College Chemistry Quiz: Properties Of The Equilibrium Constant
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Properties Of The Equilibrium ConstantQuestion 1 of 20

Consider the equilibrium CaCO3(s)⇌CaO(s)+CO2(g)CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g). Which of the following correctly represents the equilibrium constant expression?

Kc=[CaO][CO2][CaCO3]K_c = \frac{[CaO][CO_2]}{[CaCO_3]}
Kc=[CO2]K_c = [CO_2]
Kc=[CO2][CaCO3]K_c = \frac{[CO_2]}{[CaCO_3]}
Kc=[CaO][CaCO3]K_c = \frac{[CaO]}{[CaCO_3]}
Kc=[CaO]×[CO2]K_c = [CaO] \times [CO_2]
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College Chemistry Quiz

College Chemistry Quiz: Properties Of The Equilibrium Constant

Practice Properties Of The Equilibrium Constant 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 Properties Of The Equilibrium Constant, 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

Consider the equilibrium CaCO3(s)⇌CaO(s)+CO2(g)CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g). Which of the following correctly represents the equilibrium constant expression?

  1. Kc=[CaO][CO2][CaCO3]K_c = \frac{[CaO][CO_2]}{[CaCO_3]}
  2. Kc=[CO2]K_c = [CO_2] (correct answer)
  3. Kc=[CO2][CaCO3]K_c = \frac{[CO_2]}{[CaCO_3]}
  4. Kc=[CaO][CaCO3]K_c = \frac{[CaO]}{[CaCO_3]}
  5. Kc=[CaO]×[CO2]K_c = [CaO] \times [CO_2]
Explanation: When you encounter equilibrium expressions involving solids and gases, the key principle is that only species in the gas and aqueous phases appear in the equilibrium constant expression. Pure solids and pure liquids have constant concentrations and are omitted. For the reaction CaCO3(s)⇌CaO(s)+CO2(g)CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g), both CaCO3CaCO_3 and CaOCaO are pure solids, so their concentrations remain constant throughout the reaction. Only CO2CO_2 is a gas, meaning its concentration can vary. Therefore, the equilibrium expression becomes simply Kc=[CO2]K_c = [CO_2], making choice B correct. Choice A incorrectly includes both solid species in the expression. While this follows the general pattern of products over reactants, it violates the fundamental rule about excluding pure solids from equilibrium expressions. Choice C includes only CaCO3CaCO_3 as a solid but still incorrectly incorporates it in the denominator. Remember, all pure solids must be excluded, not just some of them. Choice D commits the same error as C by including the solid CaCO3CaCO_3 in the denominator, while also incorrectly including solid CaOCaO in the numerator instead of the gas CO2CO_2. Study tip: When writing equilibrium expressions, first identify the physical states of all species. Cross out any pure solids (s) and pure liquids (l) before writing your expression. Only include gases (g) and aqueous solutions (aq). This simple screening step will prevent most errors on equilibrium constant problems.

Question 2

Consider the equilibrium reaction N2(g)+3H2(g)⇌2NH3(g)N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) with Kc=0.60K_c = 0.60 at 500°C. If the equation is multiplied by 2 to give 2N2(g)+6H2(g)⇌4NH3(g)2N_2(g) + 6H_2(g) \rightleftharpoons 4NH_3(g), what is the new equilibrium constant?

  1. 0.30
  2. 0.36 (correct answer)
  3. 0.60
  4. 1.2
  5. 1.7
Explanation: When you encounter questions about manipulating chemical equilibrium expressions, remember that the equilibrium constant changes in predictable ways based on how you modify the balanced equation. For the original reaction N2(g)+3H2(g)⇌2NH3(g)N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g), the equilibrium constant is Kc=[NH3]2[N2][H2]3=0.60K_c = \frac{[NH_3]^2}{[N_2][H_2]^3} = 0.60. When you multiply an entire equation by a coefficient, you raise the equilibrium constant to that power. Since we're multiplying by 2, the new equilibrium constant becomes Kc2=(0.60)2=0.36K_c^2 = (0.60)^2 = 0.36. This happens because the new equilibrium expression is Kc,new=[NH3]4[N2]2[H2]6K_{c,new} = \frac{[NH_3]^4}{[N_2]^2[H_2]^6}, which equals ([NH3]2[N2][H2]3)2=Kc2\left(\frac{[NH_3]^2}{[N_2][H_2]^3}\right)^2 = K_c^2. Option A (0.30) represents the error of dividing the original constant by 2 instead of squaring it. Option C (0.60) incorrectly assumes the equilibrium constant doesn't change when you modify the equation. Option D (1.2) comes from multiplying by 2 instead of raising to the power of 2. The correct answer is B (0.36). Remember this key rule: when you multiply a balanced equation by a factor n, raise the equilibrium constant to the nth power. When you reverse an equation, take the reciprocal of K. These transformations are essential for solving equilibrium problems involving multiple reaction steps.

