Equilibrium
Unit 7 of AP Chemistry, worth 7–9% of the exam. 14 questions below, each with the working. Every answer was checked by a second pass before it was published.
Reversible reactions, the equilibrium constant, ICE tables, Le Chatelier's principle, reaction quotient, solubility equilibria and Ksp, free energy and equilibrium.
How this unit is tested
Start by recognizing that equilibrium problems all hinge on one idea: at equilibrium the forward and reverse rates are equal, so concentrations (or partial pressures) stop changing even though both reactions still occur. Every problem type in this unit is a variation on writing K correctly, then either solving for concentrations (ICE table), predicting a shift (Le Chatelier's or Q vs K), or connecting K to thermodynamics (ΔG°).
For any equilibrium expression, write the balanced equation first, then put products over reactants raised to their coefficients — and immediately cross out pure solids and liquids, since they don't appear in K. When a problem gives you initial amounts and asks for equilibrium amounts, build an ICE table: initial, change (in terms of x), equilibrium. Substitute the equilibrium row into the K expression and solve for x, checking whether the small-x approximation is valid (change is under about 5% of the initial value) before deciding whether you need the quadratic formula.
When a problem gives you concentrations that may or may not be at equilibrium, compute Q using the same expression as K and compare: QK means it shifts reverse, Q=K means it's already at equilibrium. Le Chatelier's principle questions are really Q vs K questions in disguise — a stress changes a concentration, pressure, or temperature, which changes Q relative to K (or changes K itself, for temperature), and the system shifts to restore equality.
For solubility equilibria, treat Ksp exactly like any other K, but remember the solid salt is omitted from the expression, so Ksp is written only in terms of dissolved ions raised to their stoichiometric coefficients. Finally, connect everything to thermodynamics: ΔG° = −RTlnK tells you that a very negative ΔG° corresponds to a large K (products favored), and ΔG = ΔG° + RTlnQ tells you which direction is spontaneous at any given composition, even away from standard conditions.
What you have to know
Law of Mass Action
For aA + bB ⇌ cC + dD, K = [C]^c[D]^d / [A]^a[B]^b, using equilibrium concentrations (or pressures). Pure solids and pure liquids are omitted because their activities equal 1.
Kc–Kp relationship
$K_p = K_c(RT)^{\Delta n}$, where Δn is moles of gaseous product minus moles of gaseous reactant, R = 0.08206 L·atm/(mol·K), and T is in kelvin.
Reaction Quotient Rule
Q has the same form as K but uses whatever concentrations exist at a given moment. If Q<K the reaction proceeds forward; if Q>K it proceeds in reverse; if Q=K the system is at equilibrium.
Le Chatelier's Principle
If a system at equilibrium is disturbed (change in concentration, pressure/volume, or temperature), the system shifts in the direction that partially counteracts the disturbance. Adding a catalyst speeds both directions equally and causes no shift.
Solubility Product Constant
For a sparingly soluble salt $M_xA_y(s) \rightleftharpoons xM^{n+}(aq) + yA^{m-}(aq)$, $K_{sp} = [M^{n+}]^x[A^{m-}]^y$. If the ion product Q exceeds Ksp, precipitation occurs; if Q<Ksp, the solution is unsaturated.
Free Energy–Equilibrium Relationship
$\Delta G^\circ = -RT\ln K$ and $\Delta G = \Delta G^\circ + RT\ln Q$. A large K corresponds to a very negative ΔG°; the reaction proceeds spontaneously in the direction that makes ΔG negative.
14 practice questions
-
A reversible reaction has reached equilibrium in a sealed container. Which statement is correct?
Show the answer
Answer. The forward and reverse reaction rates are equal, so concentrations remain constant over time.
Equilibrium is dynamic: both reactions continue to occur, but at equal rates, so no net change in concentration is observed. Concentrations of reactants and products need not be equal to each other. -
Write the Kc expression for 2NOCl(g) ⇌ 2NO(g) + Cl₂(g).
Show the answer
Answer. Kc = [NO]²[Cl₂]/[NOCl]²
Products go in the numerator and reactants in the denominator, each raised to its coefficient from the balanced equation. All species here are gases, so all appear in the expression. -
Write the Kc expression for the heterogeneous equilibrium CaCO₃(s) ⇌ CaO(s) + CO₂(g).
Show the answer
Answer. Kc = [CO₂]
Pure solids have an activity of 1 and are omitted from the equilibrium expression, so only the gaseous CO₂ concentration appears. -
For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), Kc = 0.105 at 472 K. Calculate Kp.
Show the answer
Answer. Kp ≈ 7.00×10⁻⁵
Use Kp = Kc(RT)^Δn with Δn = 2 − (1+3) = −2. RT = (0.08206)(472) = 38.73 L·atm/mol, so Kp = 0.105/(38.73)² = 0.105/1500 ≈ 7.00×10⁻⁵. -
For H₂(g) + I₂(g) ⇌ 2HI(g), K = 54.3 at 425°C. A mixture contains [H₂] = 0.10 M, [I₂] = 0.10 M, and [HI] = 0.50 M. In which direction will the reaction proceed to reach equilibrium?
Show the answer
Answer. Forward, toward more HI
Q = [HI]²/([H₂][I₂]) = (0.50)²/(0.10×0.10) = 0.25/0.01 = 25. Since Q (25) is less than K (54.3), the reaction must proceed forward to increase products and reach equilibrium. -
A 1.00 L flask contains 0.500 mol PCl₅ initially, which decomposes: PCl₅(g) ⇌ PCl₃(g) + Cl₂(g), Kc = 1.80×10⁻³. Find the equilibrium concentration of Cl₂.
