Magnitude of the Equilibrium Constant
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AP Chemistry › Magnitude of the Equilibrium Constant
At a given temperature, the equilibrium constant for $\mathrm{CO(g) + H_2O(g) \rightleftharpoons CO_2(g) + H_2(g)}$ is very small ($K \ll 1$). What does this say about the equilibrium mixture?
Reactants are favored; mostly $\mathrm{CO}$ and $\mathrm{H_2O}$ are present at equilibrium.
Products are favored; mostly $\mathrm{CO_2}$ and $\mathrm{H_2}$ are present at equilibrium.
The reaction is fast, so reactants dominate at equilibrium.
Neither side is favored; comparable amounts of all gases are present at equilibrium.
The reaction goes to completion, leaving no products 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.
For the reaction $\mathrm{2NO_2(g) \rightleftharpoons N_2O_4(g)}$ at a certain temperature, the equilibrium constant is approximately 1 ($K \approx 1$). What does this imply about product versus reactant favorability at equilibrium?
The reaction goes to completion, leaving only one species at equilibrium.
The reaction is slow, so products cannot accumulate at equilibrium.
Products are favored; mostly $\mathrm{N_2O_4}$ is present at equilibrium.
Neither side is strongly favored; both $\mathrm{NO_2}$ and $\mathrm{N_2O_4}$ are present in significant amounts at equilibrium.
Reactants are favored; mostly $\mathrm{NO_2}$ is present 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.
At a certain temperature, the equilibrium constant for $\mathrm{CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)}$ is very large ($K \gg 1$). What does this indicate about which side is favored at equilibrium?
The reaction is fast, so $\mathrm{CaCO_3}$ disappears immediately at equilibrium.
Reactants are favored; the equilibrium lies toward $\mathrm{CaCO_3}$.
Neither side is favored; reactants and products are present in comparable amounts at equilibrium.
The reaction goes to completion, leaving no solids at equilibrium.
Products are favored; the equilibrium lies toward $\mathrm{CaO}$ and $\mathrm{CO_2}$.
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.
For the reaction $\mathrm{H_2(g) + I_2(g) \rightleftharpoons 2HI(g)}$ at a certain temperature, the equilibrium constant is approximately 1 ($K \approx 1$). What does this suggest about the system at equilibrium?
The reaction goes to completion, leaving no reactants at equilibrium.
Reactants are favored; mostly $\mathrm{H_2}$ and $\mathrm{I_2}$ are present at equilibrium.
Neither side is strongly favored; appreciable amounts of reactants and products are present at equilibrium.
The reaction is slow, so reactants remain at equilibrium.
Products are favored; mostly $\mathrm{HI}$ is present 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.
At a certain temperature, the equilibrium constant for $\mathrm{Fe^{3+}(aq) + SCN^-(aq) \rightleftharpoons FeSCN^{2+}(aq)}$ is very large ($K \gg 1$). What does this indicate about the equilibrium position?
Neither side is favored; reactants and product are present in comparable amounts at equilibrium.
Reactants are favored; mostly $\mathrm{Fe^{3+}}$ and $\mathrm{SCN^-}$ remain at equilibrium.
The reaction is fast, so reactants must be favored at equilibrium.
The reaction goes to completion, leaving no reactants at equilibrium.
Products are favored; mostly $\mathrm{FeSCN^{2+}}$ is present 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.
For the reaction $\text{PCl}_5(g)\rightleftharpoons \text{PCl}_3(g)+\text{Cl}_2(g)$ at a certain temperature, $K$ is very small. What does this imply about product vs. reactant favorability at equilibrium?
Neither side is favored; all three gases are present in comparable amounts.
Reactants are favored; the equilibrium mixture contains mostly $\text{PCl}_5(g)$.
Products are favored; the equilibrium mixture contains mostly $\text{PCl}_3(g)$ and $\text{Cl}_2(g)$.
The reaction goes to completion; all $\text{PCl}_5$ decomposes at equilibrium.
The reaction is fast; $\text{PCl}_5$ decomposes quickly.
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.
For the reaction $2\text{NO}_2(g)\rightleftharpoons 2\text{NO}(g)+\text{O}_2(g)$ at a given temperature, $K$ is very large. What does this indicate about product vs. reactant favorability at equilibrium?
The reaction is fast; it reaches equilibrium quickly.
Products are favored; the equilibrium mixture contains mostly $\text{NO}(g)$ and $\text{O}_2(g)$.
Reactants are favored; the equilibrium mixture contains mostly $\text{NO}_2(g)$.
The reaction goes to completion; only products remain at equilibrium.
Neither side is favored; $\text{NO}_2$, $\text{NO}$, and $\text{O}_2$ are comparable in amount.
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.
For the aqueous reaction $\text{Ag}^+(aq)+\text{Cl}^-(aq)\rightleftharpoons \text{AgCl}(s)$ at a certain temperature, the equilibrium constant for the forward reaction is very large. What does this imply about the equilibrium mixture?
Neither side is favored; comparable amounts of ions and solid are present.
Products are favored; most silver and chloride are present as $\text{AgCl}(s)$.
The reaction is slow; little precipitate forms because the rate is low.
Reactants are favored; most silver remains as $\text{Ag}^+(aq)$ and $\text{Cl}^-(aq)$.
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.
For the reaction $\text{Fe}^{3+}(aq)+\text{SCN}^-(aq)\rightleftharpoons \text{FeSCN}^{2+}(aq)$ at a given temperature, the equilibrium constant is very large. What does this indicate about the equilibrium mixture?
Products are favored; most species are present as $\text{FeSCN}^{2+}$.
Reactants are favored; most species remain as $\text{Fe}^{3+}$ and $\text{SCN}^-$.
Neither side is favored; comparable amounts of complex and ions are present.
The reaction goes to completion; no reactant ions remain at equilibrium.
The reaction is fast; the complex forms instantly.
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.
For the reaction $\text{Br}_2(l)\rightleftharpoons \text{Br}_2(g)$ in a closed container at a certain temperature, $K\approx 1$ for the phase change as written. What does this imply about the equilibrium state?
The phase change is fast; equilibrium is reached immediately.
The phase change goes to completion; only one phase is present at equilibrium.
Neither side is strongly favored; both liquid and vapor are significant at equilibrium.
Reactants are strongly favored; almost all $\text{Br}_2$ remains liquid.
Products are strongly favored; almost all $\text{Br}_2$ is in the gas phase.
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.