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AP Chemistry

AP Chemistry Help: Reaction Quotient And Equilibrium Constant

Review real example questions for Reaction Quotient And Equilibrium Constant in AP Chemistry.

Question 1

For the reversible reaction CaCO3(s)⇌CaO(s)+CO2(g)\mathrm{CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)}CaCO3​(s)⇌CaO(s)+CO2​(g) at a given temperature, Kc=0.20K_c = 0.20Kc​=0.20. A sealed container initially has [CO2]=0.50 M[\mathrm{CO_2}] = 0.50\,\mathrm{M}[CO2​]=0.50M in the presence of solid CaCO3\mathrm{CaCO_3}CaCO3​ and solid CaO\mathrm{CaO}CaO, and the system is not at equilibrium. (Solids are not included in QcQ_cQc​.) Based on comparing QcQ_cQc​ and KcK_cKc​, in which direction will the reaction proceed to reach equilibrium?

  1. The reaction will proceed toward products.
  2. The reaction will proceed toward reactants.
  3. The system is already at equilibrium.
  4. The reaction will proceed toward products because Qc<KcQ_c < K_cQc​<Kc​.
  5. The reaction will proceed toward reactants because Qc=KcQ_c = K_cQc​=Kc​.
Explanation: This question tests the skill of reaction quotient and equilibrium constant. To determine the direction, calculate Q_c = [CO2] = 0.50 (solids omitted). Since Q_c = 0.50 is greater than K_c = 0.20, the reaction will proceed toward reactants to decrease the value of Q until it equals K. When Q is greater than K, the mixture has more products relative to reactants than at equilibrium, so the reverse reaction is favored to consume products. A common misconception is thinking that Q > K means the reaction proceeds toward products, but this is incorrect as the system needs to reduce products to reach equilibrium. Always compare Q to K first; the reaction proceeds in the direction that moves Q toward K.

Question 2

For the reversible reaction 2SO2(g)+O2(g)⇌2SO3(g)\mathrm{2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)}2SO2​(g)+O2​(g)⇌2SO3​(g) at a given temperature, Kc=1.0×102K_c = 1.0\times 10^2Kc​=1.0×102. A mixture is prepared with initial concentrations [SO2]=0.10 M[\mathrm{SO_2}] = 0.10\,\mathrm{M}[SO2​]=0.10M, [O2]=0.10 M[\mathrm{O_2}] = 0.10\,\mathrm{M}[O2​]=0.10M, and [SO3]=1.0 M[\mathrm{SO_3}] = 1.0\,\mathrm{M}[SO3​]=1.0M, and the system is not at equilibrium. Based on comparing QcQ_cQc​ and KcK_cKc​, in which direction will the reaction proceed to reach equilibrium?

  1. The reaction will proceed toward reactants.
  2. The reaction will proceed toward products.
  3. The reaction will proceed toward reactants because Qc=KcQ_c = K_cQc​=Kc​.
  4. The reaction will proceed toward products because Qc<KcQ_c < K_cQc​<Kc​.
  5. The system is already at equilibrium.
Explanation: This question tests the skill of reaction quotient and equilibrium constant. To determine the direction, calculate Q_c = [SO3]^2 / ([SO2]^2 [O2]) = (1.0)^2 / (0.10^2 × 0.10) = 1000. Since Q_c = 1000 is greater than K_c = 100, the reaction will proceed toward reactants to decrease the value of Q until it equals K. When Q is greater than K, the mixture has more products relative to reactants than at equilibrium, so the reverse reaction is favored to consume products. A common misconception is thinking that Q > K means the reaction proceeds toward products, but this is incorrect as the system needs to reduce products to reach equilibrium. Always compare Q to K first; the reaction proceeds in the direction that moves Q toward K.

Question 3

For the reversible reaction 2H2O2(aq)⇌2H2O(l)+O2(g)\mathrm{2H_2O_2(aq) \rightleftharpoons 2H_2O(l) + O_2(g)}2H2​O2​(aq)⇌2H2​O(l)+O2​(g) at a given temperature, Kc=2.0×10−3K_c = 2.0\times 10^{-3}Kc​=2.0×10−3. A mixture is prepared with initial concentrations [H2O2]=0.10 M[\mathrm{H_2O_2}] = 0.10\,\mathrm{M}[H2​O2​]=0.10M and [O2]=1.0×10−3 M[\mathrm{O_2}] = 1.0\times 10^{-3}\,\mathrm{M}[O2​]=1.0×10−3M (liquid water is not included in QcQ_cQc​), and the system is not at equilibrium. Based on comparing QcQ_cQc​ and KcK_cKc​, in which direction will the reaction proceed to reach equilibrium?

