AP Chemistry Unit 7 ReviewEquilibrium

AP Chemistry Unit 7 covers chemical equilibrium, the dynamic state where a reversible reaction's forward and reverse rates are equal, so concentrations stop changing even though both reactions keep running. The unit's single biggest idea is that the reaction quotient Q and the equilibrium constant K tell you exactly where a reaction stands and which direction it will shift to get to equilibrium. You'll also use Le Châtelier's principle to predict how systems respond to stress, and apply all of it to solubility with Ksp and the common-ion effect. Unit 7 is worth 7-9% of the AP exam.

What this unit covers

What equilibrium actually is (and isn't)

• Many processes are reversible. Water evaporates and condenses, gases absorb and desorb, salts dissolve and precipitate, acids transfer protons in both directions. When the forward and reverse rates become equal, the system is at equilibrium.
• Equilibrium is dynamic, not dead. Molecules are still reacting in both directions constantly. It just looks static from the outside because the rates cancel, so concentrations and observable properties (color, pressure, pH) hold steady.
• You can reach the same equilibrium state from either side. Start with all reactants or all products of N2 + 3H2 ⇌ 2NH3 at the same temperature and you land at the same K.
• Particulate diagrams show up here. Given pictures of particles before and at equilibrium, you should be able to count reactant and product particles and estimate whether K is large, small, or near 1.

Q and K, the unit's central tools

• For aA + bB ⇌ cC + dD, the law of mass action gives Kc = [C]^c[D]^d / ([A]^a[B]^b). For gas-phase reactions you can write the same expression with partial pressures as Kp. Pure solids and liquids never appear in the expression.
• Q has the identical mathematical form as K, but you can calculate it at any moment, not just at equilibrium. Q is a snapshot; K is the destination.
• Compare them to predict direction. If Q < K, there are too few products, so the reaction shifts forward. If Q > K, it shifts in reverse. If Q = K, the system is already at equilibrium and nothing net happens.
• The magnitude of K tells you the story at a glance. K >> 1 means the reaction goes essentially to completion (mostly products). K << 1 means it barely proceeds (mostly reactants).

Manipulating and calculating K

• K follows clean algebra rules. Reverse a reaction and K inverts (becomes 1/K). Multiply coefficients by c and K is raised to the power c. Add reactions together and the overall K is the product of the individual K values. The same rules apply to Q.
• You can calculate K directly by plugging measured equilibrium concentrations or partial pressures into the K expression.
• When you only know initial conditions, the ICE table (Initial, Change, Equilibrium) is your workhorse. Set up the change row using stoichiometry with x, plug the equilibrium row into the K expression, and solve.
• When K is very small, the "x is negligible" approximation lets you skip the quadratic formula. If K is tiny, barely any reactant converts, so initial minus x is approximately just the initial value.

Le Châtelier's principle, the qualitative side

• When a system at equilibrium gets stressed, it shifts in the direction that partially counteracts the stress. The deeper explanation is always Q versus K. A stress makes Q ≠ K, and the system shifts to bring them back into equality.
• Concentration and volume changes alter Q only, not K. Add a reactant and Q drops below K, so the reaction shifts toward products. Decrease the volume of a gas system and it shifts toward the side with fewer moles of gas.
• Temperature is the special one. It changes K itself. Treat heat like a reactant (endothermic) or product (exothermic). Raising temperature shifts an endothermic reaction forward and increases its K.
• A catalyst speeds up both directions equally. It gets you to equilibrium faster but never shifts the position. Same for adding an inert gas at constant volume, which changes nothing in the Q expression.
• Shifts have measurable consequences you can predict, like color changes, pH changes, or pressure changes.

Solubility equilibria and the common-ion effect

• Dissolving a sparingly soluble salt is an equilibrium, like AgCl(s) ⇌ Ag+(aq) + Cl-(aq), with its own constant Ksp = [Ag+][Cl-]. The solid never appears in the expression.
• Converting between Ksp and molar solubility depends on stoichiometry. For AgCl, Ksp = s². For PbCl2(s) ⇌ Pb2+ + 2Cl-, Ksp = (s)(2s)² = 4s³. Same idea, different exponents, so set up the dissolution equation first.
• Ksp values greater than 1 correspond to salts the solubility rules call soluble, so the qualitative rules from earlier in the course have a quantitative basis.
• The common-ion effect is Le Châtelier applied to dissolution. Dissolve AgCl in a solution that already contains Cl- and the equilibrium shifts toward the solid, so less AgCl dissolves. You can reason it qualitatively or calculate it with an ICE table where the common ion has a nonzero initial concentration.

Unit 7, Equilibrium at a glance

Why Unit 7, Equilibrium matters in AP Chem

Unit 7 is where AP Chem stops asking "does this reaction happen?" and starts asking "how far does it go?" That question, the extent of reaction, is one of the course's big ideas, and the Q versus K framework you build here becomes the default reasoning tool for the rest of the year.

• Equilibrium connects the macroscopic (constant color, constant pressure) to the particulate (molecules still reacting in both directions), which is the central habit of mind in AP Chem.
• Q versus K reasoning is the most transferable skill in the course. It explains acid strength, buffer behavior, solubility, and eventually the link between K and thermodynamic favorability.
• Le Châtelier's principle gives you a fast, defensible way to justify predictions in free-response answers, as long as you back it with the Q and K logic graders want to see.

