AP Chemistry Quiz: Reaction Quotient And Equilibrium Constant

20 questions · exam conditions

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Reaction Quotient And Equilibrium ConstantQuestion 1 of 20

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

The reaction will proceed toward products.

The reaction will proceed toward reactants.

The system is already at equilibrium.

The reaction will proceed toward products because

Q_c < K_c

The reaction will proceed toward reactants because

Q_c = K_c

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AP Chemistry Quiz

AP Chemistry Quiz: Reaction Quotient And Equilibrium Constant

Practice Reaction Quotient And Equilibrium Constant in AP Chemistry with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Reaction Quotient And Equilibrium Constant, 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.

All questions

Question 1

The reaction will proceed toward products.
The reaction will proceed toward reactants. (correct answer)
The system is already at equilibrium.
The reaction will proceed toward products because $Q_c < K_c$ .
The reaction will proceed toward reactants because $Q_c = K_c$ .

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 $\mathrm{2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)}$ at a given temperature, $K_c = 1.0\times 10^2$ . A mixture is prepared with initial concentrations $[\mathrm{SO_2}] = 0.10\,\mathrm{M}$ , $[\mathrm{O_2}] = 0.10\,\mathrm{M}$ , and $[\mathrm{SO_3}] = 1.0\,\mathrm{M}$ , and the system is not at equilibrium. Based on comparing $Q_c$ and $K_c$ , in which direction will the reaction proceed to reach equilibrium?

The reaction will proceed toward reactants. (correct answer)
The reaction will proceed toward products.
The reaction will proceed toward reactants because $Q_c = K_c$ .
The reaction will proceed toward products because $Q_c < K_c$ .
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 $\mathrm{2H_2O_2(aq) \rightleftharpoons 2H_2O(l) + O_2(g)}$ at a given temperature, $K_c = 2.0\times 10^{-3}$ . A mixture is prepared with initial concentrations $[\mathrm{H_2O_2}] = 0.10\,\mathrm{M}$ and $[\mathrm{O_2}] = 1.0\times 10^{-3}\,\mathrm{M}$ (liquid water is not included in $Q_c$ ), and the system is not at equilibrium. Based on comparing $Q_c$ and $K_c$ , in which direction will the reaction proceed to reach equilibrium?

The reaction will proceed toward products.
The reaction will proceed toward reactants. (correct answer)
The system is already at equilibrium.
The reaction will proceed toward reactants because $Q_c > K_c$ .
The reaction will proceed toward products because $Q_c = K_c$ .

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 $\mathrm{2NO_2(g) \rightleftharpoons N_2O_4(g)}$ has $K_c = 6.0$ . A mixture is prepared such that $[\mathrm{NO_2}] = 0.30\,\mathrm{M}$ and $[\mathrm{N_2O_4}] = 0.90\,\mathrm{M}$ , so the system is not initially at equilibrium. Based on comparing $Q_c$ (calculated from the given concentrations) and $K_c$ , in which direction will the reaction proceed to reach equilibrium?

the reaction will proceed toward products
the reaction will proceed toward reactants (correct answer)
the system is already at equilibrium
the reaction will proceed toward products until the reactants are used up
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 $\mathrm{CO(g) + Cl_2(g) \rightleftharpoons COCl_2(g)}$ at a certain temperature, $K_c = 0.25$ . A mixture is prepared with $[\mathrm{CO}] = 0.40\,\mathrm{M}$ , $[\mathrm{Cl_2}] = 0.40\,\mathrm{M}$ , and $[\mathrm{COCl_2}] = 0.020\,\mathrm{M}$ , so the system is not at equilibrium initially. Using $Q_c = \dfrac{[\mathrm{COCl_2}]}{[\mathrm{CO}][\mathrm{Cl_2}]}$ and comparing $Q_c$ to $K_c$ , in which direction will the reaction proceed to reach equilibrium?

the reaction will proceed toward products (correct answer)
the system is already at equilibrium
the reaction will proceed toward reactants because $Q_c < K_c$
the reaction will proceed toward products until all reactants are used up
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 $\mathrm{A(g) + B(g) \rightleftharpoons C(g)}$ at a certain temperature, $K_c = 2.0$ . A reaction mixture is prepared with initial concentrations $[\mathrm{A}]=1.0\,\mathrm{M}$ , $[\mathrm{B}]=1.0\,\mathrm{M}$ , and $[\mathrm{C}]=0.50\,\mathrm{M}$ , and the system is not at equilibrium. Based on comparing $Q_c$ to $K_c$ , in which direction will the reaction proceed to reach equilibrium?

the reaction will proceed toward reactants
the reaction will proceed toward products (correct answer)
the system is already at equilibrium
the reaction will proceed toward reactants because the reactant concentrations are equal
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 $\mathrm{CO(g) + H_2O(g) \rightleftharpoons CO_2(g) + H_2(g)}$ at a given temperature, $K_c = 1.0$ . A mixture is prepared with initial concentrations $[\mathrm{CO}] = 0.50\,\mathrm{M}$ , $[\mathrm{H_2O}] = 0.50\,\mathrm{M}$ , $[\mathrm{CO_2}] = 0.10\,\mathrm{M}$ , and $[\mathrm{H_2}] = 0.10\,\mathrm{M}$ , and the system is not at equilibrium. Based on comparing $Q_c$ and $K_c$ , in which direction will the reaction proceed to reach equilibrium?

