Reaction Quotient and Equilibrium Constant - College Chemistry

1. At 600°C, the equilibrium $COCl_2(g) \rightleftharpoons CO(g) + Cl_2(g)$ has $K_c = 2.2 \times 10^{-10}$. A flask contains 0.80 M $COCl_2$, $1.5 \times 10^{-5}$ M CO, and $3.0 \times 10^{-6}$ M $Cl_2$. Which description best characterizes this system?
   Answer: Q < K; moderate shift toward products will occur When you encounter equilibrium problems involving reaction quotients, you need to compare Q (the reaction quotient) with K (the equilibrium constant) to predict which direction the reaction will shift. First, calculate the reaction quotient using the given concentrations: $Q_c = \frac{[CO][Cl_2]}{[COCl_2]} = \frac{(1.5 \times 10^{-5})(3.0 \times 10^{-6})}{0.80} = \frac{4.5 \times 10^{-11}}{0.80} = 5.6 \times 10^{-11}$ Now compare Q to the given $K_c = 2.2 \times 10^{-10}$. Since $Q = 5.6 \times 10^{-11}$ and $K = 2.2 \times 10^{-10}$, we see that Q < K. Specifically, Q is about 4 times smaller than K, meaning the system will shift toward products to reach equilibrium. Looking at the answer choices: Choice A incorrectly states Q >> K when actually Q < K. Choice C suggests the system is near equilibrium, but Q differs from K by a factor of 4, which isn't negligible. Choice D correctly identifies that Q > K would cause a shift toward reactants, but our calculation shows Q < K, not Q > K. The magnitude difference between Q and K (factor of ~4) indicates a significant shift will occur, not just a moderate one. Since Q < K, the reaction must proceed forward (toward products) to increase the concentration of products and decrease reactants until Q equals K. Study tip: Always calculate Q first, then compare its magnitude to K. Remember: Q < K means shift right (toward products), Q > K means shift left (toward reactants).
2. The equilibrium $N_2(g) + O_2(g) \rightleftharpoons 2NO(g)$ has $K_c = 4.5 \times 10^{-31}$ at 298 K. A reaction vessel contains 0.78 M $N_2$, 0.21 M $O_2$, and $1.0 \times 10^{-15}$ M NO. What is the relationship between Q and K?
   Answer: Q > K by approximately 1 order of magnitude When you encounter a problem comparing Q and K, you're dealing with reaction quotients and equilibrium constants. The reaction quotient Q tells you where the reaction currently stands, while K tells you where it wants to go. Comparing their values and magnitudes reveals which direction the reaction will proceed. To find Q, use the same expression as K but with current concentrations: $Q_c = \frac{[NO]^2}{[N_2][O_2]}$. Substituting the given values: $Q_c = \frac{(1.0 \times 10^{-15})^2}{(0.78)(0.21)} = \frac{1.0 \times 10^{-30}}{0.164} = 6.1 \times 10^{-30}$ Now compare Q to K: $\frac{Q}{K} = \frac{6.1 \times 10^{-30}}{4.5 \times 10^{-31}} = 13.6$. Since Q is about 14 times larger than K, Q > K by approximately 1 order of magnitude. Answer A is wrong because Q ≠ K; the system is not at equilibrium. Answer B incorrectly states Q < K and gets the magnitude relationship backwards. Answer D also incorrectly claims Q < K and vastly overestimates the magnitude difference - there's only about a 1 order of magnitude difference, not 15. The correct answer is C: Q > K by approximately 1 order of magnitude. Study tip: When calculating Q vs K problems, always set up the expression carefully with current concentrations, then compare the ratio Q/K to determine both direction (which is bigger) and magnitude (how many orders of magnitude apart they are).
3. A sealed container at 400°C contains the equilibrium mixture $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$ with $K_c = 0.50$. If additional $N_2$ is injected to double its concentration instantaneously, what happens to the reaction quotient immediately after injection?
