AP Chemistry Equilibrium Explained: K, Q, ICE Tables, and More
AP Chemistry equilibrium explained: K expressions, Q vs K, ICE tables with worked examples, Le Chatelier's principle, and Ksp, plus how they are tested.
Chemical equilibrium is the state in which the forward and reverse reactions happen at the same rate, so the concentrations of reactants and products stop changing, though they are generally not equal. In AP Chemistry, nearly every equilibrium question comes back to one comparison: the reaction quotient Q against the equilibrium constant K. If you can write K, compare Q to K, and run an ICE table, you can handle most equilibrium questions on the exam, including the ones that show up in acid-base and solubility problems.
What is chemical equilibrium?
Picture a reversible reaction in a closed container. At first, the forward reaction is fast because reactant concentrations are high. As products build up, the reverse reaction speeds up. Eventually the two rates are equal, and the concentrations stay constant from that point on.
Two misconceptions cost students points:
- Equilibrium does not mean equal concentrations. It means constant concentrations, because the rates are equal.
- The reactions do not stop. Equilibrium is dynamic: both reactions continue at the same rate.
Equilibrium is Unit 7 of AP Chemistry, weighted at 7–9% of the exam. But its ideas run through Unit 8 (Acids and Bases, 11–15%) and Unit 9 (Thermodynamics and Electrochemistry), so it effectively accounts for much more of your score.
How do you write an equilibrium constant expression?
For a general reaction aA + bB ⇌ cC + dD, the equilibrium constant is:
K = [C]ᶜ[D]ᵈ / [A]ᵃ[B]ᵇ
Products go on top, reactants on the bottom, and each concentration is raised to its coefficient. Leave out pure solids and pure liquids, because their concentrations do not change. For gases, you can write Kp using partial pressures instead of molar concentrations.
Example: For CaCO₃(s) ⇌ CaO(s) + CO₂(g), the expression is simply Kc = [CO₂], since both solids are excluded.
What does the size of K tell you?
| Value of K | Meaning |
|---|---|
| K much greater than 1 | Products are favored at equilibrium |
| K close to 1 | Significant amounts of both reactants and products |
| K much less than 1 | Reactants are favored at equilibrium |
K tells you where the reaction ends up, not how fast it gets there. Speed is a kinetics question.
How do you manipulate K?
| Change to the reaction | New K |
|---|---|
| Reverse it | 1/K |
| Multiply coefficients by n | Kⁿ |
| Add two reactions together | K₁ × K₂ |
These rules let you find K for a target reaction from reactions with known K values, in the same way Hess's law works for enthalpy.
Q vs K: which way will the reaction shift?
The reaction quotient Q uses the same expression as K, but with the current concentrations instead of equilibrium concentrations.
| Comparison | What happens |
|---|---|
| Q < K | Too few products. The reaction proceeds forward, toward products. |
| Q > K | Too many products. The reaction proceeds in reverse, toward reactants. |
| Q = K | The system is at equilibrium. |
Q vs K reasoning is also the most rigorous way to justify a shift on a free-response question, as you will see in the Le Chatelier section below.
How to solve equilibrium problems with ICE tables
An ICE table tracks the Initial concentrations, the Change, and the Equilibrium concentrations.
Worked example 1: finding K from equilibrium data
A flask initially contains 1.00 M H₂ and 1.00 M I₂. They react according to H₂(g) + I₂(g) ⇌ 2HI(g). At equilibrium, [HI] = 1.56 M. Find Kc.
| H₂ | I₂ | HI | |
|---|---|---|---|
| Initial | 1.00 | 1.00 | 0 |
| Change | −x | −x | +2x |
| Equilibrium | 1.00 − x | 1.00 − x | 2x |
Since 2x = 1.56, x = 0.78. So [H₂] = [I₂] = 0.22 M.
Kc = (1.56)² / (0.22)(0.22) = 2.43 / 0.0484 ≈ 50
The key step is using the stoichiometric coefficients in the Change row: HI gains 2x because 2 moles of HI form for every mole of H₂ used.
Worked example 2: the small-x approximation
Find the pH of a 0.10 M solution of a weak acid HA with Ka = 1.8 × 10⁻⁵.
| HA | H⁺ | A⁻ | |
|---|---|---|---|
| Initial | 0.10 | 0 | 0 |
| Change | −x | +x | +x |
| Equilibrium | 0.10 − x | x | x |
Ka = x² / (0.10 − x) = 1.8 × 10⁻⁵
Because Ka is small, x will be small compared with 0.10, so approximate 0.10 − x ≈ 0.10:
x² = 1.8 × 10⁻⁶, so x = 1.34 × 10⁻³ M = [H⁺]
pH = −log(1.34 × 10⁻³) ≈ 2.87
Check the approximation: 1.34 × 10⁻³ is about 1.3% of 0.10, well under the commonly used 5% guideline, so the approximation holds.
