AP Chemistry · Handsworth Secondary 2026–27

Unit 7 · Equilibrium

Dr. Ras Mulinta
Handsworth Secondary
Notes Package

CED Unit 7, 7–9% of the AP exam, taught in 8 blocks. BC Chemistry 12 already gave you Le Châtelier and the idea of a reversible reaction, so we move quickly there and spend our time on the AP depth: writing K expressions, the Q-vs-K decision, ICE-table algebra, and solubility (Ksp) math. Every block is ~15 min of instruction, then you work problems while I circulate.

Blocks:  1 → A–B  ·  2 → C  ·  3 → D  ·  4 → E–F  ·  5 → G  ·  6 → H–I  ·  7 → J  ·  8 → K

Name:Block:Date:

A · Dynamic Equilibrium CED 7.1 · LO 7.1.A

Many processes are reversible, evaporation/condensation, dissolution/precipitation, proton or electron transfer. Equilibrium is not the reaction stopping; it is the forward and reverse processes running at equal rates so nothing observable changes.

Core: at equilibrium, reactants and products are both present, their concentrations (or partial pressures) stay constant, and rate_forward = rate_reverse. "Constant" does not mean "equal", the amounts are usually different, just no longer changing.
Worked example
On a concentration-vs-time graph for A → B, the [A] curve falls then levels off and the [B] curve rises then levels off, both flattening at the same time. What does the flat region mean?
Flat curves = concentrations no longer changing → the system has reached equilibrium.
Forward and reverse rates are now equal; the reaction is still occurring in both directions (dynamic), but there is no net change.

Practice A

  1. True or false: at equilibrium the forward reaction stops. Explain.
  2. Sketch rate-vs-time curves for the forward and reverse reactions of a synthesis that starts with pure reactants; mark where equilibrium is reached.
  3. Name two everyday physical processes that reach a dynamic equilibrium.

B · Reaction Direction & Relative Rates CED 7.2 · LO 7.2.A

Before equilibrium, one direction wins. Which way the net reaction runs is simply a contest between the forward rate and the reverse rate.

Core: if rate_fwd > rate_rev → net conversion of reactants to products; if rate_rev > rate_fwd → net conversion of products back to reactants; when they become equal, equilibrium is reached.
Worked example
A flask is filled with only reactants. Just after mixing, which rate is larger and which way does the net reaction proceed?
With no product present yet, the reverse rate starts at zero while the forward rate is high.
rate_fwd > rate_rev, so there is a net conversion of reactants → products until the two rates equalize.

Practice B

  1. A flask is filled with only product. Which direction does the net reaction run at the start? Why?
  2. As reactants are consumed, what happens to the forward rate, and why does that push the system toward equilibrium?
  3. Explain why a sealed equilibrium mixture shows no color change even though molecules are constantly reacting.

C · Writing Kc and Kp CED 7.3 · LO 7.3.A

The equilibrium constant is the value the reaction quotient reaches at equilibrium. For aA + bB ⇌ cC + dD, the law of mass action sets its form.

Expressions: Kc = [C]ᶜ[D]ᵈ / [A]ᵃ[B]ᵇ  ·  Kp = (P_C)ᶜ(P_D)ᵈ / (P_A)ᵃ(P_B)ᵇ. Products on top, each concentration/pressure raised to its coefficient. At equilibrium Kc = Qc and Kp = Qp.
Leave out pure solids and pure liquids, their "concentration" doesn't change, so they are not written in K or Q.
Worked example
Write Kc for  N₂(g) + 3H₂(g) ⇌ 2NH₃(g)  and Kp for  2SO₂(g) + O₂(g) ⇌ 2SO₃(g).
Kc = [NH₃]² / ([N₂][H₂]³)
Kp = (P_SO₃)² / ((P_SO₂)²(P_O₂))

Not tested: converting between Kc and Kp (the K_p = K_c(RT)^Δn formula). Just notice whether a question gives concentrations (Kc) or pressures (Kp) and use the matching expression.

Not tested: equilibria where a dissolved species is in equilibrium with that same species in the gas phase.

