Equilibrium — AP Chemistry Study Guide

For: AP Chemistry candidates sitting AP Chemistry.

Covers: This full unit overview maps the 12 core sub-topics of the AP Chemistry Equilibrium unit, from foundational dynamic equilibrium definitions through Q, K, Le Châtelier, solubility equilibria, and the unifying connection between equilibrium and thermodynamics.

You should already know: Balanced chemical reaction stoichiometry, basic thermodynamics concepts (enthalpy, entropy, free energy), fundamentals of reversible reaction kinetics.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Chemistry style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. Why This Unit Matters

Equilibrium is the fourth-largest unit on the AP Chemistry exam, accounting for 14–18% of total exam score, with concepts appearing in both multiple-choice (MCQ) and free-response (FRQ) sections, often integrated with acids and bases, thermodynamics, and solution chemistry. Unlike introductory chemistry that often frames reactions as going to completion, nearly all chemical reactions are reversible, and equilibrium provides the quantitative framework to predict how much product will form at a given set of conditions. This unit answers critical applied questions: how much of a drug will dissolve in the bloodstream? How does ocean acidification affect coral skeleton solubility? Why does changing temperature change the yield of industrial reactions like ammonia synthesis? Beyond applied problems, equilibrium unites two core themes of AP Chemistry: kinetics (how fast a reaction reaches equilibrium) and thermodynamics (how favorable the reaction is at equilibrium). Mastery of this unit is required for all subsequent study of acid-base equilibria, which builds directly on the core K and Q concepts introduced here.

2. Unit Concept Map

The Equilibrium unit builds incrementally from foundational definitions to applied, quantitative problems, with each sub-topic relying on mastery of the previous:

Foundational setup: Introduction to equilibrium defines dynamic equilibrium: the state where the rate of the forward reaction equals the rate of the reverse reaction, so macroscopic concentrations of all species remain constant, even as microscopic reactions continue. Direction of reversible reactions extends this to show that equilibrium can be reached starting from only reactants or only products, with the same final K value for a given temperature.
Core quantitative tools: Reaction quotient Q introduces the general ratio of product to reactant activities that applies at any point in a reaction, not just at equilibrium. Next, Calculating the equilibrium constant K defines K as the value of Q when a reaction reaches equilibrium at a given temperature, establishes the rules for writing K expressions, and how to manipulate K for reversed or scaled reactions. Magnitude of K teaches what K values tell us about the relative favorability of forward vs reverse reactions. Calculating equilibrium concentrations then applies K to solve for unknown equilibrium concentrations using the ICE table method.
Changing equilibrium conditions: Le Châtelier’s principle teaches how changes in concentration, pressure, volume, and temperature shift equilibrium, and which changes alter the value of K.
Applied equilibrium systems: This set of sub-topics applies core equilibrium concepts to the specific, high-yield case of ionic dissolution in water: Solubility equilibria introduces the solubility product constant $K_{s p}$ , Free energy of dissolution connects $K_{s p}$ to the thermodynamics of the dissolution process, Common ion effect applies Le Châtelier’s principle to solubility when a shared ion is already present in solution, and pH and solubility extends this to connect solution pH to the solubility of ionic compounds with basic anions.
Unifying conclusion: Free energy and equilibrium ties all prior concepts together by deriving the quantitative relationship between the standard free energy change of a reaction and its equilibrium constant, uniting thermodynamics and equilibrium into a single predictive framework.

3. A Guided Tour: How A Single Exam Problem Connects Multiple Core Sub-Topics

To see how the unit’s concepts build in a single exam-style problem, consider this question: A 1.0 L saturated solution of lead(II) chloride is prepared at 25°C. Equilibrium analysis shows $[P b^{2 +}] = 0.016 M$ . (a) Calculate the equilibrium constant for the dissolution of PbCl₂. (b) Will additional precipitation occur if 0.010 mol of solid NaCl is added to the solution? Justify your answer. (c) Calculate ΔG° for the dissolution reaction at 25°C and state if dissolution is thermodynamically favorable under standard conditions.

We work through this problem by applying sub-topics in sequence, just as the unit builds them:

First, we use the foundation from Calculating the equilibrium constant K: We write the dissolution reaction: $P b C l_{2} (s) ⇌ P b^{2 +} (a q) + 2 C l^{-} (a q)$ We omit the solid from the K (now called $K_{s p}$ for solubility) expression, so $K_{s p} = [P b^{2 +}] [C l^{-}]^{2}$ . From stoichiometry, $[C l^{-}] = 2 [P b^{2 +}] = 0.032 M$ . Calculating gives $K_{s p} = (0.016) (0.032)^{2} = 1.6 \times 1 0^{- 5}$ .
Next, we combine Reaction quotient Q and Common ion effect for part (b): After adding 0.010 mol NaCl, $[C l^{-}] = 0.032 + 0.010 = 0.042 M$ , and $[P b^{2 +}]$ remains 0.016 M immediately after addition before any shift. Calculate Q: $Q = (0.016) (0.042)^{2} = 2.8 \times 1 0^{- 5}$ . Compare Q to K: $Q > K$ , so the reaction shifts left to form more solid precipitate. This matches the prediction of the common ion effect, which tells us added common ion reduces solubility.
Finally, we use Free energy and equilibrium for part (c): The relationship is $Δ G^{\circ} = - R T ln K$ $R = 8.314 \times 1 0^{- 3} k J / (m o l \cdot K)$ , $T = 298 K$ , $ln (1.6 \times 1 0^{- 5}) = - 10.96$ . Plugging in: $Δ G^{\circ} = - (8.314 \times 1 0^{- 3}) (298) (- 10.96) = + 27 k J / m o l$ . Since ΔG° is positive, dissolution is not thermodynamically favorable under standard conditions, which matches K < 1.