  1. The reaction will proceed toward products.
  2. The reaction will proceed toward reactants.
  3. The system is already at equilibrium.
  4. The reaction will proceed toward reactants because Qc>KcQ_c > K_cQc​>Kc​.
  5. The reaction will proceed toward products because Qc=KcQ_c = K_cQc​=Kc​.
Explanation: This question tests the skill of reaction quotient and equilibrium constant. To determine the direction, calculate Q_c = [O2] / [H2O2]^2 = 1.0×10^{-3} / (0.10)^2 = 0.10. Since Q_c = 0.10 is greater than K_c = 2.0×10^{-3}, the reaction will proceed toward reactants to decrease the value of Q until it equals K. When Q is greater than K, the mixture has more products relative to reactants than at equilibrium, so the reverse reaction is favored to consume products. A common misconception is thinking that Q > K means the reaction proceeds toward products, but this is incorrect as the system needs to reduce products to reach equilibrium. Always compare Q to K first; the reaction proceeds in the direction that moves Q toward K.

Question 4

At a certain temperature, the reversible reaction 2NO2(g)⇌N2O4(g)\mathrm{2NO_2(g) \rightleftharpoons N_2O_4(g)}2NO2​(g)⇌N2​O4​(g) has Kc=6.0K_c = 6.0Kc​=6.0. A mixture is prepared such that [NO2]=0.30 M[\mathrm{NO_2}] = 0.30\,\mathrm{M}[NO2​]=0.30M and [N2O4]=0.90 M[\mathrm{N_2O_4}] = 0.90\,\mathrm{M}[N2​O4​]=0.90M, so the system is not initially at equilibrium. Based on comparing QcQ_cQc​ (calculated from the given concentrations) and KcK_cKc​, in which direction will the reaction proceed to reach equilibrium?

  1. the reaction will proceed toward products
  2. the reaction will proceed toward reactants
  3. the system is already at equilibrium
  4. the reaction will proceed toward products until the reactants are used up
  5. the reaction will proceed toward reactants until the concentrations of all species are equal
Explanation: This question tests the skill of reaction quotient and equilibrium constant. The reaction quotient Q_c is calculated using the initial concentrations in the same form as the equilibrium constant K_c, which for this reaction is Q_c = [N_2O_4] / [NO_2]^2 = 0.90 / (0.30)^2 = 10. Here, Q_c = 10 is greater than K_c = 6.0, indicating that there are more products relative to reactants than at equilibrium. Therefore, to reach equilibrium, the reaction will shift toward the reactants to decrease Q_c until it equals K_c. A common misconception is thinking that Q_c > K_c means the reaction proceeds toward products, but actually, it shifts to consume products and form reactants. Always compare Q to K first; the reaction proceeds in the direction that moves Q toward K.

Question 5

For the reversible reaction CO(g)+Cl2(g)⇌COCl2(g)\mathrm{CO(g) + Cl_2(g) \rightleftharpoons COCl_2(g)}CO(g)+Cl2​(g)⇌COCl2​(g) at a certain temperature, Kc=0.25K_c = 0.25Kc​=0.25. A mixture is prepared with [CO]=0.40 M[\mathrm{CO}] = 0.40\,\mathrm{M}[CO]=0.40M, [Cl2]=0.40 M[\mathrm{Cl_2}] = 0.40\,\mathrm{M}[Cl2​]=0.40M, and [COCl2]=0.020 M[\mathrm{COCl_2}] = 0.020\,\mathrm{M}[COCl2​]=0.020M, so the system is not at equilibrium initially. Using Qc=[COCl2][CO][Cl2]Q_c = \dfrac{[\mathrm{COCl_2}]}{[\mathrm{CO}][\mathrm{Cl_2}]}Qc​=[CO][Cl2​][COCl2​]​ and comparing QcQ_cQc​ to KcK_cKc​, in which direction will the reaction proceed to reach equilibrium?

  1. the reaction will proceed toward products
  2. the system is already at equilibrium
  3. the reaction will proceed toward reactants because Qc<KcQ_c < K_cQc​<Kc​
  4. the reaction will proceed toward products until all reactants are used up
  5. the reaction will proceed toward reactants
Explanation: This question tests understanding of reaction quotient and equilibrium constant relationships. Calculate Qc = [COCl₂]/([CO][Cl₂]) = 0.020/(0.40 × 0.40) = 0.020/0.16 = 0.125. Since Q < K (0.125 < 0.25), the system has too few products compared to equilibrium conditions. The reaction must proceed forward (toward products) to increase Q until it equals K. Choice D incorrectly reverses the logic, claiming the reaction goes backward when Q < K. The key strategy is to calculate Q using current concentrations, compare to K, and remember: Q < K means forward reaction, Q > K means reverse reaction.

