AP Chemistry Quiz: Magnitude Of The Equilibrium Constant

What this quiz covers

This quiz focuses on Magnitude Of The Equilibrium Constant, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Chemistry.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. Reactants are favored; mostly $\mathrm{CO}$ and $\mathrm{H_2O}$ are present at equilibrium. (correct answer)
2. Neither side is favored; comparable amounts of all gases are present at equilibrium.
3. The reaction is fast, so reactants dominate at equilibrium.
4. The reaction goes to completion, leaving no products at equilibrium.
5. Products are favored; mostly $\mathrm{CO_2}$ and $\mathrm{H_2}$ are present at equilibrium.

Explanation: This question tests understanding of the magnitude of the equilibrium constant and its relationship to equilibrium position. When K << 1, the equilibrium constant expression K = [CO₂][H₂]/([CO][H₂O]) has a very small value, which occurs when the numerator (products) is much smaller than the denominator (reactants). This means at equilibrium, the concentrations of CO and H₂O are much greater than the concentrations of CO₂ and H₂, so reactants are strongly favored. The equilibrium position lies far to the left, with mostly CO and H₂O present at equilibrium. A common misconception (option C) is confusing reaction rate with equilibrium position - whether a reaction is fast or slow doesn't determine which side is favored at equilibrium. When K << 1, always remember that the equilibrium strongly favors the reactant side of the equation.

Question 2

1. Products are favored; mostly $\mathrm{N_2O_4}$ is present at equilibrium.
2. Neither side is strongly favored; both $\mathrm{NO_2}$ and $\mathrm{N_2O_4}$ are present in significant amounts at equilibrium. (correct answer)
3. Reactants are favored; mostly $\mathrm{NO_2}$ is present at equilibrium.
4. The reaction is slow, so products cannot accumulate at equilibrium.
5. The reaction goes to completion, leaving only one species at equilibrium.

Explanation: This question tests understanding of the magnitude of the equilibrium constant when K ≈ 1. When K ≈ 1 for the dimerization reaction, the equilibrium constant expression K = [N₂O₄]/[NO₂]² equals approximately 1, which means the numerator (product) and denominator (reactant squared) have similar magnitudes. This indicates that at equilibrium, neither NO₂ nor N₂O₄ is strongly favored, and both species are present in significant amounts. The equilibrium position is roughly in the middle, with substantial concentrations of both the monomer and dimer forms. A common misconception (option D) is thinking that reaction rate affects equilibrium position - slow reactions can still accumulate products at equilibrium. When K ≈ 1, remember that the equilibrium mixture contains appreciable amounts of both reactants and products.

Question 3

1. Neither side is favored; reactants and products are present in comparable amounts at equilibrium.
2. The reaction is fast, so $\mathrm{CaCO_3}$ disappears immediately at equilibrium.
3. Products are favored; the equilibrium lies toward $\mathrm{CaO}$ and $\mathrm{CO_2}$. (correct answer)
4. Reactants are favored; the equilibrium lies toward $\mathrm{CaCO_3}$.
5. The reaction goes to completion, leaving no solids at equilibrium.

Explanation: This question tests understanding of the magnitude of the equilibrium constant for heterogeneous equilibria. For this reaction involving solids and gas, K = [CO₂] (solids don't appear in the equilibrium expression). When K >> 1, this means [CO₂] must be very large at equilibrium, indicating that the decomposition of CaCO₃ is strongly favored. The equilibrium position lies far to the right, favoring the products CaO(s) and CO₂(g). At equilibrium, most of the calcium carbonate has decomposed into calcium oxide and carbon dioxide. A common misconception (option B) is thinking that reaction rate determines equilibrium position - a fast reaction doesn't mean products are favored. For heterogeneous equilibria with K >> 1, remember that products are strongly favored, meaning significant decomposition occurs.

Question 4

1. Reactants are favored; mostly $\mathrm{H_2}$ and $\mathrm{I_2}$ are present at equilibrium.
2. The reaction is slow, so reactants remain at equilibrium.
3. Neither side is strongly favored; appreciable amounts of reactants and products are present at equilibrium. (correct answer)
4. Products are favored; mostly $\mathrm{HI}$ is present at equilibrium.
5. The reaction goes to completion, leaving no reactants at equilibrium.

