By Fatskills Exam Guides Team — the exam nerds behind 28,500+ quizzes and 2.1M practice questions across 500+ global exams.
Dynamic equilibrium is the state where the rate of the forward reaction equals the rate of the reverse reaction, so the concentrations of reactants and products stay constant (but not necessarily equal). This concept is essential on the AP exam—it appears in multiple-choice questions, FRQs, and lab-based questions. A real-world example: carbonated soda. When you open a bottle, CO? escapes (forward reaction), but some CO? also dissolves back into the liquid (reverse reaction). At equilibrium, the rate of CO? escaping equals the rate of it dissolving, keeping the fizz level stable (until you open the bottle and disturb it!).
Mistake: Including solids/liquids in K expressions. Correction: Only gases and aqueous solutions appear in Kc/Kp. Solids/liquids have constant concentrations and are omitted.
Mistake: Confusing Kc and Kp units. Correction: Kc and Kp are unitless (technically, activities are used, but AP ignores this). Just write the ratio.
Mistake: Assuming K changes with concentration/pressure. Correction: Only temperature changes K. Concentration/pressure shifts the equilibrium position but doesn’t alter K.
Mistake: Forgetting to raise concentrations to their coefficients in K expressions. Correction: For 2NO?-N?O?, Kc = [N?O?] / [NO?]² (not [NO?]).
Mistake: Misapplying the quadratic formula in ICE tables. Correction: If K is small (e.g., K < 10), assume x is negligible compared to initial concentrations (e.g., 1.0 - x-1.0).
For the reaction: 2SO?(g) + O?(g)-2SO?(g) (Kc = 4.0 at 25°C), which change will increase the amount of SO? at equilibrium? (A) Adding a catalyst (B) Increasing the volume of the container (C) Adding more O? (D) Increasing the temperature (assuming the reaction is exothermic)
Answer: (C) Adding more O? Explanation: Adding O? shifts the equilibrium right (toward products) to consume the excess reactant. Catalysts (A) don’t affect equilibrium, increasing volume (B) shifts to the side with more gas moles (left), and increasing temperature (D) shifts left for exothermic reactions.
The reaction CO(g) + H?O(g)-CO?(g) + H?(g) has Kc = 5.0 at 800 K. (a) Write the equilibrium expression for Kc. (b) If [CO] = 0.10 M, [H?O] = 0.10 M, [CO?] = 0.20 M, and [H?] = 0.20 M, is the system at equilibrium? If not, which way will it shift?
Answer: (a) Kc = [CO?][H?] / [CO][H?O] (b) Q = (0.20)(0.20) / (0.10)(0.10) = 4.0. Since Q < K (4.0 < 5.0), the system is not at equilibrium and will shift right to reach equilibrium.
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