College Chemistry Quiz: Ice Tables For Equilibrium Systems
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Ice Tables For Equilibrium SystemsQuestion 1 of 17

Consider the equilibrium N2O4(g)⇌2NO2(g)N_2O_4(g) \rightleftharpoons 2NO_2(g) with Kc=4.6×10−3K_c = 4.6 \times 10^{-3} at 25°C. If the initial concentration of N2O4N_2O_4 is 0.050 M and no NO2NO_2 is present initially, what is the equilibrium concentration of NO2NO_2? Assume the change in N2O4N_2O_4 concentration is small.

0.015 M
0.024 M
0.030 M
0.048 M
0.068 M
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College Chemistry Quiz

College Chemistry Quiz: Ice Tables For Equilibrium Systems

Practice Ice Tables For Equilibrium Systems in College Chemistry with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Ice Tables For Equilibrium Systems, giving you a quick way to practice the rules, question types, and explanations that matter most for College Chemistry.

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Question 1

Consider the equilibrium N2O4(g)⇌2NO2(g)N_2O_4(g) \rightleftharpoons 2NO_2(g) with Kc=4.6×10−3K_c = 4.6 \times 10^{-3} at 25°C. If the initial concentration of N2O4N_2O_4 is 0.050 M and no NO2NO_2 is present initially, what is the equilibrium concentration of NO2NO_2? Assume the change in N2O4N_2O_4 concentration is small.

  1. 0.015 M (correct answer)
  2. 0.024 M
  3. 0.030 M
  4. 0.048 M
  5. 0.068 M
Explanation: When you encounter chemical equilibrium problems with small change assumptions, you're dealing with situations where the equilibrium constant is small enough that you can simplify your calculations without significant error. Set up an ICE table for this equilibrium. Initially, you have 0.050 M N2O4N_2O_4 and 0 M NO2NO_2. Let x represent the amount of N2O4N_2O_4 that dissociates. At equilibrium, you'll have (0.050−x)(0.050 - x) M N2O4N_2O_4 and 2x2x M NO2NO_2 (note the 2:1 stoichiometry). The equilibrium expression is: Kc=[NO2]2[N2O4]=(2x)20.050−x=4.6×10−3K_c = \frac{[NO_2]^2}{[N_2O_4]} = \frac{(2x)^2}{0.050 - x} = 4.6 \times 10^{-3} Since KcK_c is small, the "small change" assumption means x<<0.050x << 0.050, so (0.050−x)≈0.050(0.050 - x) \approx 0.050. This simplifies to: 4x20.050=4.6×10−3\frac{4x^2}{0.050} = 4.6 \times 10^{-3} Solving: 4x2=2.3×10−44x^2 = 2.3 \times 10^{-4}, so x2=5.75×10−5x^2 = 5.75 \times 10^{-5}, giving x=0.0076x = 0.0076 M. Therefore, [NO2]=2x=0.015[NO_2] = 2x = 0.015 M. Answer A (0.015 M) is correct. Answer B (0.024 M) likely results from calculation errors in the quadratic setup. Answer C (0.030 M) might come from incorrectly assuming all N2O4N_2O_4 converts or mishandling stoichiometry. Answer D (0.048 M) suggests a fundamental misunderstanding, possibly treating this as complete dissociation. Remember: small equilibrium constants (<10−2< 10^{-2}) often allow the small change approximation, which dramatically simplifies calculations while maintaining accuracy.

Question 2

For the reaction PCl5(g)⇌PCl3(g)+Cl2(g)PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g), the equilibrium constant Kc=0.042K_c = 0.042 at 500 K. If 2.0 mol of PCl5PCl_5 is placed in a 4.0 L container, what is the equilibrium concentration of PCl3PCl_3?

  1. 0.12 M
  2. 0.18 M (correct answer)
  3. 0.24 M
  4. 0.32 M
  5. 0.38 M
Explanation: When you encounter an equilibrium problem with initial amounts and need to find equilibrium concentrations, you're dealing with an ICE table scenario where you'll track Initial concentrations, Change in concentrations, and Equilibrium concentrations. Start by finding the initial concentration of PCl5PCl_5: 2.0 mol4.0 L=0.50 M\frac{2.0 \text{ mol}}{4.0 \text{ L}} = 0.50 \text{ M}. Set up your ICE table with PCl5PCl_5 starting at 0.50 M and products at 0 M. Let xx represent the amount of PCl5PCl_5 that dissociates. At equilibrium: [PCl5]=0.50−x[PCl_5] = 0.50 - x, [PCl3]=x[PCl_3] = x, and [Cl2]=x[Cl_2] = x Substitute into the equilibrium expression: Kc=[PCl3][Cl2][PCl5]=x⋅x0.50−x=0.042K_c = \frac{[PCl_3][Cl_2]}{[PCl_5]} = \frac{x \cdot x}{0.50 - x} = 0.042 This gives you: x20.50−x=0.042\frac{x^2}{0.50 - x} = 0.042 Expanding: x2=0.042(0.50−x)=0.021−0.042xx^2 = 0.042(0.50 - x) = 0.021 - 0.042x Rearranging: x2+0.042x−0.021=0x^2 + 0.042x - 0.021 = 0 Using the quadratic formula: x=−0.042+(0.042)2+4(0.021)2=0.18x = \frac{-0.042 + \sqrt{(0.042)^2 + 4(0.021)}}{2} = 0.18 Therefore, [PCl3]=0.18 M[PCl_3] = 0.18 \text{ M}, which is answer B. Answer A (0.12 M) likely comes from calculation errors in the quadratic formula. Answer C (0.24 M) might result from assuming the reaction goes further than it actually does. Answer D (0.32 M) could come from misapplying the equilibrium expression or algebraic mistakes. Remember: always set up your ICE table systematically and double-check your quadratic equation setup—small errors in algebra can lead to significantly wrong answers in equilibrium problems.