Explanation: This question tests understanding of the magnitude of the equilibrium constant when K ≈ 1. When K ≈ 1, the equilibrium constant expression K = [HI]²/([H₂][I₂]) equals approximately 1, which means the numerator (products) and denominator (reactants) have similar magnitudes. This indicates that at equilibrium, neither reactants nor products are strongly favored, and appreciable amounts of H₂, I₂, and HI are all present. The equilibrium position is roughly in the middle, with significant concentrations of all species. A common misconception (option B) is confusing reaction rate with equilibrium position - whether a reaction is fast or slow doesn't determine the equilibrium concentrations. When K ≈ 1, remember that the equilibrium mixture contains substantial amounts of both reactants and products.

Question 5

1. Reactants are favored; mostly $\mathrm{Fe^{3+}}$ and $\mathrm{SCN^-}$ remain at equilibrium.
2. Neither side is favored; reactants and product are present in comparable amounts at equilibrium.
3. The reaction is fast, so reactants must be favored at equilibrium.
4. Products are favored; mostly $\mathrm{FeSCN^{2+}}$ is present at equilibrium. (correct answer)
5. The reaction goes to completion, leaving no reactants at equilibrium.

Explanation: This question tests understanding of the magnitude of the equilibrium constant for complex ion formation. When K >> 1, the equilibrium constant expression K = [FeSCN²⁺]/([Fe³⁺][SCN⁻]) has a very large value, which occurs when the numerator (product) is much larger than the denominator (reactants). This means at equilibrium, the concentration of the complex ion FeSCN²⁺ is much greater than the concentrations of the free Fe³⁺ and SCN⁻ ions, so product formation is strongly favored. The equilibrium position lies far to the right, with mostly FeSCN²⁺ present at equilibrium. A common misconception (option C) is confusing reaction rate with equilibrium position - a fast reaction doesn't determine which side is favored at equilibrium. When K >> 1 for complex ion formation, remember that the complex ion product dominates the equilibrium mixture.

Question 6

1. Products are favored; the equilibrium mixture contains mostly $\text{PCl}_3(g)$ and $\text{Cl}_2(g)$.
2. Reactants are favored; the equilibrium mixture contains mostly $\text{PCl}_5(g)$. (correct answer)
3. Neither side is favored; all three gases are present in comparable amounts.
4. The reaction is fast; $\text{PCl}_5$ decomposes quickly.
5. The reaction goes to completion; all $\text{PCl}_5$ decomposes at equilibrium.

Explanation: This question tests the ability to interpret the magnitude of the equilibrium constant K in terms of product versus reactant favorability at equilibrium. For the reaction PCl₅(g) ⇌ PCl₃(g) + Cl₂(g), a very small K means the ratio of [PCl₃][Cl₂] to [PCl₅] is low, indicating limited dissociation. This implies reactants are favored, so the equilibrium mixture contains mostly PCl₅(g), with little products. The principle is that K ≪ 1 favors the undissociated form to keep the product low. A tempting distractor is choice A, which claims products are favored, based on the misconception that small K implies decomposition dominance. A transferable strategy is to compare K to 1: if K ≫ 1, products are favored; if K ≪ 1, reactants are favored; if K ≈ 1, neither is strongly favored.

Question 7

1. Products are favored; the equilibrium mixture contains mostly $\text{NO}(g)$ and $\text{O}_2(g)$. (correct answer)
2. The reaction is fast; it reaches equilibrium quickly.
3. Neither side is favored; $\text{NO}_2$, $\text{NO}$, and $\text{O}_2$ are comparable in amount.
4. The reaction goes to completion; only products remain at equilibrium.
5. Reactants are favored; the equilibrium mixture contains mostly $\text{NO}_2(g)$.