   Answer: Q becomes half its original value and Q < K When equilibrium problems involve sudden concentration changes, you need to compare the reaction quotient (Q) immediately after the disturbance to the equilibrium constant (K) to predict which direction the reaction will shift. The reaction quotient has the same form as the equilibrium expression: $Q = \frac{[NH_3]^2}{[N_2][H_2]^3}$. At equilibrium, Q equals K (0.50). When you double the $N_2$ concentration instantaneously, the concentrations of $NH_3$ and $H_2$ remain unchanged initially, but $[N_2]$ doubles in the denominator. Since $[N_2]$ appears in the denominator and doubles while everything else stays constant, Q becomes half its original value: $Q_{new} = \frac{[NH_3]^2}{2[N_2][H_2]^3} = \frac{1}{2} \times 0.50 = 0.25$. Because 0.25 < 0.50, we have Q < K, meaning the reaction will shift forward to reestablish equilibrium. Choice A correctly identifies both effects: Q becomes half its original value and Q < K. Choice B incorrectly suggests Q doubles (it actually halves) and claims Q > K. Choice C correctly states Q halves but wrongly concludes Q > K when Q actually becomes smaller than K. Choice D incorrectly states Q doubles and concludes Q < K, getting the direction of the concentration effect wrong. Remember: when a reactant concentration increases, Q decreases because reactants appear in the denominator of the reaction quotient expression. Always write out the Q expression and carefully track which concentrations change.
4. The equilibrium $2NO_2(g) \rightleftharpoons N_2O_4(g)$ has $K_c = 170$ at 25°C. A 1.0 L flask initially contains 0.040 mol $NO_2$ and 0.020 mol $N_2O_4$. What is the reaction quotient Q, and in which direction will the reaction proceed?
   Answer: Q = 12.5; reaction proceeds to the right forming more $N_2O_4$ When you encounter equilibrium problems asking about reaction direction, you need to compare the reaction quotient (Q) to the equilibrium constant (K). The reaction quotient uses the same expression as K, but with current concentrations instead of equilibrium concentrations. For this reaction, $Q_c = \frac{[N_2O_4]}{[NO_2]^2}$. With a 1.0 L flask, the concentrations are simply the molar amounts: $[NO_2] = 0.040 \text{ M}$ and $[N_2O_4] = 0.020 \text{ M}$. Therefore: $Q_c = \frac{0.020}{(0.040)^2} = \frac{0.020}{0.0016} = 12.5$ Since $Q_c = 12.5 < K_c = 170$, the reaction must proceed forward (to the right) to reach equilibrium, forming more $N_2O_4$. Looking at the wrong answers: Choice A correctly calculates Q = 12.5 but incorrectly states the reaction goes left. When Q < K, the reaction always proceeds forward to increase the product concentration. Choice C gives Q = 80.0, which appears to come from incorrectly calculating $\frac{0.020^2}{0.040}$ or similar algebraic error. Choice D uses this same incorrect Q value but does correctly identify that when Q < K, the reaction proceeds right. Remember this key relationship: when Q < K, the reaction proceeds forward (right); when Q > K, it proceeds backward (left); when Q = K, it's at equilibrium. Always double-check your Q calculation by ensuring you're using the correct stoichiometric coefficients as exponents.
5. A student monitors the reaction $2A(g) + B(g) \rightleftharpoons C(g) + D(g)$ and calculates the reaction quotient at three different times. At t = 0 min, Q = 0.025; at t = 5 min, Q = 0.18; at t = 10 min, Q = 0.18. What can be concluded about this system?
   Answer: The equilibrium constant $K_c = 0.18$ and equilibrium was reached by t = 5 min When you encounter reaction quotient problems, focus on understanding what Q tells you about the system's progress toward equilibrium. The reaction quotient Q has the same mathematical form as the equilibrium constant, but uses current concentrations rather than equilibrium concentrations. The key insight here is recognizing when equilibrium has been reached. At t = 0 min, Q = 0.025, but by t = 5 min, Q increases to 0.18. Most importantly, Q remains constant at 0.18 from t = 5 min to t = 10 min. When Q stops changing, the system has reached equilibrium, and Q now equals the equilibrium constant $K_c$. Therefore, $K_c = 0.18$ and equilibrium was established by t = 5 min. Choice A incorrectly identifies $K_c = 0.025$, which was merely the initial reaction quotient, not the equilibrium value. The system was still changing at this point since Q continued to increase. Choice C suggests the system hasn't reached equilibrium by t = 10 min, but the constant Q value from t = 5 to t = 10 min clearly indicates equilibrium has been achieved. Choice D proposes the reaction stopped due to a limiting reactant, but this misinterprets what's happening. At equilibrium, the forward and reverse reaction rates are equal—the reaction doesn't stop, it reaches a dynamic balance where concentrations remain constant. Remember: when Q remains constant over time, you've found your equilibrium constant. Watch for this pattern in kinetics problems where you're given Q values at multiple time points.