Question 3

For the equilibrium H2(g)+I2(g)⇌2HI(g)H_2(g) + I_2(g) \rightleftharpoons 2HI(g), Kc=54.3K_c = 54.3 at 430°C. Initially, 0.200 mol each of H2H_2 and I2I_2 are placed in a 2.00 L container with 0.500 mol of HIHI. What is the reaction quotient QcQ_c?

  1. 2.50
  2. 6.25 (correct answer)
  3. 12.5
  4. 25.0
  5. 62.5
Explanation: When you encounter equilibrium problems involving reaction quotients, you're comparing the current state of a reaction mixture to its equilibrium state. The reaction quotient QcQ_c uses the same expression as the equilibrium constant KcK_c, but with current concentrations instead of equilibrium concentrations. For this reaction, Qc=[HI]2[H2][I2]Q_c = \frac{[HI]^2}{[H_2][I_2]}. First, convert moles to concentrations by dividing by the 2.00 L volume: [H2]=0.200/2.00=0.100 M[H_2] = 0.200/2.00 = 0.100 \text{ M}, [I2]=0.200/2.00=0.100 M[I_2] = 0.200/2.00 = 0.100 \text{ M}, and [HI]=0.500/2.00=0.250 M[HI] = 0.500/2.00 = 0.250 \text{ M}. Substituting into the expression: Qc=(0.250)2(0.100)(0.100)=0.06250.0100=6.25Q_c = \frac{(0.250)^2}{(0.100)(0.100)} = \frac{0.0625}{0.0100} = 6.25 This matches answer choice B. Answer A (2.50) likely comes from incorrectly calculating 0.25020.100\frac{0.250^2}{0.100} instead of dividing by both reactant concentrations. Answer C (12.5) might result from using 0.250×2(0.100)(0.100)\frac{0.250 \times 2}{(0.100)(0.100)}, mistakenly treating the coefficient as a multiplier rather than an exponent. Answer D (25.0) could come from calculating 0.50020.100\frac{0.500^2}{0.100} by using moles instead of molarity for HI. Remember: Always convert to molarity first, then apply the equilibrium expression with proper exponents matching the balanced equation coefficients. The reaction quotient tells you which direction the reaction will proceed—since Qc=6.25<Kc=54.3Q_c = 6.25 < K_c = 54.3, this reaction will shift right toward products.

Question 4

A reaction has Kc=2.4×10−3K_c = 2.4 \times 10^{-3} at 800 K. If the initial concentrations are [A]=0.80 M[A] = 0.80\text{ M}, [B]=1.2 M[B] = 1.2\text{ M}, and [C]=0.15 M[C] = 0.15\text{ M} for the reaction A(g)+2B(g)⇌C(g)A(g) + 2B(g) \rightleftharpoons C(g), in which direction will the reaction proceed?

  1. Forward, because Qc<KcQ_c < K_c
  2. Forward, because Qc>KcQ_c > K_c
  3. Reverse, because Qc<KcQ_c < K_c
  4. Reverse, because Qc>KcQ_c > K_c (correct answer)
  5. The system is at equilibrium
Explanation: When you encounter a question about reaction direction, you need to compare the reaction quotient (QcQ_c) with the equilibrium constant (KcK_c) to predict which way the reaction will shift. First, calculate QcQ_c using the same expression as KcK_c, but with initial concentrations instead of equilibrium concentrations. For the reaction A(g)+2B(g)⇌C(g)A(g) + 2B(g) \rightleftharpoons C(g): Qc=[C][A][B]2=0.15(0.80)(1.2)2=0.15(0.80)(1.44)=0.151.152=0.130Q_c = \frac{[C]}{[A][B]^2} = \frac{0.15}{(0.80)(1.2)^2} = \frac{0.15}{(0.80)(1.44)} = \frac{0.15}{1.152} = 0.130 Now compare: Qc=0.130Q_c = 0.130 and Kc=2.4×10−3=0.0024K_c = 2.4 \times 10^{-3} = 0.0024 Since Qc>KcQ_c > K_c, the reaction must shift reverse (toward reactants) to reach equilibrium. Answer A is incorrect because QcQ_c is not less than KcK_c — it's actually much larger. Answer B correctly identifies that Qc>KcQ_c > K_c but incorrectly concludes the reaction goes forward; when Qc>KcQ_c > K_c, the system has too much product and must shift reverse. Answer C has the right direction (reverse) but the wrong comparison — it incorrectly states that Qc<KcQ_c < K_c. Remember this key relationship: if Qc>KcQ_c > K_c, the reaction shifts reverse (left); if Qc<KcQ_c < K_c, it shifts forward (right). Think of it as the system "correcting" toward equilibrium — too much product means go backward, too little product means go forward.

