AP exam review verified for 2027

AP Chem Unit 7 Review: Equilibrium

Review AP Chem Unit 7 to build fluency with equilibrium constants, ICE tables, Le Chatelier's principle, and solubility equilibria. These concepts connect reversible reactions, Q vs. K reasoning, and Ksp calculations into one unified framework that appears throughout the rest of the course.

Use the topic guides, key terms, and practice questions available here to work through every concept from dynamic equilibrium to the common-ion effect.

What is AP Chem unit 7?

Chemical equilibrium describes a reversible system where the forward and reverse reactions proceed at equal rates, so concentrations stay constant even though both reactions keep happening. Unit 7 builds from that conceptual foundation through quantitative tools like K expressions, ICE tables, and Ksp calculations, then ties everything together with Le Chatelier's principle.

Equilibrium is a dynamic state where forward and reverse reaction rates are equal. The equilibrium constant K describes the ratio of product to reactant concentrations at equilibrium. Comparing Q to K tells you which direction a reaction will shift. Le Chatelier's principle predicts how a system responds to concentration, pressure, or temperature changes. Ksp extends these ideas to the dissolution of slightly soluble salts.

K and Q are the core quantitative tools

The equilibrium constant expression Kc = [C]^c[D]^d / [A]^a[B]^b (or Kp using partial pressures) describes the system at equilibrium. Q has the same form but applies at any moment. When Q is less than K the reaction shifts toward products; when Q is greater than K it shifts toward reactants. Pure solids and liquids are always omitted from both expressions.

ICE tables connect initial conditions to equilibrium

An ICE table organizes Initial concentrations, the Change in terms of a variable x, and the Equilibrium concentrations. You substitute the equilibrium row into the K expression and solve for x. When K is very small, the small-x approximation often avoids a quadratic. Always check whether the approximation is valid.

Le Chatelier's principle explains every shift

Adding a reactant, removing a product, decreasing volume in a gas-phase reaction, or changing temperature all disturb equilibrium. The system shifts to partially counteract the stress. Temperature changes are unique because they change the value of K itself, not just Q. Concentration and pressure changes shift Q without changing K.

Equilibrium is a balance of competing processes

Every equilibrium concept in Unit 7 follows from one idea: a reversible system reaches a state where opposing processes occur at equal rates. K quantifies that state, Q measures where the system is at any moment, Le Chatelier's principle describes how the system returns to balance after a disturbance, and Ksp applies the same logic to dissolving salts. Understanding this unifying idea lets you reason through any equilibrium problem rather than memorizing separate rules.

AP Chem unit 7 topics

7.1

Introduction to Equilibrium

Equilibrium is a dynamic state where forward and reverse reaction rates are equal. Concentration, partial pressure, and rate-vs-time graphs all level off at equilibrium. Examples include evaporation/condensation, dissolution/precipitation, and acid-base proton transfer.

open guide
7.2

Direction of Reversible Reactions

The direction a reversible reaction proceeds depends on whether the forward or reverse rate is larger. Net conversion of reactants to products occurs when the forward rate exceeds the reverse rate, and vice versa. Equilibrium is reached when the two rates are equal.

open guide
7.3

Reaction Quotient and Equilibrium Constant

Q and K share the same mathematical form: products over reactants, each raised to its stoichiometric coefficient. Pure solids and liquids are excluded. Comparing Q to K predicts the direction of shift: Q < K means shift toward products, Q > K means shift toward reactants.

open guide
7.4

Calculating the Equilibrium Constant

Substitute measured equilibrium concentrations or partial pressures into the K expression to calculate Kc or Kp. Stoichiometric coefficients become exponents. K is reported as a unitless number.

open guide
7.5

Magnitude of the Equilibrium Constant

K >> 1 means the equilibrium mixture is mostly products. K << 1 means mostly reactants. K near 1 means significant amounts of both. K magnitude describes composition, not reaction speed.

open guide
7.6

Properties of the Equilibrium Constant

Reversing a reaction inverts K. Multiplying coefficients by c raises K to the power c. Adding reactions multiplies their K values. These same rules apply to Q.

open guide
7.7

Calculating Equilibrium Concentrations

Use an ICE table with initial concentrations, changes expressed as multiples of x, and equilibrium values substituted into the K expression. Compare Q to K first to set the direction of x. Use the small-x approximation when K is very small and verify with the 5% rule.

