Reaction Quotient and Le Chatelier's Principle - AP Chemistry

1. In a sealed flask at constant temperature, the equilibrium $\mathrm{2NO(g) + O_2(g) \rightleftharpoons 2NO_2(g)}$ has $K_p = 6.0\times10^1$. The system is initially at equilibrium. A small amount of $\mathrm{O_2(g)}$ is removed, and immediately afterward $Q_p$ is found to be $9.0\times10^1$. As equilibrium is reestablished, what shift occurs?
   Answer: Shift toward reactants (net formation of $\mathrm{NO}$ and $\mathrm{O_2}$) 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.
2. A closed vessel at constant temperature contains an equilibrium mixture for $\mathrm{CO(g) + H_2O(g) \rightleftharpoons CO_2(g) + H_2(g)}$, where $K = 1.0$. A sample of $\mathrm{CO(g)}$ is added, and immediately afterward the reaction quotient is determined to be $Q = 0.40$. As the system returns to equilibrium, which change occurs?
   Answer: The system shifts toward products (forms more $\mathrm{CO_2}$ and $\mathrm{H_2}$). 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.
3. At constant temperature, the following reaction is at equilibrium in a closed container: $\text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s)+\text{CO}_2(g)$ with $K_p=0.80$. A stress is applied by adding $\text{CO}_2(g)$ so that immediately after the stress $P_{\text{CO}_2}=1.6\,\text{atm}$. (Assume both solids remain present.) Based on comparing $Q_p$ and $K_p$, how will the system respond to reestablish equilibrium?
   Answer: Shift toward reactants (left) 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.
4. A closed container is initially at equilibrium for $\mathrm{2NO(g) + O_2(g) \rightleftharpoons 2NO_2(g)}$. At this temperature, $K_p = 6.0$. Some $\mathrm{NO_2(g)}$ is removed, and immediately after the removal $Q_p = 0.50$. As equilibrium is reestablished, how will the system shift?
   Answer: Shift toward products (right) This question tests the skill of reaction quotient and Le Châtelier's principle. Removing NO2, a product, decreases its partial pressure, causing the reaction quotient $Q_p = \frac{[\mathrm{NO_2}]^2}{[\mathrm{NO}]^2 [\mathrm{O_2}]}$ to become less than $K_p$ since the numerator decreases. Since $Q_p < K_p$, the system will shift to increase $Q_p$ by favoring the forward reaction, which produces more NO2. This net shift toward products restores equilibrium by driving $Q_p$ back to $K_p$. A common misconception is that removing a product causes a shift left, but it actually shifts right to replace it. Always identify how the stress affects $Q$ compared to $K$, then predict the shift that drives $Q$ back toward $K$.
5. A closed container at constant temperature contains the equilibrium system $\mathrm{2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)}$ with $K_p = 3.0\times10^2$. The system is initially at equilibrium. A small amount of $\mathrm{SO_3(g)}$ is removed, and immediately afterward the reaction quotient is $Q_p = 1.0\times10^2$. As equilibrium is reestablished, how will the system shift?
   Answer: Shift toward products (net formation of $\mathrm{SO_3}$) This question tests reaction quotient and Le Châtelier's principle. Removing SO3, a product, decreases its partial pressure, which lowers the numerator in Qp, resulting in Qp = 1.0×$10^2$, less than Kp of 3.0×$10^2$. Since Qp < Kp, the system shifts toward the products to increase Qp by forming more SO3. This net shift right consumes SO2 and O2 to reestablish equilibrium. A common misconception is that removing a product makes Qp larger than Kp and shifts toward reactants (choice E), but removing a product actually decreases Qp, prompting a forward shift. To predict equilibrium shifts, identify how the stress affects Q, then predict the shift that drives Q back toward K.
6. A rigid container at constant temperature contains an equilibrium mixture for $\mathrm{NH_3(g) + HCl(g) \rightleftharpoons NH_4Cl(s)}$ with $K_p = 2.0\times10^3$. The system is initially at equilibrium. A small amount of $\mathrm{NH_4Cl(s)}$ is added, and immediately afterward the calculated reaction quotient is still $Q_p = 2.0\times10^3$. As the system returns to equilibrium, what shift occurs?
   Answer: No net shift This question tests reaction quotient and Le Châtelier's principle. Adding NH4Cl(s), a solid product, does not affect the partial pressures of the gases, so Qp remains equal to Kp at 2.0×$10^3$. Since Qp = Kp immediately after addition, the system is still at equilibrium with no perturbation. Therefore, no net shift occurs as solids are not included in Q or K expressions. A common misconception is that adding a solid product increases Qp above Kp and shifts toward reactants (choice E), but solids have activity of 1 and do not change Q. To predict equilibrium shifts, identify how the stress affects Q, then predict the shift that drives Q back toward K.
7. A flask at constant temperature contains an equilibrium mixture for $\mathrm{2NO(g) + Cl_2(g) \rightleftharpoons 2NOCl(g)}$. For this reaction, $K = 4.0$. Some $\mathrm{Cl_2(g)}$ is removed, and immediately after the removal the reaction quotient is found to be $Q = 7.0$. What will happen as equilibrium is reestablished?
   Answer: The system shifts toward reactants (forms more $\mathrm{NO}$ and $\mathrm{Cl_2}$). This question tests your understanding of reaction quotient and Le Châtelier's principle. When Cl₂ is removed from the equilibrium system, the concentration of reactants decreases, which increases Q (since [Cl₂] appears in the denominator of the Q expression). With Q = 7.0 and K = 4.0, we have Q > K, meaning there's too much product relative to equilibrium. To reestablish equilibrium, the system must shift to decrease Q back to 4.0, which occurs by converting some NOCl back to NO and Cl₂ (shift toward reactants). A common misconception (choice E) is that removing reactant makes Q < K, but removing reactant actually increases Q since reactants appear in the denominator. When Q > K, always predict a shift toward reactants to decrease Q back to K.
8. A reaction mixture is at equilibrium for $\mathrm{2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)}.$ At this temperature, $K_p = 3.0\times 10^2$. Some $\mathrm{O_2(g)}$ is removed, and immediately afterward $Q_p = 7.5\times 10^2$. As the system reestablishes equilibrium, how will it shift?
   Answer: Shift toward reactants (left) This question tests the skill of reaction quotient and Le Châtelier's principle. Removing O2, a reactant, decreases its partial pressure, causing the reaction quotient $Q_p = [SO3]^2 / ([SO2]^2 [O2])$ to become greater than $K_p$ since the denominator decreases. Since $Q_p > K_p$, the system will shift to reduce $Q_p$ by favoring the reverse reaction, which produces more SO2 and O2 while consuming SO3. This net shift toward reactants restores equilibrium by driving $Q_p$ back to $K_p$. A common misconception is that removing a reactant shifts the equilibrium to the right to 'replace' it, but it actually shifts left to produce more of the removed species. Always identify how the stress affects Q compared to K, then predict the shift that drives Q back toward K.
9. In a closed container at constant temperature, the following system is initially at equilibrium: $\text{CO}(g)+\text{H}_2\text{O}(g) \rightleftharpoons \text{CO}_2(g)+\text{H}_2(g)$ with $K_c=1.0$. A stress is applied by adding $\text{H}_2(g)$, and immediately after the addition the concentrations are $[\text{CO}]=0.30\,\text{M}$, $[\text{H}_2\text{O}]=0.30\,\text{M}$, $[\text{CO}_2]=0.30\,\text{M}$, and $[\text{H}_2]=0.60\,\text{M}$. Based on $Q_c$ versus $K_c$, how will the system respond to reestablish equilibrium?
   Answer: Shift toward reactants (left) This question tests understanding of reaction quotient and Le Châtelier's principle. When H₂ is added to the equilibrium system, we calculate $Q_c = \frac{[\text{CO}_2][\text{H}_2]}{[\text{CO}][\text{H}_2\text{O}]} = \frac{0.30 \times 0.60}{0.30 \times 0.30} = 2.0$, which is greater than $K_c = 1.0$. Since $Q_c > K_c$, the system must shift toward reactants (left) to decrease $Q_c$ back to $K_c$, consuming the excess H₂ that was added. Choice D incorrectly claims that $K_c$ changes when a substance is added—$K_c$ only changes with temperature, not concentration changes. The key strategy is to calculate Q after any stress and compare to K: when $Q > K$, the system shifts left to decrease Q back to K.
10. A closed container at constant temperature contains the equilibrium system $2\text{NO}(g)+\text{Cl}_2(g) \rightleftharpoons 2\text{NOCl}(g)$ with $K_p=1.0\times 10^{4}$. A stress is applied by removing some $\text{NOCl}(g)$. Immediately after the stress, the partial pressures are $P_{\text{NO}}=0.20\,\text{atm}$, $P_{\text{Cl}_2}=0.20\,\text{atm}$, and $P_{\text{NOCl}}=5.0\,\text{atm}$. Using $Q_p$ compared with $K_p$, which direction will the system shift to reestablish equilibrium?
    Answer: Shift toward products (right) This question tests understanding of reaction quotient and Le Châtelier's principle. After NOCl is removed, we calculate $Q_p = \frac{(P_{\text{NOCl}})^2}{(P_{\text{NO}})^2 P_{\text{Cl}_2}} = \frac{(5.0)^2}{(0.20)^2 \times 0.20} = 3125$, which is less than $K_p = 1.0\times 10^{4}$. Since $Q_p < K_p$, the system must shift toward products (right) to increase $Q_p$ back to $K_p$, producing more NOCl to replace what was removed. Choice E incorrectly suggests shifting left because a product was removed—this is backwards, as removing products always causes a rightward shift to replace them. The strategy is to calculate Q after the stress and compare to K: removing products decreases Q below K, requiring a rightward shift.
11. At constant temperature, the equilibrium $\mathrm{CO(g) + H_2O(g) \rightleftharpoons CO_2(g) + H_2(g)}$ has $K_c = 1.0$. The system is initially at equilibrium in a sealed container. A small amount of $\mathrm{CO(g)}$ is removed, and immediately afterward $Q_c$ is found to be $1.6$. As the system returns to equilibrium, what shift occurs?
    Answer: Shift toward reactants (net formation of $\mathrm{CO}$ and $\mathrm{H_2O}$) This question tests reaction quotient and Le Châtelier's principle. Removing CO, a reactant, decreases its concentration, which makes the denominator smaller in Qc, resulting in Qc = 1.6, greater than Kc of 1.0. Since Qc > Kc, the system shifts toward the reactants to decrease Qc by producing more CO and H2O. This net shift left consumes CO2 and H2 to reestablish equilibrium. A common misconception is that removing a reactant increases the forward reaction and shifts toward products (choice D), but it actually shifts toward reactants to replace the removed species. To predict equilibrium shifts, identify how the stress affects Q, then predict the shift that drives Q back toward K.
12. In a rigid container at constant temperature, the system $\mathrm{H_2(g) + I_2(g) \rightleftharpoons 2HI(g)}$ is initially at equilibrium with $K_c = 50$. A catalyst is added, and immediately afterward a measurement shows $Q_c = 50$. How will the system respond as it reestablishes equilibrium?
    Answer: No net shift This question tests reaction quotient and Le Châtelier's principle. Adding a catalyst speeds up both forward and reverse reactions equally but does not change the equilibrium constant Kc or the position of equilibrium, leaving Qc equal to Kc at 50. Since Qc = Kc immediately after addition, the system remains at equilibrium with no perturbation to concentrations. Therefore, no net shift occurs as the system is already balanced. A common misconception is that a catalyst shifts the equilibrium toward products by increasing Kc (choice D), but catalysts only affect kinetics, not thermodynamics. To predict equilibrium shifts, identify how the stress affects Q, then predict the shift that drives Q back toward K.
13. At constant temperature, a closed container contains the equilibrium system $\text{CH}_3\text{COOH}(aq) \rightleftharpoons \text{H}^+(aq)+\text{CH}_3\text{COO}^-(aq)$ with $K_c=1.8\times 10^{-5}$. A stress is applied by adding $\text{CH}_3\text{COO}^-(aq)$ (from a soluble salt). Immediately after the addition, the concentrations are $[\text{CH}_3\text{COOH}]=0.10\,\text{M}$, $[\text{H}^+]=1.8\times 10^{-3}\,\text{M}$, and $[\text{CH}_3\text{COO}^-]=0.20\,\text{M}$. Using $Q_c$ compared with $K_c$, which direction will the system shift to reestablish equilibrium?
    Answer: Shift toward reactants (left) This question tests understanding of reaction quotient and Le Châtelier's principle. After CH₃COO⁻ is added, we calculate $Q_c = \frac{[\mathrm{H}^{+}][\mathrm{CH}_3\mathrm{COO}^{-}]}{[\mathrm{CH}_3\mathrm{COOH}]} = \frac{(1.8 \times 10^{-3} \times 0.20)}{0.10} = 3.6 \times 10^{-3}$, which is greater than $K_c = 1.8 \times 10^{-5}$. Since $Q_c > K_c$, the system must shift toward reactants (left) to decrease $Q_c$ back to $K_c$, converting some H⁺ and CH₃COO⁻ back into CH₃COOH. Choice E incorrectly claims that adding a substance changes $K_c$—equilibrium constants only change with temperature, not concentration. The strategy is to calculate $Q$ after the stress: when products are added, $Q$ increases above $K$, requiring a leftward shift to decrease $Q$.
14. At $25^\circ\text{C}$, the reaction $\text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g)$ is at equilibrium in a rigid container, and $K_c = 0.20$. A small amount of $\text{NO}_2(g)$ is injected, and immediately afterward the measured concentrations are $[\text{NO}_2]=0.60\,\text{M}$ and $[\text{N}_2\text{O}_4]=0.50\,\text{M}$. Based on comparing $Q_c$ to $K_c$, how will the system respond to reestablish equilibrium?
    Answer: Shift toward reactants (left) This question tests your understanding of reaction quotient and Le Châtelier's principle. For the reaction N₂O₄(g) ⇌ 2NO₂(g), we calculate $Q_c$ = [NO₂]²/[N₂O₄] = (0.60)²/(0.50) = 0.72. Since $Q_c$ (0.72) > $K_c$ (0.20), the system has too much product relative to equilibrium, so it must shift toward reactants (left) to decrease $Q_c$ back to $K_c$. A common misconception is thinking that adding NO₂ always drives the reaction right because "adding reactant shifts right," but NO₂ is a product here, not a reactant. To solve equilibrium shift problems: calculate Q, compare it to K, then predict the shift that drives Q back toward K (if Q > K, shift left; if Q < K, shift right).
15. A sealed flask at constant temperature contains an equilibrium mixture for $\mathrm{2H_2S(g) \rightleftharpoons 2H_2(g) + S_2(g)}$, with $K = 0.10$. A small amount of $\mathrm{H_2(g)}$ is removed, and immediately afterward the reaction quotient is measured to be $Q = 0.040$. What will occur as equilibrium is reestablished?
    Answer: The system shifts toward products (forms more $\mathrm{H_2}$ and $\mathrm{S_2}$). This question tests your understanding of reaction quotient and Le Châtelier's principle. When H₂ is removed from the equilibrium system, the concentration of products decreases, which decreases Q (since [H₂]² appears in the numerator of the Q expression). With Q = 0.040 and K = 0.10, 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 0.10, which occurs by converting more H₂S into H₂ and S₂ (shift toward products). A common misconception (choice E) correctly identifies Q < K but incorrectly predicts a shift toward reactants; when Q < K, the shift is always toward products. When Q < K, always predict a shift toward products to increase Q back to K.
16. At constant temperature, a sealed flask contains an equilibrium mixture for $\mathrm{CH_3COOH(aq) \rightleftharpoons H^+(aq) + CH_3COO^-(aq)}$, with $K = 1.8\times 10^{-5}$. A small amount of $\mathrm{CH_3COO^-}$ is added, and immediately afterward the reaction quotient is found to be $Q = 3.6\times 10^{-5}$. What happens as the system returns to equilibrium?
    Answer: The system shifts toward reactants (produces more $\mathrm{CH_3COOH}$). This question tests your understanding of reaction quotient and Le Châtelier's principle. When CH₃COO⁻ is added to the equilibrium system, the concentration of products increases, which increases Q (since [CH₃COO⁻] appears in the numerator of the Q expression). With Q = 3.6×10⁻⁵ and K = 1.8×10⁻⁵, we have Q > K, meaning there's too much product relative to equilibrium. To reestablish equilibrium, the system must shift to decrease Q back to 1.8×10⁻⁵, which occurs by combining H⁺ and CH₃COO⁻ to form more CH₃COOH (shift toward reactants). A common misconception (choice E) correctly identifies Q > K but incorrectly predicts a shift toward products; when Q > K, the shift is always toward reactants. When Q > K, always predict a shift toward reactants to decrease Q back to K.
17. A closed container at constant temperature contains an equilibrium mixture for $\mathrm{Fe^{3+}(aq) + SCN^-(aq) \rightleftharpoons FeSCN^{2+}(aq)}$, with $K = 900$. Some $\mathrm{Fe^{3+}}$ is removed, and immediately after removal the reaction quotient is determined to be $Q = 1200$. As equilibrium is reestablished, what will the system do?
    Answer: The system shifts toward reactants (forms more $\mathrm{Fe^{3+}}$ and $\mathrm{SCN^-}$). This question tests your understanding of reaction quotient and Le Châtelier's principle. When Fe³⁺ is removed from the equilibrium system, the concentration of reactants decreases, which increases Q (since [Fe³⁺] appears in the denominator of the Q expression). With Q = 1200 and K = 900, we have Q > K, meaning there's too much product relative to equilibrium. To reestablish equilibrium, the system must shift to decrease Q back to 900, which occurs by converting some FeSCN²⁺ back to Fe³⁺ and SCN⁻ (shift toward reactants). A common misconception (choice E) is that removing reactant makes Q < K, but removing reactant actually increases Q since reactants appear in the denominator. When Q > K, always predict a shift toward reactants to decrease Q back to K.
18. In a closed flask at constant temperature, the system is initially at equilibrium for $\mathrm{CO(g) + H_2O(g) \rightleftharpoons CO_2(g) + H_2(g)}$. At this temperature, $K_c = 1.6$. A small amount of $\mathrm{CO_2(g)}$ is added, and immediately after the addition the reaction quotient is $Q_c = 4.0$. Which shift will occur as equilibrium is reestablished?
    Answer: Shift toward reactants (left) This question tests the skill of reaction quotient and Le Châtelier's principle. Adding CO2, a product, increases its concentration, causing the reaction quotient $Q_c = \frac{[\mathrm{CO_2}][\mathrm{H_2}]}{[\mathrm{CO}][\mathrm{H_2O}]}$ to become greater than $K_c$. Since $Q_c > K_c$, the system will shift to reduce $Q_c$ by favoring the reverse reaction, which consumes CO2 and H2 to produce CO and H2O. This net shift toward reactants restores equilibrium by driving $Q_c$ back to $K_c$. A common misconception is that adding a product always causes no shift, but it actually perturbs $Q$ and triggers a shift to rebalance. Always identify how the stress affects $Q$ compared to $K$, then predict the shift that drives $Q$ back toward $K$.
19. A container at constant temperature contains an equilibrium mixture for $\mathrm{CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)}$, where $K = 0.30$ (in terms of $P_{\mathrm{CO_2}}$). Some $\mathrm{CO_2(g)}$ is added, and immediately afterward the reaction quotient is $Q = 0.80$. As the system returns to equilibrium, what will happen?
    Answer: The system shifts toward reactants (consumes $\mathrm{CO_2}$ to form more $\mathrm{CaCO_3(s)}$). This question tests your understanding of reaction quotient and Le Châtelier's principle. For heterogeneous equilibria involving solids, K and Q are expressed only in terms of gases (here, PCO₂), since solid concentrations are constant. When CO₂ is added, its partial pressure increases, making Q = 0.80 while K = 0.30, so Q > K, meaning there's too much gaseous product relative to equilibrium. To reestablish equilibrium, the system must shift to decrease Q back to 0.30, which occurs by consuming CO₂ to form more solid CaCO₃ (shift toward reactants). A common misconception (choice C) is that solids prevent equilibrium shifts, but solids can be produced or consumed even though they don't appear in Q or K expressions. When Q > K, always predict a shift toward reactants to decrease Q back to K.
20. At constant temperature, a sealed container holds the equilibrium $\mathrm{H_2(g) + CO_2(g) \rightleftharpoons H_2O(g) + CO(g)}$ with $K_c = 0.50$. The system is initially at equilibrium. A small amount of $\mathrm{CO(g)}$ is added, and immediately afterward the reaction quotient is $Q_c = 0.80$. As equilibrium is reestablished, how will the system shift?
    Answer: Shift toward reactants (net formation of $\mathrm{H_2}$ and $\mathrm{CO_2}$) This question tests reaction quotient and Le Châtelier's principle. Adding CO, a product, increases its concentration, which raises the numerator in Qc, resulting in Qc = 0.80, greater than Kc of 0.50. Since Qc > Kc, the system shifts toward the reactants to decrease Qc by forming more H2 and CO2. This net shift left consumes H2O and CO to reestablish equilibrium. A common misconception is that adding a product makes Qc < Kc and shifts toward products (choice E), but it actually increases Qc above Kc, prompting a reverse shift. To predict equilibrium shifts, identify how the stress affects Q, then predict the shift that drives Q back toward K.

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