but reactant and product concentrations are rarely equal.
Pure Solids and Pure Liquids Omitted: Only aqueous species (aq) and gases (g) appear in equilibrium expressions because the concentrations (densities) of pure solids and pure liquids remain constant during reaction.
Dynamic Particulate Evidence: Radiotracer experiments show continuous exchange between labeled reactants and products even after macroscopic concentrations stop changing.

Key Distinction: Never write that concentrations are equal at equilibrium! Write: "Concentrations remain constant because the rate of the forward reaction equals the rate of the reverse reaction."

Topics 7.3 – 7.5 Directional Vector

Reaction Quotient (Q) vs Equilibrium Constant (K)

The reaction quotient Q has the identical mathematical form as K, but is evaluated using current non-equilibrium concentrations or partial pressures.

Directional Comparison:
• Q < K: Excess reactants present; reaction shifts Right (Forward) to generate products.
• Q = K: System is at Dynamic Equilibrium; no net shift.
• Q > K: Excess products present; reaction shifts Left (Reverse) to generate reactants.
Magnitude of K:
• K \gg 1 (> 10^3): Products strongly favored at equilibrium.
• K \ll 1 (< 10^{-3}): Reactants strongly favored; reaction barely proceeds.
Pressure Constant Relation: K_p = K_c(RT)^{\Delta n}, where \Delta n = \text{moles of gaseous products} - \text{moles of gaseous reactants}.

Alphabetical Memory Trick: Put Q and K in alphabetical order (K then Q or vice versa). Notice Q < K points right (\rightarrow shifts forward) while Q > K points left (\leftarrow shifts reverse).

Topic 7.6 Algebraic Manipulation

Properties & Algebraic Rules for Equilibrium Constants

Unlike enthalpy (which is additive under Hess's Law), equilibrium expressions are multiplicative. Manipulating chemical equations alters K exponentially or reciprocally.

Chemical Operation	Forward Example	New Equilibrium Constant (K_{new})
Reverse Equation	B \rightleftharpoons A	Invert: K_{new} = \frac{1}{K_{fwd}} = (K_{fwd})^{-1}
Multiply Coefficients by n	n A \rightleftharpoons n B	Raise to power n: K_{new} = (K_{orig})^n
Divide Coefficients by 2	\frac{1}{2} A \rightleftharpoons \frac{1}{2} B	Take square root: K_{new} = \sqrt{K_{orig}} = (K_{orig})^{1/2}
Add Multiple Reactions	Reaction 1 + Reaction 2	Multiply constants: K_{overall} = K_1 \times K_2

Do NOT Add Constants: When summing two chemical equations, multiply their K values. Adding K values is an automatic zero on AP FRQs!

Hess's Law vs K Algebra:
• Enthalpy (\Delta H): Reverse = change sign; Multiply by n = multiply by n; Add equations = add \Delta H.
• Equilibrium (K): Reverse = reciprocal (1/K); Multiply by n = power (K^n); Add equations = multiply K.

Pillar 2: ICE Tables & Le Châtelier's Principle

ICE Tables & Le Châtelier's Principle (7.7–7.10)

Equilibrium calculations quantify shifting concentrations, while Le Châtelier's principle predicts how a system responds to external disturbances.

Topics 7.7 & 7.8 Heavyweight Calculation

Calculating Equilibrium Concentrations & ICE Tables

An ICE table organizes Initial concentrations, Changes governed by stoichiometric coefficients (\pm x), and resulting Equilibrium expressions.

Change Row Stoichiometry: For \text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3, if \text{N}_2 loses -x, \text{H}_2 loses -3x, and \text{NH}_3 gains +2x.
The 5% Approximation Rule: If K is tiny (K \le 10^{-4}) and initial concentrations are relatively large (\frac{[\text{initial}]}{K} > 400), the change x is negligible compared to initial value:
[A]_{eq} = [A]_0 - x \approx [A]_0.
Validating Approximation: Verify that \frac{x}{[A]_0} \times 100\% < 5\%. If greater than 5%, the full quadratic formula must be solved.

Particulate Diagram FRQs: When counting particles in equilibrium boxes, verify that the ratio \frac{[P]^p}{[R]^r} matches K, and verify that changes in particle counts obey equation stoichiometry!

Topics 7.9 & 7.10 Core Conceptual Pillar

Le Châtelier's Principle & Stress Shifts

When a chemical system at equilibrium is subjected to a disturbance, the system counteracts the stress by shifting its position of equilibrium to re-establish Q = K.

Concentration Disturbances: Adding a substance shifts the equilibrium away from that substance; removing a substance shifts the equilibrium toward that substance to replace it. (K remains unchanged!)
Volume & Pressure Disturbances:
• Decreasing volume (increasing pressure): Shifts toward the side with fewer moles of gas.
• Increasing volume (decreasing pressure): Shifts toward the side with more moles of gas.
• Adding an inert gas (e.g., He, Ar) at constant volume increases total pressure but does NOT change partial pressures of reactants; no shift occurs!
Temperature: The ONLY Stress That Changes K:
• Endothermic (\Delta H > 0): Treat heat as a reactant. Increasing T shifts right and increases K.
• Exothermic (\Delta H < 0): Treat heat as a product. Increasing T shifts left and decreases K.

