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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 / 10

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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?

All questions

Question 1

Shift toward products (net formation of $\mathrm{NO_2}$ )
No net shift because removing $\mathrm{O_2}$ lowers both $Q_p$ and $K_p$ equally
Shift toward reactants (net formation of $\mathrm{NO}$ and $\mathrm{O_2}$ ) (correct answer)
Shift toward products because removing a reactant makes $Q_p < K_p$
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: $\text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g)$ with $K_c = 0.50$ . At equilibrium, the concentrations are $[\text{N}_2\text{O}_4]=0.40\,\text{M}$ and $[\text{NO}_2]=0.45\,\text{M}$ . A small amount of $\text{NO}_2(g)$ is suddenly removed so that immediately after the stress $[\text{NO}_2]=0.30\,\text{M}$ while $[\text{N}_2\text{O}_4]$ is unchanged at that instant. Based on comparing $Q_c$ to $K_c$ , how will the system respond to reestablish equilibrium?

Shift toward reactants (left)
No net shift
Shift toward products (right) (correct answer)
The reaction stops because equilibrium was disrupted
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

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?

Shift toward products (right)
Shift toward reactants (left) (correct answer)
No net shift
The reaction stops because solids make $Q_p$ constant
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 $\mathrm{N_2O_4(g) \rightleftharpoons 2NO_2(g)}$ . The system is initially at equilibrium with $K_c = 0.20$ . A small amount of $\mathrm{NO_2(g)}$ is then injected, and immediately after the injection the reaction quotient is $Q_c = 0.80$ . As the system reestablishes equilibrium, how will it shift?

Shift toward reactants (left) (correct answer)
No net shift
Shift toward products (right)
Shift toward products (right)
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

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?

Shift toward products (right)
Shift toward reactants (left) (correct answer)
No net shift
The reaction stops because adding $\text{H}_2$ makes $K_c$ change
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

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?

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

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

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

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

No net shift
Shift toward reactants (left)
Shift toward products (right)
No net shift
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

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?

Shift toward products (right)
Shift toward reactants (left) (correct answer)
No net shift
Shift toward products (right)
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

A reaction mixture is at equilibrium for $\mathrm{CH_3COOH(aq) \rightleftharpoons H^+(aq) + CH_3COO^-(aq)}$ . At this temperature, $K_c = 1.8\times 10^{-5}$ . A small amount of $\mathrm{CH_3COO^- (aq)}$ is added, and immediately afterward $Q_c = 9.0\times 10^{-5}$ . As the system returns to equilibrium, what net shift will occur?

Shift toward products (right)
Shift toward reactants (left) (correct answer)
No net shift
Shift toward products (right)
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 $K$ .