This tour shows how every step relies on prior concepts, which is why the unit builds incrementally the way it does.

4. Cross-Cutting Common Pitfalls (and how to avoid them)

Wrong move: Including pure solids, pure liquids, or solvents in the expression for Q or K. Why: Students confuse total amount of a species with its effective activity in the reaction; solids and pure liquids have constant activity that is already incorporated into K, so they do not change Q. Correct move: Always cross out any pure solids, pure liquids, and solvents from the numerator and denominator of Q/K before starting any calculation.
Wrong move: Comparing Q and K in reverse to predict reaction direction (e.g., saying Q > K means shift right). Why: Students mix up the definition of Q as products over reactants, leading to reversed logic. Correct move: Write the mnemonic "Q>K, too many Products, shift Left; Q<K, too many Reactants, shift Right" on your exam scratch paper before solving any equilibrium problem.
Wrong move: Changing the value of K when only concentration or pressure is changed, rather than temperature. Why: Students confuse a shift in equilibrium position with a change in the equilibrium constant. Correct move: Remember that only changes to temperature alter the value of K; all other changes (concentration, pressure, volume) only shift equilibrium to re-establish the original K.
Wrong move: Including solids in the change (x) row of an ICE table. Why: Students feel like they need to include every species from the balanced reaction, but forget that changing the amount of excess solid does not change equilibrium concentrations. Correct move: Only include aqueous and gaseous species in the ICE table; exclude solids and pure liquids entirely.
Wrong move: Reversing the sign relationship between ΔG° and K (e.g., saying K > 1 corresponds to positive ΔG°). Why: Students mix up the negative sign in the formula $Δ G^{\circ} = - R T ln K$ , leading to incorrect prediction of spontaneity. Correct move: Memorize the direct relationship: If $K > 1$ , $Δ G^{\circ} < 0$ (favorable forward); if $K < 1$ , $Δ G^{\circ} > 0$ (unfavorable forward).

5. Quick Check: Do You Know When To Use Which Sub-Topic?

For each scenario below, identify which sub-topic from the unit you would use to answer it. Check your answer after the list.

You are given initial concentrations of all reactants and products, and need to predict if the reaction will proceed forward or reverse to reach equilibrium.
You need to find the equilibrium concentration of iron(III) in a saturated solution of iron(III) hydroxide in 0.10 M sodium hydroxide.
You need to predict how increasing the temperature of a reaction changes the value of K, given that the forward reaction is endothermic.
You need to find the concentration of ammonia at equilibrium given initial concentrations of N₂ and H₂ and the value of Kc.
You need to determine if a reaction with K = 2.3 × 10⁴ is thermodynamically favorable under standard conditions.

Answers:

Reaction quotient Q (compare to K)
Common ion effect
Le Châtelier’s principle
Calculating equilibrium concentrations (ICE table method)
Free energy and equilibrium

6. Quick Reference Unit Cheatsheet

Category	Formula / Rule	Notes
Reaction Quotient ( $Q$ )	$Q = \frac{[ C ] ^{c} [ D ] ^{d}}{[ A ] ^{a} [ B ] ^{b}}$ for $a A + b B ⇌ c C + d D$	Excludes pure solids, pure liquids, and solvents; applies to any reaction point, not just equilibrium
Equilibrium Constant ( $K$ )	$K = Q$ at equilibrium, at a fixed temperature	$K_{c}$ uses molar concentrations; $K_{p}$ uses gas partial pressures (atm); K is inverted for reversed reactions, raised to a power for scaled reactions
Direction of Reaction	$Q < K \to$ shift forward; $Q = K \to$ at equilibrium; $Q > K \to$ shift reverse	Applies to all equilibrium systems
Le Châtelier Temperature Rule	Increasing $T$ shifts equilibrium in the endothermic direction; decreasing $T$ shifts in the exothermic direction	Only temperature changes alter the value of $K$
Solubility Product ( $K_{s p}$ )	$K_{s p} = [A^{n +}]^{a} [B^{m -}]^{b}$ for $A_{a} B_{b} (s) ⇌ a A^{n +} + b B^{m -}$	Excludes the solid ionic compound from the expression
Free Energy - Equilibrium Relationship	$Δ G^{\circ} = - R T ln K$	$R = 8.314 J/(mol \cdotp K)$ , $T$ in Kelvin; $Δ G^{\circ} < 0$ when $K > 1$ , $Δ G^{\circ} > 0$ when $K < 1$

7. All Sub-Topics In This Unit

Below are links to the full study guide for each sub-topic in the Equilibrium unit: