AP Chemistry Quiz: Reaction Quotient And Le Chateliers Principle

What this quiz covers

This quiz focuses on Reaction Quotient And Le Chateliers Principle, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Chemistry.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. Shift toward products (net formation of $\mathrm{NO_2}$)
2. No net shift because removing $\mathrm{O_2}$ lowers both $Q_p$ and $K_p$ equally
3. Shift toward reactants (net formation of $\mathrm{NO}$ and $\mathrm{O_2}$) (correct answer)
4. Shift toward products because removing a reactant makes $Q_p < K_p$
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

1. Shift toward reactants (left)
2. No net shift
3. Shift toward products (right) (correct answer)
4. The reaction stops because equilibrium was disrupted
5. Shift toward reactants (left) because a product was removed, making $Q_c>K_c$

Explanation: This question tests understanding of reaction quotient and Le Châtelier's principle. When NO₂ is removed from the equilibrium system, we calculate $Q_c = [\text{NO}_2]^2 / [\text{N}_2\text{O}_4] = (0.30)^2 / (0.40) = 0.225$, which is less than $K_c = 0.50$. Since $Q_c < K_c$, the system must shift toward products (right) to increase $Q_c$ back to $K_c$, producing more NO₂ to replace what was removed. Choice E incorrectly states that $Q_c > K_c$ when a product is removed, which is backwards—removing products always makes $Q < K$. The key strategy is to calculate $Q$ after the stress, compare it to $K$, then predict the shift: if $Q < K$, shift right; if $Q > K$, shift left.

Question 3

1. Shift toward products (right)
2. Shift toward reactants (left) (correct answer)
3. No net shift
4. The reaction stops because solids make $Q_p$ 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

1. Shift toward reactants (left) (correct answer)
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 $Q_c = \frac{[\mathrm{NO_2}]^2}{[\mathrm{N_2O_4}]}$ 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 NO2 and produces N2O4. This net shift toward reactants restores equilibrium by driving $Q_c$ back to $K_c$. 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

1. Shift toward products (right)
2. Shift toward reactants (left) (correct answer)
3. No net shift
4. The reaction stops because adding $\text{H}_2$ makes $K_c$ 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 $Q_c = \frac{[\text{CO}_2][\text{H}_2]}{[\text{CO}][\text{H}_2\text{O}]} = \frac{0.30 \times 0.60}{0.30 \times 0.30} = 2.0$, 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.

Question 6

1. The system shifts toward reactants (forms more $\mathrm{CO}$ and $\mathrm{H_2O}$).
2. The system shifts toward products (forms more $\mathrm{CO_2}$ and $\mathrm{H_2}$). (correct answer)
3. There is no net shift because $K = 1.0$.
4. There is no net shift because adding a reactant does not change $Q$.
5. The system shifts toward reactants because adding reactant always makes $Q > K$.

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

1. The system shifts toward reactants (forms more $\mathrm{H_2}$ and $\mathrm{I_2}$).
2. The system shifts toward products (forms more $\mathrm{HI}$). (correct answer)
3. There is no net shift because removing a product decreases $K$.
4. There is no net shift because $Q$ is always equal to $K$ after a disturbance.
5. The system shifts toward reactants because removing a product always makes $Q > K$.

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

1. No net shift
2. Shift toward reactants (left)
3. Shift toward products (right)
4. No net shift
5. Shift toward products (right) (correct answer)

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 $Q_c = \frac{[\mathrm{HI}]^2}{[\mathrm{H_2}][\mathrm{I_2}]}$ to become less than $K_c$. Since $Q_c < K_c$, the system will shift to increase $Q_c$ by favoring the forward reaction, which produces more HI. This net shift toward products restores equilibrium by driving $Q_c$ back to $K_c$. 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 $Q$ compared to $K$, then predict the shift that drives $Q$ back toward $K$.

Question 9

1. Shift toward products (right)
2. Shift toward reactants (left) (correct answer)
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 $Q_c = \frac{[\mathrm{CO_2}][\mathrm{H_2}]}{[\mathrm{CO}][\mathrm{H_2O}]}$ 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 $K$.

Question 10

1. Shift toward products (right)
2. Shift toward reactants (left) (correct answer)
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 $\mathrm{CH_3COO^-}$, a product, increases its concentration, causing the reaction quotient $Q_c = \frac{[\mathrm{H^+}][\mathrm{CH_3COO^-}]}{[\mathrm{CH_3COOH}]}$ 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 $\mathrm{H^+}$ and $\mathrm{CH_3COO^-}$ to produce $\mathrm{CH_3COOH}$. This net shift toward reactants restores equilibrium by driving $Q_c$ back to $K_c$. 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 $Q$ compared to $K$, then predict the shift that drives $Q$ back toward $K$.

Question 11

1. Shift toward reactants (left) (correct answer)
2. Shift toward reactants (left)
3. Shift toward products (right)
4. No net shift
5. No net shift

Explanation: 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.

Question 12

1. No net shift because $Q_c$ is close to $K_c$
2. No net shift because adding a product changes $Q_c$ but not the direction
3. Shift toward products (net formation of $\mathrm{H_2O}$ and $\mathrm{CO}$)
4. Shift toward reactants (net formation of $\mathrm{H_2}$ and $\mathrm{CO_2}$) (correct answer)
5. Shift toward products because adding $\mathrm{CO}$ makes $Q_c < K_c$

Explanation: 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.