Le Chatelier's principle
When a system at equilibrium is disturbed, it shifts in the direction that partially counteracts the change.
| Disturbance | Shift | Does K change? |
|---|---|---|
| Add a reactant, or remove a product | Toward products | No |
| Add a product, or remove a reactant | Toward reactants | No |
| Decrease volume (increase pressure) of gases | Toward the side with fewer moles of gas | No |
| Add an inert gas at constant volume | No shift | No |
| Increase temperature, endothermic forward reaction | Toward products | Yes, K increases |
| Increase temperature, exothermic forward reaction | Toward reactants | Yes, K decreases |
| Add a catalyst | No shift; equilibrium is reached faster | No |
Temperature is the only change in this table that changes the value of K. A useful trick: treat heat as a product of an exothermic reaction and a reactant of an endothermic one.
How to justify a shift for full credit
"The equilibrium shifts right" is a prediction, not a justification. Explain it with Q and K:
Adding H₂ increases the denominator of Q, so Q becomes less than K. To restore Q = K, the reaction proceeds in the forward direction, forming more HI.
That answer names the change, its effect on Q, the comparison with K, and the resulting direction.
Solubility equilibria and Ksp
Slightly soluble ionic compounds establish an equilibrium with their ions. For AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq), Ksp = [Ag⁺][Cl⁻].
To find molar solubility s:
| Salt type | Example | Ksp in terms of s |
|---|---|---|
| AB | AgCl | s² |
| AB₂ | CaF₂ | 4s³ |
| A₂B | Ag₂CrO₄ | 4s³ |
Common-ion effect: dissolving a salt in a solution that already contains one of its ions lowers its solubility. Adding Cl⁻ to an AgCl equilibrium raises Q above Ksp, so the reaction shifts toward the solid.
pH effects: salts with a basic anion, such as hydroxides or fluorides, become more soluble in acidic solution, because H⁺ removes the anion and pulls the equilibrium toward dissolving.
How equilibrium connects to thermodynamics
The standard free energy change and K are linked by ΔG° = −RT ln K, which appears on the equations sheet.
- If ΔG° < 0, then K > 1, and products are favored.
- If ΔG° > 0, then K < 1, and reactants are favored.
Expect questions that ask you to connect these signs without calculating anything.
Common equilibrium mistakes on the AP exam
- Including solids or pure liquids in K or Q.
- Forgetting to raise concentrations to their coefficients, or forgetting the 2x in the Change row.
- Saying a catalyst shifts equilibrium.
- Claiming that adding reactant changes K.
- Using initial concentrations in K instead of equilibrium concentrations.
- Justifying a shift with "Le Chatelier's principle" alone, without explaining Q vs K.
For the broader exam strategy, read how to get a 5 on AP Chemistry, and estimate your practice scores with the AP Chemistry score calculator.
How Ivystar can help
Equilibrium is where AP Chemistry starts to feel abstract for many students, and a tutor who can connect K, Q, and Le Chatelier's principle to clear pictures makes a real difference. Ivystar's AP Exam Prep pairs you with a tutor from schools like Harvard, Penn, Cornell, Stanford, or MIT, with a "Get an A or we work for free" guarantee on your class grade. Book a free 20-minute strategy call to get started.
Frequently asked questions
What is the difference between Q and K in AP Chemistry?+
K uses equilibrium concentrations and is constant at a given temperature, while Q uses the current concentrations at any moment. Comparing Q to K tells you which direction the reaction will proceed.
Do solids go in the equilibrium expression?+
No. Pure solids and pure liquids are left out of K and Q expressions because their concentrations do not change.
What changes the value of K?+
Only temperature. Changing concentrations, pressure, or volume shifts the equilibrium position but does not change K, and a catalyst changes neither.
When can you use the small-x approximation?+
When K is small relative to the initial concentration. After solving, check that x is less than about 5% of the initial concentration; if not, solve the full equation.
How much of the AP Chemistry exam is equilibrium?+
Unit 7, Equilibrium, is weighted at 7–9% of the exam, but equilibrium reasoning is also central to Unit 8, Acids and Bases, and to parts of Unit 9.
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