Practice C

  1. Write Kc for  2NO(g) + O₂(g) ⇌ 2NO₂(g).
  2. Write Kc for  CaCO₃(s) ⇌ CaO(s) + CO₂(g).  (Watch the solids.)
  3. Write Kp for  PCl₅(g) ⇌ PCl₃(g) + Cl₂(g).

D · Calculating K & Its Magnitude CED 7.4–7.5 · LO 7.4.A · LO 7.5.A

Plug measured equilibrium concentrations (or pressures) straight into the expression. The size of K then tells you which side the reaction favors.

Magnitude: a large K (≫1) means products dominate, the reaction goes nearly to completion. A small K (≪1) means reactants dominate, barely any reaction. K ≈ 1 means comparable amounts of both.
Worked example
For  N₂O₄(g) ⇌ 2NO₂(g)  the equilibrium concentrations are [N₂O₄] = 0.0125 M and [NO₂] = 0.0750 M. Find Kc and say which side is favored.
Kc = [NO₂]² / [N₂O₄] = (0.0750)² / 0.0125 = 0.005625 / 0.0125
Kc = 0.450  (< 1, so reactant N₂O₄ is favored).

Practice D

  1. For  2SO₃(g) ⇌ 2SO₂(g) + O₂(g)  the equilibrium pressures (atm) are P_SO₃ = 0.30, P_SO₂ = 0.10, P_O₂ = 0.20. Find Kp.
  2. Reaction X has K = 1×10⁸; reaction Y has K = 1×10⁻⁶. Which proceeds nearly to completion?
  3. If K is very small, are the equilibrium concentrations of products large or small compared with reactants?

E · Manipulating K (Multistep Reactions) CED 7.6 · LO 7.6.A

When you reverse, scale, or add reactions, K changes in predictable ways. Because Q and K share the same form, every rule below works on Q too.

Rules: reverse a reaction → K' = 1/K. Multiply all coefficients by c → K' = Kᶜ. Add reactions → K_overall = K₁ × K₂ × …
Worked example
Step 1: A ⇌ B, K₁ = 2.0.  Step 2: B ⇌ C, K₂ = 3.0.  Find K for the overall A ⇌ C, and for the reverse C ⇌ A.
Adding the steps gives A ⇌ C, so multiply: K_overall = K₁ × K₂ = 2.0 × 3.0 = 6.0
A ⇌ C: K = 6.0  ·  reverse C ⇌ A: K = 1/6.0 = 0.17

Practice E

  1. If K = 4.0 for  H₂ + I₂ ⇌ 2HI, what is K for  2HI ⇌ H₂ + I₂?
  2. For  N₂ + O₂ ⇌ 2NO  K = 0.20. Find K for  2N₂ + 2O₂ ⇌ 4NO.
  3. Reactions are added so that K₁ = 5.0 and K₂ = 0.20. What is K for the summed reaction?

F · Reaction Quotient Q vs K CED 7.7 · LO 7.7.A

Q is K's expression evaluated at any instant, not just at equilibrium. Comparing Q to K tells you which way the reaction must shift to reach equilibrium.

The decision: Q < K → too few products → shifts forward (right). Q > K → too many products → shifts reverse (left). Q = K → already at equilibrium, no net shift.
Worked example
For  N₂ + 3H₂ ⇌ 2NH₃, Kc = 0.50. A mixture has [N₂] = 0.10 M, [H₂] = 0.20 M, [NH₃] = 0.10 M. Which way does it shift?
Qc = [NH₃]² / ([N₂][H₂]³) = (0.10)² / (0.10 × (0.20)³) = 0.010 / (0.10 × 0.0080) = 0.010 / 0.00080
Qc = 12.5
Q (12.5) > K (0.50), so the reaction shifts reverse (toward reactants) to reach equilibrium.

Practice F

  1. For a reaction with K = 1.0, you find Q = 0.25. Which direction does it shift?
  2. For  H₂ + I₂ ⇌ 2HI, K = 50. A mixture has [H₂] = [I₂] = [HI] = 0.10 M. Find Q and the shift direction.
  3. Why is Q useful for a mixture that is not yet at equilibrium, while K only describes the equilibrium state?