Question 3

At 25°C, the equilibrium constant Kc=1.8×10−5K_c = 1.8 \times 10^{-5} for the reaction NH3(aq)+H2O(l)⇌NH4+(aq)+OH−(aq)NH_3(aq) + H_2O(l) \rightleftharpoons NH_4^+(aq) + OH^-(aq). What can be concluded about the position of equilibrium?

  1. The equilibrium lies far to the right since Kc>1K_c > 1
  2. The equilibrium lies far to the left since Kc<1K_c < 1 (correct answer)
  3. The equilibrium is exactly balanced since KcK_c is close to 1
  4. The position cannot be determined without concentration data
  5. The equilibrium lies slightly to the right since Kc>0K_c > 0
Explanation: When you encounter equilibrium constant problems, the key insight is that the magnitude of KcK_c directly tells you which side of the reaction is favored. The equilibrium constant compares the concentration of products to reactants at equilibrium. For this ammonia hydrolysis reaction, Kc=1.8×10−5=0.000018K_c = 1.8 \times 10^{-5} = 0.000018, which is much smaller than 1. This small value means that at equilibrium, the concentration of products (NH4+NH_4^+ and OH−OH^-) is much lower than the concentration of reactants (NH3NH_3). Therefore, the equilibrium lies far to the left, favoring the reactants. Looking at the wrong answers: Choice A incorrectly states that Kc>1K_c > 1, but 1.8×10−51.8 \times 10^{-5} is clearly much less than 1. Choice C suggests the equilibrium is balanced because KcK_c is "close to 1," but 0.0000180.000018 is actually five orders of magnitude smaller than 1 – that's nowhere near close. Choice D claims you need concentration data, but that misses the point entirely: KcK_c itself already incorporates equilibrium concentrations and tells you everything you need to know about the equilibrium position. Remember this pattern: Kc>>1K_c >> 1 means products dominate (equilibrium right), Kc<<1K_c << 1 means reactants dominate (equilibrium left), and Kc≈1K_c ≈ 1 means roughly equal amounts. Any KcK_c value with negative exponents (like 10−510^{-5}) immediately signals that reactants are heavily favored.

Question 4

For the gas-phase reaction 2SO2(g)+O2(g)⇌2SO3(g)2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g), Kc=2.8×102K_c = 2.8 \times 10^{2} at 1000 K. What is the value of KpK_p for this reaction at the same temperature? (R=0.0821 L⋅atm⋅mol−1⋅K−1R = 0.0821 \, L \cdot atm \cdot mol^{-1} \cdot K^{-1})