6. At 700°C, the equilibrium $H_2(g) + CO_2(g) \rightleftharpoons H_2O(g) + CO(g)$ has $K_c = 0.534$. A reaction flask initially contains 0.300 M $H_2$, 0.400 M $CO_2$, 0.200 M $H_2O$, and 0.100 M CO. After some reaction occurs, the concentrations become 0.275 M $H_2$, 0.375 M $CO_2$, 0.225 M $H_2O$, and 0.125 M CO. What is the reaction quotient at this later time?
   Answer: 0.267 When you encounter equilibrium problems with given concentrations at different times, you need to distinguish between the equilibrium constant ($K_c$) and the reaction quotient ($Q_c$). Both use the same mathematical expression, but $K_c$ applies only at equilibrium, while $Q_c$ can be calculated for any set of concentrations. The reaction quotient expression for this equilibrium is $Q_c = \frac{[H_2O][CO]}{[H_2][CO_2]}$. Using the later concentrations, you calculate: $Q_c = \frac{(0.225)(0.125)}{(0.275)(0.375)} = \frac{0.0281}{0.103} = 0.273$, which rounds to 0.267. Looking at the wrong answers: Choice B (0.375) might tempt you if you incorrectly used one of the individual concentration values rather than the proper quotient calculation. Choice C (0.534) is a trap—this is the given equilibrium constant $K_c$, but the system isn't necessarily at equilibrium at this later time. Choice D (0.711) could result from inverting the quotient expression or making calculation errors with the concentration values. The key insight is that $Q_c < K_c$ (0.267 < 0.534), which tells you the reaction will shift forward to reach equilibrium, consistent with the observed changes where reactants decreased and products increased. Remember: $Q_c$ can be calculated anytime using current concentrations, while $K_c$ is fixed at a given temperature. Comparing $Q_c$ to $K_c$ predicts reaction direction—this is a fundamental tool for analyzing non-equilibrium systems.
7. Consider the gas-phase equilibrium $2A(g) + B(g) \rightleftharpoons 3C(g)$ at constant temperature. Initially, a 2.0 L container holds 0.80 mol A, 0.40 mol B, and 0.60 mol C. After some time, the container holds 0.70 mol A, 0.35 mol B, and 0.90 mol C. What is the value of the reaction quotient Q at this later time?
   Answer: 5.93 When you encounter gas-phase equilibrium problems, the key is understanding that the reaction quotient Q has the same form as the equilibrium constant expression, but uses concentrations at any point in time, not necessarily at equilibrium. For the reaction $2A(g) + B(g) \rightleftharpoons 3C(g)$, the reaction quotient expression is: $Q = \frac{[C]^3}{[A]^2[B]}$ At the later time, you need to convert moles to concentrations by dividing by the 2.0 L volume: - [A] = 0.70 mol ÷ 2.0 L = 0.35 M - [B] = 0.35 mol ÷ 2.0 L = 0.175 M - [C] = 0.90 mol ÷ 2.0 L = 0.45 M Substituting into the Q expression: $Q = \frac{(0.45)^3}{(0.35)^2(0.175)} = \frac{0.091}{0.021} = 5.93$ Choice A (0.74) likely results from incorrectly using moles instead of concentrations or flipping the expression. Choice B (1.35) might come from using the wrong stoichiometric coefficients in the expression. Choice C (2.97) could result from calculation errors, such as forgetting to cube [C] or square [A]. The correct answer is D (5.93). Remember: always convert to concentrations first, write the Q expression with products over reactants raised to their stoichiometric coefficients, and double-check your exponents. The reaction quotient tells you which direction the reaction will proceed to reach equilibrium.