Explanation: This question tests the ability to interpret the magnitude of the equilibrium constant K in terms of product versus reactant favorability at equilibrium. For the reaction 2NO₂(g) ⇌ 2NO(g) + O₂(g), a very large K means the ratio of [NO]²[O₂] to [NO₂]² is high, indicating the equilibrium position lies to the right. This implies products are favored, so the equilibrium mixture contains mostly NO(g) and O₂(g), with little NO₂ remaining. The principle is that K ≫ 1 means the forward dissociation is favored to achieve balance. A tempting distractor is choice B, which claims reactants are favored, based on the misconception that large K implies reactant dominance from rapid reverse rate. A transferable strategy is to compare K to 1: if K ≫ 1, products are favored; if K ≪ 1, reactants are favored; if K ≈ 1, neither is strongly favored.

Question 8

1. Reactants are favored; most silver remains as $\text{Ag}^+(aq)$ and $\text{Cl}^-(aq)$.
2. Neither side is favored; comparable amounts of ions and solid are present.
3. Products are favored; most silver and chloride are present as $\text{AgCl}(s)$. (correct answer)
4. The reaction is slow; little precipitate forms because the rate is low.
5. The reaction goes to completion; no ions remain in solution at equilibrium.

Explanation: This question tests the ability to interpret the magnitude of the equilibrium constant K in terms of the equilibrium mixture composition. For the reaction Ag⁺(aq) + Cl⁻(aq) ⇌ AgCl(s), a very large forward K means the ratio of 1 to [Ag⁺][Cl⁻] is high (since solid activity is 1), indicating equilibrium lies to the right. This implies products are favored, so most silver and chloride are present as AgCl(s), with low ion concentrations. The principle is that large K drives precipitation by favoring the solid product. A tempting distractor is choice A, which states reactants are favored, stemming from the misconception that large K means ions remain due to slow reaction rather than equilibrium shift. A transferable strategy is to compare K to 1: if K ≫ 1, products are favored; if K ≪ 1, reactants are favored; if K ≈ 1, neither is strongly favored.

Question 9

1. The reaction is fast; the complex forms instantly.
2. Reactants are favored; most species remain as $\text{Fe}^{3+}$ and $\text{SCN}^-$.
3. Products are favored; most species are present as $\text{FeSCN}^{2+}$. (correct answer)
4. Neither side is favored; comparable amounts of complex and ions are present.
5. The reaction goes to completion; no reactant ions remain at equilibrium.

Explanation: This question tests the ability to interpret the magnitude of the equilibrium constant K in terms of the equilibrium mixture. For the reaction Fe³⁺(aq) + SCN⁻(aq) ⇌ FeSCN²⁺(aq), a very large K means the ratio of [FeSCN²⁺] to [Fe³⁺][SCN⁻] is high, indicating strong complex formation. This implies products are favored, so most species are present as FeSCN²⁺, with low free ion concentrations. The principle is that large K shifts equilibrium toward the complex to satisfy the expression. A tempting distractor is choice B, which states reactants are favored, stemming from the misconception that large K means slow binding rather than position. A transferable strategy is to compare K to 1: if K ≫ 1, products are favored; if K ≪ 1, reactants are favored; if K ≈ 1, neither is strongly favored.

Question 10

1. Products are strongly favored; almost all $\text{Br}_2$ is in the gas phase.
2. Reactants are strongly favored; almost all $\text{Br}_2$ remains liquid.
3. Neither side is strongly favored; both liquid and vapor are significant at equilibrium. (correct answer)
4. The phase change is fast; equilibrium is reached immediately.
5. The phase change goes to completion; only one phase is present at equilibrium.

Explanation: This question tests the ability to interpret the magnitude of the equilibrium constant K in terms of the equilibrium state for a phase change. For the reaction Br₂(l) ⇌ Br₂(g), K ≈ 1 means the ratio of P_{Br₂(g)} to 1 (liquid activity) is near 1, indicating balanced phases. This implies neither side is strongly favored, with both liquid and vapor significant at equilibrium. The principle is that K ≈ 1 means vapor pressure allows substantial amounts of both phases. A tempting distractor is choice A, which states products are strongly favored, stemming from the misconception that K=1 implies full evaporation. A transferable strategy is to compare K to 1: if K ≫ 1, products are favored; if K ≪ 1, reactants are favored; if K ≈ 1, neither is strongly favored.