  1. 3.4×1003.4 \times 10^{0} (correct answer)
  2. 2.8×1022.8 \times 10^{2}
  3. 1.2×1041.2 \times 10^{4}
  4. 2.3×1042.3 \times 10^{4}
  5. 6.7×1066.7 \times 10^{6}
Explanation: When you encounter equilibrium constant problems involving both KcK_c and KpK_p, you need to understand the relationship between these two expressions. KcK_c uses molar concentrations while KpK_p uses partial pressures, and they're connected by the equation: Kp=Kc(RT)ΔnK_p = K_c(RT)^{\Delta n}, where Δn\Delta n is the change in moles of gas. First, calculate Δn\Delta n by finding the difference between moles of gaseous products and reactants: Δn=(2 mol SO3)−(2 mol SO2+1 mol O2)=2−3=−1\Delta n = (2 \text{ mol } SO_3) - (2 \text{ mol } SO_2 + 1 \text{ mol } O_2) = 2 - 3 = -1. Now substitute into the equation: Kp=(2.8×102)(0.0821×1000)−1=(2.8×102)(82.1)−1=28082.1=3.4K_p = (2.8 \times 10^2)(0.0821 \times 1000)^{-1} = (2.8 \times 10^2)(82.1)^{-1} = \frac{280}{82.1} = 3.4 This gives us Kp=3.4×100K_p = 3.4 \times 10^0, which is answer A. Looking at the wrong answers: B (2.8×1022.8 \times 10^2) assumes Kp=KcK_p = K_c, which only occurs when Δn=0\Delta n = 0. C (1.2×1041.2 \times 10^4) results from incorrectly using Δn=+1\Delta n = +1 instead of −1-1. D (2.3×1042.3 \times 10^4) appears to involve calculation errors or misapplication of the formula. Study tip: Always calculate Δn\Delta n carefully by counting gas molecules on each side of the equation. Remember that when Δn\Delta n is negative, KpK_p will be smaller than KcK_c, and when positive, KpK_p will be larger than KcK_c.

Question 5

For a general reaction aA+bB⇌cC+dDaA + bB \rightleftharpoons cC + dD, which expression correctly represents the equilibrium constant KcK_c?

  1. Kc=[A]a[B]b[C]c[D]dK_c = \frac{[A]^a[B]^b}{[C]^c[D]^d}
  2. Kc=[C]c[D]d[A]a[B]bK_c = \frac{[C]^c[D]^d}{[A]^a[B]^b} (correct answer)
  3. Kc=[C][D][A][B]K_c = \frac{[C][D]}{[A][B]}
  4. Kc=[C]c[D]d−[A]a[B]bK_c = [C]^c[D]^d - [A]^a[B]^b
  5. Kc=c[C]+d[D]a[A]+b[B]K_c = \frac{c[C] + d[D]}{a[A] + b[B]}
Explanation: When you encounter equilibrium constant expressions, remember that they follow a universal pattern: products over reactants, with each concentration raised to its stoichiometric coefficient as an exponent. For any equilibrium reaction, the equilibrium constant KcK_c is constructed by placing the concentrations of products in the numerator and reactants in the denominator. Each concentration is raised to the power of its coefficient from the balanced equation. For the reaction aA+bB⇌cC+dDaA + bB \rightleftharpoons cC + dD, products C and D go on top, while reactants A and B go on bottom, giving us Kc=[C]c[D]d[A]a[B]bK_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}. This is answer choice B. Choice A reverses this relationship, incorrectly placing reactants over products. This is a common mistake that would give you the expression for 1Kc\frac{1}{K_c} rather than KcK_c itself. Choice C includes the correct products-over-reactants setup but omits the crucial exponents, ignoring the stoichiometric coefficients entirely. This fundamentally misrepresents how equilibrium responds to concentration changes. Choice D uses subtraction instead of division, which has no basis in equilibrium theory and wouldn't produce a meaningful constant value. The key study tip: Always remember "products over reactants, coefficients become exponents." This pattern applies to every equilibrium constant expression you'll encounter, whether it's KcK_c, KpK_p, or KeqK_{eq}. The stoichiometric coefficients are essential—they reflect how the equilibrium position responds to concentration changes according to Le Chatelier's principle.

Question 6

The equilibrium constant Kc=4.2×10−3K_c = 4.2 \times 10^{-3} for the reaction N2O4(g)⇌2NO2(g)N_2O_4(g) \rightleftharpoons 2NO_2(g) at 298 K. If the initial concentration of N2O4N_2O_4 is 0.50 M and no NO2NO_2 is initially present, which statement is correct about the equilibrium concentrations?