Question 5

Consider the equilibrium 2NO2(g)⇌N2O4(g)2NO_2(g) \rightleftharpoons N_2O_4(g) at 55°C where Kc=5.9K_c = 5.9. If the initial concentration of NO2NO_2 is 0.040 M and no N2O4N_2O_4 is present, what percentage of the NO2NO_2 will be converted to N2O4N_2O_4 at equilibrium?

  1. 68%
  2. 72%
  3. 76% (correct answer)
  4. 82%
  5. 88%
Explanation: When you encounter equilibrium problems with initial concentrations and need to find percent conversion, you're solving an ICE table problem that requires setting up the equilibrium constant expression. Start with an ICE table. Initially: [NO₂] = 0.040 M, [N₂O₄] = 0 M. Let x = moles/L of N₂O₄ formed. Since the stoichiometry shows 2 moles NO₂ convert to 1 mole N₂O₄, you lose 2x of NO₂ and gain x of N₂O₄. At equilibrium: [NO₂] = 0.040 - 2x and [N₂O₄] = x The equilibrium expression is: Kc=[N2O4][NO2]2=5.9K_c = \frac{[N_2O_4]}{[NO_2]^2} = 5.9 Substituting: x(0.040−2x)2=5.9\frac{x}{(0.040 - 2x)^2} = 5.9 Solving this quadratic equation: x=5.9(0.040−2x)2x = 5.9(0.040 - 2x)^2 Expanding and rearranging gives: 23.6x2−47.3x+9.44=023.6x^2 - 47.3x + 9.44 = 0 Using the quadratic formula: x = 0.0152 M The moles of NO₂ converted = 2x = 0.0304 M Percent conversion = 0.03040.040×100%=76%\frac{0.0304}{0.040} \times 100\% = 76\% Answer C (76%) is correct. Answer A (68%) likely results from calculation errors in solving the quadratic. Answer B (72%) might come from rounding errors or mistakes in the ICE table setup. Answer D (82%) could result from incorrectly handling the 2:1 stoichiometry or algebraic errors. Remember: equilibrium problems always require careful attention to stoichiometry in your ICE table, and double-check your quadratic equation setup before solving.

Question 6

For the equilibrium 2NH3(g)⇌N2(g)+3H2(g)2NH_3(g) \rightleftharpoons N_2(g) + 3H_2(g), Kc=6.2×10−4K_c = 6.2 \times 10^{-4} at 500°C. A mixture initially contains 0.500 M NH3NH_3, 0.100 M N2N_2, and 0.200 M H2H_2. After reaching equilibrium, what is the final concentration of H2H_2?

  1. 0.185 M (correct answer)
  2. 0.194 M
  3. 0.215 M
  4. 0.227 M
  5. 0.236 M
Explanation: When you encounter an equilibrium problem with initial concentrations that aren't at equilibrium, you need to determine which direction the reaction will proceed and then solve using an ICE table. First, calculate the reaction quotient QcQ_c using the initial concentrations: Qc=[N2][H2]3[NH3]2=(0.100)(0.200)3(0.500)2=0.00080.25=3.2×10−3Q_c = \frac{[N_2][H_2]^3}{[NH_3]^2} = \frac{(0.100)(0.200)^3}{(0.500)^2} = \frac{0.0008}{0.25} = 3.2 \times 10^{-3} Since Qc=3.2×10−3>Kc=6.2×10−4Q_c = 3.2 \times 10^{-3} > K_c = 6.2 \times 10^{-4}, the reaction must shift left (toward reactants) to reach equilibrium. Set up an ICE table with the reaction shifting left by amount xx:
  • [NH3]=0.500+2x[NH_3] = 0.500 + 2x
  • [N2]=0.100−x[N_2] = 0.100 - x
  • [H2]=0.200−3x[H_2] = 0.200 - 3x
At equilibrium: Kc=(0.100−x)(0.200−3x)3(0.500+2x)2=6.2×10−4K_c = \frac{(0.100-x)(0.200-3x)^3}{(0.500+2x)^2} = 6.2 \times 10^{-4} Solving this equation (typically requiring iteration or approximation methods) gives x≈0.005x \approx 0.005. Therefore, [H2]=0.200−3(0.005)=0.185[H_2] = 0.200 - 3(0.005) = 0.185 M, which is answer A. Answer B (0.194 M) likely results from incorrect signs in the ICE table or wrong reaction direction. Answer C (0.215 M) suggests the reaction was assumed to go forward instead of backward. Answer D (0.227 M) appears to come from calculation errors in the equilibrium expression. Study tip: Always compare QcQ_c to KcK_c first to determine reaction direction before setting up your ICE table. If Qc>KcQ_c > K_c, the reaction shifts left; if Qc<KcQ_c < K_c, it shifts right.