The reaction will proceed toward products. (correct answer)
The reaction will proceed toward reactants.
The system is already at equilibrium.
The reaction will proceed toward reactants because $Q_c < K_c$ .
The reaction will proceed toward products because $Q_c = K_c$ .

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 $\mathrm{N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)}$ at a given temperature, $K_c = 4.0\times 10^2$ . A mixture is prepared with initial concentrations $[\mathrm{N_2}] = 0.50\,\mathrm{M}$ , $[\mathrm{H_2}] = 0.50\,\mathrm{M}$ , and $[\mathrm{NH_3}] = 0.010\,\mathrm{M}$ , and the system is not at equilibrium. Based on comparing $Q_c$ and $K_c$ , in which direction will the reaction proceed to reach equilibrium?

The reaction will proceed toward reactants.
The system is already at equilibrium.
The reaction will proceed toward products. (correct answer)
The reaction will proceed toward reactants because $Q_c < K_c$ .
The reaction will proceed toward products because $Q_c > K_c$ .

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 $\mathrm{2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)}$ at a certain temperature, where $K_c = 1.0\times10^2$ . A reaction mixture is prepared with $Q_c = 0.50$ , so the system is not initially at equilibrium. Based on the comparison of $Q_c$ and $K_c$ , in which direction will the reaction proceed to reach equilibrium?

the system is already at equilibrium
the reaction will proceed toward products until the reactants are used up
the reaction will proceed toward reactants until the concentrations of all species are equal
the reaction will proceed toward reactants
the reaction will proceed toward products (correct answer)

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 $\mathrm{CH_3COOH(aq) \rightleftharpoons H^+(aq) + CH_3COO^-(aq)}$ at a given temperature, $K_c = 1.8\times 10^{-5}$ . A solution is prepared with initial concentrations $[\mathrm{CH_3COOH}] = 0.10\,\mathrm{M}$ , $[\mathrm{H^+}] = 1.0\times 10^{-5}\,\mathrm{M}$ , and $[\mathrm{CH_3COO^-}] = 1.0\times 10^{-5}\,\mathrm{M}$ , and the system is not at equilibrium. Based on comparing $Q_c$ and $K_c$ , in which direction will the reaction proceed to reach equilibrium?

The reaction will proceed toward reactants.
The reaction will proceed toward products. (correct answer)
The system is already at equilibrium.
The reaction will proceed toward products because $Q_c > K_c$ .
The reaction will proceed toward reactants because $Q_c < K_c$ .

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.

Question 11

For the reversible reaction $\mathrm{PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)}$ at a certain temperature, $K_c = 0.40$ . A reaction mixture is prepared such that $Q_c = 3.0$ , so the system is not initially at equilibrium. Based on the comparison of $Q_c$ and $K_c$ , in which direction will the reaction proceed to reach equilibrium?

the reaction will proceed toward reactants until the concentrations of all species are equal
the reaction will proceed toward products
the system is already at equilibrium
the reaction will proceed toward reactants (correct answer)
the reaction will proceed toward products until the reactants are used up

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 = [PCl_3][Cl_2] / [PCl_5]. Here, Q_c = 3.0 is greater than K_c = 0.40, 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 12

For the reversible reaction $\mathrm{CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)}$ at a certain temperature, $K_c = 0.020$ . A sealed container initially has $[\mathrm{CO_2}] = 0.50\,\mathrm{M}$ (solids are present), so the system is not at equilibrium initially. Using $Q_c = [\mathrm{CO_2}]$ for this reaction and comparing $Q_c$ to $K_c$ , in which direction will the reaction proceed to reach equilibrium?

the reaction will proceed toward products
the system is already at equilibrium
the reaction will proceed toward reactants (correct answer)
the reaction will proceed toward products because solids are not included in $Q_c$
the reaction will proceed toward reactants because $K_c$ is less than 1

Explanation: This question tests understanding of reaction quotient and equilibrium constant for heterogeneous equilibria. For reactions involving pure solids, only gases and aqueous species appear in Q and K expressions, so Qc = [CO₂] = 0.50. Since Q > K (0.50 > 0.020), the system has too much CO₂ gas compared to equilibrium. The reaction must shift backward (toward reactants) to decrease CO₂ concentration until Q equals K. Choice D correctly notes that solids aren't in Q but incorrectly predicts forward reaction when Q > K always means reverse. To solve heterogeneous equilibria problems, write Q using only gases/aqueous species, then apply the standard rule: Q > K means reaction goes toward reactants.

