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AP Chemistry

AP Chemistry Help: Reaction Quotient And Le Chateliers Principle

Review real example questions for Reaction Quotient And Le Chateliers Principle in AP Chemistry.

Question 1

In a sealed flask at constant temperature, the equilibrium 2NO(g)+O2(g)⇌2NO2(g)\mathrm{2NO(g) + O_2(g) \rightleftharpoons 2NO_2(g)}2NO(g)+O2​(g)⇌2NO2​(g) has Kp=6.0×101K_p = 6.0\times10^1Kp​=6.0×101. The system is initially at equilibrium. A small amount of O2(g)\mathrm{O_2(g)}O2​(g) is removed, and immediately afterward QpQ_pQp​ is found to be 9.0×1019.0\times10^19.0×101. As equilibrium is reestablished, what shift occurs?

  1. Shift toward products (net formation of NO2\mathrm{NO_2}NO2​)
  2. No net shift because removing O2\mathrm{O_2}O2​ lowers both QpQ_pQp​ and KpK_pKp​ equally
  3. Shift toward reactants (net formation of NO\mathrm{NO}NO and O2\mathrm{O_2}O2​)
  4. Shift toward products because removing a reactant makes Qp<KpQ_p < K_pQp​<Kp​
  5. No net shift because the system was initially at equilibrium
Explanation: This question tests reaction quotient and Le Châtelier's principle. Removing O2, a reactant, decreases its partial pressure, which makes the denominator smaller in Qp, resulting in Qp = 9.0×10^1, greater than Kp of 6.0×10^1. Since Qp > Kp, the system shifts toward the reactants to decrease Qp by forming more NO and O2. This net shift left consumes NO2 to reestablish equilibrium. A common misconception is that removing a reactant makes Qp < Kp and shifts toward products (choice E), but it actually increases Qp, prompting a reverse shift. To predict equilibrium shifts, identify how the stress affects Q, then predict the shift that drives Q back toward K.

Question 2

A sealed rigid container at constant temperature contains the system at equilibrium: N2O4(g)⇌2NO2(g)\text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g)N2​O4​(g)⇌2NO2​(g) with Kc=0.50K_c = 0.50Kc​=0.50. At equilibrium, the concentrations are [N2O4]=0.40 M[\text{N}_2\text{O}_4]=0.40\,\text{M}[N2​O4​]=0.40M and [NO2]=0.45 M[\text{NO}_2]=0.45\,\text{M}[NO2​]=0.45M. A small amount of NO2(g)\text{NO}_2(g)NO2​(g) is suddenly removed so that immediately after the stress [NO2]=0.30 M[\text{NO}_2]=0.30\,\text{M}[NO2​]=0.30M while [N2O4][\text{N}_2\text{O}_4][N2​O4​] is unchanged at that instant. Based on comparing QcQ_cQc​ to KcK_cKc​, how will the system respond to reestablish equilibrium?

  1. Shift toward reactants (left)
  2. No net shift
  3. Shift toward products (right)
  4. The reaction stops because equilibrium was disrupted
  5. Shift toward reactants (left) because a product was removed, making Qc>KcQ_c>K_cQc​>Kc​
Explanation: This question tests understanding of reaction quotient and Le Châtelier's principle. When NO₂ is removed from the equilibrium system, we calculate Qc=[NO2]2/[N2O4]=(0.30)2/(0.40)=0.225Q_c = [\text{NO}_2]^2 / [\text{N}_2\text{O}_4] = (0.30)^2 / (0.40) = 0.225Qc​=[NO2​]2/[N2​O4​]=(0.30)2/(0.40)=0.225, which is less than Kc=0.50K_c = 0.50Kc​=0.50. Since Qc<KcQ_c < K_cQc​<Kc​, the system must shift toward products (right) to increase QcQ_cQc​ back to KcK_cKc​, producing more NO₂ to replace what was removed. Choice E incorrectly states that Qc>KcQ_c > K_cQc​>Kc​ when a product is removed, which is backwards—removing products always makes Q<KQ < KQ<K. The key strategy is to calculate QQQ after the stress, compare it to KKK, then predict the shift: if Q<KQ < KQ<K, shift right; if Q>KQ > KQ>K, shift left.

