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This deck focuses on Representations Of Equilibrium, giving you a quick way to review the definitions, rules, and examples that matter most for AP Chemistry.
Study Representations Of Equilibrium in AP Chemistry with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.
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What is the relationship between Kp and Kc for a gas-phase reaction?
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Kp=Kc(RT)Δn, where Δn is the change in moles of gas. Accounts for the pressure difference when converting between concentration and pressure equilibrium constants.
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This deck focuses on Representations Of Equilibrium, giving you a quick way to review the definitions, rules, and examples that matter most for AP Chemistry.
Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.
Answer: Kp=Kc(RT)Δn, where Δn is the change in moles of gas. Accounts for the pressure difference when converting between concentration and pressure equilibrium constants.
Answer: The system is at equilibrium. When the reaction quotient equals the equilibrium constant, rates are balanced.
Answer: M^{-2}. Units are M⋅M3M2=M−2 based on the concentration terms.
Answer: Q=[A]2[B][C]=221×3=0.75. Using Q=[A]2[B][C] with the given concentrations.
Answer: Le Chatelier's Principle. States that systems respond to stress by shifting to counteract the disturbance.
Answer: The reaction shifts left towards reactants. When Q>Kc, there are too many products relative to equilibrium.
Answer: Higher Ksp indicates greater solubility. Larger Ksp values correspond to more soluble compounds at equilibrium.
Answer: M^{-2}. Units are M⋅M3M2=M−2 based on the concentration terms.
Answer: No change in equilibrium position. At constant volume, adding inert gas doesn't change partial pressures of reactants.
Answer: Equilibrium shifts toward the side with fewer moles of gas. System reduces stress by shifting to the side that occupies less volume.
Answer: Shifts equilibrium to the left (toward reactants). Heat is treated as a reactant in endothermic reactions, so removing heat shifts left.
Answer: Products are favored at equilibrium. Large equilibrium constants mean the forward reaction is thermodynamically favorable.
Answer: Neither reactants nor products are favored. Equal concentrations of reactants and products at equilibrium when Keq=1.
Answer: A catalyst does not affect the position of equilibrium. Catalysts only speed up the rate of reaching equilibrium but don't change the final position.
Answer: No effect if H2O is not a reactant or product in gaseous form. Liquid water doesn't affect gas-phase equilibria unless it participates as a gas.
Answer: Equilibrium shifts right, producing more NH3. Adding reactants increases their concentration, driving the reaction forward.
Answer: Shifts towards the side with fewer moles of gas. System responds to pressure increase by reducing the total number of gas molecules.
Answer: Shifts the equilibrium to the left (toward reactants). Heat is treated as a product in exothermic reactions, so adding heat shifts left.
Answer: Q=[A]2[B][C]=221×3=0.75. Using Q=[A]2[B][C] with the given concentrations.
Answer: Pressure changes do not affect Kc.. Only temperature changes affect Kc; pressure and concentration changes don't.
Answer: Kc decreases. For exothermic reactions, higher temperature makes products less stable, decreasing Kc.
Answer: Kc=[H2][I2][HI]2. Products raised to stoichiometric coefficients over reactants raised to their coefficients.
Answer: Solubility decreases due to the common ion effect. Increased concentration of a common ion shifts equilibrium toward the undissolved solid.
Answer: The reaction shifts left towards reactants. When Q>Kc, there are too many products relative to equilibrium.
Answer: Ksp=[Ca2+][F−]2. Products of ion concentrations raised to their stoichiometric coefficients for the dissolution.
Answer: Equilibrium shifts right, producing more products. Increasing reactant concentration makes the forward reaction more favorable.
Answer: Equilibrium shifts right to produce more product. Removing products decreases Q, making the forward reaction more favorable.
Answer: Kp=Kc(RT)Δn, where Δn is the change in moles of gas. Accounts for the pressure difference when converting between concentration and pressure equilibrium constants.
