Reaction Quotient and Equilibrium Constant - AP Chemistry

1. 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?
   Answer: the reaction will proceed toward reactants 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.
2. 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?
   Answer: The reaction will proceed toward products. 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.
3. Consider the reversible reaction $\mathrm{CH_3COOH(aq) \rightleftharpoons H^+(aq) + CH_3COO^-(aq)}$ at a certain temperature, where $K_c = 1.8 \times 10^{-5}$. A solution is prepared with $[\mathrm{CH_3COOH}] = 0.10\,\mathrm{M}$, $[\mathrm{H^+}] = 1.0 \times 10^{-4}\,\mathrm{M}$, and $[\mathrm{CH_3COO^-}] = 1.0 \times 10^{-4}\,\mathrm{M}$, so the system is not at equilibrium initially. Using $Q_c = \dfrac{[\mathrm{H^+}][\mathrm{CH_3COO^-}]}{[\mathrm{CH_3COOH}]}$ and comparing $Q_c$ to $K_c$, in which direction will the reaction proceed to reach equilibrium?
   Answer: the reaction will proceed toward products This problem tests understanding of reaction quotient and equilibrium constant for a weak acid dissociation. Calculate Qc = [H⁺][CH₃COO⁻]/[CH₃COOH] = (1.0×10⁻⁴)(1.0×10⁻⁴)/(0.10) = 1.0×10⁻⁷. Since Q < K (1.0×10⁻⁷ < 1.8×10⁻⁵), the system has too few products (ions) compared to equilibrium. The reaction must proceed forward (toward products) to increase the ion concentrations until Q = K. Choice E incorrectly suggests that K being small means the reaction goes forward, but the direction depends only on the Q/K comparison. Remember: calculate Q using current concentrations, then compare to K to determine if the reaction shifts forward (Q < K) or backward (Q > K).
4. 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?
   Answer: The reaction will proceed toward reactants. 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.
5. 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?
   Answer: The reaction will proceed toward products. 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.
6. For the reversible reaction $\mathrm{2\,SO_2(g) + O_2(g) \rightleftharpoons 2\,SO_3(g)}$ at a certain temperature, $K_c = 1.0\times10^{2}$. A reaction mixture is prepared with $[\mathrm{SO_2}] = 0.20\,\mathrm{M}$, $[\mathrm{O_2}] = 0.10\,\mathrm{M}$, and $[\mathrm{SO_3}] = 0.10\,\mathrm{M}$, 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?
   Answer: The reaction will proceed toward products. This question tests understanding of reaction quotient and equilibrium constant. For the reaction 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), we calculate $Q_c$ = [SO₃]²/([SO₂]²[O₂]) = (0.10)²/((0.20)²(0.10)) = 2.5. Since $Q_c$ (2.5) < $K_c$ (100), 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 reactions proceed until all reactants are consumed, but equilibrium is a dynamic state where both reactants and products remain present. To solve equilibrium problems: calculate Q using initial concentrations with proper stoichiometric exponents, compare to K, and remember that Q < K means shift right (toward products).
7. 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?
   Answer: The reaction will proceed toward reactants. 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.
8. Consider the reversible reaction $\mathrm{PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)}$ at a certain temperature, where $K_c = 0.10$. A reaction mixture is prepared with $Q_c = 0.50$, 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?
   Answer: the reaction will proceed toward reactants This problem involves comparing reaction quotient and equilibrium constant for a decomposition reaction. Given that Qc = 0.50 and Kc = 0.10, we see that Q > K. When Q exceeds K, the system has too many products (PCl₃ and Cl₂) relative to reactant (PCl₅) compared to equilibrium. The reaction must shift backward (toward reactants) to decrease Q until it equals K. Choice D incorrectly suggests that Q being less than 1 determines the direction, but only the Q/K comparison matters, not the absolute value of Q. To solve equilibrium direction problems, always compare Q to K first: if Q > K, the reaction proceeds toward reactants.
9. 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?
   Answer: the reaction will proceed toward reactants 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.
10. For the reversible reaction $\mathrm{CH_3COOH(aq) \rightleftharpoons H^+(aq) + CH_3COO^-(aq)}$ at $25^\circ\mathrm{C}$, $K_a = 1.8\times10^{-5}$. A solution is prepared with initial concentrations $[\mathrm{CH_3COOH}]=0.10\,\mathrm{M}$, $[\mathrm{H^+}]=1.0\times10^{-5}\,\mathrm{M}$, and $[\mathrm{CH_3COO^-}]=1.0\times10^{-5}\,\mathrm{M}$, and the system is not at equilibrium. Based on comparing $Q$ to $K_a$, in which direction will the reaction proceed to reach equilibrium?
    Answer: the reaction will proceed toward products This question tests understanding of reaction quotient and equilibrium constant. To determine the direction of reaction, we calculate Q = [H⁺][CH₃COO⁻]/[CH₃COOH] = (1.0×10⁻⁵)(1.0×10⁻⁵)/(0.10) = 1.0×10⁻¹⁰/0.10 = 1.0×10⁻⁹. Since Q (1.0×10⁻⁹) < $K_a$ (1.8×10⁻⁵), the system has too little product relative to equilibrium, so the reaction must shift toward products (more ionization) to increase Q until it equals $K_a$. Choice E incorrectly claims the reaction proceeds toward products because $K_a$ < 1, but the direction depends on Q vs K comparison, not on K's magnitude. 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.