AP Free Response Trap: Catalysts and inert gases at constant volume NEVER cause an equilibrium shift! Catalysts speed up forward and reverse rates equally without changing equilibrium yield or K.

Pillar 3: Solubility Product & Common-Ion Effect

Solubility Product (K_{sp}), Common-Ion Effect & pH (7.11–7.14)

Heterogeneous equilibria between undissolved ionic solids and dissolved aqueous ions dictate precipitation and selective dissolution.

Topic 7.11 Solubility Math

Solubility Product (K_{sp}) & Molar Solubility (s)

The solubility product constant K_{sp} governs saturated solutions of slightly soluble salts. Molar solubility (s) is the moles of solid that dissolve per liter of saturated solution.

1:1 Salt (\text{AgCl}(s) \rightleftharpoons \text{Ag}^+ + \text{Cl}^-):
K_{sp} = [s][s] = s^2 \implies s = \sqrt{K_{sp}}.
1:2 Salt (\text{PbI}_2(s) \rightleftharpoons \text{Pb}^{2+} + 2\text{I}^-):
K_{sp} = [s][2s]^2 = 4s^3 \implies s = \sqrt[3]{\frac{K_{sp}}{4}}.
Precipitation Prediction via Q_{sp}:
• Q_{sp} < K_{sp}: Unsaturated; no precipitate forms.
• Q_{sp} = K_{sp}: Exactly saturated at equilibrium.
• Q_{sp} > K_{sp}: Supersaturated; precipitate will form until Q_{sp} = K_{sp}.

Comparing Solubilities: You can only compare molar solubilities directly by comparing K_{sp} values if the salts have the same ion stoichiometry (e.g., both 1:1 salts). If comparing a 1:1 salt (\text{AgCl}) to a 1:2 salt (\text{Ag}_2\text{CrO}_4), you MUST calculate molar solubility s!

Topics 7.12 – 7.14 Coupled Equilibria

Common-Ion Effect, pH & Free Energy of Dissolution

Adding a soluble source of an ion already present in a saturated equilibrium suppresses solid dissolution according to Le Châtelier's principle.

Common-Ion Effect: Dissolving \text{AgCl} into 0.10\text{ M NaCl} shifts \text{AgCl}(s) \rightleftharpoons \text{Ag}^+ + \text{Cl}^- to the left, drastically reducing [\text{Ag}^+] and solid solubility.
pH and Solubility:
• If a salt contains a basic anion (e.g., \text{OH}^-, \text{F}^-, \text{CO}_3^{2-}), adding acid (\text{H}^+) removes the anion via neutralization, shifting equilibrium forward and increasing solubility.
• Salts of conjugate bases of strong acids (e.g., \text{Cl}^-, \text{Br}^-, \text{NO}_3^-) are unaffected by pH because these anions do not react with \text{H}^+!
Thermodynamic Relation: \Delta G^\circ = -RT \ln K_{sp}. A very small K_{sp} corresponds to a positive standard free energy change (\Delta G^\circ > 0).

Acid Solubility Question: "Will \text{CaF}_2 dissolve better in 1\text{ M HCl} or pure water?" Answer: In HCl! \text{F}^- is the conjugate base of weak acid \text{HF}; \text{H}^+ consumes \text{F}^-, driving dissolution forward.

⚖️ Equilibrium Matrix: The Le Châtelier Architect

Master stress shifts, reaction quotients (Q vs K), and K_sp precipitation. Free & interactive—no sign-up required.

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Equilibrium Trial 1

Target Reaction Scheme

Disturbance scenario description...

Haber-Bosch Q vs K Reaction Vector Arena

Reaction: \text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g) \quad (K_c = 0.50 \text{ at } 400^\circ\text{C})
Adjust non-equilibrium initial concentrations to calculate Q_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3} and predict the directional shift.

[N₂] Concentration: 1.00 M

[H₂] Concentration: 1.00 M

[NH₃] Concentration: 1.00 M

Solubility Product & Common-Ion Simulator

Equilibrium: \text{AgCl}(s) \rightleftharpoons \text{Ag}^+(aq) + \text{Cl}^-(aq) \quad (K_{sp} = 1.8 \times 10^{-10})
Observe how adding dissolved sodium chloride (a common ion source of \text{Cl}^-) suppresses the molar solubility of silver chloride.

Added NaCl Concentration: 0.00 M (Pure Water)

Microscopic Mechanics: Adding \text{Cl}^- increases ion collisions with \text{Ag}^+, driving the reverse precipitation rate higher than the forward dissolution rate until a new equilibrium with suppressed [\text{Ag}^+] is reached.

Continue Your AP Chemistry Mastery Route

Navigate directly to related foundational and advanced curriculum modules:

Unit 1: Atomic Structure Unit 2: Molecular & Ionic Bonding Unit 3: Intermolecular Forces Unit 4: Chemical Reactions Unit 5: Kinetics Unit 6: Thermodynamics Unit 7: Equilibrium (Active) Unit 8: Acids and Bases Unit 9: Applications of Thermo

Master Equilibrium: The Equilibrium Constant, ICE Tables, Le Châtelier's Principle, and K_sp.