College Chemistry Quiz: Direction Of Reversible Reactions
18 questions · exam conditions
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Direction Of Reversible ReactionsQuestion 1 of 18

For the reaction N2O4(g)⇌2NO2(g)N_2O_4(g) ⇌ 2NO_2(g), Kc=4.63×10−3K_c = 4.63 × 10^{-3} at 25°C. A reaction mixture contains 0.0236 M N2O4N_2O_4 and 0.0318 M NO2NO_2. What is the direction of the net reaction?

The reaction will proceed in the reverse direction
The reaction will proceed in the forward direction
The reaction is at equilibrium
The direction cannot be determined from the given information
The reaction will proceed in both directions equally
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College Chemistry Quiz

College Chemistry Quiz: Direction Of Reversible Reactions

Practice Direction Of Reversible Reactions in College 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 Direction Of Reversible Reactions, giving you a quick way to practice the rules, question types, and explanations that matter most for College 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

For the reaction N2O4(g)⇌2NO2(g)N_2O_4(g) ⇌ 2NO_2(g), Kc=4.63×10−3K_c = 4.63 × 10^{-3} at 25°C. A reaction mixture contains 0.0236 M N2O4N_2O_4 and 0.0318 M NO2NO_2. What is the direction of the net reaction?

  1. The reaction will proceed in the reverse direction (correct answer)
  2. The reaction will proceed in the forward direction
  3. The reaction is at equilibrium
  4. The direction cannot be determined from the given information
  5. The reaction will proceed in both directions equally
Explanation: When you encounter equilibrium problems asking about reaction direction, you need to compare the reaction quotient (Q) with the equilibrium constant (K). This comparison tells you which way the reaction must shift to reach equilibrium. First, calculate the reaction quotient using the same expression as the equilibrium constant, but with current concentrations instead of equilibrium concentrations: Qc=[NO2]2[N2O4]=(0.0318)2(0.0236)=0.001010.0236=0.0428Q_c = \frac{[NO_2]^2}{[N_2O_4]} = \frac{(0.0318)^2}{(0.0236)} = \frac{0.00101}{0.0236} = 0.0428 Now compare Q to the given Kc=4.63×10−3=0.00463K_c = 4.63 × 10^{-3} = 0.00463: Since Qc=0.0428>Kc=0.00463Q_c = 0.0428 > K_c = 0.00463, the reaction quotient is greater than the equilibrium constant. This means there's too much product (NO2NO_2) relative to reactant (N2O4N_2O_4) compared to the equilibrium ratio. The reaction must shift left (reverse direction) to decrease the numerator and increase the denominator until Q equals K. Choice A is correct because Q > K indicates reverse reaction direction. Choice B is wrong because Q > K never indicates forward direction (that only happens when Q < K). Choice C is incorrect because the reaction is at equilibrium only when Q = K, which isn't the case here. Choice D is wrong because comparing Q and K always determines reaction direction when both values and concentrations are known. Study tip: Remember the Q vs K rule: Q < K means forward, Q > K means reverse, Q = K means equilibrium. Always calculate Q first, then compare to make your prediction.

Question 2

For the equilibrium 2NO2(g)⇌N2O4(g)2NO_2(g) ⇌ N_2O_4(g), Kp=6.8K_p = 6.8 at 45°C. A reaction mixture has partial pressures PNO2=0.35P_{NO_2} = 0.35 atm and PN2O4=0.82P_{N_2O_4} = 0.82 atm. To predict the direction of reaction, a student calculates Qp=0.82(0.35)2=6.7Q_p = \frac{0.82}{(0.35)^2} = 6.7. What conclusion should the student draw?

  1. Since Q ≈ K, the system is essentially at equilibrium with minimal net reaction (correct answer)
  2. Since Q < K, the reaction will proceed strongly in the forward direction
  3. Since Q > K, the reaction will proceed strongly in the reverse direction
  4. The calculation is incorrect because temperature was not included
  5. Since Q < K, the reaction will proceed strongly in the reverse direction
Explanation: When you encounter equilibrium problems involving reaction quotients, you're comparing the current state of a system to its equilibrium state to predict which direction the reaction will proceed. The student correctly calculated the reaction quotient: Qp=PN2O4PNO22=0.82(0.35)2=6.7Q_p = \frac{P_{N_2O_4}}{P_{NO_2}^2} = \frac{0.82}{(0.35)^2} = 6.7. Since Kp=6.8K_p = 6.8 at this temperature, we have Qp=6.7Q_p = 6.7 and Kp=6.8K_p = 6.8. These values are essentially equal (within experimental uncertainty), indicating the system is very close to equilibrium with minimal driving force in either direction. Answer A correctly identifies this situation. Answer B is wrong because Q<KQ < K would normally drive the forward reaction, but the difference here (6.7 vs 6.8) is negligible—certainly not "strongly" in either direction. Answer C incorrectly states that Q>KQ > K, which isn't true since 6.7 < 6.8. Even if it were true, the tiny difference wouldn't cause a "strong" reverse reaction. Answer D is incorrect because temperature doesn't appear in the QpQ_p calculation itself—temperature affects the value of KpK_p, but once you know KpK_p at a given temperature, you can compare it directly to QpQ_p. Remember that when QpQ_p and KpK_p are very close in value (typically within 5-10% of each other), treat the system as essentially at equilibrium rather than predicting significant reaction in either direction. Focus on the magnitude of the difference, not just which is larger.