Question 3

At constant temperature, the following reaction is at equilibrium in a closed container: CaCO3(s)⇌CaO(s)+CO2(g)\text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s)+\text{CO}_2(g)CaCO3​(s)⇌CaO(s)+CO2​(g) with Kp=0.80K_p=0.80Kp​=0.80. A stress is applied by adding CO2(g)\text{CO}_2(g)CO2​(g) so that immediately after the stress PCO2=1.6 atmP_{\text{CO}_2}=1.6\,\text{atm}PCO2​​=1.6atm. (Assume both solids remain present.) Based on comparing QpQ_pQp​ and KpK_pKp​, how will the system respond to reestablish equilibrium?

  1. Shift toward products (right)
  2. Shift toward reactants (left)
  3. No net shift
  4. The reaction stops because solids make QpQ_pQp​ constant
  5. Shift toward products (right) because adding gas increases pressure
Explanation: This question tests understanding of reaction quotient and Le Châtelier's principle. For this heterogeneous equilibrium, solids don't appear in the equilibrium expression, so Q_p = P_CO₂ = 1.6 atm, which is greater than K_p = 0.80 atm. Since Q_p > K_p, the system must shift toward reactants (left) to decrease the CO₂ pressure back to equilibrium, converting some CO₂ back into CaCO₃. Choice D incorrectly claims that solids make Q_p constant—solids are omitted from Q and K expressions, but Q still varies with gas pressures. The key insight is that for heterogeneous equilibria, only gases and aqueous species appear in Q and K expressions.

Question 4

A sealed container at constant temperature contains the equilibrium system N2O4(g)⇌2NO2(g)\mathrm{N_2O_4(g) \rightleftharpoons 2NO_2(g)}N2​O4​(g)⇌2NO2​(g). The system is initially at equilibrium with Kc=0.20K_c = 0.20Kc​=0.20. A small amount of NO2(g)\mathrm{NO_2(g)}NO2​(g) is then injected, and immediately after the injection the reaction quotient is Qc=0.80Q_c = 0.80Qc​=0.80. As the system reestablishes equilibrium, how will it shift?

  1. Shift toward reactants (left)
  2. No net shift
  3. Shift toward products (right)
  4. Shift toward products (right)
  5. No net shift
Explanation: This question tests the skill of reaction quotient and Le Châtelier's principle. Injecting NO2, a product, increases its concentration, causing the reaction quotient Qc=[NO2]2[N2O4]Q_c = \frac{[\mathrm{NO_2}]^2}{[\mathrm{N_2O_4}]}Qc​=[N2​O4​][NO2​]2​ to become greater than KcK_cKc​. Since Qc>KcQ_c > K_cQc​>Kc​, the system will shift to reduce QcQ_cQc​ by favoring the reverse reaction, which consumes NO2 and produces N2O4. This net shift toward reactants restores equilibrium by driving QcQ_cQc​ back to KcK_cKc​. A common misconception is that adding any substance always shifts the equilibrium to the right, but it actually depends on whether the added species is a reactant or product and its impact on Q. Always identify how the stress affects Q compared to K, then predict the shift that drives Q back toward K.

Question 5

In a closed container at constant temperature, the following system is initially at equilibrium: CO(g)+H2O(g)⇌CO2(g)+H2(g)\text{CO}(g)+\text{H}_2\text{O}(g) \rightleftharpoons \text{CO}_2(g)+\text{H}_2(g)CO(g)+H2​O(g)⇌CO2​(g)+H2​(g) with Kc=1.0K_c=1.0Kc​=1.0. A stress is applied by adding H2(g)\text{H}_2(g)H2​(g), and immediately after the addition the concentrations are [CO]=0.30 M[\text{CO}]=0.30\,\text{M}[CO]=0.30M, [H2O]=0.30 M[\text{H}_2\text{O}]=0.30\,\text{M}[H2​O]=0.30M, [CO2]=0.30 M[\text{CO}_2]=0.30\,\text{M}[CO2​]=0.30M, and [H2]=0.60 M[\text{H}_2]=0.60\,\text{M}[H2​]=0.60M. Based on QcQ_cQc​ versus KcK_cKc​, how will the system respond to reestablish equilibrium?

