Many chemical reactions are reversible — they can proceed in both the forward and reverse directions. When the rate of the forward reaction equals the rate of the reverse reaction, the system has reached dynamic equilibrium.
At equilibrium:
• Rateforward = Ratereverse
• Concentrations of reactants and products remain constant (but NOT necessarily equal).
• The reaction has NOT stopped — both forward and reverse reactions continue at equal rates. This is why it is called dynamic equilibrium.
• Equilibrium can be reached from either direction (starting with all reactants OR all products).
Concentrations of reactants and products approach constant values at equilibrium. The forward and reverse rates become equal. (Wikimedia Commons)
Conditions for Equilibrium
The system must be closed (no matter enters or leaves).
Temperature must be constant (changing T changes K).
Both forward and reverse reactions must be occurring.
Macroscopic properties (color, pressure, concentration) remain constant over time.
7.2 — Direction of Reversible Reactions
A reversible reaction can shift toward products (forward) or toward reactants (reverse) depending on conditions. The system always moves toward equilibrium.
How to know which direction a reaction shifts:
• If the system has too many reactants relative to equilibrium → the reaction shifts forward (toward products) to reach equilibrium.
• If the system has too many products relative to equilibrium → the reaction shifts reverse (toward reactants) to reach equilibrium.
• We quantify this using the reaction quotient Q compared to the equilibrium constant K.
At the molecular level, the shift occurs because reaction rates are concentration-dependent. If [reactants] is high, the forward rate exceeds the reverse rate, consuming reactants and forming products until rates equalize again.
7.3 — Reaction Quotient and Equilibrium Constant
The equilibrium constant (K) is the ratio of product concentrations to reactant concentrations at equilibrium, each raised to the power of its stoichiometric coefficient.
Relationship: Kp = Kc(RT)Δn where Δn = (moles of gaseous products) − (moles of gaseous reactants)
The Reaction Quotient (Q)
Q has the same formula as K, but uses concentrations at any point in time (not necessarily at equilibrium).
Comparing Q to K:
• Q < K: Too few products. Reaction shifts forward (toward products) to reach equilibrium.
• Q = K: System is at equilibrium. No net change.
• Q > K: Too many products. Reaction shifts reverse (toward reactants) to reach equilibrium.
What to Include and Exclude in K Expressions
Include: Aqueous species (aq) and gases (g).
Exclude: Pure solids (s) and pure liquids (l). Their concentrations are constant and are incorporated into K.
Given equilibrium concentrations, you can calculate K by substituting directly into the expression. If initial concentrations and one equilibrium concentration are given, use an ICE table.
Example — Direct calculation:
N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
At equilibrium: [N₂] = 0.50 M, [H₂] = 0.30 M, [NH₃] = 0.20 M
The value of K tells you the extent to which a reaction proceeds at a given temperature.
Value of K
Meaning
Position of Equilibrium
K >> 1 (e.g., 10³ or larger)
Products are strongly favored
Equilibrium lies far to the right (mostly products)
K ≈ 1
Neither side is strongly favored
Significant amounts of both reactants and products
K << 1 (e.g., 10⁻³ or smaller)
Reactants are strongly favored
Equilibrium lies far to the left (mostly reactants)
Important: K only tells you the position of equilibrium, NOT the rate at which equilibrium is reached. A reaction can have a very large K but still be extremely slow (e.g., diamond converting to graphite has K > 1 but occurs incredibly slowly at room temperature).
7.6 — Properties of the Equilibrium Constant
The equilibrium constant can be manipulated mathematically when you change the way a reaction is written.
Rules for manipulating K:
1. Reverse the reaction: Knew = 1 / Koriginal
If A ⇌ B has K = 100, then B ⇌ A has K = 1/100 = 0.01
2. Multiply the reaction by a factor n: Knew = Koriginaln
If A ⇌ B has K = 100, then 2A ⇌ 2B has K = 100² = 10,000
3. Add two reactions: Koverall = K₁ × K₂
If reaction 1 has K₁ and reaction 2 has K₂, the combined reaction has K = K₁ × K₂
Example:
Given: N₂ + O₂ ⇌ 2NO K₁ = 4.0 × 10⁻³¹
Find K for: NO ⇌ ½N₂ + ½O₂
This topic extends ICE table calculations to more complex scenarios, including cases where you need to solve a quadratic equation or use approximations.
The Small-x Approximation
When K is very small (K < ~10⁻³ relative to initial concentrations), the change in concentration x is negligible compared to the initial concentration.