G · ICE Tables: Solving for Equilibrium CED 7.7 · LO 7.7.A

Given a balanced reaction, the initial amounts, and K, an ICE table (Initial, Change, Equilibrium) lets you solve for every equilibrium concentration. Let x be the amount reacted; use coefficients for the Change row.

Shortcut: when K is very small, the change x is tiny next to the initial concentration, so you may approximate (c₀ − x) ≈ c₀. Validity check: the approximation is fine if x is < 5% of c₀.
Worked example, small K, approximation
For  A ⇌ B + C, Kc = 1.6×10⁻⁵, starting with [A]₀ = 0.10 M and no B or C. Find the equilibrium concentrations.
ICE: [A] = 0.10 − x, [B] = x, [C] = x. So Kc = x² / (0.10 − x).
Assume x ≪ 0.10:  x² / 0.10 ≈ 1.6×10⁻⁵ → x² = 1.6×10⁻⁶ → x = 1.3×10⁻³.
Check: 1.3×10⁻³ / 0.10 = 1.3% < 5% ✓ approximation valid.
[A] ≈ 0.099 M, [B] = [C] = 1.3×10⁻³ M.
Worked example, K not small, quadratic
For  PCl₅(g) ⇌ PCl₃(g) + Cl₂(g), Kc = 0.040, starting with [PCl₅]₀ = 0.50 M. Find the equilibrium concentrations.
ICE: [PCl₅] = 0.50 − x, [PCl₃] = [Cl₂] = x.  Kc = x² / (0.50 − x) = 0.040.
x² + 0.040x − 0.020 = 0 → x = (−0.040 + √(0.0016 + 0.080)) / 2 = (−0.040 + 0.2857) / 2.
x = 0.123 (take the positive root).
[PCl₅] = 0.50 − 0.123 = 0.377 M,  [PCl₃] = [Cl₂] = 0.123 M.

Practice G

  1. For  A ⇌ 2B, Kc = 4.0×10⁻⁶, [A]₀ = 0.20 M. Find x and [B] using the small-x approximation; verify it is < 5%.
  2. For  H₂ + I₂ ⇌ 2HI, Kc = 50.5, starting [H₂]₀ = [I₂]₀ = 1.00 M. (Hint: this one is a perfect square, take √K of both sides.) Find [HI] at equilibrium.
  3. Why must you check the 5% condition after using the small-x approximation?

H · Particulate Models & Le Châtelier CED 7.8–7.9 · LO 7.8.A · LO 7.9.A

A particulate diagram shows the relative counts of reactant and product particles, many products vs few reactants means a large K. Le Châtelier's principle predicts how a system at equilibrium responds when you disturb it.

Le Châtelier: a system at equilibrium shifts to partly counteract a stress. Add a species → shift away from it. Remove a species → shift toward it. Decrease volume (raise pressure) → shift toward the side with fewer gas moles. For temperature, treat heat as a reagent: heating an endothermic reaction (heat on the left) shifts it right.
Worked example
For the exothermic  N₂(g) + 3H₂(g) ⇌ 2NH₃(g) + heat, predict the shift when you (a) add H₂, (b) decrease the volume, (c) raise the temperature.
(a) Adding H₂ is a stress on the left → shift right (makes more NH₃).
(b) Smaller volume favors fewer gas moles: left = 4 mol, right = 2 mol → shift right.
(c) Heat is a product (exothermic); adding heat → shift left (less NH₃, K decreases).

Not tested: adding an inert gas at constant volume does nothing, it changes total pressure but not any partial pressure or concentration, so there is no shift.

Practice H

  1. For  2SO₂ + O₂ ⇌ 2SO₃ (exothermic), predict the shift when O₂ is removed.
  2. For  N₂O₄(g) ⇌ 2NO₂(g), which way does increasing the volume shift the equilibrium? Why?
  3. A particulate box at equilibrium shows 9 product particles and 1 reactant particle. Is K large or small? Explain.

I · Stresses, Q, K, and the New Equilibrium CED 7.10 · LO 7.10.A

Le Châtelier is the qualitative story; Q vs K is the quantitative engine behind it. A stress changes Q away from K (or changes K itself), and the system redistributes until Q = K again.