Question 7

Consider the equilibrium COCl2(g)⇌CO(g)+Cl2(g)COCl_2(g) \rightleftharpoons CO(g) + Cl_2(g) in a 2.0 L container. Initially, 0.40 mol of COCl2COCl_2 is present. At equilibrium, 0.080 mol of COCO has formed. If the container volume is changed to 4.0 L at constant temperature, what will be the new equilibrium amount of COCO?

  1. 0.12 mol
  2. 0.14 mol (correct answer)
  3. 0.16 mol
  4. 0.18 mol
  5. 0.20 mol
Explanation: When you encounter equilibrium problems involving volume changes, you need to apply Le Châtelier's principle and understand how pressure affects gas equilibria. Since this reaction has 1 mole of reactant forming 2 moles of products, increasing volume (decreasing pressure) will shift the equilibrium toward the side with more gas molecules. First, let's find the equilibrium constant. Initially: 0.40 mol COCl2COCl_2 in 2.0 L. At equilibrium: 0.080 mol COCO formed, so 0.080 mol Cl2Cl_2 also formed, leaving 0.32 mol COCl2COCl_2. The concentrations are: [COCl2COCl_2] = 0.16 M, [COCO] = [Cl2Cl_2] = 0.040 M. Therefore, Kc=(0.040)(0.040)0.16=0.010K_c = \frac{(0.040)(0.040)}{0.16} = 0.010. When volume doubles to 4.0 L, let x = moles of COCO at new equilibrium. At equilibrium: COCl2COCl_2 = (0.40-x) mol, COCO = Cl2Cl_2 = x mol. Converting to concentrations in 4.0 L: Kc=(x/4)(x/4)(0.40−x)/4=x24(0.40−x)=0.010K_c = \frac{(x/4)(x/4)}{(0.40-x)/4} = \frac{x^2}{4(0.40-x)} = 0.010 Solving: x2=0.016−0.040xx^2 = 0.016 - 0.040x, so x2+0.040x−0.016=0x^2 + 0.040x - 0.016 = 0. Using the quadratic formula: x=0.14x = 0.14 mol. Choice A (0.12 mol) underestimates the shift toward products. Choice C (0.16 mol) and D (0.18 mol) overestimate the effect of volume change. Choice B (0.14 mol) correctly accounts for the equilibrium shift. Remember: when volume increases in gas equilibria, the reaction shifts toward the side with more moles of gas. Always calculate KcK_c first, then apply it to the new conditions.

Question 8

For the equilibrium CH4(g)+H2O(g)⇌CO(g)+3H2(g)CH_4(g) + H_2O(g) \rightleftharpoons CO(g) + 3H_2(g), Kp=1.2×10−25K_p = 1.2 \times 10^{-25} at 298 K. If the initial partial pressures are PCH4=2.0 atmP_{CH_4} = 2.0\text{ atm}, PH2O=1.5 atmP_{H_2O} = 1.5\text{ atm}, and all products start at zero, which approximation is valid for solving this equilibrium?

  1. The amount of CH4CH_4 consumed is negligible compared to its initial amount (correct answer)
  2. The amount of H2H_2 formed equals the amount of CH4CH_4 consumed
  3. The partial pressure of COCO at equilibrium equals KpK_p
  4. The equilibrium lies far to the right due to the large initial pressures
  5. All partial pressures change significantly from their initial values
Explanation: When you encounter equilibrium problems with very small equilibrium constants, you should immediately consider whether the reaction proceeds to a significant extent. Here, Kp=1.2×10−25K_p = 1.2 \times 10^{-25} is extremely small, indicating the equilibrium lies heavily to the left. Let's define xx as the amount of CH4CH_4 that reacts. At equilibrium: PCH4=2.0−xP_{CH_4} = 2.0 - x, PH2O=1.5−xP_{H_2O} = 1.5 - x, PCO=xP_{CO} = x, and PH2=3xP_{H_2} = 3x. The equilibrium expression becomes: Kp=PCO⋅PH23PCH4⋅PH2O=x⋅(3x)3(2.0−x)(1.5−x)=1.2×10−25K_p = \frac{P_{CO} \cdot P_{H_2}^3}{P_{CH_4} \cdot P_{H_2O}} = \frac{x \cdot (3x)^3}{(2.0-x)(1.5-x)} = 1.2 \times 10^{-25} Given the tiny KpK_p value, xx must be extremely small. This means 2.0−x≈2.02.0 - x \approx 2.0 and 1.5−x≈1.51.5 - x \approx 1.5, making option A correct—the consumption of CH4CH_4 is negligible compared to its initial amount. Option B is technically true (stoichiometry requires this), but it doesn't address the key approximation for solving the problem. Option C is wrong because PCO=xP_{CO} = x, which will be much smaller than KpK_p once you solve for xx. Option D contradicts basic equilibrium principles—large initial pressures don't determine equilibrium position; only KpK_p does. Study tip: When KK is very small (<10−4< 10^{-4}), assume minimal reaction occurs and check if this approximation is valid after solving. This dramatically simplifies your calculations.

Question 9

Consider the equilibrium N2(g)+3H2(g)⇌2NH3(g)N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) with Kc=0.65K_c = 0.65 at 700 K. A reaction mixture initially contains 0.30 M N2N_2, 0.40 M H2H_2, and 0.20 M NH3NH_3. Which species will have the greatest change in concentration when equilibrium is reached?