Question 3

The water-gas shift reaction CO(g)+H2O(g)⇌CO2(g)+H2(g)CO(g) + H_2O(g) ⇌ CO_2(g) + H_2(g) reaches equilibrium with Kc=0.64K_c = 0.64 at 800°C. If a reactor initially contains 2.0 M CO, 2.0 M H2OH_2O, 1.0 M CO2CO_2, and 1.0 M H2H_2, what happens immediately after startup?

  1. The reaction proceeds forward because Q = 0.25 < K = 0.64 (correct answer)
  2. The reaction proceeds reverse because Q = 4.0 > K = 0.64
  3. The reaction is at equilibrium because all concentrations are equal
  4. The reaction proceeds reverse because Q = 0.25 < K = 0.64
  5. The reaction proceeds forward because Q = 4.0 > K = 0.64
Explanation: When you encounter equilibrium problems with given concentrations, you need to determine the reaction direction by comparing the reaction quotient (Q) to the equilibrium constant (K). First, calculate Q using the same expression as the equilibrium constant: Qc=[CO2][H2][CO][H2O]Q_c = \frac{[CO_2][H_2]}{[CO][H_2O]}. Substituting the initial concentrations: Qc=(1.0)(1.0)(2.0)(2.0)=1.04.0=0.25Q_c = \frac{(1.0)(1.0)}{(2.0)(2.0)} = \frac{1.0}{4.0} = 0.25. Since Q = 0.25 and K = 0.64, we have Q < K. When Q < K, the reaction must proceed forward to reach equilibrium, converting more reactants into products until Q equals K. Looking at the answer choices: Choice A correctly identifies that Q = 0.25 < K = 0.64, leading to a forward reaction. Choice B incorrectly calculates Q as 4.0 – this happens when you flip the fraction and put reactants in the numerator instead of products. Choice C wrongly assumes equal concentrations mean equilibrium, but equilibrium depends on the specific ratio defined by K, not equal amounts. Choice D correctly calculates Q = 0.25 < K = 0.64 but incorrectly concludes the reaction goes in reverse – when Q < K, the reaction always proceeds forward. Remember this pattern: Q < K means forward reaction, Q > K means reverse reaction, and Q = K means equilibrium. Always double-check your Q expression matches the K expression with products over reactants.

Question 4

A student is studying the equilibrium H2(g)+I2(g)⇌2HI(g)H_2(g) + I_2(g) ⇌ 2HI(g) with Kc=54.3K_c = 54.3 at 700 K. The student prepares a mixture with [H2H_2] = 0.100 M, [I2I_2] = 0.200 M, and [HI] = 1.50 M, then predicts the direction using Qc=(1.50)2(0.100)(0.200)=112.5Q_c = \frac{(1.50)^2}{(0.100)(0.200)} = 112.5. What should the student conclude?

  1. Since Q > K, the reverse reaction is favored and [HI] will decrease (correct answer)
  2. Since Q > K, the forward reaction is favored and [HI] will increase
  3. Since Q < K, the reverse reaction is favored and [HI] will decrease
  4. The calculation contains an error because temperature was not considered
  5. Since Q < K, the forward reaction is favored and [HI] will increase
Explanation: When you encounter equilibrium problems involving reaction quotients, you're being tested on your ability to predict which direction a reaction will proceed to reach equilibrium. The student correctly calculated the reaction quotient: Qc=[HI]2[H2][I2]=(1.50)2(0.100)(0.200)=112.5Q_c = \frac{[HI]^2}{[H_2][I_2]} = \frac{(1.50)^2}{(0.100)(0.200)} = 112.5. Since Qc=112.5>Kc=54.3Q_c = 112.5 > K_c = 54.3, the system has too much product (HI) relative to reactants compared to the equilibrium ratio. To reach equilibrium, the reaction must shift left (reverse direction) to consume HI and produce more H2H_2 and I2I_2. This means [HI] will decrease. Looking at the wrong answers: Choice B incorrectly states that when Q > K, the forward reaction is favored. This is backwards—when Q > K, there's excess product, so the reverse reaction must be favored. Choice C correctly identifies that the reverse reaction is favored and [HI] will decrease, but wrongly claims Q < K. The calculation clearly shows Q > K. Choice D suggests a calculation error involving temperature, but temperature is already incorporated into the given KcK_c value of 54.3 at 700 K, and the student's math is correct. Remember this key relationship: when Q > K, the reaction shifts left (reverse); when Q < K, it shifts right (forward). Always compare your calculated Q value to the given K value first, then determine the direction of shift based on whether you have excess products or reactants.

Question 5

The equilibrium C(s)+CO2(g)⇌2CO(g)C(s) + CO_2(g) ⇌ 2CO(g) has Kp=1.52K_p = 1.52 at 700°C. A reaction vessel contains solid carbon with PCO2=0.50P_{CO_2} = 0.50 atm and PCO=1.10P_{CO} = 1.10 atm. What change will occur as the system approaches equilibrium?