If [A]₀ = 0.50 M and K is very small: (0.50 − x) ≈ 0.50
This simplifies the algebra considerably.
Validity check: After solving, verify that x is < 5% of the initial concentration. If x > 5%, use the quadratic formula instead.
Example with small-x approximation:
N₂O₄(g) ⇌ 2NO₂(g) Kc = 4.6 × 10⁻³
Initially: [N₂O₄] = 0.500 M, [NO₂] = 0
N₂O₄
2NO₂
I
0.500
0
C
−x
+2x
E
0.500 − x
2x
K = (2x)² / (0.500 − x) ≈ 4x² / 0.500 = 4.6 × 10⁻³
x² = 5.75 × 10⁻⁴ → x = 0.0240
Check: 0.0240 / 0.500 = 4.8% < 5% — approximation is valid!
[NO₂] = 2(0.0240) = 0.0480 M, [N₂O₄] = 0.500 − 0.024 = 0.476 M
7.8 — Representations of Equilibrium
Equilibrium can be represented in multiple ways on the AP exam. You must be comfortable with all of them:
Concentration vs. time graphs: Concentrations change initially, then flatten out at constant values when equilibrium is reached.
Particulate diagrams: Show the number of reactant and product particles at equilibrium. You should be able to calculate K from the particle counts.
Rate vs. time graphs: Forward rate decreases and reverse rate increases until they meet (are equal) at equilibrium.
Reading particulate diagrams for K:
Count the number of each species in the diagram. Convert counts to concentrations (if volume is given) or use counts directly as relative amounts.
Example: A 1.0 L container at equilibrium contains 3 molecules of A and 6 molecules of B for A ⇌ 2B:
K = [B]² / [A] = (6)² / (3) = 36/3 = 12 (using particle counts as concentrations in 1.0 L)
7.9 — Introduction to Le Chatelier’s Principle
Le Chatelier’s Principle: If a stress is applied to a system at equilibrium, the system will shift to partially counteract the stress and establish a new equilibrium.
Stress Applied
Direction of Shift
Effect on K
Add reactant
Shifts right (toward products)
K unchanged
Remove reactant
Shifts left (toward reactants)
K unchanged
Add product
Shifts left (toward reactants)
K unchanged
Remove product
Shifts right (toward products)
K unchanged
Decrease volume (increase pressure)
Shifts toward the side with fewer moles of gas
K unchanged
Increase volume (decrease pressure)
Shifts toward the side with more moles of gas
K unchanged
Add inert gas (at constant V)
No shift (concentrations unchanged)
K unchanged
Increase temperature
Shifts in the endothermic direction
K changes
Decrease temperature
Shifts in the exothermic direction
K changes
Add catalyst
No shift (reaches equilibrium faster)
K unchanged
Critical point: Only temperature changes K. Changes in concentration, pressure, and volume shift the position of equilibrium but do NOT change the value of K. A catalyst speeds up both forward and reverse reactions equally — it does NOT change K or the equilibrium position.
Temperature and K:
Think of heat as a “reactant” or “product”:
• Exothermic reaction: A + B ⇌ C + heat. Increasing T shifts left, K decreases.
• Endothermic reaction: A + B + heat ⇌ C. Increasing T shifts right, K increases.
7.10 — Reaction Quotient and Le Chatelier’s Principle
Le Chatelier’s Principle can be understood quantitatively through Q vs. K analysis. When a stress is applied, Q changes, and the system shifts to return Q to K.
How stresses change Q:
• Add reactant: Denominator of Q increases → Q decreases → Q < K → shifts right.
• Add product: Numerator of Q increases → Q increases → Q > K → shifts left.
• Decrease volume: All concentrations increase. If Δngas ≠ 0, Q changes. The system shifts toward fewer moles of gas to reduce Q back toward K.
Example: For N₂(g) + 3H₂(g) ⇌ 2NH₃(g) at equilibrium, if we add more N₂:
Q = [NH₃]² / ([N₂][H₂]³). Adding N₂ increases the denominator → Q < K → the reaction shifts right to produce more NH₃ until Q = K again.
7.11 — Introduction to Solubility Equilibria
When a sparingly soluble ionic compound is placed in water, it partially dissolves until a dynamic equilibrium is established between the solid and its dissolved ions. The equilibrium constant for this process is the solubility product constant (Ksp).
For: AmBn(s) ⇌ mAn+(aq) + nBm−(aq)
Ksp = [An+]m[Bm−]n
The solid is NOT included in the expression (pure solid).