Question 6

For the reversible reaction A(g)+B(g)⇌C(g)\mathrm{A(g) + B(g) \rightleftharpoons C(g)}A(g)+B(g)⇌C(g) at a certain temperature, Kc=2.0K_c = 2.0Kc​=2.0. A reaction mixture is prepared with initial concentrations [A]=1.0 M[\mathrm{A}]=1.0\,\mathrm{M}[A]=1.0M, [B]=1.0 M[\mathrm{B}]=1.0\,\mathrm{M}[B]=1.0M, and [C]=0.50 M[\mathrm{C}]=0.50\,\mathrm{M}[C]=0.50M, and the system is not at equilibrium. Based on comparing QcQ_cQc​ to KcK_cKc​, in which direction will the reaction proceed to reach equilibrium?

  1. the reaction will proceed toward reactants
  2. the reaction will proceed toward products
  3. the system is already at equilibrium
  4. the reaction will proceed toward reactants because the reactant concentrations are equal
  5. the reaction will proceed toward products until all reactants are consumed
Explanation: This question tests understanding of reaction quotient and equilibrium constant. To determine the direction of reaction, we calculate Q_c = [C]/([A][B]) = 0.50/((1.0)(1.0)) = 0.50/1.0 = 0.50. Since Q_c (0.50) < K_c (2.0), the system has too little product relative to equilibrium, so the reaction must shift toward products to increase Q_c until it equals K_c. Choice E incorrectly suggests the reaction will consume all reactants, but equilibrium reactions never go to completion - they reach a balance where both reactants and products are present. The key strategy is to calculate Q, compare it to K, and remember that when Q < K, the reaction shifts right (toward products) to reach equilibrium.

Question 7

For the reversible reaction CO(g)+H2O(g)⇌CO2(g)+H2(g)\mathrm{CO(g) + H_2O(g) \rightleftharpoons CO_2(g) + H_2(g)}CO(g)+H2​O(g)⇌CO2​(g)+H2​(g) at a given temperature, Kc=1.0K_c = 1.0Kc​=1.0. A mixture is prepared with initial concentrations [CO]=0.50 M[\mathrm{CO}] = 0.50\,\mathrm{M}[CO]=0.50M, [H2O]=0.50 M[\mathrm{H_2O}] = 0.50\,\mathrm{M}[H2​O]=0.50M, [CO2]=0.10 M[\mathrm{CO_2}] = 0.10\,\mathrm{M}[CO2​]=0.10M, and [H2]=0.10 M[\mathrm{H_2}] = 0.10\,\mathrm{M}[H2​]=0.10M, and the system is not at equilibrium. Based on comparing QcQ_cQc​ and KcK_cKc​, in which direction will the reaction proceed to reach equilibrium?

  1. The reaction will proceed toward products.
  2. The reaction will proceed toward reactants.
  3. The system is already at equilibrium.
  4. The reaction will proceed toward reactants because Qc<KcQ_c < K_cQc​<Kc​.
  5. The reaction will proceed toward products because Qc=KcQ_c = K_cQc​=Kc​.
Explanation: This question tests the skill of reaction quotient and equilibrium constant. To determine the direction, calculate Q_c = [CO2][H2] / ([CO][H2O]) = (0.10 × 0.10) / (0.50 × 0.50) = 0.04. Since Q_c = 0.04 is less than K_c = 1.0, the reaction will proceed toward products to increase the value of Q until it equals K. When Q is less than K, the mixture has fewer products relative to reactants than at equilibrium, so the forward reaction is favored to produce more products. A common misconception is believing that Q < K means the reaction proceeds toward reactants, but this is incorrect because the system actually needs to generate more products to approach equilibrium. Always compare Q to K first; the reaction proceeds in the direction that moves Q toward K.

Question 8

For the reversible reaction N2(g)+3H2(g)⇌2NH3(g)\mathrm{N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)}N2​(g)+3H2​(g)⇌2NH3​(g) at a given temperature, Kc=4.0×102K_c = 4.0\times 10^2Kc​=4.0×102. A mixture is prepared with initial concentrations [N2]=0.50 M[\mathrm{N_2}] = 0.50\,\mathrm{M}[N2​]=0.50M, [H2]=0.50 M[\mathrm{H_2}] = 0.50\,\mathrm{M}[H2​]=0.50M, and [NH3]=0.010 M[\mathrm{NH_3}] = 0.010\,\mathrm{M}[NH3​]=0.010M, and the system is not at equilibrium. Based on comparing QcQ_cQc​ and KcK_cKc​, in which direction will the reaction proceed to reach equilibrium?