  1. PCOP_{CO} will increase and PCO2P_{CO_2} will decrease as the reaction proceeds forward
  2. PCOP_{CO} will decrease and PCO2P_{CO_2} will increase as the reaction proceeds reverse (correct answer)
  3. Both pressures will remain constant because solid carbon buffers the system
  4. PCOP_{CO} will increase and PCO2P_{CO_2} will increase as more carbon reacts
  5. The pressures will oscillate until dynamic equilibrium is reached
Explanation: When you encounter equilibrium problems, you need to compare the reaction quotient (Q) with the equilibrium constant (K) to predict which direction the reaction will proceed. First, calculate the reaction quotient using the given pressures. For this reaction, Qp=PCO2PCO2Q_p = \frac{P_{CO}^2}{P_{CO_2}} (note that solid carbon doesn't appear in the expression). Substituting the values: Qp=(1.10)20.50=1.210.50=2.42Q_p = \frac{(1.10)^2}{0.50} = \frac{1.21}{0.50} = 2.42 Since Qp=2.42>Kp=1.52Q_p = 2.42 > K_p = 1.52, the system has too much product relative to reactant compared to equilibrium. The reaction must shift left (reverse direction) to reach equilibrium, consuming CO and producing CO2CO_2. Answer B correctly identifies this: PCOP_{CO} will decrease and PCO2P_{CO_2} will increase as the reaction proceeds in reverse. Answer A describes the forward reaction, which would occur if Qp<KpQ_p < K_p, but that's not the case here. Answer C incorrectly suggests that solid carbon "buffers" the system—while solids don't affect equilibrium expressions, they don't prevent pressure changes in the gas phase. Answer D is chemically impossible since both gas pressures cannot increase simultaneously in this closed system without violating mass balance. Study tip: Always calculate Q first and compare it to K. If Q > K, the reaction shifts left (reverse); if Q < K, it shifts right (forward). Remember that solids and pure liquids don't appear in equilibrium expressions, but they don't prevent the reaction from proceeding.

Question 6

For the equilibrium 3A(g)+B(g)⇌2C(g)+D(g)3A(g) + B(g) ⇌ 2C(g) + D(g), Kc=0.85K_c = 0.85 at 450 K. A reaction mixture has [A] = 1.2 M, [B] = 0.80 M, [C] = 0.60 M, and [D] = 0.40 M. A student needs to predict whether the concentrations of products will increase or decrease. What is the correct prediction?

  1. Product concentrations will increase because Q = 0.104 < K (correct answer)
  2. Product concentrations will decrease because Q = 0.104 > K
  3. Product concentrations will increase because Q = 1.44 > K
  4. Product concentrations will decrease because Q = 1.44 > K
  5. Product concentrations will remain constant because Q ≈ K
Explanation: When you encounter equilibrium problems asking about concentration changes, you need to compare the reaction quotient (Q) to the equilibrium constant (K) to determine which direction the reaction will proceed. First, calculate Q using the same expression as K, but with current concentrations instead of equilibrium concentrations. For this reaction, Q=[C]2[D][A]3[B]Q = \frac{[C]^2[D]}{[A]^3[B]}. Substituting the given values: Q=(0.60)2(0.40)(1.2)3(0.80)=0.1441.382=0.104Q = \frac{(0.60)^2(0.40)}{(1.2)^3(0.80)} = \frac{0.144}{1.382} = 0.104 Since Q = 0.104 and K = 0.85, we have Q < K. This means the reaction must shift right (toward products) to reach equilibrium, so product concentrations will increase. Looking at the wrong answers: Option B incorrectly states that Q > K when Q is actually less than K. Options C and D both calculate Q incorrectly as 1.44 - this likely comes from flipping the Q expression upside down, putting reactants in the numerator and products in the denominator. Even if Q were 1.44, option C would be wrong because when Q > K, the reaction shifts left (decreasing products), not right. The correct answer is A: product concentrations will increase because Q = 0.104 < K. Remember this key relationship: when Q < K, the reaction shifts right toward products; when Q > K, it shifts left toward reactants. Always double-check your Q expression matches the equilibrium constant format with products over reactants, each raised to their stoichiometric coefficients.

Question 7

A researcher examines the equilibrium CH4(g)+H2O(g)⇌CO(g)+3H2(g)CH_4(g) + H_2O(g) ⇌ CO(g) + 3H_2(g) at 1000 K, where Kp=3.2K_p = 3.2. The partial pressures are PCH4=0.80P_{CH_4} = 0.80 atm, PH2O=1.2P_{H_2O} = 1.2 atm, PCO=0.45P_{CO} = 0.45 atm, and PH2=1.5P_{H_2} = 1.5 atm. Based on these conditions, what will happen to the partial pressure of H2H_2?

  1. PH2P_{H_2} will increase because Q = 1.6 < K (correct answer)
  2. PH2P_{H_2} will decrease because Q = 1.6 > K
  3. PH2P_{H_2} will increase because Q = 3.5 > K
  4. PH2P_{H_2} will decrease because Q = 3.5 > K
  5. PH2P_{H_2} will remain constant because Q ≈ K
Explanation: When you encounter equilibrium problems asking about the direction of reaction, you need to compare the reaction quotient Q to the equilibrium constant K to predict which way the system will shift. First, calculate Q using the same expression as K, but with current partial pressures instead of equilibrium values. For this reaction, Qp=PCO×PH23PCH4×PH2OQ_p = \frac{P_{CO} \times P_{H_2}^3}{P_{CH_4} \times P_{H_2O}}. Substituting the given values: Qp=(0.45)(1.5)3(0.80)(1.2)=(0.45)(3.375)0.96=1.58≈1.6Q_p = \frac{(0.45)(1.5)^3}{(0.80)(1.2)} = \frac{(0.45)(3.375)}{0.96} = 1.58 ≈ 1.6. Since Q = 1.6 < K = 3.2, the reaction must shift forward (to the right) to reach equilibrium. This means more products will form, including more H2H_2, so the partial pressure of H2H_2 will increase. This confirms answer A is correct. Looking at the wrong answers: B incorrectly states that Q > K when we calculated Q < K. C and D both claim Q = 3.5, which represents a calculation error—likely from miscalculating PH23=(1.5)3=3.375P_{H_2}^3 = (1.5)^3 = 3.375 somewhere in the arithmetic. Additionally, D combines this wrong Q value with the wrong prediction about pressure change direction. Remember this key relationship: when Q < K, the reaction shifts forward (toward products), and when Q > K, it shifts backward (toward reactants). Always double-check your Q calculation, especially when dealing with coefficients greater than 1—they become exponents in the expression.