Question 13

Consider the reversible reaction $\mathrm{NH_3(g) + HCl(g) \rightleftharpoons NH_4Cl(s)}$ at a certain temperature, where $K_p = 2.0\times10^{4}$ . A mixture is prepared with $P_{\mathrm{NH_3}} = 0.10\,\mathrm{atm}$ and $P_{\mathrm{HCl}} = 0.10\,\mathrm{atm}$ , and the system is not initially at equilibrium. Based on comparing $Q_p$ to $K_p$ , in which direction will the reaction proceed to reach equilibrium?

The reaction will proceed toward reactants.
The system is already at equilibrium.
The reaction will proceed toward products. (correct answer)
The reaction will proceed toward reactants until the concentrations of all species are equal.
The reaction will proceed toward products until all reactants are consumed.

Explanation: This question tests understanding of reaction quotient and equilibrium constant. For the reaction NH₃(g) + HCl(g) ⇌ NH₄Cl(s), we calculate Q_p = 1/(P_NH₃ × P_HCl) = 1/((0.10)(0.10)) = 100, noting that solids don't appear in Q. Since Q_p (100) < K_p (2.0×10⁴ = 20,000), the system has too little product relative to equilibrium. When Q < K, the reaction must shift toward products to increase Q until it equals K. A common misconception is including solids in the Q expression, but only gases and aqueous species appear in equilibrium expressions. To solve heterogeneous equilibrium problems: write Q excluding pure solids/liquids, compare to K, and remember that very large K values indicate strong product formation.

Question 14

Consider the reversible reaction $\mathrm{H_2(g) + I_2(g) \rightleftharpoons 2\,HI(g)}$ at a certain temperature, where $K_c = 50$ . A reaction mixture is prepared with $[\mathrm{H_2}] = 0.10\,\mathrm{M}$ , $[\mathrm{I_2}] = 0.10\,\mathrm{M}$ , and $[\mathrm{HI}] = 0.20\,\mathrm{M}$ , and the system is not initially at equilibrium. Based on comparing $Q_c$ to $K_c$ , in which direction will the reaction proceed to reach equilibrium?

The reaction will proceed toward reactants.
The reaction will proceed toward products. (correct answer)
The system is already at equilibrium.
The reaction will proceed toward reactants until the concentrations of all species are equal.
The reaction will proceed toward products until all reactants are consumed.

Explanation: This question tests understanding of reaction quotient and equilibrium constant. For the reaction H₂(g) + I₂(g) ⇌ 2HI(g), we calculate Q_c = [HI]²/([H₂][I₂]) = (0.20)²/((0.10)(0.10)) = 4.0. Since Q_c (4.0) < K_c (50), the system has too little product relative to equilibrium. When Q < K, the reaction must shift toward products to increase Q until it equals K. A common error is assuming equal concentrations mean equilibrium, but equilibrium depends on the K value, not concentration equality. To predict reaction direction: always calculate Q first, compare it to K, and remember the reaction proceeds in the direction that moves Q toward K.

Question 15

For the reversible reaction $\mathrm{2\,NO(g) + Cl_2(g) \rightleftharpoons 2\,NOCl(g)}$ at a certain temperature, $K_c = 1.0\times10^{4}$ . A reaction mixture is prepared with $[\mathrm{NO}] = 0.10\,\mathrm{M}$ , $[\mathrm{Cl_2}] = 0.10\,\mathrm{M}$ , and $[\mathrm{NOCl}] = 0.10\,\mathrm{M}$ , and the system is not initially at equilibrium. Based on comparing $Q_c$ to $K_c$ , in which direction will the reaction proceed to reach equilibrium?

The reaction will proceed toward reactants.
The reaction will proceed toward products. (correct answer)
The system is already at equilibrium.
The reaction will proceed toward reactants until the concentrations of all species are equal.
The reaction will proceed toward products until all reactants are consumed.

Explanation: This question tests understanding of reaction quotient and equilibrium constant. For the reaction 2NO(g) + Cl₂(g) ⇌ 2NOCl(g), we calculate Q_c = [NOCl]²/([NO]²[Cl₂]) = (0.10)²/((0.10)²(0.10)) = 100. Since Q_c (100) < K_c (1.0×10⁴ = 10,000), the system has too little product relative to equilibrium. When Q < K, the reaction must shift toward products to increase Q until it equals K. A common error is thinking equal initial concentrations mean equilibrium, but the large K value indicates products are strongly favored at equilibrium. To predict reaction direction: calculate Q with proper stoichiometry, compare to K, and recognize that very large K values mean significant product formation is needed.