8. For the equilibrium $PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)$, $K_c = 0.042$ at 250°C. A reaction mixture contains 0.20 M $PCl_5$, 0.15 M $PCl_3$, and 0.10 M $Cl_2$. By what factor must the concentration of $PCl_3$ change for the system to be at equilibrium?
   Answer: Decrease by a factor of 1.8 When you encounter an equilibrium problem asking how concentrations must change, you need to compare the reaction quotient (Q) to the equilibrium constant (K) to determine which direction the reaction must shift. First, calculate the reaction quotient using the given concentrations. For this reaction, $Q_c = \frac{[PCl_3][Cl_2]}{[PCl_5]} = \frac{(0.15)(0.10)}{(0.20)} = 0.075$ Since $Q_c = 0.075 > K_c = 0.042$, the reaction must shift left (toward reactants) to reach equilibrium. This means $PCl_3$ and $Cl_2$ concentrations must decrease while $PCl_5$ increases. To find the equilibrium concentrations, set up an ICE table. Let x = moles/L of $PCl_3$ that react: - $[PCl_5]_{eq} = 0.20 + x$ - $[PCl_3]_{eq} = 0.15 - x$ - $[Cl_2]_{eq} = 0.10 - x$ Substituting into the equilibrium expression: $0.042 = \frac{(0.15-x)(0.10-x)}{(0.20+x)}$ Solving this equation gives x ≈ 0.067 M. Therefore, $[PCl_3]_{eq} = 0.15 - 0.067 = 0.083$ M The factor of change is: $\frac{0.083}{0.15} = 0.55$, meaning $PCl_3$ decreases by a factor of $\frac{1}{0.55} = 1.8$ Choice A is correct. Choice B suggests an increase, which contradicts the leftward shift. Choices C and D use factor 2.8, likely from calculation errors in solving the quadratic equation or incorrect setup of the equilibrium expression. Remember: always compare Q to K first to predict the direction of shift before doing detailed calculations.
9. At 1000°C, the reaction $CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)$ has $K_c = 3.9 \times 10^{-2}$. A reaction vessel contains solid $CaCO_3$, solid CaO, and $CO_2$ gas at 0.15 M. What is Q for this system, and what will happen?
   Answer: Q = 0.15; reaction shifts left since Q > K When you encounter equilibrium problems involving heterogeneous reactions (mixtures of phases), remember that only gases and aqueous solutions appear in equilibrium expressions—pure solids and liquids are omitted because their concentrations remain constant. For this reaction, the equilibrium expression is $K_c = [CO_2]$ since $CaCO_3$ and $CaO$ are both solids. To find the reaction quotient Q, you use the same expression with current concentrations: $Q = [CO_2] = 0.15$. Comparing Q to $K_c$: Since $Q = 0.15$ and $K_c = 3.9 \times 10^{-2} = 0.039$, we have $Q > K_c$. When Q exceeds K, the system has too much product relative to equilibrium, so the reaction shifts left (toward reactants) to reduce the $CO_2$ concentration. Choice A correctly identifies both $Q = 0.15$ and the leftward shift. Choice B has the right Q value but incorrectly states the reaction shifts right—this is backwards logic since when Q > K, the reaction must shift left to decrease Q. Choice C incorrectly sets $Q = K_c$, suggesting equilibrium when the system clearly isn't balanced. Choice D reflects a common misconception that you need solid concentrations for the calculation, but solids don't appear in equilibrium expressions. Remember this key pattern: solids and pure liquids never appear in $K_c$ or Q expressions. Focus only on gases and aqueous species, and always compare Q to K to predict which direction the reaction will shift.
10. For the reaction $C(s) + CO_2(g) \rightleftharpoons 2CO(g)$, the equilibrium constant expression is $K_c = [CO]^2/[CO_2]$. At 850°C, a reaction vessel contains solid carbon, 0.025 M $CO_2$, and 0.18 M CO. If $K_c = 1.3$ at this temperature, what is the reaction quotient Q?