Question 8

For the equilibrium 2NO2(g)⇌N2O4(g)2NO_2(g) ⇌ N_2O_4(g), Kp=6.7K_p = 6.7 at 45°C. A reaction vessel initially contains only NO2NO_2 at 0.50 atm partial pressure. After some time, PNO2=0.32P_{NO_2} = 0.32 atm and PN2O4=0.09P_{N_2O_4} = 0.09 atm. What is occurring in this system at this moment?

  1. The reaction is proceeding toward equilibrium in the forward direction (correct answer)
  2. The reaction is proceeding toward equilibrium in the reverse direction
  3. The reaction has reached equilibrium and the rates are equal
  4. The reaction will stop because insufficient NO2NO_2 remains
  5. The system is oscillating around the equilibrium position
Explanation: When you encounter equilibrium problems asking about the direction of reaction, you need to compare the reaction quotient (Q) to the equilibrium constant (K) to determine which way the reaction will proceed. First, calculate the reaction quotient using the given partial pressures. For 2NO2(g)⇌N2O4(g)2NO_2(g) ⇌ N_2O_4(g), the expression is Qp=PN2O4(PNO2)2Q_p = \frac{P_{N_2O_4}}{(P_{NO_2})^2}. Substituting the values: Qp=0.09(0.32)2=0.090.1024=0.88Q_p = \frac{0.09}{(0.32)^2} = \frac{0.09}{0.1024} = 0.88 Now compare Q to the given Kp=6.7K_p = 6.7. Since Qp=0.88<Kp=6.7Q_p = 0.88 < K_p = 6.7, the reaction must shift forward to reach equilibrium, producing more N2O4N_2O_4 and consuming more NO2NO_2. Looking at the wrong answers: Answer B suggests the reverse direction, but since Q < K, the forward reaction is favored. Answer C claims equilibrium has been reached, but this would only be true if Q equaled K (6.7), not 0.88. Answer D incorrectly assumes the reaction stops due to insufficient reactant, but equilibrium reactions continue in both directions even when reactant concentrations are low. The correct answer is A because when Q < K, the reaction proceeds forward until Q increases to equal K. Study tip: Remember the Q vs K rule: Q < K means go forward, Q > K means go reverse, Q = K means equilibrium. Always calculate Q first, then compare to the given K value to predict reaction direction.

Question 9

A student is investigating the equilibrium H2S(g)+I2(s)⇌2HI(g)+S(s)H_2S(g) + I_2(s) ⇌ 2HI(g) + S(s) at 60°C, where Kc=1.34×10−5K_c = 1.34 × 10^{-5}. The concentrations are [H2SH_2S] = 0.045 M and [HI] = 0.0012 M, with solid I2I_2 and S present. What will happen to the [HI] as the system approaches equilibrium?

  1. [HI] will increase because Q = 3.2 × 10^{-5} > K
  2. [HI] will decrease because Q = 3.2 × 10^{-5} > K (correct answer)
  3. [HI] will increase because Q = 3.2 × 10^{-5} < K
  4. [HI] will decrease because Q = 8.0 × 10^{-6} < K
  5. [HI] will increase because Q = 8.0 × 10^{-6} < K
Explanation: When you encounter equilibrium problems asking about concentration changes, you need to compare the reaction quotient (Q) to the equilibrium constant (K) to predict the direction of change. First, calculate Q using the same expression as K, but with current concentrations. For this reaction, Qc=[HI]2[H2S]Q_c = \frac{[HI]^2}{[H_2S]} (solids are excluded from equilibrium expressions). Substituting the given values: Qc=(0.0012)20.045=1.44×10−60.045=3.2×10−5Q_c = \frac{(0.0012)^2}{0.045} = \frac{1.44 × 10^{-6}}{0.045} = 3.2 × 10^{-5} Since Qc=3.2×10−5Q_c = 3.2 × 10^{-5} is greater than Kc=1.34×10−5K_c = 1.34 × 10^{-5}, the reaction must shift left (toward reactants) to reach equilibrium. This means [HI] will decrease as the system converts HI back to reactants. Option A incorrectly states that [HI] will increase despite correctly calculating Q > K. When Q > K, the reaction always shifts left, decreasing product concentrations. Option C makes the opposite error, correctly predicting an increase in [HI] but falsely claiming Q < K. Option D uses an incorrect Q calculation (8.0×10−68.0 × 10^{-6}) that's smaller than K, leading to the wrong directional prediction. Remember this key relationship: when Q > K, the reaction shifts left (products decrease); when Q < K, the reaction shifts right (products increase). Always calculate Q carefully using the correct equilibrium expression, excluding any solids or pure liquids from your calculation.