open guide
7.8

Representations of Equilibrium

Particulate diagrams show the relative numbers of reactant and product particles at equilibrium. More product particles indicate a larger K. Atom conservation must be maintained in any particle diagram.

open guide
7.9

Introduction to Le Chatelier's Principle

A system at equilibrium shifts to partially counteract a stress. Stresses include adding or removing a species, changing volume or pressure in a gas-phase system, changing temperature, or diluting the system. Observable properties like color and pH can signal the direction of shift.

open guide
7.10

Reaction Quotient and Le Chatelier's Principle

Any disturbance changes Q relative to K. Concentration and pressure changes affect Q only; temperature changes affect K itself. In both cases, the system shifts until Q equals K again.

open guide
7.11

Introduction to Solubility Equilibria

Dissolution of a slightly soluble salt is modeled as an equilibrium with constant Ksp. Molar solubility is calculated from Ksp using an ICE table, but the math depends on the stoichiometry of the dissolution equation. Comparing Qsp to Ksp predicts precipitation.

open guide
7.12

Common-Ion Effect

A salt dissolves less in a solution already containing one of its ions. The common ion shifts the dissolution equilibrium back toward the solid, reducing molar solubility. Ksp itself does not change; only the equilibrium position shifts.

open guide
7.14

7.14 Free Energy of Dissolution

Review AP Chemistry free energy of dissolution, including entropy, enthalpy, Gibbs free energy, Delta G = Delta H - T Delta S, salt dissolution, solvent reorganization, ion-solvent attractions, thermodynamic favorability, and Ksp.

open guide
practice snapshot

Hardest AP Chemistry unit 7 topics

This snapshot uses Fiveable practice activity to show where students tend to miss questions and which review moves are worth prioritizing first.

58%average MCQ accuracy

Across 8.9k multiple-choice practice attempts for this unit.

8.9kMCQ attempts

Practice activity included in this snapshot.

60%average FRQ score

Across 25 scored free-response attempts for this unit.

Hardest topics in unit 7

MCQ miss rate
7.12

Review Common-Ion Effect with attention to how the concept appears in AP-style source and evidence questions.

49%655 tries
7.3

Review Reaction Quotient and Equilibrium Constant with attention to how the concept appears in AP-style source and evidence questions.

48%1,001 tries
7.5

Review Magnitude of the Equilibrium Constant with attention to how the concept appears in AP-style source and evidence questions.

41%604 tries
7.2

Review Direction of Reversible Reactions with attention to how the concept appears in AP-style source and evidence questions.

35%1,124 tries

Unit 7 review notes

7.1

Dynamic Equilibrium and Reaction Direction

Equilibrium is reached when the forward and reverse reaction rates become equal. At that point, concentrations and partial pressures stop changing, but both reactions continue. On a concentration-vs-time or rate-vs-time graph, equilibrium appears where the curves level off. The direction a reversible reaction proceeds depends entirely on which rate is currently larger.

  • Dynamic equilibrium: The forward and reverse reactions occur simultaneously at equal rates, so no net change in concentrations is observed.
  • Reversible reaction: A reaction that can proceed in both the forward and reverse directions, represented by a double arrow.
  • Forward reaction rate > reverse rate: Net conversion of reactants to products occurs until the rates equalize.
  • Concentration-vs-time graph: Reactant concentrations decrease and product concentrations increase until both level off at equilibrium.
Sketch a concentration-vs-time graph for a reversible reaction and label the point where equilibrium is established. Explain why the concentrations are constant but the reaction has not stopped.
7.3

Writing and Calculating K and Q

The reaction quotient Q and equilibrium constant K share the same mathematical form: products over reactants, each raised to its stoichiometric coefficient. For aA + bB = cC + dD, Kc = [C]^c[D]^d / [A]^a[B]^b and Kp uses partial pressures. Pure solids and pure liquids are excluded from both expressions. To calculate K experimentally, substitute measured equilibrium concentrations or partial pressures directly into the expression.