  1. Shift toward products (right)
  2. Shift toward reactants (left)
  3. No net shift
  4. The reaction stops because adding H2\text{H}_2H2​ makes KcK_cKc​ change
  5. Shift toward products (right) because a product was added
Explanation: This question tests understanding of reaction quotient and Le Châtelier's principle. When H₂ is added to the equilibrium system, we calculate Qc=[CO2][H2][CO][H2O]=0.30×0.600.30×0.30=2.0Q_c = \frac{[\text{CO}_2][\text{H}_2]}{[\text{CO}][\text{H}_2\text{O}]} = \frac{0.30 \times 0.60}{0.30 \times 0.30} = 2.0Qc​=[CO][H2​O][CO2​][H2​]​=0.30×0.300.30×0.60​=2.0, which is greater than Kc=1.0K_c = 1.0Kc​=1.0. Since Qc>KcQ_c > K_cQc​>Kc​, the system must shift toward reactants (left) to decrease QcQ_cQc​ back to KcK_cKc​, consuming the excess H₂ that was added. Choice D incorrectly claims that KcK_cKc​ changes when a substance is added—KcK_cKc​ only changes with temperature, not concentration changes. The key strategy is to calculate Q after any stress and compare to K: when Q>KQ > KQ>K, the system shifts left to decrease Q back to K.

Question 6

A closed vessel at constant temperature contains an equilibrium mixture for 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), where K=1.0K = 1.0K=1.0. A sample of CO(g)\mathrm{CO(g)}CO(g) is added, and immediately afterward the reaction quotient is determined to be Q=0.40Q = 0.40Q=0.40. As the system returns to equilibrium, which change occurs?

  1. The system shifts toward reactants (forms more CO\mathrm{CO}CO and H2O\mathrm{H_2O}H2​O).
  2. The system shifts toward products (forms more CO2\mathrm{CO_2}CO2​ and H2\mathrm{H_2}H2​).
  3. There is no net shift because K=1.0K = 1.0K=1.0.
  4. There is no net shift because adding a reactant does not change QQQ.
  5. The system shifts toward reactants because adding reactant always makes Q>KQ > KQ>K.
Explanation: This question tests your understanding of reaction quotient and Le Châtelier's principle. When CO is added to the equilibrium system, the concentration of reactants increases, which decreases Q (since [CO] appears in the denominator of the Q expression). With Q = 0.40 and K = 1.0, we have Q < K, meaning there's too little product relative to equilibrium. To reestablish equilibrium, the system must shift to increase Q back to 1.0, which occurs by converting more CO and H₂O into CO₂ and H₂ (shift toward products). A common misconception (choice D) is that adding a reactant doesn't change Q, but adding reactant decreases Q since reactants appear in the denominator. When Q < K, always predict a shift toward products to increase Q back to K.

Question 7

A rigid container holds an equilibrium mixture for H2(g)+I2(g)⇌2HI(g)\mathrm{H_2(g) + I_2(g) \rightleftharpoons 2HI(g)}H2​(g)+I2​(g)⇌2HI(g) at constant temperature. For this system, K=50K = 50K=50. Some HI(g)\mathrm{HI(g)}HI(g) is removed, and immediately afterward the reaction quotient is found to be Q=20Q = 20Q=20. As equilibrium is reestablished, what will the system do?

  1. The system shifts toward reactants (forms more H2\mathrm{H_2}H2​ and I2\mathrm{I_2}I2​).
  2. The system shifts toward products (forms more HI\mathrm{HI}HI).
  3. There is no net shift because removing a product decreases KKK.
  4. There is no net shift because QQQ is always equal to KKK after a disturbance.
  5. The system shifts toward reactants because removing a product always makes Q>KQ > KQ>K.
Explanation: This question tests your understanding of reaction quotient and Le Châtelier's principle. When HI is removed from the equilibrium system, the concentration of products decreases, which decreases Q (since [HI]² appears in the numerator). With Q = 20 and K = 50, we have Q < K, meaning there's too little product relative to equilibrium. To reestablish equilibrium, the system must shift to increase Q back to 50, which occurs by converting more H₂ and I₂ into HI (shift toward products). A common misconception (choice E) is that removing product makes Q > K, but removing product actually decreases Q since products appear in the numerator. When Q < K, always predict a shift toward products to increase Q back to K.

Question 8

A mixture of gases in a rigid container is at equilibrium for the reaction H2(g)+I2(g)⇌2HI(g)\mathrm{H_2(g) + I_2(g) \rightleftharpoons 2HI(g)}H2​(g)+I2​(g)⇌2HI(g). At this temperature, Kc=50K_c = 50Kc​=50. Some HI(g)\mathrm{HI(g)}HI(g) is removed, and immediately afterward Qc=8Q_c = 8Qc​=8. As the system returns to equilibrium, what net shift will occur?