Question 10

At 1000 K, the equilibrium constant for CO(g)+H2O(g)⇌CO2(g)+H2(g)CO(g) + H_2O(g) ⇌ CO_2(g) + H_2(g) is Kc=1.56K_c = 1.56. If a reaction vessel contains 0.100 M CO, 0.100 M H2OH_2O, 0.200 M CO2CO_2, and 0.200 M H2H_2, which statement correctly describes the system?

  1. Q = 4.00, so the reaction will proceed to the left (correct answer)
  2. Q = 0.25, so the reaction will proceed to the right
  3. Q = 1.56, so the reaction is at equilibrium
  4. Q = 4.00, so the reaction will proceed to the right
  5. Q = 0.25, so the reaction will proceed to the left
Explanation: When you encounter equilibrium problems with given concentrations, you need to compare the reaction quotient (Q) to the equilibrium constant (K) to determine which direction the reaction will proceed. First, calculate Q using the same expression as the equilibrium constant: Qc=[CO2][H2][CO][H2O]Q_c = \frac{[CO_2][H_2]}{[CO][H_2O]}. Substituting the given concentrations: Qc=(0.200)(0.200)(0.100)(0.100)=0.04000.0100=4.00Q_c = \frac{(0.200)(0.200)}{(0.100)(0.100)} = \frac{0.0400}{0.0100} = 4.00. Since Qc=4.00Q_c = 4.00 is greater than Kc=1.56K_c = 1.56, the system has too much product relative to reactants compared to equilibrium. To reach equilibrium, the reaction must shift left (toward reactants) to decrease the numerator and increase the denominator until Q equals K. Looking at the answer choices: A correctly identifies Q = 4.00 and the leftward shift. B incorrectly calculates Q as 0.25 (likely from inverting the expression) and wrongly predicts a rightward shift. C correctly notes that when Q = K the system is at equilibrium, but Q ≠ 1.56 here. D correctly calculates Q = 4.00 but incorrectly concludes the reaction proceeds right—when Q > K, the reaction always shifts left to reduce the product-to-reactant ratio. Remember this key relationship: when Q > K, shift left; when Q < K, shift right; when Q = K, you're at equilibrium. Always double-check your Q calculation by ensuring products are in the numerator and reactants in the denominator, just like the K expression.

Question 11

Consider the equilibrium Fe3+(aq)+SCN−(aq)⇌FeSCN2+(aq)Fe^{3+}(aq) + SCN^-(aq) ⇌ FeSCN^{2+}(aq) with Kc=138K_c = 138 at room temperature. A solution is prepared by mixing equal volumes of 0.002 M Fe3+Fe^{3+} and 0.002 M SCN−SCN^-. Immediately after mixing, what is the direction of the net reaction?

  1. Forward, because no product is initially present so Q = 0 < K (correct answer)
  2. Reverse, because the reactant concentrations are too low
  3. No net reaction because the reactants are in stoichiometric proportion
  4. Forward, because Q = 1 < K due to equal initial concentrations
  5. The direction cannot be determined without knowing the final equilibrium concentrations
Explanation: When you encounter equilibrium problems asking about reaction direction, you need to compare the reaction quotient Q to the equilibrium constant K. The reaction will proceed in whichever direction brings Q closer to K. To find Q, you use the same expression as K but with initial concentrations. Here, Qc=[FeSCN2+][Fe3+][SCN−]Q_c = \frac{[FeSCN^{2+}]}{[Fe^{3+}][SCN^-]}. Since equal volumes are mixed, each concentration is halved: both [Fe3+][Fe^{3+}] and [SCN−][SCN^-] become 0.001 M, while [FeSCN2+]=0[FeSCN^{2+}] = 0 initially. Therefore: Qc=0(0.001)(0.001)=0Q_c = \frac{0}{(0.001)(0.001)} = 0 Since Qc=0<Kc=138Q_c = 0 < K_c = 138, the reaction must proceed forward to increase Q until it equals K. Answer A correctly identifies that Q = 0 because no product is present initially, and since Q < K, the forward reaction occurs. Answer B incorrectly suggests the reaction goes reverse - but when Q < K, reactions always go forward, regardless of concentration magnitude. Answer C wrongly assumes stoichiometric proportions prevent net reaction - but equilibrium position depends on K, not initial ratios. Answer D makes an error in calculating Q, claiming Q = 1 due to equal concentrations, which ignores that the product concentration is zero. Remember: always calculate Q first using initial concentrations, then compare to K. If Q < K, the reaction goes forward; if Q > K, it goes reverse. The key insight is that Q = 0 whenever products are initially absent.

Question 12

A student analyzes the equilibrium SO2(g)+12O2(g)⇌SO3(g)SO_2(g) + \frac{1}{2}O_2(g) ⇌ SO_3(g) with Kp=128K_p = 128 at 627°C. The partial pressures are PSO2=0.40P_{SO_2} = 0.40 atm, PO2=0.60P_{O_2} = 0.60 atm, and PSO3=25P_{SO_3} = 25 atm. The student calculates Qp=25(0.40)(0.60)0.5=250.310=80.6Q_p = \frac{25}{(0.40)(0.60)^{0.5}} = \frac{25}{0.310} = 80.6. What should the student conclude about the reaction direction?