  • Kc: Equilibrium constant expressed in terms of molar concentrations of aqueous or gaseous species.
  • Kp: Equilibrium constant expressed in terms of partial pressures of gaseous species.
  • Omitting pure solids and liquids: Their concentrations do not change with amount, so they are excluded from K and Q expressions.
  • Stoichiometric coefficient as exponent: Each concentration or partial pressure in the K expression is raised to the power of its coefficient in the balanced equation.
Write the Kc expression for 2SO2(g) + O2(g) = 2SO3(g). Then write the Kp expression. Explain why you would omit a solid like CaCO3 if it appeared in the reaction.
7.5

Magnitude and Algebraic Properties of K

The size of K tells you the composition of the equilibrium mixture. K much greater than 1 means the mixture is mostly products; K much less than 1 means mostly reactants. K values also follow predictable algebraic rules when reactions are manipulated: reversing a reaction inverts K, multiplying coefficients by a factor c raises K to the power c, and adding reactions together multiplies their K values. These rules apply equally to Q.

  • K >> 1: Equilibrium lies far to the right; the mixture is predominantly products.
  • K << 1: Equilibrium lies far to the left; the mixture is predominantly reactants.
  • Reversing a reaction: K for the reverse reaction equals 1 divided by K for the forward reaction.
  • Multiplying coefficients by c: The new K equals the original K raised to the power c.
  • Adding reactions: The overall K equals the product of the individual K values, analogous to Hess's law for enthalpy.
Given K1 for A = B and K2 for B = C, write the K expression for 2A = 2C. Show each algebraic step.
ManipulationEffect on K
Reverse the reactionK becomes 1/K
Multiply coefficients by factor cK becomes K^c
Add two reactionsK becomes K1 x K2
No change to reactionK stays the same
7.7

Calculating Equilibrium Concentrations and Particulate Models

ICE tables are the standard method for finding equilibrium concentrations when you know initial conditions and K. Set up Initial, Change (as +x or -x based on stoichiometry), and Equilibrium rows, then substitute into the K expression. When K is very small, the small-x approximation simplifies the algebra. Particulate diagrams represent equilibrium visually: the ratio of product particles to reactant particles reflects the magnitude of K.

  • ICE table: Organizes Initial concentrations, Change in terms of x, and Equilibrium concentrations to set up the K expression.
  • Small-x approximation: When K is very small, x is negligible compared to initial concentrations, avoiding a quadratic equation. Verify with the 5% rule.
  • Q vs. K before setting up ICE: Compare Q to K first to determine the direction of the shift and the sign of x.
  • Particulate diagram: A molecular-level drawing showing relative numbers of reactant and product particles; more product particles indicate a larger K.
For N2(g) + 3H2(g) = 2NH3(g) with given initial concentrations and Kc, set up the ICE table and identify whether you need the quadratic formula or can use the small-x approximation.
7.9

Le Chatelier's Principle and Q vs. K

Le Chatelier's principle states that a system at equilibrium shifts to partially counteract any applied stress. Adding a reactant or removing a product shifts the reaction toward products; the reverse is also true. Decreasing volume in a gas-phase reaction increases pressure and shifts toward the side with fewer moles of gas. Temperature changes are fundamentally different: they change the value of K itself. For an endothermic reaction, increasing temperature increases K; for an exothermic reaction, increasing temperature decreases K. Every shift can be explained quantitatively by noting that the stress changes Q relative to K, and the system moves to restore Q = K.

  • Concentration stress: Adding or removing a species changes Q but not K; the reaction shifts to restore Q = K.
  • Pressure/volume stress: Decreasing volume increases all partial pressures and shifts toward the side with fewer moles of gas.
  • Temperature stress: Changes K itself; treat heat as a reactant (endothermic) or product (exothermic) to predict the direction of shift.
  • Inert gas addition at constant volume: Does not change partial pressures of reactants or products, so no shift occurs.
  • Q vs. K after a stress: If Q < K after the stress, the reaction shifts toward products; if Q > K, it shifts toward reactants.
For the exothermic reaction 2NO2(g) = N2O4(g), predict the direction of shift when (a) NO2 is added, (b) volume is decreased, and (c) temperature is increased. Justify each answer using Q vs. K reasoning.
StressEffect on QEffect on KDirection of shift
Add reactantQ decreasesNo changeToward products
Remove productQ decreasesNo changeToward products
Decrease volume (gas)Q changes based on moles of gasNo changeToward fewer moles of gas
Increase temperature (endothermic)No immediate changeK increasesToward products
Increase temperature (exothermic)No immediate changeK decreasesToward reactants
7.11

Solubility Equilibria and Ksp

The dissolution of a slightly soluble salt is a reversible process with its own equilibrium constant, Ksp, called the solubility product. For a salt MxAy dissolving into M^y+ and A^x- ions, the Ksp expression includes only the ion concentrations raised to their stoichiometric coefficients; the solid is omitted. Molar solubility s can be calculated from Ksp using an ICE table, but the relationship between s and Ksp depends on the stoichiometry of the dissolution. Comparing the ion product Qsp to Ksp predicts whether a precipitate will form.