11. For the reversible reaction $\mathrm{Fe^{3+}(aq) + SCN^-(aq) \rightleftharpoons FeSCN^{2+}(aq)}$ at a certain temperature, $K_c = 1.0 \times 10^2$. A mixture is prepared with $[\mathrm{Fe^{3+}}] = 0.10\,\mathrm{M}$, $[\mathrm{SCN^-}] = 0.10\,\mathrm{M}$, and $[\mathrm{FeSCN^{2+}}] = 5.0\,\mathrm{M}$, so the system is not at equilibrium initially. Using $Q_c = \dfrac{[\mathrm{FeSCN^{2+}}]}{[\mathrm{Fe^{3+}}][\mathrm{SCN^-}]}$ and comparing $Q_c$ to $K_c$, in which direction will the reaction proceed to reach equilibrium?
    Answer: the reaction will proceed toward reactants This question involves comparing reaction quotient and equilibrium constant for a complexation reaction. Calculate Qc = [FeSCN²⁺]/([Fe³⁺][SCN⁻]) = 5.0/(0.10 × 0.10) = 5.0/0.01 = 500. Since Q > K (500 > 100), the system has too much product (complex ion) compared to equilibrium conditions. The reaction must shift backward (toward reactants) to decrease the complex concentration until Q equals K. Choice D incorrectly claims that a large Q means the reaction goes forward, when actually Q > K always means reverse reaction. To determine equilibrium shift direction, always calculate Q first, then compare: if Q > K, the reaction proceeds toward reactants.
12. 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?
    Answer: the reaction will proceed toward products 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.
13. 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?
    Answer: The reaction will proceed toward products. 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.
14. For the reversible reaction $\mathrm{2NO_2(g) \rightleftharpoons N_2O_4(g)}$ at a certain temperature, $K_c = 6.0$. A reaction mixture is prepared with initial concentrations $[\mathrm{NO_2}]=0.50\,\mathrm{M}$ and $[\mathrm{N_2O_4}]=2.0\,\mathrm{M}$, 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?
    Answer: the reaction will proceed toward reactants This question tests understanding of reaction quotient and equilibrium constant. To determine the direction of reaction, we calculate $Q_c$ = [N₂O₄]/[NO₂]² = 2.0/(0.50)² = 2.0/0.25 = 8.0. Since $Q_c$ (8.0) > $K_c$ (6.0), the system has too much product relative to equilibrium, so the reaction must shift toward reactants to decrease $Q_c$ until it equals $K_c$. Choice D incorrectly relates the direction to the number of moles of gas, but equilibrium direction depends only on Q vs K comparison, not on stoichiometry. The key strategy is to calculate Q, compare it to K, and remember that when Q > K, the reaction shifts left (toward reactants) to reach equilibrium.
15. 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?
    Answer: the reaction will proceed toward products 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.
16. 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?
    Answer: The reaction will proceed toward products. 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.
17. For the reversible reaction $\mathrm{PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)}$ at a certain temperature, $K_c = 0.040$. A reaction mixture is prepared with initial concentrations $[\mathrm{PCl_5}]=0.50\,\mathrm{M}$, $[\mathrm{PCl_3}]=0.10\,\mathrm{M}$, and $[\mathrm{Cl_2}]=0.10\,\mathrm{M}$, 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?
    Answer: the reaction will proceed toward products This question tests understanding of reaction quotient and equilibrium constant. To determine the direction of reaction, we calculate $Q_c$ = [PCl₃][Cl₂]/[PCl₅] = (0.10)(0.10)/(0.50) = 0.01/0.50 = 0.020. Since $Q_c$ (0.020) < $K_c$ (0.040), 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 D incorrectly relates the direction to $K_c$ being less than 1, but the direction depends on comparing Q to K, not on K's magnitude relative to 1. 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.
18. 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?
    Answer: the reaction will proceed toward reactants 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.
19. 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?
    Answer: the reaction will proceed toward products 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.
20. For the reversible reaction $\mathrm{N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)}$ at a certain temperature, $K_c = 4.0 \times 10^2$. A mixture is prepared such that the reaction quotient is $Q_c = 2.0 \times 10^1$, 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?
    Answer: the reaction will proceed toward products This question tests understanding of reaction quotient and equilibrium constant. When Q < K (here, Qc = 20 while Kc = 400), the system has too few products relative to reactants compared to the equilibrium state. To reach equilibrium, the reaction must produce more products, so it proceeds in the forward direction (toward products). The system will shift to increase Q until Q = K. A common misconception is reversing the rule and thinking that because Q < K the reaction goes backward (choice C), but Q < K means there are too few products, so the reaction proceeds forward. To solve these problems, always compare Q to K: if Q < K, the reaction goes forward; if Q > K, the reaction goes backward.

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