  1. The reaction will proceed forward because Q < K (correct answer)
  2. The reaction will proceed reverse because Q > K
  3. The reaction will proceed reverse because Q < K
  4. The system is near equilibrium because Q ≈ K
  5. The calculation is invalid because fractional exponents cannot be used
Explanation: When you encounter equilibrium problems involving reaction quotient (Q) and equilibrium constant (K), you're determining which direction a reaction will proceed to reach equilibrium. The key principle is comparing Q to K. The student correctly calculated Qp=80.6Q_p = 80.6 using the equilibrium expression Qp=PSO3PSO2⋅PO20.5Q_p = \frac{P_{SO_3}}{P_{SO_2} \cdot P_{O_2}^{0.5}}. Since Qp=80.6Q_p = 80.6 and Kp=128K_p = 128, we have Q < K. When Q < K, the reaction quotient is smaller than it should be at equilibrium, meaning there's too little product relative to reactants. The reaction must shift forward (toward products) to increase Q until it equals K. Looking at the wrong answers: Choice B incorrectly states that Q > K, which is mathematically false since 80.6 < 128. Choice C makes the right comparison (Q < K) but draws the wrong conclusion about reaction direction—when Q < K, the reaction proceeds forward, not reverse. Choice D suggests the system is near equilibrium, but with Q and K differing by nearly 50 units (128 - 80.6 = 47.4), this represents a significant deviation from equilibrium. Choice A correctly identifies both the relationship (Q < K) and the proper conclusion (forward reaction). Study tip: Remember the Q vs K rule: Q < K means "not enough product" so the reaction goes forward; Q > K means "too much product" so the reaction goes reverse. The reaction always shifts to make Q equal to K.

Question 13

For the equilibrium 2SO3(g)⇌2SO2(g)+O2(g)2SO_3(g) ⇌ 2SO_2(g) + O_2(g), Kp=1.8×10−5K_p = 1.8 × 10^{-5} at 600°C. A mixture has PSO3=2.0P_{SO_3} = 2.0 atm, PSO2=0.050P_{SO_2} = 0.050 atm, and PO2=0.025P_{O_2} = 0.025 atm. What is the direction of the net reaction?

  1. Forward, because Q = 1.56 × 10^{-5} < K (correct answer)
  2. Reverse, because Q = 1.56 × 10^{-5} > K
  3. Forward, because Q = 6.4 × 10^{-6} < K
  4. Reverse, because Q = 6.4 × 10^{-6} > K
  5. At equilibrium, because Q ≈ K
Explanation: When you encounter equilibrium problems asking about reaction direction, you need to compare the reaction quotient (Q) to the equilibrium constant (K). This tells you whether the system is at equilibrium or which way it will shift to reach equilibrium. First, calculate Q using the same expression as K, but with current pressures instead of equilibrium pressures. For this reaction, Qp=PSO22×PO2PSO32Q_p = \frac{P_{SO_2}^2 \times P_{O_2}}{P_{SO_3}^2} Substituting the given values: Qp=(0.050)2×(0.025)(2.0)2=0.0025×0.0254.0=6.25×10−54.0=1.56×10−5Q_p = \frac{(0.050)^2 \times (0.025)}{(2.0)^2} = \frac{0.0025 \times 0.025}{4.0} = \frac{6.25 \times 10^{-5}}{4.0} = 1.56 \times 10^{-5} Since Qp=1.56×10−5<Kp=1.8×10−5Q_p = 1.56 \times 10^{-5} < K_p = 1.8 \times 10^{-5}, the reaction will proceed forward to increase Q until it equals K. Looking at the wrong answers: Answer B incorrectly states that 1.56×10−5>K1.56 \times 10^{-5} > K, which is mathematically wrong. Answer C gets the wrong Q value (6.4×10−66.4 \times 10^{-6}), likely from calculation errors in the numerator or denominator. Answer D combines both mistakes—the wrong Q value and incorrect comparison. Study tip: Remember the Q vs. K rules: when Q < K, the reaction goes forward; when Q > K, it goes reverse; when Q = K, it's at equilibrium. Double-check your Q calculation by ensuring you're using the correct stoichiometric coefficients as exponents and placing products over reactants.

Question 14

Consider the equilibrium COCl2(g)⇌CO(g)+Cl2(g)COCl_2(g) ⇌ CO(g) + Cl_2(g) with Kp=8.3×10−15K_p = 8.3 × 10^{-15} at 25°C. A container initially holds 0.50 atm COCl2COCl_2, 0.10 atm CO, and 0.10 atm Cl2Cl_2. What is the direction of the spontaneous reaction?