  • Ksp: The solubility product constant; the equilibrium constant for a slightly soluble salt dissolving into its ions in a saturated solution.
  • Molar solubility: The number of moles of a salt that dissolve per liter of solution before the solution becomes saturated.
  • ICE table for Ksp: Set initial ion concentrations to zero (or a common-ion value), express change as multiples of s, and substitute into the Ksp expression.
  • Qsp vs. Ksp: If Qsp > Ksp, the solution is supersaturated and a precipitate forms; if Qsp < Ksp, more solid can dissolve.
Write the Ksp expression for Ca3(PO4)2. Set up an ICE table and express Ksp in terms of molar solubility s. Identify how the stoichiometry affects the relationship between s and Ksp.
7.12

Common-Ion Effect

When a salt is dissolved in a solution that already contains one of its ions, the solubility of the salt decreases. This is the common-ion effect. It follows directly from Le Chatelier's principle: the extra ion shifts the dissolution equilibrium back toward the solid. Quantitatively, place the initial concentration of the common ion in the ICE table and solve for the new molar solubility. The Ksp value itself does not change; only the solubility changes.

  • Common-ion effect: The reduction in solubility of a salt when dissolved in a solution already containing one of the salt's ions.
  • Ksp is constant at constant temperature: Adding a common ion does not change Ksp; it changes the equilibrium position and therefore the molar solubility.
  • ICE table with common ion: Place the initial concentration of the common ion in the Initial row; the change x will be much smaller than that initial value.
  • Le Chatelier explanation: The added ion increases Qsp above Ksp, so the system shifts toward the solid, reducing solubility.
Calculate the molar solubility of AgCl in 0.10 M NaCl solution given Ksp = 1.8 x 10^-10. Compare your answer to the molar solubility in pure water and explain the difference.

Practice AP Chem unit 7 questions

Try stimulus-based AP practice questions and written prompts after you review the notes.

Example stimulus-based MCQs

open all practice
graph

Stimulus-based practice question

A(g)+B(g)2C(g)A(g)+B(g)\rightleftharpoons 2C(g)

The graph shows the concentrations of the gases versus time at constant temperature. The system is initially at equilibrium. At t=10 mint=10\ \text{min}, additional A(g)A(g) is added to the rigid container.

Question

Based on the graph, what is the new equilibrium concentration of CC ?

4.8 M

4.4 M

5.2 M

5.6 M

diagram

Stimulus-based practice question

The diagram represents the equilibrium mixture for the reaction 2SO2(g)+O2(g)2SO3(g)2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g) in a sealed flask. Each particle represents 0.01 mol0.01 \text{ mol} of gas. A student wants to calculate the equilibrium constant KpK_p for this reaction.

Question

Which of the following quantities is required to calculate KpK_p from the provided particulate model?

The total pressure of the equilibrium mixture

The equilibrium concentrations of all gaseous species

The initial number of reactant particles and the volume of the flask

The number of moles of each gas at equilibrium and the temperature

Example FRQs

open all FRQs
SAQ

Silver chromate solubility and ionic compound dissolution

7. A student investigates the properties of silver chromate, Ag₂CrO₄. A Lewis diagram for the chromate ion, CrO₄²⁻, is shown in Figure 1.

Figure 1. Lewis diagram for the chromate ion, CrO₄²⁻, showing which oxygen atoms carry formal charge −1.

Figure 1
A.

On the Lewis diagram in Figure 1, circle one oxygen atom that carries a formal charge of -1.

The dissolution of Ag₂CrO₄(s) in pure water is represented by the equation below. At 25°C, the concentration of Ag⁺ in a saturated solution is determined to be 1.3 × 10⁻⁴ M.