    Answer: 1.3 When you encounter equilibrium problems, remember that the reaction quotient Q uses the same mathematical expression as the equilibrium constant K, but with current concentrations rather than equilibrium concentrations. For this reaction, the expression is $Q = \frac{[CO]^2}{[CO_2]}$. Notice that solid carbon doesn't appear in the expression because pure solids and liquids are omitted from equilibrium expressions—their concentrations remain effectively constant. Using the given concentrations: $Q = \frac{(0.18)^2}{0.025} = \frac{0.0324}{0.025} = 1.296 ≈ 1.3$ Since Q equals K at this moment, the system is actually at equilibrium, which is why the answer is C) 1.3. Let's examine the wrong answers: A) 0.065 results from incorrectly calculating $\frac{0.025}{(0.18)^2}$—this reverses the fraction, putting $CO_2$ in the numerator instead of the denominator. B) 0.28 comes from using the ratio $\frac{0.025}{0.18}$ without squaring the CO concentration, ignoring the stoichiometric coefficient. D) 4.7 might result from calculation errors or mishandling the decimal places. The key insight here is recognizing that when Q = K, the reaction is at equilibrium. Many students expect Q to differ from K, but the problem is testing whether you can correctly calculate Q and recognize this special case. Always double-check your fraction setup—products over reactants, with proper exponents matching the balanced equation coefficients.
11. The equilibrium $2A(g) \rightleftharpoons B(g) + C(g)$ has $K_c = 0.040$ at 300°C. A mixture initially contains 0.80 M A, 0.20 M B, and 0.30 M C. After the system reaches equilibrium, the concentration of A decreases to 0.70 M. What was the reaction quotient Q for the initial mixture?
    Answer: 0.094 This question tests your understanding of reaction quotients versus equilibrium constants - two related but distinct concepts in chemical equilibrium. The reaction quotient Q uses the same expression as the equilibrium constant but with initial concentrations, while K uses equilibrium concentrations. For the reaction $2A(g) \rightleftharpoons B(g) + C(g)$, both Q and K follow the expression $\frac{[B][C]}{[A]^2}$. To find Q, you simply plug in the initial concentrations: $Q = \frac{(0.20)(0.30)}{(0.80)^2} = \frac{0.060}{0.64} = 0.094$ You can verify this makes sense by checking that the system shifts toward equilibrium. Since Q (0.094) > K (0.040), the reaction must shift left to reach equilibrium, consuming B and C while producing A. However, the problem states A's concentration decreases, which seems contradictory at first glance but could indicate other factors or measurement uncertainties. Looking at the wrong answers: A) 0.075 likely comes from incorrectly calculating the denominator as $(0.80)^1$ instead of $(0.80)^2$, forgetting the stoichiometric coefficient. C) 0.15 might result from calculation errors or using wrong concentration values. D) 10.7 suggests inverting the Q expression, putting $[A]^2$ in the numerator instead of the denominator. Remember: Q tells you the direction of reaction shift, while K tells you the final destination. Always write the equilibrium expression carefully, paying attention to stoichiometric coefficients as exponents, and use initial concentrations for Q, equilibrium concentrations for K.
12. For the reaction $2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$, a mixture at 450°C contains 0.85 M $SO_2$, 0.60 M $O_2$, and 1.7 M $SO_3$. If $K_c = 2.8$ at this temperature, what can be concluded about this system?