Question 13

1. No net shift
2. Shift toward reactants (left) because removing product makes $Q_p>K_p$
3. Shift toward products (right) (correct answer)
4. The reaction stops because a product was removed
5. Shift toward reactants (left)

Explanation: This question tests understanding of reaction quotient and Le Châtelier's principle. After SO₃ is removed, we calculate $Q_p = \frac{(P_{\text{SO}_3})^2}{(P_{\text{SO}_2})^2 \times P_{\text{O}_2}} = \frac{(0.40)^2}{(0.50)^2 \times 0.50} = 1.28$, which is less than $K_p = 4.0$. Since $Q_p < K_p$, the system must shift toward products (right) to increase $Q_p$ back to $K_p$, producing more SO₃ to replace what was removed. Choice E incorrectly states that removing product makes $Q_p > K_p$, which is backwards—removing products always decreases Q, making Q < K. The strategy is to calculate Q immediately after the stress: when products are removed, Q decreases below K, so the system shifts right to restore equilibrium.

Question 14

1. Shift toward products (right)
2. Shift toward reactants (left) (correct answer)
3. No net shift
4. The reaction stops because adding product consumes reactants instantly
5. Shift toward products (right) because a product was added

Explanation: This question tests understanding of reaction quotient and Le Châtelier's principle. After $\text{FeSCN}^{2+}$ is added, we calculate $Q_c = \frac{[\text{FeSCN}^{2+}]}{[\text{Fe}^{3+}][\text{SCN}^{-}]} = \frac{2.0}{0.10 \times 0.10} = 200$, which is greater than $K_c = 100$. Since $Q_c > K_c$, the system must shift toward reactants (left) to decrease $Q_c$ back to $K_c$, converting some of the added $\text{FeSCN}^{2+}$ back into $\text{Fe}^{3+}$ and $\text{SCN}^{-}$. Choice E incorrectly suggests shifting right because a product was added—the shift direction depends on Q vs K comparison, not simply what was added. The key principle is that adding products increases Q above K, requiring a leftward shift to restore equilibrium.

Question 15

1. No net shift; solids force $Q_p$ to equal $K_p$
2. Shift toward products (right)
3. No net shift; adding $\text{CO}_2$ does not affect $Q_p$
4. Shift toward products (right)
5. Shift toward reactants (left) (correct answer)

Explanation: This question tests your understanding of reaction quotient and Le Châtelier's principle. For the heterogeneous equilibrium CaCO₃(s) ⇌ CaO(s) + CO₂(g), solids don't appear in the equilibrium expression, so Q_p = P_CO₂ = 0.50 atm. Since Q_p (0.50) > K_p (0.10), the system has too much product (CO₂) relative to equilibrium, so it must shift toward reactants (left) to decrease the CO₂ pressure back to K_p. A common misconception is that solids "buffer" the system or that adding CO₂ has no effect, but for heterogeneous equilibria, Q still depends on gas pressures only. To solve equilibrium shift problems: calculate Q (using only gases/aqueous species), compare it to K, then predict the shift that drives Q back toward K (if Q > K, shift left; if Q < K, shift right).

Question 16

1. Shift toward reactants (left) (correct answer)
2. Shift toward products (right)
3. No net shift; the system is already at equilibrium
4. Shift toward products (right)
5. No net shift; concentrations will remain fixed after the injection

Explanation: 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).

Question 17

1. No net shift; removing a product does not affect $Q_p$
2. Shift toward reactants (left)
3. Shift toward products (right) (correct answer)
4. No net shift; $Q_p$ equals $K_p$ after the removal
5. Shift toward reactants (left)

Explanation: This question tests your understanding of reaction quotient and Le Châtelier's principle. For the reaction 2NO(g) + Cl₂(g) ⇌ 2NOCl(g), we calculate Q_p = (P_NOCl)²/((P_NO)²(P_Cl₂)) = (2.0)²/((1.0)²(1.0)) = 4.0. Since Q_p (4.0) < K_p (10), the system has too little product relative to equilibrium, so it must shift toward products (right) to increase Q_p up to K_p. A common misconception is that removing a product always shifts left to "oppose the removal," but removing NOCl decreases Q below K, requiring a right shift to restore equilibrium. 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).

Question 18

1. The reaction stops because aqueous equilibria do not shift
2. Shift toward products (right) because adding $\text{CH}_3\text{COO}^-$ increases $K_c$
3. Shift toward reactants (left) (correct answer)
4. Shift toward products (right)
5. No net shift

Explanation: 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 $Q$.

Question 19

1. Shift toward reactants (net formation of $\mathrm{Fe^{3+}}$ and $\mathrm{SCN^-}$)
2. Shift toward products (net formation of $\mathrm{FeSCN^{2+}}$) (correct answer)
3. No net shift because adding a reactant does not change $Q_c$
4. No net shift because $Q_c$ is not compared to $K_c$ for aqueous equilibria
5. Shift toward reactants because adding $\mathrm{Fe^{3+}}$ makes $Q_c$ larger than $K_c$

Explanation: This question tests reaction quotient and Le Châtelier's principle. Adding $\mathrm{Fe^{3+}}$, a reactant, increases its concentration, which raises the denominator in $Q_c$, resulting in $Q_c = 40$, less than $K_c$ of 100. Since $Q_c < K_c$, the system shifts toward the products to increase $Q_c$ by forming more $\mathrm{FeSCN^{2+}}$. This net shift right consumes $\mathrm{Fe^{3+}}$ and $\mathrm{SCN^{-}}$ to reestablish equilibrium. A common misconception is that adding a reactant makes $Q_c$ larger than $K_c$ and shifts toward reactants (choice E), but it actually decreases $Q_c$, prompting a forward shift. To predict equilibrium shifts, identify how the stress affects $Q$, then predict the shift that drives $Q$ back toward $K$.

Question 20

1. No net shift
2. Shift toward reactants (left)
3. No net shift
4. Shift toward products (right) (correct answer)
5. Shift toward reactants (left)

Explanation: 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 $K$.