B.
i.

Calculate the equilibrium concentration of CrO₄²⁻ in the saturated solution at 25°C.

ii.

Using your answer to part B(i), calculate the value of the solubility product constant, Ksp, for Ag₂CrO₄ at 25°C.

The student compares the solubility of silver chromate with that of silver sulfate, Ag₂SO₄. The solubility product constant for Ag₂SO₄ is provided in Table 1.

Table 1. Solubility product constant for silver sulfate

Substance

Ksp at 25°C

Ag₂SO₄

1.2 × 10⁻⁵

C.

A student claims that Ag₂SO₄ is less soluble than Ag₂CrO₄ at 25°C. Do you agree or disagree? Justify your answer using the data in Table 1 and your result from part B(ii).

SAQ

Magnesium hydroxide solubility equilibrium and common ion effect

6. A student investigates the solubility of magnesium hydroxide, Mg(OH)2\text{Mg(OH)}_2, at 298 K. The student prepares a saturated solution by adding excess solid Mg(OH)2\text{Mg(OH)}_2 to distilled water, as represented in Figure 1. Relevant solubility product constants (KspK_{sp}) are listed in Table 1.

Table 1. Solubility Product Constants at 298 K

Substance

Ksp at 298 K

Mg(OH)2\text{Mg(OH)}_2

5.6×10125.6 × 10^{-12}

Ca(OH)2\text{Ca(OH)}_2

5.0×1065.0 × 10^{-6}

Mn(OH)2\text{Mn(OH)}_2

1.9×10131.9 × 10^{-13}

Zn(OH)2\text{Zn(OH)}_2

3.0×10173.0 × 10^{-17}

Figure 1. Particle diagram of a saturated Mg(OH)2 solution at equilibrium with excess Mg(OH)2(s) (water molecules not shown).

Figure 1
A.

Write the expression for the solubility product constant, KspK_{sp}, for the dissolution of Mg(OH)2\text{Mg(OH)}_2.

B.

Calculate the molar solubility of Mg(OH)2\text{Mg(OH)}_2 in pure water at 298 K.

C.

The student prepares a solution that is 0.10 M0.10\ M in Mg(NO3)2\text{Mg(NO}_3)_2 and adds excess solid Mg(OH)2\text{Mg(OH)}_2. Is the molar solubility of Mg(OH)2\text{Mg(OH)}_2 in this solution greater than, less than, or equal to the molar solubility in pure water? Justify your answer with a calculation.

D.

The student wants to prepare a saturated solution with the highest possible concentration of hydroxide ions, [OH][\text{OH}^-]. The student has access to the salts listed in Table 1. Calculate the maximum [OH][\text{OH}^-], in MM, that can be produced at 298 K using one of these salts.

FRQ

Silver chromate solubility and common ion effect

2. Answer the following questions about silver chromate, Ag₂CrO₄.

Silver chromate is a sparingly soluble red solid that dissolves in water according to the equation shown in part A. A student investigates the solubility of Ag₂CrO₄ at 25°C.

A.

The dissolution of solid silver chromate in water is represented by a balanced chemical equation.

i.

Write the balanced equation for the dissolution of Ag₂CrO₄(s) in water. Include states of matter.

ii.

Write the expression for the solubility product constant, KspK_{sp}, for Ag₂CrO₄.

Figure 1. Concentration of silver ions, Ag⁺, in a solution containing Ag₂CrO₄(s) versus time at 25°C (equilibrium plateau at 1.3 × 10⁻⁴ M).

Figure 1
B.

A student adds excess solid Ag₂CrO₄ to pure water at 25°C and stirs. The concentration of Ag⁺(aq) is monitored over time, as shown in Figure 1.

i.

Determine the equilibrium concentration of Ag⁺(aq) from the graph in Figure 1.

ii.

Calculate the value of the solubility product constant, KspK_{sp}, for Ag₂CrO₄ at 25°C.

C.

The student adds a small amount of solid AgNO₃ to the saturated solution prepared in part B. Does the concentration of dissolved chromate ion, CrO₄²⁻(aq), increase, decrease, or remain the same? Justify your answer.