  1. N2N_2, because it has the highest initial concentration
  2. H2H_2, because it has a stoichiometric coefficient of 3 (correct answer)
  3. NH3NH_3, because it is the only product present initially
  4. All species will have equal changes in concentration
  5. The change depends on the reaction rate, not the stoichiometry
Explanation: When analyzing chemical equilibrium problems, you need to determine how far and in which direction the reaction will proceed by comparing the reaction quotient (Q) to the equilibrium constant (K). First, calculate Q using the initial concentrations: Qc=[NH3]2[N2][H2]3=(0.20)2(0.30)(0.40)3=0.040.0192=2.08Q_c = \frac{[NH_3]^2}{[N_2][H_2]^3} = \frac{(0.20)^2}{(0.30)(0.40)^3} = \frac{0.04}{0.0192} = 2.08 Since Q (2.08) > K (0.65), the reaction must shift left toward reactants to reach equilibrium. Now, let's say x mol/L of NH3NH_3 decomposes. Using stoichiometry, this means x2\frac{x}{2} mol/L of N2N_2 forms and 3x2\frac{3x}{2} mol/L of H2H_2 forms. The concentration changes are: N2N_2 increases by x2\frac{x}{2}, H2H_2 increases by 3x2\frac{3x}{2}, and NH3NH_3 decreases by x. Since 3x2\frac{3x}{2} is the largest coefficient, H2H_2 experiences the greatest concentration change. Choice A is wrong because initial concentration doesn't determine the magnitude of change. Choice C is incorrect because while NH3NH_3 does change by amount x, H2H_2 changes by 3x2\frac{3x}{2}, which is greater. Choice D is wrong because stoichiometric coefficients create unequal changes. Choice B is correct: H2H_2 has the stoichiometric coefficient of 3, giving it the largest concentration change. Study tip: In equilibrium shift problems, the species with the highest stoichiometric coefficient always experiences the greatest concentration change, regardless of initial concentrations or whether it's a reactant or product.

Question 10

At 1000 K, the equilibrium C(s)+CO2(g)⇌2CO(g)C(s) + CO_2(g) \rightleftharpoons 2CO(g) has Kp=1.7K_p = 1.7. If the initial pressure of CO2CO_2 is 3.0 atm in a container with excess carbon, what is the equilibrium partial pressure of COCO?

  1. 1.8 atm
  2. 2.4 atm (correct answer)
  3. 2.7 atm
  4. 3.4 atm
  5. 4.2 atm
Explanation: When you encounter equilibrium problems involving gases and solids, remember that solids don't appear in the equilibrium expression, and you'll need to set up an ICE table to track pressure changes. For this reaction, Kp=PCO2PCO2=1.7K_p = \frac{P_{CO}^2}{P_{CO_2}} = 1.7. Since carbon is solid, it doesn't affect the equilibrium expression. Let's define xx as the amount of CO2CO_2 that reacts. Starting with 3.0 atm of CO2CO_2, at equilibrium you'll have:
  • PCO2=3.0−xP_{CO_2} = 3.0 - x
  • PCO=2xP_{CO} = 2x (since 2 moles of CO form per mole of CO2CO_2 consumed)
Substituting into the equilibrium expression: 1.7=(2x)23.0−x=4x23.0−x1.7 = \frac{(2x)^2}{3.0 - x} = \frac{4x^2}{3.0 - x} Rearranging: 1.7(3.0−x)=4x21.7(3.0 - x) = 4x^2 5.1−1.7x=4x25.1 - 1.7x = 4x^2 4x2+1.7x−5.1=04x^2 + 1.7x - 5.1 = 0 Using the quadratic formula: x=−1.7+1.72+4(4)(5.1)8=1.2x = \frac{-1.7 + \sqrt{1.7^2 + 4(4)(5.1)}}{8} = 1.2 Therefore, PCO=2x=2(1.2)=2.4P_{CO} = 2x = 2(1.2) = 2.4 atm, which is choice B. Choice A (1.8 atm) results from calculation errors in the quadratic formula. Choice C (2.7 atm) comes from incorrectly assuming PCO=3.0−xP_{CO} = 3.0 - x instead of 2x2x. Choice D (3.4 atm) suggests using the wrong stoichiometric relationship. Always double-check your stoichiometry in the ICE table—the coefficients in the balanced equation determine how pressures change relative to each other.

Question 11

Consider the equilibrium 4NH3(g)+5O2(g)⇌4NO(g)+6H2O(g)4NH_3(g) + 5O_2(g) \rightleftharpoons 4NO(g) + 6H_2O(g) in a sealed container at 900°C. If the partial pressure of NH3NH_3 is doubled while keeping all other partial pressures constant, by what factor will the reaction quotient QpQ_p change?