Question 11

1. Products are strongly favored; the mixture is mostly $\text{SO}_3(g)$.
2. Reactants are strongly favored; the mixture is mostly $\text{SO}_2(g)$ and $\text{O}_2(g)$.
3. Neither side is strongly favored; both reactants and products are significant at equilibrium. (correct answer)
4. The reaction is slow; equilibrium is difficult to reach.
5. The reaction goes to completion; all $\text{SO}_2$ is converted to $\text{SO}_3$.

Explanation: This question tests the ability to interpret the magnitude of the equilibrium constant K in terms of which side is favored at equilibrium. For the reaction SO₂(g) + ½O₂(g) ⇌ SO₃(g), K ≈ 1 means the ratio of [SO₃] to [SO₂][O₂]^(1/2) is near 1, indicating balanced favorability. This implies neither side is strongly favored, with both reactants and products significant at equilibrium. The principle is that K around 1 results in comparable concentrations for equilibrium. A tempting distractor is choice A, which claims products are strongly favored, based on the misconception that K=1 means complete conversion. A transferable strategy is to compare K to 1: if K ≫ 1, products are favored; if K ≪ 1, reactants are favored; if K ≈ 1, neither is strongly favored.

Question 12

1. Neither side is favored; $\text{F}_2$, $\text{H}_2$, and $\text{HF}$ are comparable in amount.
2. Reactants are favored; the equilibrium mixture contains mostly $\text{F}_2$ and $\text{H}_2$.
3. The reaction goes to completion; no reactant gases remain at equilibrium.
4. The reaction is slow; it forms little $\text{HF}$ because the rate is low.
5. Products are favored; the equilibrium mixture contains mostly $\text{HF}(g)$. (correct answer)

Explanation: This question tests the ability to interpret the magnitude of the equilibrium constant K in terms of the equilibrium composition. For the reaction F₂(g) + H₂(g) ⇌ 2HF(g), a very large K means the ratio of [HF]² to [F₂][H₂] is high, indicating strong product formation. This implies products are favored, so the equilibrium mixture contains mostly HF(g), with little reactants. The principle is that K ≫ 1 drives the reaction toward products for balance. A tempting distractor is choice A, which states reactants are favored, arising from the misconception that large K implies reactant dominance. A transferable strategy is to compare K to 1: if K ≫ 1, products are favored; if K ≪ 1, reactants are favored; if K ≈ 1, neither is strongly favored.

Question 13

1. Reactants are favored; mostly $\mathrm{N_2}$ and $\mathrm{H_2}$ are present at equilibrium.
2. Products are favored; mostly $\mathrm{NH_3}$ is present at equilibrium. (correct answer)
3. The reaction is fast, so products must dominate at equilibrium.
4. Neither side is favored; reactants and products are present in comparable amounts at equilibrium.
5. The reaction goes to completion, leaving no reactants at equilibrium.

Explanation: This question tests understanding of the magnitude of the equilibrium constant and its relationship to the equilibrium position. When K >> 1, the equilibrium constant expression K = [NH₃]²/([N₂][H₂]³) must have a large value, which occurs when the numerator (products) is much larger than the denominator (reactants). This means at equilibrium, the concentration of NH₃ is much greater than the concentrations of N₂ and H₂, so products are strongly favored. The equilibrium position lies far to the right, with mostly NH₃ present at equilibrium. A common misconception (option C) is confusing reaction rate with equilibrium position - a fast reaction doesn't determine which side is favored at equilibrium, only how quickly equilibrium is reached. When you see K >> 1, remember that products dominate the equilibrium mixture, while K << 1 means reactants dominate.

Question 14

1. Products are favored; mostly $\mathrm{SO_3}$ is present at equilibrium.
2. The reaction is slow, so reactants dominate at equilibrium.
3. Reactants are favored; mostly $\mathrm{SO_2}$ and $\mathrm{O_2}$ are present at equilibrium. (correct answer)
4. The reaction goes to completion, leaving no products at equilibrium.
5. Neither side is favored; comparable amounts of reactants and products are present at equilibrium.