Key distinction: changing a concentration or pressure changes Q only: K is unchanged, and the system shifts to restore Q = K. Changing temperature changes K itself, creating a genuinely new equilibrium constant.
Worked example
A system sits at equilibrium (Q = K). You suddenly add more product. Explain, using Q and K, what happens.
Adding product increases the numerator of Q, so now Q > K.
To restore equality the system consumes products and makes reactants, a reverse shift.
K is unchanged (no temperature change); the concentrations redistribute until Q falls back to K.

Practice I

  1. You remove a reactant from an equilibrium mixture. Is Q now > K or < K, and which way does the system shift?
  2. Cooling an exothermic reaction makes more product. Did Q change first, or did K change? Explain.
  3. Why does adding a catalyst leave the equilibrium position (and K) unchanged?

J · Solubility Equilibria: Ksp & Molar Solubility CED 7.11 · LO 7.11.A

Dissolving a slightly soluble salt is just a reversible reaction, so it has an equilibrium constant, the solubility product Ksp. From Ksp you can get molar solubility, and vice versa.

Set-up: for MₓXᵧ(s) ⇌ x Mⁿ⁺ + y X^(m−),  Ksp = [Mⁿ⁺]ˣ[X^(m−)]ʸ (the solid is left out). Let molar solubility = s; build the ion concentrations from s using the coefficients. Larger Ksp generally means more soluble (for salts of the same ion ratio).
Worked example
Find the molar solubility of PbI₂ given Ksp = 7.1×10⁻⁹.
PbI₂(s) ⇌ Pb²⁺ + 2I⁻. If s mol/L dissolve: [Pb²⁺] = s, [I⁻] = 2s.
Ksp = (s)(2s)² = 4s³ = 7.1×10⁻⁹ → s³ = 1.775×10⁻⁹ → s = ∛(1.775×10⁻⁹).
s = 1.2×10⁻³ M  (so [Pb²⁺] = 1.2×10⁻³ M and [I⁻] = 2.4×10⁻³ M).

Practice J

  1. Find the molar solubility of AgCl, Ksp = 1.8×10⁻¹⁰. (1:1 salt, so Ksp = s².)
  2. Find the molar solubility of CaF₂, Ksp = 3.9×10⁻¹¹. (Use Ksp = 4s³.)
  3. The molar solubility of a 1:1 salt MX is 1.0×10⁻⁴ M. Find its Ksp.

K · The Common-Ion Effect CED 7.12 · LO 7.12.A

If a solution already contains one of the salt's ions, the salt is less soluble. This is just Le Châtelier applied to dissolving: the extra ion pushes the dissolution equilibrium back toward the solid.

Method: the common ion's starting concentration goes into the Ksp expression. Because the salt dissolves only slightly, the small amount it adds is usually negligible next to the common ion already present.
Worked example
Find the molar solubility of PbI₂ (Ksp = 7.1×10⁻⁹) in 0.10 M NaI, and compare it to the value in pure water.
Now [I⁻] starts at 0.10 M. Let s = molar solubility: [Pb²⁺] = s, [I⁻] ≈ 0.10 (the 2s added is negligible).
Ksp = (s)(0.10)² = 7.1×10⁻⁹ → s = 7.1×10⁻⁹ / 0.010 = 7.1×10⁻⁷ M.
s = 7.1×10⁻⁷ M, about 1700× less soluble than the 1.2×10⁻³ M found in pure water (Section J). The common ion suppresses solubility.

Practice K

  1. Find the molar solubility of AgCl (Ksp = 1.8×10⁻¹⁰) in 0.10 M NaCl. Compare to pure water (1.3×10⁻⁵ M).
  2. Find the molar solubility of CaF₂ (Ksp = 3.9×10⁻¹¹) in 0.10 M NaF. (Common ion is F⁻; use [F⁻] ≈ 0.10.)
  3. Use Le Châtelier to explain, without numbers, why adding NaCl makes AgCl less soluble.
AP Chemistry · Unit 7, Equilibrium · Dr. Ras Mulinta · Handsworth Secondary 2026–27. Pegged to the College Board AP Chemistry CED (topics 7.1–7.12) and BC Chemistry 12. Ksp values from standard reference tables; atomic masses from the IUPAC periodic table.