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Unit 7 · Equilibrium Unit Hub Flashcards Cheat Sheet Essentials Visual Review MC Practice SAQ Practice

AP Chemistry Unit 7 Visual Review

A topic-by-topic visual walkthrough of Equilibrium — K and Q, calculating equilibrium, Le Chatelier's principle, and solubility equilibria.

← Back to Unit 7 hub
TOPIC 7.1 Introduction to Equilibrium RATES APPROACHING EQUILIBRIUM forward rate ↓ reverse rate ↑ equal rates rate time → Dynamic equilibrium Equilibrium is reached when the FORWARD and REVERSE reaction rates become EQUAL. Both reactions still occur — it's "dynamic," not stopped. Concentrations stay CONSTANT At equilibrium, the amounts of reactants and products no longer change — but they are NOT necessarily equal to each other. Only a reversible reaction (⇌) can reach it. Reaching equilibrium from either direction A reaction reaches the SAME equilibrium state whether you start with all reactants or all products. The double arrow (⇌) signals reversibility. A closed system is required — nothing can enter or leave. At equilibrium, forward rate = reverse rate — concentrations stay constant, but the reactions continue. The Review Hub · AP Chemistry Unit 7 TOPIC 7.2 Direction of Reversible Reactions Reversible reactions run both ways In a reversible reaction, products can re-form reactants. Which way the net reaction PROCEEDS depends on the current concentrations relative to the equilibrium position — measured by comparing Q to K (Topic 7.3, 7.10). If a system isn't yet at equilibrium, it shifts toward it. Q < K Too many reactants. Shifts FORWARD → (makes more products) to the right Q = K At equilibrium. No net shift — concentrations stable. rates are equal Q > K Too many products. Shifts REVERSE ← (re-forms reactants) to the left Think of it as "chasing" balance: the system always shifts in the direction that moves Q toward K. Compare Q to K: Q < K shifts forward, Q > K shifts reverse, Q = K is equilibrium. The Review Hub · AP Chemistry Unit 7 TOPIC 7.3 Reaction Quotient (Q) & Equilibrium Constant (K) for aA + bB ⇌ cC + dD: K = [C]ᶜ[D]ᵈ ÷ [A]ᵃ[B]ᵇ Products over reactants Multiply product concentrations (raised to their coefficients) and divide by reactant concentrations. Leave OUT pure solids (s) and pure liquids (l). Their concentration is constant, so they don't appear in K. Q vs. K — same formula, different time Q (reaction quotient) uses concentrations at ANY moment. K (equilibrium constant) uses them AT equilibrium. Comparing Q to K tells you which way the reaction will shift (Topic 7.2). At equilibrium, Q = K. Kc vs. Kp Kc uses molar concentrations (mol/L). Kp uses partial pressures of gases (atm). Use whichever fits the data. K has no units on the AP exam. A reaction and its reverse have reciprocal K values (K_reverse = 1/K_forward). If you multiply an equation by a factor n, the new K is the old K raised to the power n (K → Kⁿ). Adding reactions multiplies their K values together. K = products ÷ reactants (each to its coefficient), leaving out solids & liquids; Q is the same but any time. The Review Hub · AP Chemistry Unit 7 TOPIC 7.4 Calculating the Equilibrium Constant Use an ICE table An ICE table (Initial, Change, Equilibrium) tracks concentrations as a reaction moves to equilibrium. The Change row uses the mole ratios (with x); the Equilibrium row = Initial + Change. ICE TABLE for N₂ + 3H₂ ⇌ 2NH₃ N₂H₂NH₃ 1.03.00 −x−3x+2x 1.0−x3.0−3x2x Initial Change Equil. Then solve for K (or x) Plug the Equilibrium row into the K expression: K = [NH₃]² ÷ ([N₂][H₂]³) If K is known, solve for x, then find each equilibrium concentration. If K is small, the "x is negligible" approximation often simplifies the math. Use an ICE table (Initial, Change, Equilibrium) to organize concentrations, then plug into the K expression. The Review Hub · AP Chemistry Unit 7 TOPIC 7.5 Magnitude of the Equilibrium Constant K ≫ 1 (large) Equilibrium favors the PRODUCTS. At equilibrium, mostly products are present. The reaction goes "nearly to completion." The bigger K is, the further right the reaction lies. e.g., K = 10⁶ → almost all product. K ≪ 1 (small) Equilibrium favors the