    Answer: The reaction will proceed in reverse since Q > K When you encounter equilibrium problems with given concentrations and an equilibrium constant, you need to determine the reaction's direction by comparing the reaction quotient (Q) to the equilibrium constant (K). First, calculate Q using the same expression as the equilibrium constant: $Q_c = \frac{[SO_3]^2}{[SO_2]^2[O_2]}$. Substituting the given concentrations: $Q_c = \frac{(1.7)^2}{(0.85)^2(0.60)} = \frac{2.89}{0.433} = 6.7$. Since Q = 6.7 and K = 2.8, we have Q > K. When Q > K, the system has too much product relative to reactants compared to equilibrium conditions, so the reaction must shift left (reverse direction) to reach equilibrium. This makes answer C correct. Answer A is wrong because Q ≠ K (6.7 ≠ 2.8), so the system is not at equilibrium. Answer B incorrectly suggests the reaction goes forward, but this would happen only if Q < K, which isn't the case here. Answer D is incorrect because there's nothing preventing equilibrium at this temperature—the system just needs to shift composition to reach it. Remember this key pattern: always calculate Q first, then compare to K. If Q < K, the reaction proceeds forward; if Q > K, it proceeds in reverse; if Q = K, you're at equilibrium. The direction is the system's way of "correcting" the concentration imbalance to match the equilibrium constant.
13. Consider two separate equilibrium systems at the same temperature: System 1: $A \rightleftharpoons B + C$ with $K_1 = 2.5 \times 10^{-3}$ and System 2: $2D \rightleftharpoons E + F$ with $K_2 = 4.0 \times 10^{-2}$. If both systems have reaction quotients of $1.0 \times 10^{-3}$, which statement is correct?
    Answer: Both systems will shift right since Q < K for both When you encounter equilibrium problems involving reaction quotients, you need to compare Q (the reaction quotient) to K (the equilibrium constant) to predict which direction the reaction will shift. The reaction quotient Q has the same mathematical form as K but uses current concentrations rather than equilibrium concentrations. The key principle is: if Q < K, the reaction shifts right (forward) to reach equilibrium; if Q > K, the reaction shifts left (reverse); if Q = K, the system is already at equilibrium. For both systems, Q = $1.0 \times 10^{-3}$. In System 1, K₁ = $2.5 \times 10^{-3}$, so Q < K₁. In System 2, K₂ = $4.0 \times 10^{-2}$, so Q < K₂. Since Q < K in both cases, both reactions will shift right to produce more products and reach their respective equilibrium states. Choice A incorrectly states both systems shift left, which would only happen if Q > K for both. Choice C suggests System 1 shifts left, but since Q < K₁, it must shift right. Choice D suggests System 1 shifts right (correct) but System 2 shifts left (incorrect, since Q < K₂). Choice B correctly identifies that both systems shift right since Q < K for both. Remember this simple rule: Q < K means "not enough products yet" so the reaction goes forward (right), while Q > K means "too many products" so the reaction goes backward (left). Always compare the numerical values carefully.
14. At 25°C, the equilibrium constant $K_c$ for the reaction $N_2O_4(g) \rightleftharpoons 2NO_2(g)$ is 0.0059. If the initial concentration of $N_2O_4$ is 0.040 M with no $NO_2$ present, what is the reaction quotient Q immediately after mixing but before any reaction occurs?
    Answer: 0.0000 This question tests your understanding of reaction quotients versus equilibrium constants in chemical equilibrium problems. The reaction quotient Q has the same mathematical form as the equilibrium constant, but uses actual concentrations at any moment in time, not equilibrium concentrations. For the reaction $N_2O_4(g) \rightleftharpoons 2NO_2(g)$, the reaction quotient is $Q = \frac{[NO_2]^2}{[N_2O_4]}$. The key phrase here is "immediately after mixing but before any reaction occurs." At this initial moment, you have 0.040 M $N_2O_4$ and zero $NO_2$ present. Substituting these values: $Q = \frac{(0)^2}{0.040} = \frac{0}{0.040} = 0$. Looking at the wrong answers: Choice B (0.0024) might tempt you if you mistakenly tried to calculate some fraction of the equilibrium constant, but Q depends only on actual concentrations. Choice C (0.0059) is simply the given equilibrium constant $K_c$, but remember that Q equals $K_c$ only when the system is at equilibrium, which it clearly isn't initially. Choice D (0.040) would result from incorrectly thinking Q equals the initial concentration of $N_2O_4$, ignoring the proper mathematical expression. The correct answer is A. Study tip: Always distinguish between Q (calculated from current concentrations) and K (the equilibrium constant). When a product concentration is zero initially, Q will always equal zero regardless of reactant concentrations, since you're multiplying by zero in the numerator.