Example — Calculating molar solubility from Ksp:
AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq) Ksp = 1.8 × 10⁻¹⁰
Let s = molar solubility (mol/L that dissolves):
[Ag⁺] = s, [Cl⁻] = s
Ksp = s × s = s² = 1.8 × 10⁻¹⁰
s = √(1.8 × 10⁻¹⁰) = 1.3 × 10⁻⁵ M
Compare the ion product Qsp to Ksp:
• Qsp < Ksp: Unsaturated solution — no precipitate forms. More solid can dissolve.
• Qsp = Ksp: Saturated solution — at equilibrium.
• Qsp > Ksp: Supersaturated — precipitate forms until Q = Ksp.
7.12 — Common-Ion Effect
The common-ion effect occurs when a soluble salt provides an ion that is already present in the equilibrium. By Le Chatelier’s Principle, adding a common ion shifts the equilibrium to the left, decreasing the solubility of the sparingly soluble salt.
Example: What is the solubility of AgCl in 0.10 M NaCl?
AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq) Ksp = 1.8 × 10⁻¹⁰
NaCl provides 0.10 M Cl⁻ already in solution (common ion).
[Ag⁺] = s, [Cl⁻] = 0.10 + s ≈ 0.10 (since s is very small)
Ksp = s(0.10) = 1.8 × 10⁻¹⁰
s = 1.8 × 10⁻¹⁰ / 0.10 = 1.8 × 10⁻⁹ M
Compare: In pure water, s = 1.3 × 10⁻⁵ M. The common ion reduced solubility by a factor of ~7000!
7.13 — pH and Solubility
The solubility of certain salts is affected by the pH of the solution. This is especially true for salts whose anions are the conjugate bases of weak acids.
Salts that are MORE soluble in acidic solution:
If the anion is a weak base (conjugate base of a weak acid), adding H⁺ will react with the anion and remove it from the equilibrium, shifting it to the right and increasing solubility.
• CaCO₃(s) ⇌ Ca²⁺(aq) + CO₃²⁻(aq)
• In acid: CO₃²⁻ + 2H⁺ → H₂CO₃ → H₂O + CO₂(g)
• Removing CO₃²⁻ shifts the dissolution equilibrium right → more CaCO₃ dissolves.
Salts NOT affected by pH:
Salts with anions that are conjugate bases of strong acids (Cl⁻, Br⁻, I⁻, NO₃⁻) are unaffected because these anions do not react with H⁺ in any meaningful way.
AP Exam tip: Common anions whose solubility IS affected by pH: OH⁻, S²⁻, CO₃²⁻, PO₄³⁻, F⁻, CH₃COO⁻, C₂O₄²⁻. These are all weak bases that react with H⁺.
7.14 — Free Energy of Dissolution
The spontaneity of dissolution depends on both enthalpy and entropy changes. The Gibbs free energy change (ΔG) determines whether dissolution is spontaneous.
ΔG = ΔH − TΔS
• ΔHdissolution: Can be positive (endothermic, e.g., NH₄NO₃) or negative (exothermic, e.g., NaOH).
• ΔSdissolution: Usually positive (ions or molecules become more disordered in solution than in the solid lattice).
• If ΔG < 0: Dissolution is spontaneous.
• If ΔG > 0: Dissolution is not spontaneous (the compound is insoluble or sparingly soluble).
Connecting ΔG to Ksp:
ΔG° = −RT ln Ksp
• If Ksp is large (very soluble): ΔG° is negative (spontaneous).
• If Ksp is very small (sparingly soluble): ΔG° is positive (not spontaneous under standard conditions).
Example: If Ksp = 1.8 × 10⁻¹⁰ at 298 K:
ΔG° = −(8.314)(298) ln(1.8 × 10⁻¹⁰) = −(2478)(−22.4) = +55.5 kJ/mol
Positive ΔG° confirms AgCl is sparingly soluble.
Entropy-Driven Dissolution
Some endothermic dissolutions ARE spontaneous because the large positive ΔS (increased disorder of ions in solution) overcomes the positive ΔH. Example: NH₄NO₃ dissolves endothermically (ΔH > 0, solution feels cold), but the large ΔS makes ΔG negative at room temperature.
📝 Practice Multiple-Choice Questions
Test your knowledge of Unit 7. Click “Show Answer” to reveal the correct choice and explanation.