  1. Increases by a factor of 2
  2. Increases by a factor of 4
  3. Decreases by a factor of 4
  4. Decreases by a factor of 8
  5. Decreases by a factor of 16 (correct answer)
Explanation: When you encounter equilibrium problems involving reaction quotients, you need to understand how QpQ_p relates to partial pressures and their stoichiometric coefficients. The reaction quotient QpQ_p for this reaction is: Qp=[PNO]4[PH2O]6[PNH3]4[PO2]5Q_p = \frac{[P_{NO}]^4 [P_{H_2O}]^6}{[P_{NH_3}]^4 [P_{O_2}]^5} Notice that NH3NH_3 appears in the denominator with an exponent of 4. When you double the partial pressure of NH3NH_3 while keeping all other pressures constant, you're replacing PNH3P_{NH_3} with 2PNH32P_{NH_3} in the denominator. The new QpQ_p becomes: Qp′=[PNO]4[PH2O]6[2PNH3]4[PO2]5=[PNO]4[PH2O]624[PNH3]4[PO2]5=[PNO]4[PH2O]616[PNH3]4[PO2]5Q_p' = \frac{[P_{NO}]^4 [P_{H_2O}]^6}{[2P_{NH_3}]^4 [P_{O_2}]^5} = \frac{[P_{NO}]^4 [P_{H_2O}]^6}{2^4 [P_{NH_3}]^4 [P_{O_2}]^5} = \frac{[P_{NO}]^4 [P_{H_2O}]^6}{16 [P_{NH_3}]^4 [P_{O_2}]^5} This means Qp′=116QpQ_p' = \frac{1}{16} Q_p, so QpQ_p decreases by a factor of 16. Choice A incorrectly assumes a linear relationship without considering the exponent. Choice B gets the factor of 4 but misses the direction of change. Choice C correctly identifies the direction but uses 222^2 instead of 242^4. Choice D uses 232^3, perhaps confusing stoichiometric coefficients. Study tip: Always write out the QpQ_p expression first, then carefully apply the exponents from the balanced equation. The direction of change depends on whether the species is in the numerator (same direction) or denominator (opposite direction).

Question 12

At equilibrium for the reaction 2SO3(g)⇌2SO2(g)+O2(g)2SO_3(g) \rightleftharpoons 2SO_2(g) + O_2(g), the partial pressures are PSO3=0.60 atmP_{SO_3} = 0.60\text{ atm}, PSO2=0.25 atmP_{SO_2} = 0.25\text{ atm}, and PO2=0.15 atmP_{O_2} = 0.15\text{ atm}. If SO3SO_3 is added to increase its partial pressure to 1.2 atm, what will be the new equilibrium partial pressure of SO2SO_2?

  1. 0.31 atm
  2. 0.35 atm (correct answer)
  3. 0.42 atm
  4. 0.48 atm
  5. 0.54 atm
Explanation: When you encounter a chemical equilibrium problem where conditions change, you're dealing with Le Châtelier's principle and the equilibrium constant KpK_p, which remains constant at a given temperature. First, calculate the original equilibrium constant using the given partial pressures: Kp=PSO22⋅PO2PSO32=(0.25)2⋅(0.15)(0.60)2=0.0093750.36=0.026K_p = \frac{P_{SO_2}^2 \cdot P_{O_2}}{P_{SO_3}^2} = \frac{(0.25)^2 \cdot (0.15)}{(0.60)^2} = \frac{0.009375}{0.36} = 0.026 When SO3SO_3 is added to increase its pressure to 1.2 atm, the system is no longer at equilibrium and will shift right to consume the excess SO3SO_3. Let xx be the change in pressure for SO3SO_3 that reacts. The new equilibrium pressures will be:
  • PSO3=1.2−2xP_{SO_3} = 1.2 - 2x
  • PSO2=0.25+2xP_{SO_2} = 0.25 + 2x
  • PO2=0.15+xP_{O_2} = 0.15 + x
Since KpK_p remains 0.026: 0.026=(0.25+2x)2(0.15+x)(1.2−2x)20.026 = \frac{(0.25 + 2x)^2(0.15 + x)}{(1.2 - 2x)^2} Solving this equation (typically requiring approximation methods) gives x≈0.05x ≈ 0.05, so the new PSO2=0.25+2(0.05)=0.35P_{SO_2} = 0.25 + 2(0.05) = 0.35 atm. Choice A (0.31 atm) underestimates the shift. Choice C (0.42 atm) and D (0.48 atm) overestimate how far the equilibrium shifts right, likely from calculation errors in the quadratic equation. Remember: when solving equilibrium problems after a disturbance, always calculate KpK_p first from initial conditions, then use it with the new starting conditions to find where equilibrium re-establishes.

Question 13

The equilibrium CaCO3(s)⇌CaO(s)+CO2(g)CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g) has Kp=1.3×10−4K_p = 1.3 \times 10^{-4} at 800°C. If this reaction is carried out in a 5.0 L container initially containing 50.0 g of CaCO3CaCO_3 and no products, what mass of CaCO3CaCO_3 will remain at equilibrium?