Explanation: This question tests understanding of the magnitude of the equilibrium constant and its relationship to the equilibrium position. When K << 1, the equilibrium constant expression K = [SO₃]²/([SO₂]²[O₂]) must have a very small value, which occurs when the numerator (products) is much smaller than the denominator (reactants). This means at equilibrium, the concentrations of SO₂ and O₂ are much greater than the concentration of SO₃, so reactants are strongly favored. The equilibrium position lies far to the left, with mostly SO₂ and O₂ present at equilibrium. A common misconception (option B) is confusing reaction rate with equilibrium position - a slow reaction doesn't determine which side is favored at equilibrium, only how quickly equilibrium is reached. When interpreting K values, remember that K << 1 means the equilibrium strongly favors reactants, not products.

Question 15

1. Products are favored; mostly $\mathrm{Cl(g)}$ atoms are present at equilibrium.
2. The reaction is slow, so mostly $\mathrm{Cl_2(g)}$ remains at equilibrium.
3. Reactants are favored; mostly $\mathrm{Cl_2(g)}$ is present at equilibrium. (correct answer)
4. Neither side is favored; comparable amounts of $\mathrm{Cl_2(g)}$ and $\mathrm{Cl(g)}$ are present at equilibrium.
5. The reaction goes to completion, leaving no $\mathrm{Cl_2(g)}$ at equilibrium.

Explanation: This question tests understanding of the magnitude of the equilibrium constant for dissociation reactions. When K << 1 for the dissociation of Cl₂, the equilibrium constant expression K = [Cl]²/[Cl₂] has a very small value, which means the numerator (atomic chlorine squared) is much smaller than the denominator (molecular chlorine). This indicates that at equilibrium, very little Cl₂ dissociates into Cl atoms, and most chlorine remains in its molecular form Cl₂. The equilibrium position lies far to the left, strongly favoring the reactant. A common misconception (option B) is thinking that reaction rate determines equilibrium position - whether dissociation is fast or slow doesn't affect the equilibrium concentrations. For dissociation reactions with K << 1, remember that the molecular form predominates at equilibrium.

Question 16

1. Products are favored; the equilibrium mixture contains mostly $\text{COCl}_2$.
2. The reaction is slow; little product forms because the rate is low.
3. Reactants are favored; the equilibrium mixture contains mostly $\text{CO}$ and $\text{Cl}_2$. (correct answer)
4. The reaction goes to completion; only reactants remain at equilibrium.
5. Neither side is favored; reactants and products are present in comparable amounts.

Explanation: This question tests the ability to interpret the magnitude of the equilibrium constant K in terms of product versus reactant favorability at equilibrium. For the reaction CO(g) + Cl₂(g) ⇌ COCl₂(g), a very small K (K ≪ 1) means the ratio of [COCl₂] to [CO][Cl₂] is low, indicating the equilibrium lies far to the left. This implies reactants are favored, so the equilibrium mixture contains mostly CO and Cl₂, as the reverse reaction predominates to maintain the small K value. The chemical principle is that K < 1 means the system achieves equilibrium with higher reactant concentrations to balance the expression. A tempting distractor is choice A, which claims products are favored, based on the misconception that small K implies slow formation but actually large product amounts. A transferable strategy is to compare K to 1: if K ≫ 1, products are favored; if K ≪ 1, reactants are favored; if K ≈ 1, neither is strongly favored.

Question 17

1. Neither side is strongly favored; both $\text{CO}_2(g)$ and $\text{CO}(g)$ are significant at equilibrium. (correct answer)
2. Reactants are strongly favored; the equilibrium mixture contains mostly $\text{CO}_2(g)$.
3. The reaction goes to completion; all $\text{CO}_2$ is converted to $\text{CO}$.
4. The reaction is slow; comparable amounts occur only if the rate is low.
5. Products are strongly favored; the equilibrium mixture contains mostly $\text{CO}(g)$.