1. At equilibrium, which of the following is true?
(A) The concentrations of reactants and products are equal
(B) The reaction has stopped
(C) The rates of the forward and reverse reactions are equal
(D) The value of Q is zero
Answer: (C)
At dynamic equilibrium, the rates of the forward and reverse reactions are equal, NOT the concentrations. The reaction continues in both directions; concentrations remain constant but are not necessarily equal.
2. For the reaction 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), which is the correct equilibrium expression?
(A) K = [SO₃]² / [SO₂]²[O₂]
(B) K = [SO₂]²[O₂] / [SO₃]²
(C) K = 2[SO₃] / (2[SO₂] + [O₂])
(D) K = [SO₃] / [SO₂][O₂]
Answer: (A)
K = products over reactants, each raised to the stoichiometric coefficient: [SO₃]² / ([SO₂]²[O₂]). Coefficients become exponents, not multipliers.
3. If Q > K for a reaction, the system will:
(A) Shift right to form more products
(B) Shift left to form more reactants
(C) Remain at equilibrium
(D) Increase in temperature
Answer: (B)
When Q > K, there are too many products relative to equilibrium. The reaction shifts left (reverse) to consume products and form reactants until Q decreases to equal K.
4. If K = 3.2 × 10⁸ for a reaction, which statement is correct?
(A) At equilibrium, mostly reactants are present
(B) The reaction is very slow
(C) At equilibrium, mostly products are present
(D) The reaction is endothermic
Answer: (C)
K >> 1 means the equilibrium lies far to the right — products are strongly favored. K says nothing about the rate (B) or whether the reaction is endo/exothermic (D).
5. If the reaction A ⇌ 2B has K = 25, what is K for 2B ⇌ A?
(A) 25
(B) −25
(C) 0.040
(D) 625
Answer: (C)
Reversing a reaction takes the reciprocal of K: Kreverse = 1/K = 1/25 = 0.040. K values are never negative.
6. For the exothermic reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g), which change will increase the value of K?
(A) Increasing the pressure
(B) Decreasing the temperature
(C) Adding a catalyst
(D) Adding more N₂
Answer: (B)
Only temperature changes K. For an exothermic reaction, decreasing the temperature shifts equilibrium right and increases K. Pressure, catalysts, and concentration changes shift the position but do NOT change K.
7. Adding an inert gas (such as Ar) to an equilibrium mixture at constant volume will:
(A) Shift the equilibrium toward products
(B) Shift the equilibrium toward reactants
(C) Have no effect on the equilibrium
(D) Change the value of K
Answer: (C)
Adding an inert gas at constant volume increases total pressure but does NOT change the partial pressures or concentrations of the reacting gases. Since Q is unchanged, there is no shift in equilibrium.
8. The Ksp of PbI₂ is 9.8 × 10⁻⁹. What is the molar solubility of PbI₂ in pure water?
9. The solubility of CaF₂ (Ksp = 3.9 × 10⁻¹¹) would be LOWEST in which solution?
(A) Pure water
(B) 0.10 M NaF
(C) 0.10 M NaCl
(D) 0.010 M NaF
Answer: (B)
F⁻ is a common ion. By the common-ion effect, having the highest [F⁻] (0.10 M NaF) suppresses the dissolution of CaF₂ the most, giving the lowest solubility. NaCl has no common ion effect.
10. CaCO₃ is more soluble in acidic solution than in pure water because:
(A) Acid increases the temperature of the solution
(B) H⁺ reacts with CO₃²⁻, removing it from equilibrium and shifting dissolution right
(C) Acid changes the Ksp of CaCO₃
(D) Ca²⁺ reacts with the acid
Answer: (B)
CO₃²⁻ is a weak base that reacts with H⁺: CO₃²⁻ + 2H⁺ → H₂O + CO₂. This removes CO₃²⁻ from the dissolution equilibrium, shifting CaCO₃(s) ⇌ Ca²⁺ + CO₃²⁻ to the right, increasing solubility. Ksp itself does not change (only temperature changes Ksp).
💡 Study Tip for Unit 7
Equilibrium is one of the most tested topics on the AP Chemistry exam. Master writing K expressions (remember: no solids or liquids!), ICE tables, and Q vs. K analysis. Le Chatelier’s Principle appears on almost every exam — memorize the table of stresses and shifts, and remember that only temperature changes K. For solubility equilibria, practice Ksp calculations with 1:1 and non-1:1 salts, the common-ion effect, and predicting precipitation. The connection between Ksp, ΔG, and pH-dependent solubility ties together multiple units.