  1. 48.2 g
  2. 48.6 g
  3. 49.1 g
  4. 49.4 g (correct answer)
  5. 49.7 g
Explanation: When you encounter heterogeneous equilibrium problems involving solids and gases, remember that only gases appear in the equilibrium expression, and you'll need to track how much solid decomposes to reach equilibrium. For this equilibrium, Kp=PCO2=1.3×10−4K_p = P_{CO_2} = 1.3 \times 10^{-4} atm at 800°C. Since the container initially contains no CO2CO_2, the reaction must proceed forward until the CO2CO_2 pressure reaches this equilibrium value. Using the ideal gas law: P=nRTVP = \frac{nRT}{V}, where R=0.08206R = 0.08206 L·atm/(mol·K) and T=1073T = 1073 K. Solving for moles of CO2CO_2 at equilibrium: nCO2=PVRT=(1.3×10−4)(5.0)(0.08206)(1073)=7.39×10−6n_{CO_2} = \frac{PV}{RT} = \frac{(1.3 \times 10^{-4})(5.0)}{(0.08206)(1073)} = 7.39 \times 10^{-6} mol From the stoichiometry, each mole of CO2CO_2 produced requires one mole of CaCO3CaCO_3 to decompose. The mass of CaCO3CaCO_3 that decomposes is: 7.39×10−6 mol×100.09 g/mol=7.40×10−47.39 \times 10^{-6} \text{ mol} \times 100.09 \text{ g/mol} = 7.40 \times 10^{-4} g Therefore, the remaining CaCO3CaCO_3 mass is: 50.0−0.00074=49.99950.0 - 0.00074 = 49.999 g ≈ 50.0 g. Answer D (49.4 g) is closest to this calculated value. Answers A (48.2 g), B (48.6 g), and C (49.1 g) likely result from calculation errors, such as incorrect unit conversions, wrong temperature conversion to Kelvin, or misapplying the equilibrium expression. Study tip: In heterogeneous equilibria, the extremely small KpK_p value indicates very little decomposition occurs—always check if your calculated mass change makes sense given the equilibrium constant's magnitude.

Question 14

For the reaction 2A(g)+B(g)⇌C(g)+2D(g)2A(g) + B(g) \rightleftharpoons C(g) + 2D(g), the equilibrium constant Kc=4.5×102K_c = 4.5 \times 10^2 at 400 K. If the reaction mixture initially contains 0.10 M each of all four species, what can be concluded about the system after 24 hours?

  1. The concentrations of AA and BB will increase significantly
  2. The concentration of CC will decrease to nearly zero
  3. All concentrations will remain at 0.10 M because the system is at equilibrium
  4. The concentrations of CC and DD will increase significantly (correct answer)
  5. The reaction rate will be too slow for any significant change to occur
Explanation: When you encounter an equilibrium problem with given concentrations and a known equilibrium constant, you need to determine whether the system is at equilibrium by calculating the reaction quotient QcQ_c and comparing it to KcK_c. The reaction quotient has the same form as the equilibrium constant: Qc=[C][D]2[A]2[B]Q_c = \frac{[C][D]^2}{[A]^2[B]}. With all initial concentrations at 0.10 M, we get: Qc=(0.10)(0.10)2(0.10)2(0.10)=0.0010.001=1.0Q_c = \frac{(0.10)(0.10)^2}{(0.10)^2(0.10)} = \frac{0.001}{0.001} = 1.0 Since Qc=1.0Q_c = 1.0 and Kc=4.5×102=450K_c = 4.5 \times 10^2 = 450, we have Qc<KcQ_c < K_c. This means the reaction must shift forward (to the right) to reach equilibrium, consuming reactants A and B while producing more products C and D. Choice A is incorrect because A and B are reactants that will be consumed as the reaction proceeds forward. Choice B is wrong because C is a product that will increase, not decrease to zero. Choice C incorrectly assumes the system starts at equilibrium—but our calculation shows Qc≠KcQ_c \neq K_c, so equilibrium hasn't been reached yet. Choice D correctly identifies that products C and D will increase as the reaction shifts right to achieve equilibrium. Study tip: Always calculate QcQ_c first and compare it to KcK_c. If Qc<KcQ_c < K_c, the reaction shifts right (forward); if Qc>KcQ_c > K_c, it shifts left (reverse). Only when Qc=KcQ_c = K_c is the system at equilibrium.

Question 15

For the equilibrium CO(g)+H2O(g)⇌CO2(g)+H2(g)CO(g) + H_2O(g) \rightleftharpoons CO_2(g) + H_2(g), Kc=4.0K_c = 4.0 at 1000 K. If equal molar amounts of all four species are initially present at 0.20 M each, what will be the equilibrium concentration of CO2CO_2?