Question 16

The reversible reaction $\mathrm{CH_3COOH(aq) \rightleftharpoons H^+(aq) + CH_3COO^-(aq)}$ has $K_c = 1.8\times10^{-5}$ at a certain temperature. A solution is prepared with $[\mathrm{H^+}] = 1.0\times10^{-3}\,\mathrm{M}$ , $[\mathrm{CH_3COO^-}] = 1.0\times10^{-3}\,\mathrm{M}$ , and $[\mathrm{CH_3COOH}] = 1.0\times10^{-2}\,\mathrm{M}$ , so the system is not initially at equilibrium. Based on comparing $Q_c$ (calculated from the given concentrations) and $K_c$ , in which direction will the reaction proceed to reach equilibrium?

the reaction will proceed toward products until the reactants are used up
the reaction will proceed toward reactants (correct answer)
the reaction will proceed toward reactants until the concentrations of all species are equal
the system is already at equilibrium
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 = [H^+][CH_3COO^-] / [CH_3COOH] = ( $1.0×10^{-3}$ )( $1.0×10^{-3}$ ) / ( $1.0×10^{-2}$ ) = 1.0×10^{-4}. Here, Q_c = 1.0×10^{-4} is greater than K_c = 1.8×10^{-5}, 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 17

The reversible reaction $\mathrm{CO(g) + Cl_2(g) \rightleftharpoons COCl_2(g)}$ has $K_c = 8.0$ at a certain temperature. A mixture is prepared with $Q_c = 0.20$ , so the system is not initially at equilibrium. Based on the comparison of $Q_c$ and $K_c$ , in which direction will the reaction proceed to reach equilibrium?

the reaction will proceed toward products until the reactants are used up
the reaction will proceed toward reactants
the reaction will proceed toward products (correct answer)
the system is already at equilibrium
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 = [COCl_2] / ([CO][ $Cl_2$ ]). Here, Q_c = 0.20 is less than K_c = 8.0, 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 18

For the reversible reaction $\mathrm{N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)}$ at a certain temperature, $K_c = 4.0\times10^2$ . A mixture is prepared such that the reaction quotient is $Q_c = 2.0\times10^3$ , so the system is not initially at equilibrium. Based on the comparison of $Q_c$ and $K_c$ , in which direction will the reaction proceed to reach equilibrium?

the system is already at equilibrium
the reaction will proceed toward products
the reaction will proceed toward products until the reactants are used up
the reaction will proceed toward reactants until the concentrations of all species are equal
the reaction will proceed toward reactants (correct answer)

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 = [NH_3]^2 / ([ $N_2$ ][ $H_2$ ]^3). Here, Q_c = 2.0×10^3 is greater than K_c = 4.0×10^2, 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 19

For the reversible reaction $\mathrm{CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)}$ at a certain temperature, $K_c = 0.80$ (solids are omitted from $Q_c$ ). A sealed container is prepared with $[\mathrm{CO_2}] = 0.20\,\mathrm{M}$ , so the system is not initially at equilibrium. Based on comparing $Q_c$ (calculated from the given concentration) and $K_c$ , in which direction will the reaction proceed to reach equilibrium?

the reaction will proceed toward products (correct answer)
the reaction will proceed toward products until the reactants are used up
the reaction will proceed toward reactants
the reaction will proceed toward reactants until the concentrations of all species are equal
the system is already at equilibrium

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 = [CO_2] = 0.20 (omitting solids). Here, Q_c = 0.20 is less than K_c = 0.80, 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 20

For the reversible reaction $\mathrm{2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)}$ at a certain temperature, $K_c = 1.0 \times 10^3$ . A reaction mixture is prepared with $Q_c = 5.0 \times 10^3$ , so the system is not at equilibrium initially. Based on the comparison of $Q_c$ and $K_c$ , in which direction will the reaction proceed to reach equilibrium?

the system is already at equilibrium
the reaction will proceed toward reactants until all products are consumed
the reaction will proceed toward products
the reaction will proceed toward products because $Q_c > K_c$
the reaction will proceed toward reactants (correct answer)

Explanation: This question requires comparing reaction quotient and equilibrium constant to predict reaction direction. When Q > K (here, Qc = 5000 while Kc = 1000), the system has too many products relative to reactants compared to equilibrium. To reach equilibrium, some products must convert back to reactants, so the reaction proceeds in the reverse direction (toward reactants). The reaction will continue until Q decreases to equal K. Choice D incorrectly states the reaction goes forward when Q > K, which is a common misconception about the Q/K relationship. Remember the strategy: when Q > K, the reaction shifts left (toward reactants); when Q < K, the reaction shifts right (toward products).