  1. [NO2]>[N2O4][NO_2] > [N_2O_4] at equilibrium because the reaction proceeds to completion
  2. [NO2]<[N2O4][NO_2] < [N_2O_4] at equilibrium because Kc<1K_c < 1 indicates reactants are favored (correct answer)
  3. [NO2]=[N2O4][NO_2] = [N_2O_4] at equilibrium because the stoichiometry is 1:2
  4. [NO2]>[N2O4][NO_2] > [N_2O_4] at equilibrium because two moles of product form per mole of reactant
  5. The concentrations cannot be predicted without solving the equilibrium expression
Explanation: When you encounter equilibrium problems with given KcK_c values, the magnitude of the equilibrium constant tells you which species will predominate at equilibrium. Since Kc=4.2×10−3<1K_c = 4.2 \times 10^{-3} < 1, the equilibrium lies to the left, favoring reactants over products. To verify this quantitatively, set up an ICE table. Starting with 0.50 M N2O4N_2O_4 and 0 M NO2NO_2, let xx be the amount of N2O4N_2O_4 that dissociates. At equilibrium: [N2O4]=0.50−x[N_2O_4] = 0.50 - x and [NO2]=2x[NO_2] = 2x. Substituting into the equilibrium expression: Kc=[NO2]2[N2O4]=(2x)20.50−x=4.2×10−3K_c = \frac{[NO_2]^2}{[N_2O_4]} = \frac{(2x)^2}{0.50 - x} = 4.2 \times 10^{-3} Solving this equation yields x≈0.023x \approx 0.023 M, giving equilibrium concentrations of approximately [N2O4]=0.48[N_2O_4] = 0.48 M and [NO2]=0.046[NO_2] = 0.046 M. Clearly, [NO2]<[N2O4][NO_2] < [N_2O_4], confirming answer B. Choice A is wrong because reactions with small KcK_c values don't proceed to completion—they reach equilibrium with mostly unreacted starting material. Choice C incorrectly assumes stoichiometry determines concentration ratios at equilibrium; stoichiometry only affects the rate at which concentrations change. Choice D makes the error of thinking that producing more moles of product automatically means higher product concentration at equilibrium. Remember: Kc<1K_c < 1 means reactants dominate at equilibrium, while Kc>1K_c > 1 means products dominate. The actual numerical value tells you how strongly the equilibrium favors one side.

Question 7

For the reaction PCl5(g)⇌PCl3(g)+Cl2(g)PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g) at 250°C, Kc=0.042K_c = 0.042. A reaction mixture at equilibrium contains 0.15 M PCl3PCl_3 and 0.15 M Cl2Cl_2. What is the equilibrium concentration of PCl5PCl_5?

  1. 0.054 M
  2. 0.28 M
  3. 0.54 M (correct answer)
  4. 1.9 M
  5. 3.6 M
Explanation: When you encounter equilibrium problems involving KcK_c, you're working with the equilibrium constant expression that relates product and reactant concentrations at equilibrium. For this reaction, Kc=[PCl3][Cl2][PCl5]K_c = \frac{[PCl_3][Cl_2]}{[PCl_5]}. To find the equilibrium concentration of PCl5PCl_5, substitute the known values into the equilibrium expression: 0.042=(0.15)(0.15)[PCl5]0.042 = \frac{(0.15)(0.15)}{[PCl_5]}. Solving for [PCl5][PCl_5]: [PCl5]=(0.15)(0.15)0.042=0.02250.042=0.54 M[PCl_5] = \frac{(0.15)(0.15)}{0.042} = \frac{0.0225}{0.042} = 0.54 \text{ M}. This confirms answer C is correct. Let's examine why the other options are wrong. Answer A (0.054 M) results from incorrectly multiplying KcK_c by the product concentrations instead of dividing: 0.042×0.15×0.15=0.009450.042 \times 0.15 \times 0.15 = 0.00945, then somehow getting 0.054—this shows a fundamental misunderstanding of the KcK_c expression. Answer B (0.28 M) might come from using 0.0420.15=0.28\frac{0.042}{0.15} = 0.28, incorrectly treating the reaction as if only one product exists. Answer D (1.9 M) could result from inverting the calculation: 0.042×0.15×0.150.0422\frac{0.042 \times 0.15 \times 0.15}{0.042^2} or similar algebraic errors. Remember this key strategy: always write out the KcK_c expression first, then substitute known values methodically. Double-check that products go in the numerator and reactants in the denominator, with proper stoichiometric coefficients as exponents. This systematic approach prevents the algebraic mix-ups that create most wrong answers in equilibrium problems.