REACTANTS. At equilibrium, mostly reactants remain. The reaction barely proceeds. The smaller K is, the further left it lies. e.g., K = 10⁻⁶ → almost all reactant. What K does and does NOT tell you K tells you the EXTENT of a reaction — how far it goes (the ratio of products to reactants at equilibrium). K says NOTHING about the RATE — a reaction can have a huge K but be extremely slow (that's kinetics, Unit 5). K ≈ 1 means significant amounts of BOTH reactants and products are present at equilibrium. K depends only on TEMPERATURE — it does not change with concentration, pressure, or a catalyst. Changing the temperature is the ONLY way to change the value of K. K ≫ 1 favors products, K ≪ 1 favors reactants — magnitude shows extent, not rate. The Review Hub · AP Chemistry Unit 7 TOPIC 7.6 Properties of the Equilibrium Constant Manipulating equations changes K Reverse the reaction → K becomes 1/K Multiply coefficients by n → K becomes Kⁿ Add two reactions → multiply their K's (K₁ × K₂) These rules let you build a new K from known ones — the equilibrium version of Hess's law. Only temperature changes K K is a fixed number for a given reaction AT a given temperature. Adding reactant, changing pressure, or a catalyst shifts the POSITION but never changes K. Raising T increases K for endothermic reactions and decreases it for exothermic ones. Which species appear in K • Gases and aqueous species (their concentration or pressure varies) → INCLUDED in the K expression. • Pure solids and pure liquids (constant "concentration") → OMITTED from K. This is why adding more solid to a reaction doesn't shift the equilibrium — the solid isn't in K. The value of K sets the ratio, but many different equilibrium concentrations can satisfy the same K. Remember: a catalyst helps a system REACH equilibrium faster but leaves K (and the final position) unchanged. Reverse → 1/K, scale → Kⁿ, add → multiply K's; only temperature changes K itself. The Review Hub · AP Chemistry Unit 7 TOPIC 7.7 Calculating Equilibrium Concentrations Given K and initial amounts, solve for equilibrium Set up an ICE table, write the equilibrium row in terms of x, substitute into the K expression, and solve for x. Then plug x back into each equilibrium expression to get the final concentrations. If x appears squared, you may need the quadratic formula — unless the approximation applies. The small-x approximation If K is very small (≤ ~10⁻³), only a tiny amount reacts, so "initial − x ≈ initial." This avoids the quadratic. Valid when x is < ~5% of the initial concentration. Always check the 5% rule after solving to confirm it was valid. Worked idea For A ⇌ B + C starting with 1.0 M A and small K: K = x² ÷ (1.0 − x) ≈ x² ÷ 1.0 → x = √K, giving [B] = [C] = √K. The approximation turns a quadratic into a simple square root. If Q ≠ K, first determine the direction of the shift, then define x accordingly (which side loses vs. gains). ICE table + K expression → solve for x; when K is small, the x-approximation simplifies the math. The Review Hub · AP Chemistry Unit 7 TOPIC 7.8 Representations of Equilibrium PARTICLE COUNTS STAY CONSTANT AT EQUILIBRIUM time 1 time 2 (equil.) same numbers of each species — even as individuals react Particulate view At equilibrium, the COUNTS of reactant and product particles stay constant over time — although individual molecules keep reacting back and forth. Graphical view On a concentration-vs-time graph, the curves LEVEL OFF (become flat) once equilibrium is reached. The ratio of the flat values matches K. Reading equilibrium representations The AP exam shows equilibrium as particle diagrams, tables, or graphs. In all of them, equilibrium is where amounts stop changing. Use the relative amounts of products vs. reactants to estimate whether K is large or small. At equilibrium, particle counts hold steady and graphs level off — even as reactions continue. The Review Hub · AP Chemistry Unit 7 TOPIC 7.9 Le Châtelier's Principle A system at equilibrium resists change If you disturb (stress) a system at equilibrium, it SHIFTS in the direction that partially counteracts the stress and re-establishes equilibrium. Three stresses to know: concentration, pressure/volume, and temperature. Concentration Add a substance → shifts AWAY from it (to use it up). Remove one → shifts TOWARD it (to replace it). Pressure / volume Increase pressure (↓ volume) → shifts toward the side with FEWER moles of gas. Only affects gaseous equilibria. Temperature Treat heat as a reactant or product. Add heat → shifts in the ENDOthermic direction. This is the ONLY stress that changes K. A catalyst and adding an inert gas (at constant volume) do NOT shift the equilibrium. Stress an equilibrium and it shifts to counteract the change — only temperature changes K. The Review Hub · AP Chemistry Unit 7 TOPIC 7.10 Reaction Quotient & Le Châtelier's Principle Q explains WHY Le Châtelier shifts happen A stress momentarily changes Q, making it no longer equal to K. The system then shifts to bring Q back to K — which is exactly the direction Le Châtelier's principle predicts. Q is the quantitative reason behind the rule. Q ≠ K means "not at equilibrium," so a net reaction occurs until Q = K again. Adding a reactant (example) Adding reactant increases the DENOMINATOR of Q, so Q DROPS below K (Q < K). The system shifts FORWARD to rebuild products until Q rises back to K. Temperature is the special case Concentration/pressure stresses change Q but NOT K — the system returns to the SAME K. Temperature changes K ITSELF, so the equilibrium position moves to satisfy the new K. Bottom line: the system always moves in whichever direction makes Q equal to K again. A stress makes Q ≠ K; the reaction shifts to restore Q = K — that's Le Châtelier explained by Q. The Review Hub · AP Chemistry Unit 7 TOPIC 7.11 Solubility Equilibria (Ksp) AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq) Ksp = [Ag⁺][Cl⁻] The solubility product, Ksp For a "slightly soluble" salt, Ksp is the equilibrium constant for it dissolving into its ions. The SOLID is left out (as always). A SMALLER Ksp means LESS soluble. Molar solubility from Ksp Let x = molar solubility (mol dissolved per liter). Write each ion's concentration in terms of x, then solve Ksp = x·x. For AgCl: Ksp = x² → x = √Ksp Watch coefficients: for CaF₂, Ksp = (x)(2x)² = 4x³. Will a precipitate form? Compare Q to Ksp Calculate Q (the ion product) with the current ion concentrations, then compare: • Q < Ksp → unsaturated, no precipitate • Q = Ksp → saturated (at equilibrium) • Q > Ksp → supersaturated → a PRECIPITATE forms until Q drops back to Ksp This mirrors Q vs. K for any equilibrium — Ksp is just the solubility version. Ksp is the equilibrium constant for a salt dissolving; compare Q to Ksp to predict precipitation. The Review Hub · AP Chemistry Unit 7 TOPIC 7.12 The Common-Ion Effect A shared ion decreases solubility Adding an ion that's ALREADY part of a solubility equilibrium (a "common ion") pushes the equilibrium toward the SOLID (Le Châtelier), so LESS of the salt dissolves. Ksp is unchanged, but solubility goes down. It's just Le Châtelier's principle applied to a dissolving equilibrium. Worked idea AgCl(s) ⇌ Ag⁺ + Cl⁻. What happens if you add NaCl (a source of Cl⁻)? The added Cl⁻ raises [Cl⁻], so the equilibrium shifts LEFT (toward solid AgCl) to keep [Ag⁺][Cl⁻] = Ksp. Result: [Ag⁺] drops and less AgCl dissolves — AgCl is LESS soluble in a chloride solution than in pure water. The same idea explains why a weak acid is less ionized in a solution that already contains its conjugate base (Unit 8 buffers). Key takeaway: a common ion suppresses ionization/dissolving. Ksp (and K) stay the same — only the position shifts. A common ion decreases solubility — Le Châtelier pushes the equilibrium back toward the solid. The Review Hub · AP Chemistry Unit 7
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How to use the visual review

Spend 30 seconds per slide before clicking next. Look at the diagram, then ask yourself: "Could I draw this from memory and explain it?"

Use the fullscreen button () on desktop for the best experience. Use arrow keys to navigate. Tap "Show all slides" to jump around.

This is great for review the night before the exam — fast, visual, and covers everything you need to remember about Unit 7's equilibrium content.