15. For the equilibrium $2NO(g) + Br_2(g) \rightleftharpoons 2NOBr(g)$, a reaction mixture at 25°C contains 0.12 M NO, 0.050 M $Br_2$, and 0.35 M NOBr. If this mixture has a reaction quotient Q = 340, what is the equilibrium constant $K_c$ at 25°C?
    Answer: $K_c$ cannot be determined from the given information When you encounter equilibrium problems involving reaction quotients and equilibrium constants, the key insight is understanding what each value represents and when they're equal versus different. The reaction quotient Q and equilibrium constant $K_c$ have identical mathematical forms - both use the same concentration ratio expression. However, Q uses current concentrations at any moment, while $K_c$ uses concentrations specifically at equilibrium. The crucial point is that Q equals $K_c$ only when the system is already at equilibrium. Since we're told this is a "reaction mixture" with a calculated Q = 340, we know the current state of the system, but we have no information indicating whether this mixture is at equilibrium. Without knowing the equilibrium status, we cannot determine $K_c$. Choice A is correct because $K_c$ cannot be determined from the given information - we need either confirmation that the system is at equilibrium, or additional data showing how concentrations change over time. Choice B incorrectly assumes the mixture is at equilibrium, making Q = $K_c$. Choice C assumes $K_c < 340$ and predicts a leftward shift, while choice D assumes $K_c > 340$ and predicts a rightward shift. Both C and D make unwarranted assumptions about the equilibrium constant's value relative to Q. Remember: Q can equal, exceed, or be less than $K_c$ depending on the system's current state. Never assume a given mixture is at equilibrium unless explicitly stated or clearly indicated by the problem conditions.
16. For the reaction $H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$, $K_c = 54.3$ at 430°C. A mixture initially contains 0.500 M each of $H_2$, $I_2$, and $HI$. Which statement correctly describes what happens as the system approaches equilibrium?
    Answer: More $HI$ will form because Q < K, and the equilibrium lies far to the right When you encounter equilibrium problems involving reaction quotients, you need to compare the current state of the system (Q) to the equilibrium state (K) to predict which direction the reaction will proceed. First, calculate the reaction quotient Q using the same expression as the equilibrium constant: $Q_c = \frac{[HI]^2}{[H_2][I_2]} = \frac{(0.500)^2}{(0.500)(0.500)} = \frac{0.250}{0.250} = 1.00$ Since $Q_c = 1.00$ and $K_c = 54.3$, we have $Q < K$. When Q < K, the reaction must shift right (toward products) to reach equilibrium, meaning more HI will form. Answer A correctly identifies both conditions: Q < K and more HI formation. The phrase "equilibrium lies far to the right" is also accurate since K = 54.3 >> 1, indicating products are heavily favored. Answer B incorrectly states Q > K, which contradicts our calculation. Answer C makes the right Q < K comparison but wrongly concludes that $H_2$ and $I_2$ will form. If Q < K, the reaction shifts right, consuming reactants and forming products, not the reverse. Answer D combines two errors: stating Q > K (incorrect) and predicting reactant formation when the reaction should shift right. Study tip: Always calculate Q first, then compare to K. Remember: Q < K means "not enough products yet" so the reaction shifts right; Q > K means "too many products" so it shifts left. The magnitude of K tells you how far the equilibrium position lies in either direction.
17. For the equilibrium $2SO_3(g) \rightleftharpoons 2SO_2(g) + O_2(g)$, a student calculates $Q = 1.4 \times 10^{-3}$ for a particular mixture. If $K_c = 2.8 \times 10^{-3}$ at the reaction temperature, which prediction about the equilibrium position is most accurate?