Table 1. Experimental conditions for mixing trials

Trial

Volume of AgNO₃(aq)

Concentration of AgNO₃(aq)

Volume of K₂CrO₄(aq)

Concentration of K₂CrO₄(aq)

1

50.0 mL

4.0 × 10⁻⁴ M

50.0 mL

4.0 × 10⁻⁴ M

D.

In a separate experiment, the student mixes solutions of silver nitrate and potassium chromate as described in Table 1. Assume that volumes are additive.

i.

Calculate the value of the reaction quotient, QQ, for the mixture in Trial 1 immediately after mixing but before any reaction occurs.

ii.

Based on your calculation in part D(i), predict whether a precipitate of Ag₂CrO₄ will form. Justify your answer.

E.

The student adds a strong acid to a saturated solution of Ag₂CrO₄ in contact with solid Ag₂CrO₄. Explain how the addition of acid affects the solubility of Ag₂CrO₄. The chromate ion reacts with protons in acidic solution according to the following equilibrium equation:
2CrO42(aq)+2H+(aq)Cr2O72(aq)+H2O(l)2CrO_4^{2-}(aq) + 2H^+(aq) \rightleftharpoons Cr_2O_7^{2-}(aq) + H_2O(l)

Key terms

TermDefinition
Chemical EquilibriumA dynamic state in which the forward and reverse reactions occur at equal rates, so the concentrations of all species remain constant over time.
Dynamic EquilibriumThe condition where forward and reverse reaction rates are equal and both reactions continue, resulting in no net change in concentrations.
Reversible ReactionA reaction that can proceed in both the forward and reverse directions, represented by a double arrow and capable of reaching equilibrium.
Equilibrium Constant ExpressionThe ratio of product concentrations to reactant concentrations at equilibrium, each raised to its stoichiometric coefficient; pure solids and liquids are excluded.
K ValueThe numerical value of the equilibrium constant; K >> 1 indicates a product-favored equilibrium, K << 1 indicates a reactant-favored equilibrium.
Partial PressureThe pressure exerted by a single gas in a mixture; used in Kp expressions for gas-phase equilibria.
Particulate DiagramA molecular-level drawing showing the relative numbers of reactant and product particles at equilibrium; the particle ratio reflects the magnitude of K.
Molar SolubilityThe number of moles of a slightly soluble salt that dissolve per liter of solution to form a saturated solution; calculated from Ksp using an ICE table.
Solubility product constant (Ksp)The equilibrium constant for a slightly soluble salt dissolving into its constituent ions; equals the product of ion concentrations each raised to their stoichiometric coefficients.
PrecipitationThe formation of an insoluble solid when the ion product Qsp exceeds Ksp, driving the dissolution equilibrium back toward the solid.
Stoichiometric CoefficientsThe numbers in a balanced equation that become exponents in the K expression and determine how molar solubility relates to ion concentrations in Ksp problems.
DissociationThe separation of an ionic compound into its constituent ions when dissolved in water; the basis for writing Ksp expressions.

Common unit 7 mistakes

Including pure solids or liquids in K expressions

Pure solids and pure liquids have concentrations that do not depend on amount, so they are always excluded from Kc, Kp, and Ksp expressions. A common error is including the dissolving solid in a Ksp expression or including water in an aqueous equilibrium.

Confusing a change in Q with a change in K

Only a temperature change alters the value of K. Adding or removing a species, changing pressure, or diluting the solution changes Q but leaves K the same. Saying that adding a reactant increases K is incorrect.

Setting up the ICE table in the wrong direction

Always compare Q to K before writing the change row. If Q > K, x represents a decrease in products and an increase in reactants. Skipping this step leads to a negative equilibrium concentration, which signals an error.

Applying the small-x approximation without checking it

The approximation that x is negligible compared to the initial concentration is only valid when K is very small relative to the initial concentration. Always verify using the 5% rule: x divided by the initial concentration must be less than 0.05.

Ignoring stoichiometry when relating molar solubility to Ksp

For a salt like Ag2CrO4 that dissolves as Ag2CrO4(s) = 2Ag+(aq) + CrO4^2-(aq), [Ag+] = 2s, not s. Forgetting the coefficient leads to a Ksp expression of s^3 instead of the correct 4s^3.

How this unit shows up on the AP exam

Justify equilibrium shifts with explicit Q vs. K reasoning

AP Chemistry free-response questions frequently ask you to explain why a system shifts in a particular direction after a stress. A complete answer names the stress, states how it changes Q relative to K, and concludes with the direction of shift. Saying only that Le Chatelier's principle predicts a shift is not sufficient without the Q vs. K justification.

