Magnitude of the Equilibrium Constant
Help Questions
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?
The reaction goes to completion, leaving no products at equilibrium.
Neither side is favored; comparable amounts of all gases 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.
Reactants are favored; mostly $\mathrm{CO}$ and $\mathrm{H_2O}$ 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.
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.
Neither side is strongly favored; both $\mathrm{NO_2}$ and $\mathrm{N_2O_4}$ are present in significant amounts at equilibrium.
Products are favored; mostly $\mathrm{N_2O_4}$ is present at equilibrium.
Reactants are favored; mostly $\mathrm{NO_2}$ is present at equilibrium.
The reaction is slow, so products cannot accumulate 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 goes to completion, leaving no solids at equilibrium.
Reactants are favored; the equilibrium lies toward $\mathrm{CaCO_3}$.
Products are favored; the equilibrium lies toward $\mathrm{CaO}$ and $\mathrm{CO_2}$.
The reaction is fast, so $\mathrm{CaCO_3}$ disappears immediately at equilibrium.
Neither side is favored; reactants and products are present in comparable amounts 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.
For the gas-phase reaction $\mathrm{N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)}$ at a certain temperature, the equilibrium constant is reported to be very large ($K \gg 1$). What does this imply about the mixture at equilibrium?
The reaction goes to completion, leaving no reactants at equilibrium.
Products are favored; mostly $\mathrm{NH_3}$ is present at equilibrium.
Reactants are favored; mostly $\mathrm{N_2}$ and $\mathrm{H_2}$ are present at equilibrium.
Neither side is favored; reactants and products are present in comparable amounts at equilibrium.
The reaction is fast, so products must dominate 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.
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.
The reaction is slow, so reactants remain at equilibrium.
Neither side is strongly favored; appreciable amounts of reactants and products are present 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?
Reactants are favored; mostly $\mathrm{Fe^{3+}}$ and $\mathrm{SCN^-}$ remain at equilibrium.
Neither side is favored; reactants and product are present in comparable amounts at equilibrium.
Products are favored; mostly $\mathrm{FeSCN^{2+}}$ is present at equilibrium.
The reaction goes to completion, leaving no reactants at equilibrium.
The reaction is fast, so reactants must be favored 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 $\mathrm{2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)}$ at a certain temperature, the equilibrium constant is very small ($K \ll 1$). What does this indicate about the composition at equilibrium?
The reaction is slow, so reactants dominate at equilibrium.
The reaction goes to completion, leaving no products at equilibrium.
Reactants are favored; mostly $\mathrm{SO_2}$ and $\mathrm{O_2}$ are present at equilibrium.
Neither side is favored; comparable amounts of reactants and products are present at equilibrium.
Products are favored; mostly $\mathrm{SO_3}$ is 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.
For the reaction $\mathrm{Cl_2(g) \rightleftharpoons 2Cl(g)}$ at a certain temperature, the equilibrium constant is very small ($K \ll 1$). What does this imply about the equilibrium mixture?
The reaction is slow, so mostly $\mathrm{Cl_2(g)}$ remains at equilibrium.
The reaction goes to completion, leaving no $\mathrm{Cl_2(g)}$ at equilibrium.
Products are favored; mostly $\mathrm{Cl(g)}$ atoms are present at equilibrium.
Reactants are favored; mostly $\mathrm{Cl_2(g)}$ is present at equilibrium.
Neither side is favored; comparable amounts of $\mathrm{Cl_2(g)}$ and $\mathrm{Cl(g)}$ are present 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.
For the reaction $\text{F}_2(g)+\text{H}_2(g)\rightleftharpoons 2\text{HF}(g)$ at a certain temperature, the equilibrium constant is very large. What does this indicate about the equilibrium composition?
The reaction is slow; it forms little $\text{HF}$ because the rate is low.
Reactants are favored; the equilibrium mixture contains mostly $\text{F}_2$ and $\text{H}_2$.
Products are favored; the equilibrium mixture contains mostly $\text{HF}(g)$.
The reaction goes to completion; no reactant gases remain at equilibrium.
Neither side is favored; $\text{F}_2$, $\text{H}_2$, and $\text{HF}$ are comparable in amount.
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.
For the reaction $\text{CO}(g)+\text{Cl}_2(g)\rightleftharpoons \text{COCl}_2(g)$ at a certain temperature, the equilibrium constant is very small ($K\ll 1$). What does this magnitude of $K$ imply about product vs. reactant favorability at equilibrium?
Neither side is favored; reactants and products are present in comparable amounts.
The reaction goes to completion; only reactants remain at equilibrium.
Reactants are favored; the equilibrium mixture contains mostly $\text{CO}$ and $\text{Cl}_2$.
Products are favored; the equilibrium mixture contains mostly $\text{COCl}_2$.
The reaction is slow; little product forms because the rate is low.
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.