Show the answer
Answer. [Cl₂] ≈ 0.0291 M
ICE gives Kc = x²/(0.500−x) = 1.80×10⁻³. Solving the resulting quadratic x² + 0.00180x − 0.000900 = 0 gives x ≈ 0.0291 M, which is [Cl₂] at equilibrium (x is about 5.8% of 0.500, too large to safely approximate as negligible). -
For the exothermic synthesis N₂(g) + 3H₂(g) ⇌ 2NH₃(g), what happens if the container's volume is decreased at constant temperature?
Show the answer
Answer. The equilibrium shifts toward NH₃ (fewer moles of gas), increasing the amount of NH₃ present
Decreasing volume increases pressure. By Le Chatelier's principle, the system shifts toward the side with fewer moles of gas to partially relieve the increased pressure — here, that's the product side (2 mol vs 4 mol of gas). -
A chemist adds a catalyst to a reaction already at equilibrium. What is the effect on the position of equilibrium and on K?
Show the answer
Answer. No effect on either — the catalyst speeds up both the forward and reverse rates equally, so the system reaches the same equilibrium faster but K and the equilibrium concentrations are unchanged.
Catalysts lower the activation energy for both directions equally, so they do not favor products or reactants; they only shorten the time needed to reach the same equilibrium state. -
For an exothermic reaction A(g) ⇌ B(g) + heat, what happens to K if temperature is increased?
Show the answer
Answer. K decreases, because increasing temperature shifts equilibrium toward reactants (favoring the endothermic reverse direction) for an exothermic reaction.
Treating heat as a product, increasing temperature is like adding a product, which shifts equilibrium in reverse. Unlike concentration or pressure changes, a temperature change actually alters the numerical value of K itself. -
The molar solubility of AgCl in pure water at 25°C is 1.3×10⁻⁵ M. Calculate Ksp for AgCl.
Show the answer
Answer. Ksp ≈ 1.7×10⁻¹⁰
AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq), so Ksp = [Ag⁺][Cl⁻] = s² = (1.3×10⁻⁵)² ≈ 1.7×10⁻¹⁰. -
Using Ksp(AgCl) = 1.8×10⁻¹⁰, calculate the molar solubility of AgCl in a 0.10 M NaCl solution.
Show the answer
Answer. s ≈ 1.8×10⁻⁹ M
With a common Cl⁻ ion already present, Ksp = s(0.10 + s) ≈ s(0.10), since s is negligible compared to 0.10. Solving gives s = Ksp/0.10 = 1.8×10⁻⁹ M, far less than the solubility in pure water — this is the common ion effect. -
Two solutions are mixed to give [Pb²⁺] = 1.0×10⁻³ M and [Cl⁻] = 2.0×10⁻³ M. Given Ksp(PbCl₂) = 1.7×10⁻⁵, will a precipitate form?
Show the answer
Answer. No precipitate forms, because Q < Ksp
Q = [Pb²⁺][Cl⁻]² = (1.0×10⁻³)(2.0×10⁻³)² = 4.0×10⁻⁹, which is far smaller than Ksp = 1.7×10⁻⁵. Since Q < Ksp, the solution remains unsaturated and no PbCl₂ precipitates. -
A reaction has K = 5.0×10³ at 298 K. Calculate ΔG° for the reaction.
Show the answer
Answer. ΔG° ≈ −21.1 kJ/mol
Use $\Delta G^\circ = -RT\ln K$ with R = 8.314 J/(mol·K), T = 298 K, and lnK = ln(5000) ≈ 8.517. ΔG° = −(8.314)(298)(8.517) ≈ −21,100 J/mol ≈ −21.1 kJ/mol. The large positive K corresponds to a substantially negative ΔG°, meaning products are strongly favored at standard conditions. -
A reaction mixture has a reaction quotient Q that is greater than its equilibrium constant K. What can be said about ΔG for the reaction as written, and which direction is spontaneous?
Show the answer
Answer. ΔG is positive for the forward direction as written, so the reverse reaction (toward reactants) is spontaneous until Q decreases to equal K.
From ΔG = ΔG° + RTlnQ (equivalently, using ΔG = RTln(Q/K)), when Q > K, ΔG is positive for the forward reaction, meaning the system must shift in reverse to reach equilibrium, where Q equals K and ΔG becomes zero.
What people get wrong
- Including pure solids or liquids in the K or Ksp expression. Instead, cross them out immediately after writing the balanced equation — only gases and aqueous species appear.
- Assuming the small-x approximation always works. Instead, check whether x is under about 5% of the initial concentration after solving; if not, redo the problem with the quadratic formula.
- Confusing Q with K by plugging in initial concentrations when the problem actually wants equilibrium concentrations, or vice versa. Instead, always identify whether the given concentrations are at equilibrium before deciding whether you're computing Q or K.
- Treating a catalyst or an inert gas added at constant volume as a stress that shifts equilibrium. Instead, remember a catalyst changes only the rate to reach equilibrium, and an inert gas at constant volume doesn't change any partial pressure of the reacting species, so K and the position of equilibrium are unaffected.
- Forgetting that Ksp expressions require raising each ion concentration to its stoichiometric coefficient, not just multiplying them. Instead, write out the dissociation equation first and match exponents to coefficients before calculating molar solubility.
- Mixing up ΔG° (fixed value tied to K under standard conditions) with ΔG (variable, depends on Q at a given moment). Instead, use ΔG° = −RTlnK only for the standard-state constant, and ΔG = ΔG° + RTlnQ to judge spontaneity at any other composition.
Drill this unit until it sticks
These questions come back on a schedule built from what you get wrong, alongside the rest of AP Chemistry. Free, and no account needed to start.