  1. Forward, because Q = 2.0 × 10^{-2} < K
  2. Reverse, because Q = 2.0 × 10^{-2} > K (correct answer)
  3. Forward, because Q = 5.0 × 10^{-2} > K
  4. Reverse, because Q = 5.0 × 10^{-2} > K
  5. At equilibrium, because the partial pressures are constant
Explanation: When you encounter an equilibrium problem asking about reaction direction, you need to compare the reaction quotient Q to the equilibrium constant K. The reaction will proceed in whichever direction brings Q closer to K. First, calculate the reaction quotient using the same expression as the equilibrium constant, but with initial pressures instead of equilibrium pressures: Qp=PCO×PCl2PCOCl2=(0.10)(0.10)0.50=0.0100.50=2.0×10−2Q_p = \frac{P_{CO} \times P_{Cl_2}}{P_{COCl_2}} = \frac{(0.10)(0.10)}{0.50} = \frac{0.010}{0.50} = 2.0 × 10^{-2} Now compare Q to K: Qp=2.0×10−2=0.02Q_p = 2.0 × 10^{-2} = 0.02 while Kp=8.3×10−15K_p = 8.3 × 10^{-15}. Since Q is much larger than K, the reaction must shift in the reverse direction to decrease Q until it equals K. Choice B correctly identifies both the calculated Q value and the reverse direction. Choice A has the right Q calculation but wrong direction—when Q > K, the reaction goes reverse, not forward. Choice C uses an incorrect Q calculation (5.0×10−25.0 × 10^{-2}), likely from a math error in the division. Choice D has the same incorrect Q value as C, though it does correctly identify the reverse direction. Remember this pattern: Q > K means reverse reaction, Q < K means forward reaction, Q = K means equilibrium. Also, always double-check your Q calculation—make sure products are in the numerator and reactants in the denominator, each raised to their stoichiometric coefficients.

Question 15

The equilibrium CaCO3(s)⇌CaO(s)+CO2(g)CaCO_3(s) ⇌ CaO(s) + CO_2(g) has Kp=1.16K_p = 1.16 atm at 900°C. A sealed container holds solid CaCO3CaCO_3 and solid CaO with PCO2=2.3P_{CO_2} = 2.3 atm. What will occur when the system reaches equilibrium?

  1. CO2CO_2 pressure will decrease as CaCO3CaCO_3 forms from CaO (correct answer)
  2. CO2CO_2 pressure will increase as more CaCO3CaCO_3 decomposes
  3. CO2CO_2 pressure will remain constant because solids don't affect equilibrium
  4. The system is already at equilibrium because solids are present
  5. CO2CO_2 pressure will decrease to zero as all CaCO3CaCO_3 decomposes
Explanation: When you encounter equilibrium problems involving both solids and gases, focus on comparing the reaction quotient (Q) to the equilibrium constant (K) to predict which direction the reaction will proceed. Here, the equilibrium constant Kp=1.16K_p = 1.16 atm represents the CO₂ pressure when the system is at equilibrium. Since only gases appear in the equilibrium expression for heterogeneous equilibria, Kp=PCO2K_p = P_{CO_2}. Currently, PCO2=2.3P_{CO_2} = 2.3 atm, which is greater than Kp=1.16K_p = 1.16 atm. When the actual pressure exceeds the equilibrium pressure, the system must shift left (toward reactants) to reduce the CO₂ pressure. This means CO₂ will react with CaO to form more CaCO₃, decreasing the CO₂ pressure until it reaches 1.16 atm. Looking at the wrong answers: Choice B suggests the opposite direction, but since PCO2>KpP_{CO_2} > K_p, decomposition would move the system further from equilibrium. Choice C incorrectly assumes solids prevent equilibrium changes—while solids don't appear in the equilibrium expression, the reaction can still proceed in either direction. Choice D misunderstands equilibrium; having all species present doesn't guarantee equilibrium—you must compare Q to K. Remember this pattern: when PCO2>KpP_{CO_2} > K_p, the reverse reaction occurs to reduce gas pressure; when PCO2<KpP_{CO_2} < K_p, the forward reaction increases gas pressure. The equilibrium constant tells you exactly what the final gas pressure must be.

Question 16

The equilibrium 2IBr(g)⇌I2(g)+Br2(g)2IBr(g) ⇌ I_2(g) + Br_2(g) has Kc=8.5×10−3K_c = 8.5 × 10^{-3} at 150°C. In a sealed flask, [IBr] = 0.25 M, [I2I_2] = 0.10 M, and [Br2Br_2] = 0.080 M. Which statement correctly describes the immediate behavior of this system?

  1. The system will shift toward reactants because Q = 1.28 × 10^{-2} > K (correct answer)
  2. The system will shift toward products because Q = 1.28 × 10^{-2} < K
  3. The system will shift toward reactants because Q = 3.2 × 10^{-2} > K
  4. The system will shift toward products because Q = 3.2 × 10^{-2} > K
  5. The system is at equilibrium because Q equals K within experimental error
Explanation: When you encounter an equilibrium problem asking about system behavior, you need to compare the reaction quotient (Q) to the equilibrium constant (K) to predict which direction the reaction will proceed. First, calculate the reaction quotient Q using the same expression as the equilibrium constant, but with the current concentrations: Qc=[I2][Br2][IBr]2=(0.10)(0.080)(0.25)2=0.0080.0625=1.28×10−2Q_c = \frac{[I_2][Br_2]}{[IBr]^2} = \frac{(0.10)(0.080)}{(0.25)^2} = \frac{0.008}{0.0625} = 1.28 × 10^{-2} Since Qc=1.28×10−2Q_c = 1.28 × 10^{-2} and Kc=8.5×10−3K_c = 8.5 × 10^{-3}, we have Q > K. When Q > K, the system has too much product relative to equilibrium, so it shifts toward reactants to decrease Q until it equals K. Looking at the answer choices: Choice A correctly identifies both the Q value (1.28 × 10−210^{-2}) and the proper conclusion that the system shifts toward reactants because Q > K. Choice B uses the correct Q value but incorrectly states Q < K, leading to the wrong directional prediction. Choice C calculates Q incorrectly as 3.2 × 10^{-2} (likely from arithmetic errors) but reaches the correct directional conclusion by chance. Choice D combines the incorrect Q calculation with an impossible conclusion—it states Q > K but predicts a shift toward products, which contradicts equilibrium principles. Remember: Q > K means shift left (toward reactants), Q < K means shift right (toward products). Always double-check your Q calculation since small arithmetic errors can lead to wrong predictions about equilibrium direction.