Set up and solve multi-step equilibrium calculations

Calculation tasks in Unit 7 often require writing a correct K expression, setting up an ICE table, and solving for an unknown concentration or Ksp. Errors in the K expression, such as including a solid or using the wrong exponent, propagate through the entire calculation. Show each step clearly and include units where appropriate.

Interpret particulate diagrams and connect representations

The AP Chemistry exam uses multiple representations of equilibrium, including graphs, particulate diagrams, and symbolic equations. You may be asked to identify the equilibrium state from a concentration-vs-time graph, draw a particulate model consistent with a given K value, or connect a particle-level description to a macroscopic observation such as a color change or pH shift.

Final unit 7 review checklist

  • Write correct K and Q expressionsGiven any balanced equation, write Kc or Kp correctly: products over reactants, coefficients as exponents, pure solids and liquids excluded. Practice with heterogeneous equilibria like CaCO3(s) = CaO(s) + CO2(g).
  • Use Q vs. K to predict reaction directionGiven initial concentrations or partial pressures, calculate Q, compare it to K, and state whether the reaction shifts toward products, toward reactants, or is already at equilibrium.
  • Set up and solve ICE tablesOrganize initial, change, and equilibrium rows correctly. Choose the sign of x based on Q vs. K. Decide whether to use the small-x approximation or the quadratic formula, and verify the approximation if used.
  • Apply K algebraic rulesReverse a reaction, scale coefficients, or combine reactions and adjust K correctly in each case. Show each step when combining multiple equilibria into an overall K.
  • Predict Le Chatelier shifts for all stress typesDistinguish between concentration stresses (change Q, not K), pressure/volume stresses (affect gas-phase reactions based on moles of gas), and temperature stresses (change K itself). Justify each prediction with Q vs. K reasoning.
  • Calculate molar solubility from KspWrite the dissolution equation, set up an ICE table with s as the variable, and solve. Account for stoichiometry: for CaF2, [F-] = 2s, so Ksp = (s)(2s)^2 = 4s^3.
  • Apply the common-ion effect quantitativelyPlace the initial common-ion concentration in the ICE table, solve for the new molar solubility, and explain the reduction using Le Chatelier's principle. Confirm that Ksp is unchanged.

How to study unit 7

Start with dynamic equilibrium and Q vs. KRead the topic guides for 7.1 and 7.2 to build the conceptual foundation, then work through 7.3 and 7.4 to practice writing and calculating K and Q expressions. Focus on getting the form of the expression right before moving to calculations.
Build fluency with K magnitude and algebraic rulesReview 7.5 and 7.6 together. Practice interpreting K values qualitatively and manipulating K algebraically by reversing reactions, scaling coefficients, and combining equilibria. Use the comparison table in the review notes as a quick reference.
Practice ICE tables and particulate modelsWork through 7.7 and 7.8 by setting up ICE tables for a variety of reactions. Practice both the small-x approximation and the quadratic approach. Sketch particulate diagrams and check that your particle ratios match the K value.
Apply Le Chatelier's principle to all stress typesReview 7.9 and 7.10 together. For each stress type, concentration, pressure/volume, and temperature, practice predicting the shift and justifying it with Q vs. K reasoning. Use the comparison table in the review notes to check your logic.
Finish with solubility equilibria and the common-ion effectWork through 7.11 and 7.12 by writing Ksp expressions, calculating molar solubility with ICE tables, and solving common-ion problems. Pay close attention to stoichiometry in the dissolution equation. Use the AP score calculator to estimate how your practice performance maps to an AP score.

More ways to review

Topic study guides

Open the individual guides for Unit 7 when you want a closer review of one topic.

browse guides

FRQ practice

Practice free-response reasoning and compare your answer with scoring guidance.

practice FRQs

Cram archive videos

Watch past review streams filtered to Unit 7 when you want a video walkthrough.

open videos

Cheatsheets

Use unit cheatsheets for a quick visual review after you work through the notes.

open cheatsheets

Score calculator

Estimate your broader AP score goal after you review the course and exam format.

open calculator

Frequently Asked Questions

What topics are covered in AP Chem Unit 7?