  1. The reaction will proceed toward reactants.
  2. The system is already at equilibrium.
  3. The reaction will proceed toward products.
  4. The reaction will proceed toward reactants because Qc<KcQ_c < K_cQc​<Kc​.
  5. The reaction will proceed toward products because Qc>KcQ_c > K_cQc​>Kc​.
Explanation: This question tests the skill of reaction quotient and equilibrium constant. To determine the direction, calculate Q_c = [NH3]^2 / ([N2][H2]^3) = (0.010)^2 / (0.50 × 0.50^3) = 0.0016. Since Q_c = 0.0016 is less than K_c = 400, the reaction will proceed toward products to increase the value of Q until it equals K. When Q is less than K, the mixture has fewer products relative to reactants than at equilibrium, so the forward reaction is favored to produce more products. A common misconception is believing that Q < K means the reaction proceeds toward reactants, but this is incorrect because the system actually needs to generate more products to approach equilibrium. Always compare Q to K first; the reaction proceeds in the direction that moves Q toward K.

Question 9

Consider the reversible reaction 2SO2(g)+O2(g)⇌2SO3(g)\mathrm{2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)}2SO2​(g)+O2​(g)⇌2SO3​(g) at a certain temperature, where Kc=1.0×102K_c = 1.0\times10^2Kc​=1.0×102. A reaction mixture is prepared with Qc=0.50Q_c = 0.50Qc​=0.50, so the system is not initially at equilibrium. Based on the comparison of QcQ_cQc​ and KcK_cKc​, in which direction will the reaction proceed to reach equilibrium?

  1. the system is already at equilibrium
  2. the reaction will proceed toward products until the reactants are used up
  3. the reaction will proceed toward reactants until the concentrations of all species are equal
  4. the reaction will proceed toward reactants
  5. the reaction will proceed toward products
Explanation: This question tests the skill of reaction quotient and equilibrium constant. The reaction quotient Q_c is calculated using the initial concentrations in the same form as the equilibrium constant K_c, which for this reaction is Q_c = [SO_3]^2 / ([SO_2]^2 [O_2]). Here, Q_c = 0.50 is less than K_c = 1.0×10^2, indicating that there are fewer products relative to reactants than at equilibrium. Therefore, to reach equilibrium, the reaction will shift toward the products to increase Q_c until it equals K_c. A common misconception is thinking that Q_c < K_c means the reaction proceeds toward reactants, but actually, it shifts to form more products. Always compare Q to K first; the reaction proceeds in the direction that moves Q toward K.

Question 10

For the reversible reaction CH3COOH(aq)⇌H+(aq)+CH3COO−(aq)\mathrm{CH_3COOH(aq) \rightleftharpoons H^+(aq) + CH_3COO^-(aq)}CH3​COOH(aq)⇌H+(aq)+CH3​COO−(aq) at a given temperature, Kc=1.8×10−5K_c = 1.8\times 10^{-5}Kc​=1.8×10−5. A solution is prepared with initial concentrations [CH3COOH]=0.10 M[\mathrm{CH_3COOH}] = 0.10\,\mathrm{M}[CH3​COOH]=0.10M, [H+]=1.0×10−5 M[\mathrm{H^+}] = 1.0\times 10^{-5}\,\mathrm{M}[H+]=1.0×10−5M, and [CH3COO−]=1.0×10−5 M[\mathrm{CH_3COO^-}] = 1.0\times 10^{-5}\,\mathrm{M}[CH3​COO−]=1.0×10−5M, and the system is not at equilibrium. Based on comparing QcQ_cQc​ and KcK_cKc​, in which direction will the reaction proceed to reach equilibrium?

  1. The reaction will proceed toward reactants.
  2. The reaction will proceed toward products.
  3. The system is already at equilibrium.
  4. The reaction will proceed toward products because Qc>KcQ_c > K_cQc​>Kc​.
  5. The reaction will proceed toward reactants because Qc<KcQ_c < K_cQc​<Kc​.
Explanation: This question tests the skill of reaction quotient and equilibrium constant. To determine the direction, calculate Q_c = [H+][CH3COO-] / [CH3COOH] = (1.0×10^{-5} × 1.0×10^{-5}) / 0.10 = 1.0×10^{-9}. Since Q_c = 1.0×10^{-9} is less than K_c = 1.8×10^{-5}, the reaction will proceed toward products to increase the value of Q until it equals K. When Q is less than K, the mixture has fewer products relative to reactants than at equilibrium, so the forward reaction is favored to produce more products. A common misconception is believing that Q < K means the reaction proceeds toward reactants, but this is incorrect because the system actually needs to generate more products to approach equilibrium. Always compare Q to K first; the reaction proceeds in the direction that moves Q toward K.