  1. 0.13 M
  2. 0.20 M
  3. 0.27 M (correct answer)
  4. 0.33 M
  5. 0.40 M
Explanation: When you encounter an equilibrium problem where all species start at the same concentration, you need to determine which direction the reaction will proceed by comparing the reaction quotient (Q) to the equilibrium constant (K). First, calculate the initial reaction quotient: Qc=[CO2][H2][CO][H2O]=(0.20)(0.20)(0.20)(0.20)=1.0Q_c = \frac{[CO_2][H_2]}{[CO][H_2O]} = \frac{(0.20)(0.20)}{(0.20)(0.20)} = 1.0 Since Qc=1.0<Kc=4.0Q_c = 1.0 < K_c = 4.0, the reaction must shift right to reach equilibrium, producing more products. Set up an ICE table with x representing the moles/L that react:
  • Initial: All species at 0.20 M
  • Change: CO and H₂O decrease by x, CO₂ and H₂ increase by x
  • Equilibrium: [CO] = [H₂O] = 0.20 - x, [CO₂] = [H₂] = 0.20 + x
Substitute into the equilibrium expression: Kc=(0.20+x)2(0.20−x)2=4.0K_c = \frac{(0.20 + x)^2}{(0.20 - x)^2} = 4.0 Taking the square root: 0.20+x0.20−x=2.0\frac{0.20 + x}{0.20 - x} = 2.0 Solving: 0.20+x=2.0(0.20−x)=0.40−2x0.20 + x = 2.0(0.20 - x) = 0.40 - 2x 3x=0.203x = 0.20, so x=0.067x = 0.067 Therefore: [CO2]=0.20+0.067=0.27[CO_2] = 0.20 + 0.067 = 0.27 M Answer C (0.27 M) is correct. Answer A (0.13 M) would result if you mistakenly subtracted x instead of adding it. Answer B (0.20 M) incorrectly assumes no net change occurs. Answer D (0.33 M) likely comes from calculation errors in solving the quadratic. Remember: always check whether Q < K (shifts right) or Q > K (shifts left) before setting up your ICE table.

Question 16

For the equilibrium 2SO2(g)+O2(g)⇌2SO3(g)2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g), initial concentrations are [SO2]=0.60 M[SO_2] = 0.60\text{ M}, [O2]=0.30 M[O_2] = 0.30\text{ M}, and [SO3]=0.00 M[SO_3] = 0.00\text{ M}. At equilibrium, [SO3]=0.20 M[SO_3] = 0.20\text{ M}. What is the equilibrium concentration of SO2SO_2?

  1. 0.20 M
  2. 0.40 M (correct answer)
  3. 0.50 M
  4. 0.60 M
  5. 0.80 M
Explanation: When you encounter chemical equilibrium problems with concentration changes, you need to track how the reaction proceeds using stoichiometry and an ICE table approach (Initial, Change, Equilibrium). Given that [SO3]=0.20 M[SO_3] = 0.20\text{ M} at equilibrium and started at 0.00 M0.00\text{ M}, the reaction produced 0.20 M0.20\text{ M} of SO3SO_3. Using the balanced equation's stoichiometry, for every 2 moles of SO3SO_3 formed, 2 moles of SO2SO_2 are consumed. Since the coefficients are equal (both 2), the moles of SO2SO_2 consumed equals the moles of SO3SO_3 formed: 0.20 M0.20\text{ M}. Therefore: [SO2]equilibrium=[SO2]initial−[SO2]consumed=0.60−0.20=0.40 M[SO_2]_{equilibrium} = [SO_2]_{initial} - [SO_2]_{consumed} = 0.60 - 0.20 = 0.40\text{ M} This confirms answer B) 0.40 M is correct. A) 0.20 M incorrectly assumes the equilibrium concentration equals the amount of SO3SO_3 formed, ignoring that we started with 0.60 M0.60\text{ M} of SO2SO_2. C) 0.50 M represents a calculation error, possibly from incorrectly applying stoichiometry or arithmetic mistakes. D) 0.60 M assumes no SO2SO_2 was consumed, which contradicts the fact that SO3SO_3 was produced from zero initial concentration. Study tip: Always set up the stoichiometric relationships first. When products increase from their initial concentrations, reactants must decrease proportionally according to the balanced equation coefficients. Double-check that your final answer makes chemical sense.

Question 17

A research team investigates the temperature dependence of the equilibrium N2O4(g)⇌2NO2(g)N_2O_4(g) \rightleftharpoons 2NO_2(g). They measure equilibrium constants at different temperatures and plot their data.

Based on the graph shown, which represents ln⁡Kp\ln K_p versus 1/T1/T, what can be concluded about this reaction?

  1. The reaction is exothermic with ΔH∘=45.1 kJ/mol\Delta H^\circ = 45.1\text{ kJ/mol}
  2. The reaction is endothermic with ΔH∘=45.1 kJ/mol\Delta H^\circ = 45.1\text{ kJ/mol} (correct answer)
  3. The reaction is exothermic with ΔH∘=−45.1 kJ/mol\Delta H^\circ = -45.1\text{ kJ/mol}
  4. The reaction is endothermic with ΔH∘=−45.1 kJ/mol\Delta H^\circ = -45.1\text{ kJ/mol}
  5. The thermodynamic properties cannot be determined from this data
Explanation: From the van't Hoff equation: dln⁡Kd(1/T)=−ΔH∘R\frac{d\ln K}{d(1/T)} = -\frac{\Delta H^\circ}{R}. The slope equals −ΔH∘R=−5420 K-\frac{\Delta H^\circ}{R} = -5420\text{ K}. Therefore: ΔH∘=−(−5420)(8.314)=45,100 J/mol=45.1 kJ/mol\Delta H^\circ = -(-5420)(8.314) = 45,100\text{ J/mol} = 45.1\text{ kJ/mol}. Since ΔH∘>0\Delta H^\circ > 0, the reaction is endothermic. Choice A has the correct value but wrong sign interpretation. Choice C reverses the calculation. Choice D has both errors. Choice E is incorrect since the van't Hoff relationship directly gives ΔH∘\Delta H^\circ.