Question 8

Which of the following statements about equilibrium constants is FALSE?

  1. The equilibrium constant is independent of the initial concentrations of reactants and products
  2. The equilibrium constant changes with temperature for a given reaction
  3. The equilibrium constant expression includes concentration terms for pure solids and liquids (correct answer)
  4. The equilibrium constant for the reverse reaction is the reciprocal of the forward reaction constant
  5. The equilibrium constant is dimensionless when all concentrations are expressed in molarity
Explanation: When you encounter questions about equilibrium constants, focus on the fundamental rules that govern how these constants are defined and calculated. The key issue here involves what gets included in equilibrium constant expressions. For any equilibrium reaction, the expression K=[products][reactants]K = \frac{[products]}{[reactants]} only includes species whose concentrations can actually change during the reaction. Pure solids and pure liquids have constant concentrations that don't change as the reaction proceeds, so they're omitted from the equilibrium expression and assigned an activity of 1. This makes option C false – pure solids and liquids are specifically excluded from equilibrium constant expressions. Let's examine why the other options are true. Option A is correct because KK depends only on the ratio of concentrations at equilibrium, not on where you started. You can begin with any initial concentrations and still reach the same equilibrium constant value. Option B is accurate since temperature changes affect the forward and reverse reaction rates differently, shifting the equilibrium position and changing KK. Option D reflects a mathematical relationship: if Kforward=[products][reactants]K_{forward} = \frac{[products]}{[reactants]}, then Kreverse=[reactants][products]=1KforwardK_{reverse} = \frac{[reactants]}{[products]} = \frac{1}{K_{forward}}. Remember this pattern: equilibrium expressions only include gases and aqueous solutions – never pure solids, pure liquids, or solvents. When writing KK expressions, mentally cross out any pure phases before setting up your fraction.

Question 9

For the equilibrium 2NO2(g)⇌N2O4(g)2NO_2(g) \rightleftharpoons N_2O_4(g), Kc=170K_c = 170 at 298 K. If the concentrations at a particular moment are [NO2]=0.10 M[NO_2] = 0.10 \, M and [N2O4]=0.50 M[N_2O_4] = 0.50 \, M, what is the value of the reaction quotient QcQ_c?

  1. 0.020
  2. 0.50
  3. 5.0
  4. 50 (correct answer)
  5. 170
Explanation: When you encounter equilibrium problems, it's crucial to distinguish between the equilibrium constant (KcK_c) and the reaction quotient (QcQ_c). While KcK_c uses equilibrium concentrations, QcQ_c uses any given concentrations to determine which direction the reaction will proceed. The reaction quotient has the same mathematical form as the equilibrium constant. For the reaction 2NO2(g)⇌N2O4(g)2NO_2(g) \rightleftharpoons N_2O_4(g), the expression is: Qc=[N2O4][NO2]2Q_c = \frac{[N_2O_4]}{[NO_2]^2} Notice that NO2NO_2 is squared because its coefficient is 2. Substituting the given concentrations: Qc=0.50(0.10)2=0.500.01=50Q_c = \frac{0.50}{(0.10)^2} = \frac{0.50}{0.01} = 50 Looking at the wrong answers: Choice A (0.020) represents the reciprocal of the correct answer, suggesting you inverted the fraction. Choice B (0.50) is simply the concentration of N2O4N_2O_4, indicating you forgot to include the denominator entirely. Choice C (5.0) results from forgetting to square the NO2NO_2 concentration, calculating 0.500.10\frac{0.50}{0.10} instead. The correct answer is D (50). Study tip: Always write out the QcQ_c expression first, paying careful attention to stoichiometric coefficients becoming exponents. Remember that Qc<KcQ_c < K_c means the reaction proceeds forward, Qc>KcQ_c > K_c means it proceeds backward, and Qc=KcQ_c = K_c indicates equilibrium. In this case, since Qc=50Q_c = 50 is much less than Kc=170K_c = 170, the forward reaction is favored.