AP Chem Unit 7 covers 12 topics on equilibrium: Introduction to Equilibrium, Direction of Reversible Reactions, Reaction Quotient and Equilibrium Constant, Calculating the Equilibrium Constant, Magnitude of the Equilibrium Constant, Properties of the Equilibrium Constant, Calculating Equilibrium Concentrations, Representations of Equilibrium, Introduction to Le Châtelier's Principle, Reaction Quotient and Le Châtelier's Principle, Introduction to Solubility Equilibria, and Common-Ion Effect. The unit builds from the concept of dynamic equilibrium up through predicting how systems shift under stress. You'll work with K expressions, ICE tables, Q vs. K comparisons, and Ksp calculations. See AP Chem Unit 7 for topic-by-topic breakdowns.

How much of the AP Chem exam is Unit 7?

Unit 7 makes up 7-9% of the AP Chem exam. That weight covers everything from writing equilibrium constant expressions and calculating K to applying Le Châtelier's principle and solving solubility equilibria problems. It's a focused unit, but the concepts show up in calculation-heavy multiple-choice and free-response questions.

What's on the AP Chem Unit 7 progress check (MCQ and FRQ)?

The AP Chem Unit 7 progress check includes MCQ and FRQ parts drawn from all 12 equilibrium topics. MCQ questions test your ability to interpret Q vs. K, predict reaction direction, and identify how Le Châtelier's principle applies to concentration, temperature, and volume changes. FRQ questions typically ask you to set up ICE tables, calculate equilibrium concentrations, write Ksp expressions, or explain shifts using Le Châtelier's principle. The progress check pulls heavily from topics like Reaction Quotient and Equilibrium Constant (7.3), Calculating Equilibrium Concentrations (7.7), Introduction to Le Châtelier's Principle (7.9), and Introduction to Solubility Equilibria (7.11). Practicing those topics first gives you the most progress check coverage. Head to AP Chem Unit 7 for matched practice on each topic.

How do I practice AP Chem Unit 7 FRQs?

AP Chem Unit 7 FRQs most often come from three areas: calculating equilibrium concentrations using ICE tables, applying Le Châtelier's principle to explain system shifts, and solving solubility equilibria problems with Ksp. Questions usually ask you to show your setup, not just a final answer, so writing out every step of an ICE table or Ksp calculation matters. To practice effectively, work through past College Board FRQs that involve equilibrium constant expressions and reaction quotient comparisons. Focus on explaining your reasoning in full sentences when the prompt says 'justify' or 'explain.' You can find topic-aligned FRQ practice at AP Chem Unit 7.

Where can I find AP Chem Unit 7 practice questions?

The best place to find AP Chem Unit 7 practice questions, including multiple-choice and practice test sets, is AP Chem Unit 7. That page organizes practice by topic so you can target equilibrium constant calculations, reaction quotient problems, Le Châtelier's principle scenarios, and solubility equilibria separately. For MCQ practice, look for questions that give you a reaction and ask whether Q is greater than, less than, or equal to K. For a practice test experience, work through a full set of Unit 7 questions timed, then review any ICE table or Ksp problems you missed.

How should I study AP Chem Unit 7?

Start AP Chem Unit 7 by building a solid understanding of what equilibrium means physically: a dynamic state where forward and reverse reaction rates are equal. Once that clicks, the math follows more naturally. Here's a practical study order: 1. **Learn K expressions first.** Practice writing equilibrium constant expressions for gases and solutions before touching calculations. 2. **Master the Q vs. K comparison.** Knowing whether the reaction quotient Q is less than, greater than, or equal to K tells you which direction a reaction shifts. This shows up constantly. 3. **Work ICE tables by hand.** Set up and solve at least 10 ICE table problems for Calculating Equilibrium Concentrations (7.7). Speed and accuracy here pay off on FRQs. 4. **Study Le Châtelier's principle with real examples.** Practice predicting shifts for concentration changes, temperature changes, and volume changes separately. 5. **Finish with solubility equilibria.** Ksp problems and the Common-Ion Effect (7.12) build on everything above, so save them for last. Visit AP Chem Unit 7 for topic-by-topic resources that follow this order.

Ready to review Unit 7?Start with the notes, check the topic cards, and use the practice or resource links when they are available for this course.