AP Chemistry, AP exams, 7–9%

Unit 7: Equilibrium

Unit 7 develops the reaction quotient and equilibrium constant, the calculations that follow from them, and Le Chatelier's principle as a qualitative prediction tool. Solubility equilibria, the common-ion effect, and the relationship between pH and solubility close the unit. The reasoning built here is reused directly in Units 8 and 9.

Position: 7 of 9 units
Published weighting: 7–9% in the official framework
Topics covered: 6 topics in this guide

What is covered

Topics inside this unit.

These topics describe the scope students most often review for this part of AP Chemistry. Treat them as a study checklist rather than an exhaustive list of what a single exam form can sample.

Reaction quotient and equilibrium constant: Writing expressions, comparing Q with K, and predicting the direction of net change.
Calculating equilibrium concentrations: Setting up and solving an ICE table, including when a simplifying assumption is justified.
Magnitude and properties of K: What a very large or very small K implies, and how K changes when a reaction is reversed or scaled.
Le Chatelier's principle: Predicting the response to changes in concentration, pressure, volume, and temperature.
Solubility equilibria: Solubility product expressions, molar solubility calculations, and predicting precipitation from Q versus Ksp.
Common-ion effect and pH: How a shared ion or a change in pH shifts a solubility equilibrium, with the reasoning behind it.

Study tips

How to practice Equilibrium.

Section-specific guidance that follows from how this material is usually tested.

Always compare Q with K before predicting a shift; the habit prevents most Le Chatelier mistakes.
Practice ICE tables until the setup is mechanical, then focus effort on justifying the approximation you make.
Connect solubility equilibria back to Unit 4 solubility rules so the two are one topic rather than two.

Where this fits

Unit 7: Equilibrium is 7 of 9 units in the AP Chemistry guide, and it carries a published weighting of 7–9% in the official framework. Use the full guide to balance it against everything else you have to review.

Back to the AP Chemistry guide Official exam information

Sample Equilibrium questions

Practice questions, not yet expert-reviewed

Original practice questions written to the published framework for this section. No expert has reviewed them, and they are not taken from any official exam. Try each one before you open the answer.

Question 1Foundational
What is the equilibrium constant expression Kc for the reaction N2(g) + 3 H2(g) <-> 2 NH3(g)?
1. A([N2][H2]^3) / [NH3]^2
2. B[NH3]^2 / ([N2][H2]^3)
3. C[NH3]^2 / ([N2][H2])
4. D(2[NH3]) / ([N2] + 3[H2])
Show answer
Answer: B. [NH3]^2 / ([N2][H2]^3)
Products go in the numerator and reactants in the denominator, and each concentration is raised to its coefficient from the balanced equation. That gives ammonia squared over nitrogen times hydrogen cubed. Writing the reciprocal is the tempting error and produces the constant for the reverse reaction instead.
Key insight: Coefficients become exponents in an equilibrium expression, never multipliers.
Question 2Foundational
For the equilibrium 2 SO2(g) + O2(g) <-> 2 SO3(g), what happens to the amount of SO3 when the container volume is decreased at constant temperature?
1. AIt decreases
2. BIt stays exactly the same
3. CIt increases
4. DIt falls to zero
Show answer
Answer: C. It increases
Reducing the volume raises the total pressure, and the system relieves that stress by shifting toward the side with fewer moles of gas. The reactant side has three moles of gas and the product side has two, so the equilibrium shifts right and more SO3 forms. The equilibrium constant itself is unchanged, because only temperature can change K.
Key insight: Compress a gaseous equilibrium and it shifts toward whichever side has fewer moles of gas.
Question 3Intermediate
A reaction mixture has a reaction quotient Q that is smaller than the equilibrium constant K. In which direction does the net reaction proceed?
1. AToward the reactants
2. BIt is already at equilibrium
3. CThe direction cannot be determined from Q and K
4. DToward the products
Show answer
Answer: D. Toward the products
Q has the same form as K but uses current concentrations, so Q smaller than K means the mixture holds too few products relative to equilibrium. The net reaction therefore runs forward until Q rises to equal K. Reversing the comparison is the standard error, so always ask which quantity has to grow.
Key insight: Q must move toward K, so a small Q means the reaction runs forward to make more product.
Question 4Intermediate
The molar solubility of AgCl in pure water is 1.3 x 10^-5 mol/L. What is the solubility product Ksp for AgCl?
1. A1.7 x 10^-10
2. B1.3 x 10^-5
3. C1.7 x 10^-5
4. D2.6 x 10^-5
Show answer
Answer: A. 1.7 x 10^-10
Dissolving AgCl releases one silver ion and one chloride ion, so each is present at 1.3 x 10^-5 mol/L and Ksp equals that value squared, about 1.7 x 10^-10. Answering 2.6 x 10^-5 doubles the solubility instead of squaring it, which mistakes the number of ions for an exponent. Solids never appear in the expression, so AgCl itself is left out.
Key insight: Ksp is a product of ion concentrations raised to their coefficients, so a 1 to 1 salt gives solubility squared.
Question 5Advanced
For the gas-phase reaction A <-> B, Kc equals 4.0. If a sealed flask initially contains A at 1.00 mol/L and no B, what is the concentration of B at equilibrium?
1. A0.20 mol/L
2. B0.40 mol/L
3. C0.67 mol/L
4. D0.80 mol/L
Show answer
Answer: D. 0.80 mol/L
Let x be the amount of A converted, so at equilibrium B is x and A is 1.00 minus x, and the expression x divided by (1.00 minus x) equals 4.0. Solving gives x equals 4.00 minus 4.00x, so 5.00x equals 4.00 and x is 0.80 mol/L. The value 0.20 mol/L is the equilibrium concentration of A, the tempting answer for anyone who solves for the wrong species.
Key insight: Because K is well above 1 here, most of the starting material must end up as product.

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Unit 7: Equilibrium