Question 17

For the gas-phase reaction A2(g)+3B2(g)⇌2AB3(g)A_2(g) + 3B_2(g) ⇌ 2AB_3(g), Kp=4.5×104K_p = 4.5 × 10^4 at 400 K. A reaction vessel contains PA2=0.10P_{A_2} = 0.10 atm, PB2=0.30P_{B_2} = 0.30 atm, and PAB3=15P_{AB_3} = 15 atm. A student claims the reaction will proceed in the reverse direction. Is this claim correct?

  1. No, because Q = 1.85 × 10^4 < K, so the reaction proceeds forward
  2. Yes, because Q = 8.3 × 10^4 > K, so the reaction proceeds reverse (correct answer)
  3. No, because the high product pressure indicates forward reaction is favored
  4. Yes, because Q = 1.85 × 10^4 > K, so the reaction proceeds reverse
  5. No, because Q = 5.4 × 10^4 < K, so the reaction proceeds forward
Explanation: When you encounter equilibrium problems asking about reaction direction, you need to compare the reaction quotient (Q) to the equilibrium constant (K). This tells you which way the reaction will shift to reach equilibrium. First, calculate Q using the same expression as K, but with the given pressures instead of equilibrium pressures: Qp=(PAB3)2(PA2)(PB2)3=(15)2(0.10)(0.30)3=225(0.10)(0.027)=2250.0027=8.3×104Q_p = \frac{(P_{AB_3})^2}{(P_{A_2})(P_{B_2})^3} = \frac{(15)^2}{(0.10)(0.30)^3} = \frac{225}{(0.10)(0.027)} = \frac{225}{0.0027} = 8.3 × 10^4 Since Qp=8.3×104>Kp=4.5×104Q_p = 8.3 × 10^4 > K_p = 4.5 × 10^4, the reaction quotient exceeds the equilibrium constant. This means there are too many products relative to reactants compared to the equilibrium position, so the reaction must proceed in the reverse direction to reach equilibrium. Answer A incorrectly calculates Q as 1.85×1041.85 × 10^4 (likely from computational error) and draws the wrong conclusion about reaction direction. Answer C ignores the quantitative relationship between Q and K, making a qualitative guess based on "high product pressure" without proper calculation. Answer D gets the wrong Q value (same error as A) but would have the correct reasoning if the calculation were right. Remember: when Q > K, the reaction shifts left (reverse); when Q < K, it shifts right (forward); when Q = K, it's at equilibrium. Always calculate Q carefully using the correct stoichiometric coefficients as exponents.

Question 18

The decomposition reaction 2HI(g)⇌H2(g)+I2(g)2HI(g) ⇌ H_2(g) + I_2(g) has Kc=1.26×10−3K_c = 1.26 × 10^{-3} at 500°C. A sealed flask initially contains only 0.500 M HI. After some time, analysis shows [HI] = 0.482 M, [H2H_2] = 0.009 M, and [I2I_2] = 0.009 M. What is happening in this system?

  1. The reaction is proceeding toward equilibrium in the forward direction (correct answer)
  2. The reaction is proceeding toward equilibrium in the reverse direction
  3. The reaction has reached equilibrium
  4. The reaction will stop because there is insufficient HI remaining
  5. The reaction is proceeding in both directions at different rates
Explanation: When you encounter equilibrium problems involving concentration changes, you need to determine whether the system has reached equilibrium by comparing the reaction quotient (Q) to the equilibrium constant (K). First, calculate the reaction quotient using the given concentrations: Qc=[H2][I2][HI]2=(0.009)(0.009)(0.482)2=8.1×10−50.232=3.49×10−4Q_c = \frac{[H_2][I_2]}{[HI]^2} = \frac{(0.009)(0.009)}{(0.482)^2} = \frac{8.1 × 10^{-5}}{0.232} = 3.49 × 10^{-4} Since Qc=3.49×10−4Q_c = 3.49 × 10^{-4} is less than Kc=1.26×10−3K_c = 1.26 × 10^{-3}, the reaction must continue in the forward direction to reach equilibrium. When Q < K, the system shifts right to increase the numerator (products) and decrease the denominator (reactants) until Q equals K. Answer A is correct because Q < K indicates the forward reaction will continue. Answer B is wrong because the reverse direction would occur if Q > K, which isn't the case here. Answer C is incorrect because Q ≠ K, so equilibrium hasn't been reached. Answer D reflects a common misconception—reactions don't "stop" due to insufficient reactants unless one is completely consumed. The system will continue shifting until equilibrium is established, regardless of how much starting material remains. Remember this key relationship: when Q < K, the reaction proceeds forward; when Q > K, it proceeds in reverse; when Q = K, you're at equilibrium. Always calculate Q first when analyzing whether an equilibrium system is at rest or still shifting.