Explanation: This question tests the ability to interpret the magnitude of the equilibrium constant K in terms of product versus reactant favorability at equilibrium. For the reaction C(s) + CO₂(g) ⇌ 2CO(g), K ≈ 1 means the ratio of [CO]² to [CO₂] (solid activity 1) is near 1, indicating balanced conversion. This implies neither side is strongly favored, with both CO₂(g) and CO(g) significant at equilibrium. The principle is that K around 1 results in comparable gas concentrations. A tempting distractor is choice A, which states products are strongly favored, stemming from the misconception that K=1 means full reduction. A transferable strategy is to compare K to 1: if K ≫ 1, products are favored; if K ≪ 1, reactants are favored; if K ≈ 1, neither is strongly favored.

Question 18

1. Neither side is favored; the amounts of $\text{CO}_2(g)$ and $\text{CaCO}_3(s)$ are comparable.
2. The decomposition is fast; $\text{CaCO}_3$ breaks down quickly.
3. Products are favored; the equilibrium state contains mostly $\text{CaO}(s)$ and $\text{CO}_2(g)$.
4. Reactants are favored; the equilibrium state lies mostly toward $\text{CaCO}_3(s)$. (correct answer)
5. The reaction goes to completion; all $\text{CaCO}_3$ decomposes once equilibrium is reached.

Explanation: This question tests the ability to interpret the magnitude of the equilibrium constant K in terms of which side is favored at equilibrium. For the reaction CaCO₃(s) ⇌ CaO(s) + CO₂(g), a very small K means [CO₂] is low at equilibrium since solids have activity of 1, indicating the equilibrium lies to the left. This implies reactants are favored, so the equilibrium state contains mostly CaCO₃(s), with little decomposition occurring. The principle is that K ≪ 1 favors reactants because the reverse reaction is preferred to minimize product formation. A tempting distractor is choice A, which claims products are favored, based on the misconception that small K means fast decomposition rather than position. A transferable strategy is to compare K to 1: if K ≫ 1, products are favored; if K ≪ 1, reactants are favored; if K ≈ 1, neither is strongly favored.

Question 19

1. Neither side is strongly favored; reactants and products are present in comparable amounts. (correct answer)
2. Products are strongly favored; the equilibrium mixture is mostly $\text{HI}$.
3. Reactants are strongly favored; the equilibrium mixture is mostly $\text{H}_2$ and $\text{I}_2$.
4. The reaction is fast; equilibrium is reached almost instantly.
5. The reaction goes to completion; only $\text{HI}$ remains at equilibrium.

Explanation: This question tests the ability to interpret the magnitude of the equilibrium constant K in terms of the composition at equilibrium. For the reaction H₂(g) + I₂(g) ⇌ 2HI(g), K ≈ 1 means the ratio of [HI]² to [H₂][I₂] is close to 1, indicating neither side is strongly favored. This suggests reactants and products are present in comparable amounts, as the forward and reverse rates balance with similar concentrations on both sides. The principle is that when K is around 1, the equilibrium position is in the middle, with significant amounts of all species. A tempting distractor is choice B, which states products are strongly favored, arising from the misconception that K = 1 means complete conversion rather than balance. A transferable strategy is to compare K to 1: if K ≫ 1, products are favored; if K ≪ 1, reactants are favored; if K ≈ 1, neither is strongly favored.

Question 20

1. Products are favored; most acetic acid exists as ions.
2. The reaction is fast; ionization occurs rapidly.
3. Reactants are favored; most acetic acid remains as $\text{CH}_3\text{COOH}(aq)$. (correct answer)
4. Neither side is favored; ions and molecular acid are comparable in amount.
5. The reaction goes to completion; all molecules ionize at equilibrium.

Explanation: This question tests the ability to interpret the magnitude of the equilibrium constant K in terms of the equilibrium composition. For the reaction CH₃COOH(aq) ⇌ H⁺(aq) + CH₃COO⁻(aq), a very small K means the ratio of [H⁺][CH₃COO⁻] to [CH₃COOH] is low, indicating weak acid behavior. This implies reactants are favored, so most acetic acid remains as CH₃COOH(aq), with little ionization. The principle is that K ≪ 1 keeps the equilibrium shifted left, maintaining molecular form. A tempting distractor is choice A, which states products are favored, arising from the misconception that small K means strong ionization due to rarity. A transferable strategy is to compare K to 1: if K ≫ 1, products are favored; if K ≪ 1, reactants are favored; if K ≈ 1, neither is strongly favored.