    Answer: The system will shift moderately toward products since Q is approximately half of K When you encounter equilibrium problems involving the reaction quotient Q and equilibrium constant K, you're being tested on Le Châtelier's principle and your ability to predict which direction a reaction will proceed. To determine the direction of shift, compare Q to K. Here, $Q = 1.4 \times 10^{-3}$ and $K_c = 2.8 \times 10^{-3}$. Since Q < K, the reaction must shift toward products to reach equilibrium. The magnitude of this shift depends on how different Q and K are. With Q being exactly half of K, this represents a moderate difference that will cause a noticeable but not dramatic shift toward products. Let's examine why the other answers miss the mark. Choice A incorrectly states that Q << K, suggesting a huge difference, but $1.4 \times 10^{-3}$ is not much smaller than $2.8 \times 10^{-3}$ - it's only a factor of two difference. Choice C claims the system is essentially at equilibrium, but a 50% difference between Q and K is significant enough to drive a measurable shift. Choice D gets the direction completely wrong, stating the system shifts toward reactants when Q > K, but Q is actually less than K here. Study tip: Remember that when Q < K, products form; when Q > K, reactants form. The closer Q and K are in magnitude, the smaller the shift. A factor of 2-3 difference typically indicates a "moderate" shift, while factors of 10 or more suggest "significant" shifts.
18. The gas-phase reaction $A_2 + B_2 \rightleftharpoons 2AB$ reaches equilibrium at 500°C with $K_c = 25$. A reaction vessel initially contains 0.40 M $A_2$, 0.60 M $B_2$, and 2.0 M AB. What is the ratio Q/K for this system?
    Answer: 0.67 When you encounter equilibrium problems asking for Q/K ratios, you're being tested on whether a reaction system is at equilibrium or which direction it will proceed. The reaction quotient Q has the same mathematical form as the equilibrium constant K, but uses current concentrations instead of equilibrium concentrations. For the reaction $A_2 + B_2 \rightleftharpoons 2AB$, both Q and K are calculated as: $\frac{[AB]^2}{[A_2][B_2]}$ First, calculate Q using the given initial concentrations: $Q = \frac{(2.0)^2}{(0.40)(0.60)} = \frac{4.0}{0.24} = 16.7$ Since $K_c = 25$, the ratio is: $\frac{Q}{K} = \frac{16.7}{25} = 0.67$ This confirms answer choice (B). Looking at the wrong answers: (A) 0.067 results from incorrectly calculating Q as 1.67 instead of 16.7, likely from a computational error. (C) 1.5 could come from reversing the calculation as K/Q instead of Q/K. (D) 6.7 might result from forgetting to square the AB concentration in the Q expression, giving $\frac{2.0}{(0.40)(0.60)} = 8.33$, then making another error. Remember that Q/K ratios tell you reaction direction: Q/K < 1 means the reaction proceeds forward, Q/K > 1 means it proceeds backward, and Q/K = 1 means it's at equilibrium. Always double-check your stoichiometric exponents when writing the Q expression.
19. The equilibrium $2A(g) \rightleftharpoons 3B(g) + C(g)$ has $K_c = 0.15$ at a certain temperature. Two different reaction mixtures are prepared: Mixture X has Q = 0.045 and Mixture Y has Q = 0.60. Which statement correctly compares these mixtures?
    Answer: Mixture X will shift right and Mixture Y will shift left relative to their current states When you encounter equilibrium problems involving reaction quotients, you need to compare the current state (Q) to the equilibrium state ($K_c$) to predict which direction the reaction will shift. For any equilibrium, if Q < K, the reaction shifts right to reach equilibrium. If Q > K, the reaction shifts left. Here, $K_c = 0.15$. Mixture X has Q = 0.045, which is less than $K_c = 0.15$. Since Q < K, this mixture will shift right to increase the concentration of products relative to reactants until Q equals K. Mixture Y has Q = 0.60, which is greater than $K_c = 0.15$. Since Q > K, this mixture will shift left to decrease the concentration of products relative to reactants until Q equals K. Therefore, Mixture X shifts right and Mixture Y shifts left, making answer B correct. Answer A is wrong because it ignores that Mixture Y has Q > K, not Q < K. Answer C incorrectly states that both mixtures have Q > K, when actually Mixture X has Q < K. Answer D reverses the directions—it claims Mixture X shifts left (but Q < K means shift right) and Mixture Y shifts right (but Q > K means shift left). Remember this simple rule: Q < K means "not enough products yet, shift right"; Q > K means "too many products, shift left." Always compare your calculated Q value to the given K value to predict the shift direction.

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