Answer: Products are favored at equilibrium. Large equilibrium constants mean the forward reaction is thermodynamically favorable.
Answer: Shifts towards the side with more moles of dissolved species. Dilution favors the side that produces more particles to maintain equilibrium.
Answer: The reaction favors reactants at equilibrium. Small Kc values mean the equilibrium position lies far to the left.
Answer: Equilibrium shifts toward the side with fewer moles of gas. System reduces stress by shifting to the side that occupies less volume.
Answer: No effect on the equilibrium position. Solid concentrations don't appear in equilibrium expressions and don't affect position.
Answer: Keq>1 indicates a spontaneous reaction. Large equilibrium constants indicate thermodynamically favorable forward reactions.
Answer: Shifts towards the side with fewer moles of gas. System responds to pressure increase by reducing the total number of gas molecules.
Answer: Shifts towards the side with more moles of gas. Increasing volume decreases pressure, favoring the side with more gas molecules.
Answer: No change in equilibrium position. When Δn=0, pressure changes don't affect equilibrium position.
Answer: Shifts left, towards reactants. When Q>Kc, there are excess products, so the reaction proceeds backward.
Answer: Kc=[H2][I2][HI]2. Products raised to stoichiometric coefficients over reactants raised to their coefficients.
Answer: Ksp=[Ca2+][F−]2. Products of ion concentrations raised to their stoichiometric coefficients for the dissolution.
Answer: Neither reactants nor products are favored. Equal concentrations of reactants and products at equilibrium when Keq=1.
Answer: No effect if H2O is not a reactant or product in gaseous form. Liquid water doesn't affect gas-phase equilibria unless it participates as a gas.
Answer: No effect on the equilibrium position. Inert gases don't participate in the reaction and don't change partial pressures.
Answer: A catalyst does not affect the position of equilibrium. Catalysts only speed up the rate of reaching equilibrium but don't change the final position.
Answer: The system is at equilibrium. When the reaction quotient equals the equilibrium constant, rates are balanced.
Answer: Shifts left, towards reactants. When Q>Kc, there are excess products, so the reaction proceeds backward.
Answer: Kp=Kc(RT)Δn, where Δn is the change in moles of gas. Accounts for the pressure difference when converting between concentration and pressure equilibrium constants.
Answer: Shifts equilibrium to the left (toward reactants). Heat is treated as a reactant in endothermic reactions, so removing heat shifts left.
Answer: No effect on the equilibrium position. Inert gases don't participate in the reaction and don't change partial pressures.
Answer: Solubility decreases due to the common ion effect. Increased concentration of a common ion shifts equilibrium toward the undissolved solid.
Answer: No effect on the equilibrium position. Solid concentrations don't appear in equilibrium expressions and don't affect position.
Answer: Kc changes with temperature. Temperature is the only factor that changes the equilibrium constant value.
Answer: Equilibrium shifts right to produce more product. Removing products decreases Q, making the forward reaction more favorable.
Answer: Equilibrium shifts right, producing more products. Increasing reactant concentration makes the forward reaction more favorable.
Answer: Pressure changes do not affect Kc.. Only temperature changes affect Kc; pressure and concentration changes don't.
Answer: Neither reactants nor products are favored. Equal concentrations of reactants and products at equilibrium when Keq=1.
Answer: Higher Ksp indicates greater solubility. Larger Ksp values correspond to more soluble compounds at equilibrium.
Answer: Kc changes with temperature. Temperature is the only factor that changes the equilibrium constant value.
Answer: Equilibrium shifts right to produce more product. Removing products decreases Q, making the forward reaction more favorable.
Answer: No change in equilibrium position. At constant volume, adding inert gas doesn't change partial pressures of reactants.
Answer: Kc changes with temperature. Temperature is the only factor that changes the equilibrium constant value.
Answer: Equilibrium shifts right to produce more product. Removing products decreases Q, making the forward reaction more favorable.