  1. No net shift
  2. Shift toward reactants (left)
  3. Shift toward products (right)
  4. No net shift
  5. Shift toward products (right)
Explanation: This question tests the skill of reaction quotient and Le Châtelier's principle. Removing HI, a product, decreases its concentration, causing the reaction quotient Qc=[HI]2[H2][I2]Q_c = \frac{[\mathrm{HI}]^2}{[\mathrm{H_2}][\mathrm{I_2}]}Qc​=[H2​][I2​][HI]2​ to become less than KcK_cKc​. Since Qc<KcQ_c < K_cQc​<Kc​, the system will shift to increase QcQ_cQc​ by favoring the forward reaction, which produces more HI. This net shift toward products restores equilibrium by driving QcQ_cQc​ back to KcK_cKc​. A common misconception is that removing a product causes a shift to the left, but actually, it shifts right to replace the removed species. Always identify how the stress affects QQQ compared to KKK, then predict the shift that drives QQQ back toward KKK.

Question 9

In a closed flask at constant temperature, the system is initially at equilibrium for 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 this temperature, Kc=1.6K_c = 1.6Kc​=1.6. A small amount of CO2(g)\mathrm{CO_2(g)}CO2​(g) is added, and immediately after the addition the reaction quotient is Qc=4.0Q_c = 4.0Qc​=4.0. Which shift will occur as equilibrium is reestablished?

  1. Shift toward products (right)
  2. Shift toward reactants (left)
  3. No net shift
  4. Shift toward products (right)
  5. No net shift
Explanation: This question tests the skill of reaction quotient and Le Châtelier's principle. Adding CO2, a product, increases its concentration, causing the reaction quotient Qc=[CO2][H2][CO][H2O]Q_c = \frac{[\mathrm{CO_2}][\mathrm{H_2}]}{[\mathrm{CO}][\mathrm{H_2O}]}Qc​=[CO][H2​O][CO2​][H2​]​ to become greater than KcK_cKc​. Since Qc>KcQ_c > K_cQc​>Kc​, the system will shift to reduce QcQ_cQc​ by favoring the reverse reaction, which consumes CO2 and H2 to produce CO and H2O. This net shift toward reactants restores equilibrium by driving QcQ_cQc​ back to KcK_cKc​. A common misconception is that adding a product always causes no shift, but it actually perturbs QQQ and triggers a shift to rebalance. Always identify how the stress affects QQQ compared to KKK, then predict the shift that drives QQQ back toward KKK.

Question 10

A reaction mixture is at equilibrium for 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 this temperature, Kc=1.8×10−5K_c = 1.8\times 10^{-5}Kc​=1.8×10−5. A small amount of CH3COO−(aq)\mathrm{CH_3COO^- (aq)}CH3​COO−(aq) is added, and immediately afterward Qc=9.0×10−5Q_c = 9.0\times 10^{-5}Qc​=9.0×10−5. As the system returns to equilibrium, what net shift will occur?

  1. Shift toward products (right)
  2. Shift toward reactants (left)
  3. No net shift
  4. Shift toward products (right)
  5. No net shift
Explanation: This question tests the skill of reaction quotient and Le Châtelier's principle. Adding CH3COO−\mathrm{CH_3COO^-}CH3​COO−, a product, increases its concentration, causing the reaction quotient Qc=[H+][CH3COO−][CH3COOH]Q_c = \frac{[\mathrm{H^+}][\mathrm{CH_3COO^-}]}{[\mathrm{CH_3COOH}]}Qc​=[CH3​COOH][H+][CH3​COO−]​ to become greater than KcK_cKc​. Since Qc>KcQ_c > K_cQc​>Kc​, the system will shift to reduce QcQ_cQc​ by favoring the reverse reaction, which consumes H+\mathrm{H^+}H+ and CH3COO−\mathrm{CH_3COO^-}CH3​COO− to produce CH3COOH\mathrm{CH_3COOH}CH3​COOH. This net shift toward reactants restores equilibrium by driving QcQ_cQc​ back to KcK_cKc​. A common misconception is that adding a common ion has no effect on weak acid equilibrium, but it actually suppresses dissociation via Le Châtelier's principle. Always identify how the stress affects QQQ compared to KKK, then predict the shift that drives QQQ back toward KKK.