How this unit connects across the course

• Kinetics (Unit 5) is the prerequisite idea. Equilibrium happens when the forward and reverse rates are equal, so the rate concepts you just learned literally define this unit. Catalysts speed the approach to equilibrium without moving it, a direct kinetics-equilibrium handoff.
• Thermochemistry (Unit 6) feeds the temperature rule. Whether a reaction is endothermic or exothermic determines how heating or cooling changes K, so you need ΔH in hand to apply Le Châtelier to temperature.
• Acids and Bases (Unit 8) is Unit 7 with new constants. Ka, Kb, and Kw are just equilibrium constants for proton transfer, and every weak-acid pH problem is an ICE table. If your Unit 7 skills are solid, Unit 8 is mostly applying them.
• Thermodynamics and Electrochemistry (Unit 9) closes the loop. There you'll connect K to ΔG° through ΔG° = -RT ln K, finally explaining why K has the value it does.
• The solubility rules from Chemical Reactions (Unit 4) get their quantitative explanation here. "Insoluble" salts are just salts with very small Ksp values.

Key equations and processes

• Kc = [C]^c[D]^d / ([A]^a[B]^b) for aA + bB ⇌ cC + dD. The law of mass action; pure solids and liquids are excluded.
• Kp, the same expression written with partial pressures, used for gas-phase equilibria.
• Q, the identical expression evaluated at any moment. Compare Q to K to predict the direction of shift.
• Q < K shifts forward, Q > K shifts reverse, Q = K means equilibrium. The decision rule behind every Le Châtelier answer.
• Manipulation rules. Reversed reaction gives 1/K, coefficients times c gives K^c, summed reactions give the product of the K values.
• ICE table process. Fill in Initial, Change (in terms of x with stoichiometric coefficients), and Equilibrium rows, then substitute the equilibrium row into the K expression and solve for x.
• Ksp expressions and molar solubility. For AB, Ksp = s²; for AB2, Ksp = 4s³. The relationship between Ksp and solubility depends on the dissolution stoichiometry.
• Common-ion calculation. Run the Ksp ICE table with a nonzero initial concentration for the shared ion; solubility comes out lower than in pure water.

Unit 7, Equilibrium on the AP exam

Unit 7 is 7-9% of the AP exam, and its reasoning style shows up well beyond its own questions because Unit 8 acid-base problems lean on the same skills.

• Multiple-choice questions test Q versus K direction predictions, the effect of a specific stress on an equilibrium system, interpreting particulate diagrams of mixtures before and at equilibrium, and quick K manipulations (reversing, scaling, adding reactions).
• Free-response questions love full ICE-table calculations. Expect to be given initial concentrations or pressures plus K (or equilibrium data and asked to find K), and to show every step of the setup.
• Justification is graded as much as the answer. "The equilibrium shifts left" earns little by itself. Explain it through Q and K, for example "adding product increases Q above K, so the reaction proceeds in reverse to decrease Q until Q = K."
• Solubility questions ask you to calculate molar solubility from Ksp (watch the stoichiometry), rank salts by solubility, and explain or compute the common-ion effect.
• Lab-flavored stimuli appear too, like predicting a color or pH change when a stress is applied to an observable equilibrium system.

Essential questions

• Why do concentrations stop changing at equilibrium even though reactions never stop?
• How can a single number, K, summarize how far a reaction goes, and how does Q tell you which way it still needs to move?
• Why does temperature change K itself while concentration and volume changes only change Q?
• How does treating dissolution as an equilibrium explain why some salts barely dissolve and why a common ion makes them dissolve even less?

Key terms to know

• Dynamic equilibrium: The state where forward and reverse reactions occur at equal rates, so concentrations stay constant with no net change.
• Reversible reaction: A reaction that can proceed in both the forward and reverse directions, shown with the ⇌ arrow.
• Equilibrium constant (K): The value of the mass-action expression at equilibrium; its magnitude indicates whether products or reactants are favored.
• Reaction quotient (Q): The same mass-action expression evaluated at any point in a reaction, used to predict which direction the system will shift.
• Law of mass action: The rule that the equilibrium expression equals products over reactants, each raised to its stoichiometric coefficient.
• Kp: The equilibrium constant written in terms of partial pressures for gas-phase reactions.
• ICE table: An Initial, Change, Equilibrium organizer used to solve for unknown equilibrium concentrations or for K.
• Le Châtelier's principle: The prediction that a system at equilibrium shifts to partially counteract an applied stress.
• Ksp (solubility product constant): The equilibrium constant for the dissolution of a sparingly soluble salt.
• Molar solubility: The number of moles of a salt that dissolve per liter of solution, related to Ksp through the dissolution stoichiometry.
• Common-ion effect: The decrease in a salt's solubility when the solution already contains one of the salt's ions.
• Particulate representation: A diagram of individual particles used to show relative amounts of reactants and products before and at equilibrium.

Common mix-ups

• Q changes versus K changes. Adding or removing species, diluting, or changing volume changes Q only. Temperature is the only stress in this unit that changes K itself.
• A bigger Ksp does not always mean more soluble. You can only compare Ksp values directly for salts with the same dissolution stoichiometry. Otherwise convert to molar solubility first.
• Catalysts and inert gases do not shift equilibrium. A catalyst speeds both directions equally, and an inert gas added at constant volume changes no concentration in the Q expression.
• Constant concentrations do not mean the reaction stopped. Both reactions are still running at equal rates. Saying the reaction "stops" at equilibrium loses points on free response.
• Don't put solids or pure liquids in K expressions. AgCl(s) and H2O(l) never appear; only gases and aqueous species do.