Answer: Shifts the equilibrium to the left (toward reactants). Heat is treated as a product in exothermic reactions, so adding heat shifts left.
Answer: Equilibrium shifts toward the side with fewer moles of gas. System reduces stress by shifting to the side that occupies less volume.
Answer: Ksp=[Ca2+][F−]2. Products of ion concentrations raised to their stoichiometric coefficients for the dissolution.
Answer: The system is at equilibrium. When the reaction quotient equals the equilibrium constant, rates are balanced.
Answer: Solubility decreases due to the common ion effect. Increased concentration of a common ion shifts equilibrium toward the undissolved solid.
Answer: Shifts towards the side with more moles of gas. Increasing volume decreases pressure, favoring the side with more gas molecules.
Answer: Kc changes with temperature. Temperature is the only factor that changes the equilibrium constant value.
Answer: Le Chatelier's Principle. States that systems respond to stress by shifting to counteract the disturbance.
Answer: No effect on the equilibrium position. Inert gases don't participate in the reaction and don't change partial pressures.
Answer: Shifts equilibrium to the left (toward reactants). Heat is treated as a reactant in endothermic reactions, so removing heat shifts left.
Answer: Ksp=[Ca2+][F−]2. Products of ion concentrations raised to their stoichiometric coefficients for the dissolution.
Answer: Equilibrium shifts right, producing more products. Increasing reactant concentration makes the forward reaction more favorable.
Answer: M^{-2}. Units are M⋅M3M2=M−2 based on the concentration terms.
Answer: Higher Ksp indicates greater solubility. Larger Ksp values correspond to more soluble compounds at equilibrium.
Answer: The system is at equilibrium. When the reaction quotient equals the equilibrium constant, rates are balanced.
Answer: Pressure changes do not affect Kc.. Only temperature changes affect Kc; pressure and concentration changes don't.
Answer: Equilibrium shifts toward the side with fewer moles of gas. System reduces stress by shifting to the side that occupies less volume.
Answer: Kc decreases. For exothermic reactions, higher temperature makes products less stable, decreasing Kc.
Answer: Equilibrium shifts right, producing more NH3. Adding reactants increases their concentration, driving the reaction forward.
Answer: The reaction favors reactants at equilibrium. Small Kc values mean the equilibrium position lies far to the left.
Answer: Solubility decreases due to the common ion effect. Increased concentration of a common ion shifts equilibrium toward the undissolved solid.
Answer: Keq>1 indicates a spontaneous reaction. Large equilibrium constants indicate thermodynamically favorable forward reactions.
Answer: Kp=Kc(RT)Δn, where Δn is the change in moles of gas. Accounts for the pressure difference when converting between concentration and pressure equilibrium constants.
Answer: Shifts towards the side with more moles of dissolved species. Dilution favors the side that produces more particles to maintain equilibrium.
Answer: Equilibrium shifts right, producing more NH3. Adding reactants increases their concentration, driving the reaction forward.
Answer: M^{-2}. Units are M⋅M3M2=M−2 based on the concentration terms.
Answer: Pressure changes do not affect Kc.. Only temperature changes affect Kc; pressure and concentration changes don't.
Answer: Shifts towards the side with more moles of dissolved species. Dilution favors the side that produces more particles to maintain equilibrium.
Answer: Shifts towards the side with more moles of dissolved species. Dilution favors the side that produces more particles to maintain equilibrium.
Answer: Shifts equilibrium to the left (toward reactants). Heat is treated as a reactant in endothermic reactions, so removing heat shifts left.
Answer: No effect on the equilibrium position. Inert gases don't participate in the reaction and don't change partial pressures.
Answer: The reaction favors reactants at equilibrium. Small Kc values mean the equilibrium position lies far to the left.
Answer: Equilibrium shifts right, producing more products. Increasing reactant concentration makes the forward reaction more favorable.
Answer: The reaction favors reactants at equilibrium. Small Kc values mean the equilibrium position lies far to the left.