Question 10

At a certain temperature, Kc=4.0×10−2K_c = 4.0 \times 10^{-2} for the reaction 2HI(g)⇌H2(g)+I2(g)2HI(g) \rightleftharpoons H_2(g) + I_2(g). What is KcK_c for the reaction 12H2(g)+12I2(g)⇌HI(g)\frac{1}{2}H_2(g) + \frac{1}{2}I_2(g) \rightleftharpoons HI(g)?

  1. 0.020
  2. 0.20
  3. 2.0
  4. 5.0 (correct answer)
  5. 25
Explanation: When you encounter questions about equilibrium constants and reaction manipulation, remember that the equilibrium constant changes predictably when you alter the chemical equation. The key is understanding how reversing reactions and changing stoichiometric coefficients affect KcK_c. The original reaction has Kc=4.0×10−2K_c = 4.0 \times 10^{-2} for 2HI(g)⇌H2(g)+I2(g)2HI(g) \rightleftharpoons H_2(g) + I_2(g). The target reaction is 12H2(g)+12I2(g)⇌HI(g)\frac{1}{2}H_2(g) + \frac{1}{2}I_2(g) \rightleftharpoons HI(g). Notice two changes: the reaction is reversed (products become reactants), and all coefficients are divided by 4 (since the original has 2 HI, and the new reaction has 1 HI). When you reverse a reaction, the new KcK_c becomes 1Koriginal\frac{1}{K_{original}}. When you divide all coefficients by a factor, you raise KcK_c to the power equal to that factor. Here, reversing gives 14.0×10−2=25\frac{1}{4.0 \times 10^{-2}} = 25. Then dividing coefficients by 4 means taking the fourth root: Kc=(25)1/4=(52)1/4=51/2=25=5.0K_c = (25)^{1/4} = (5^2)^{1/4} = 5^{1/2} = \sqrt{25} = 5.0. Choice A (0.020) incorrectly keeps the original KcK_c value. Choice B (0.20) appears to reverse the reaction but ignores the coefficient change. Choice C (2.0) might result from taking the square root of the reversed KcK_c instead of the fourth root. Remember this pattern: reversed reaction means Knew=1KoriginalK_{new} = \frac{1}{K_{original}}, and coefficient changes mean raising KcK_c to the power of the scaling factor. Always identify both transformations separately, then apply them sequentially.

Question 11

For the heterogeneous equilibrium 2C(s)+O2(g)⇌2CO(g)2C(s) + O_2(g) \rightleftharpoons 2CO(g), which change will NOT affect the value of the equilibrium constant KcK_c?

  1. Increasing the temperature from 298 K to 398 K
  2. Decreasing the temperature from 398 K to 298 K
  3. Adding more solid carbon to the reaction vessel
  4. Adding a platinum catalyst to speed the reaction
  5. Both (c) and (d) (correct answer)
Explanation: When you encounter equilibrium constant questions, remember that KcK_c depends only on temperature for a given reaction. The equilibrium constant is a fundamental thermodynamic property that remains fixed at a specific temperature, regardless of other reaction conditions. Looking at this heterogeneous equilibrium, the value of KcK_c will only change if temperature changes. Both options A and B involve temperature changes (increasing from 298 K to 398 K, and decreasing from 398 K to 298 K respectively), so both would alter the equilibrium constant value. Since the reaction involves breaking and forming bonds, KcK_c is temperature-dependent according to the van't Hoff equation. Option C might seem like it wouldn't affect KcK_c since pure solids don't appear in equilibrium expressions. While adding more carbon won't change the equilibrium constant's numerical value, this isn't the best answer because it's not the primary concept being tested here. Option D represents the correct principle: catalysts speed up both forward and reverse reactions equally, helping the system reach equilibrium faster without changing the equilibrium position or the equilibrium constant. A platinum catalyst would decrease activation energy but leave KcK_c unchanged. The key insight is distinguishing between factors that shift equilibrium position (concentration changes, pressure changes) and factors that actually change the equilibrium constant itself (only temperature). Catalysts are unique because they affect reaction rate but never affect equilibrium thermodynamics. Study tip: Remember "Only T affects K" - temperature is the sole factor that changes equilibrium constants, while catalysts only affect how quickly equilibrium is reached.

Question 12

At 25°C, the equilibrium constant Kc=5.6×10−12K_c = 5.6 \times 10^{-12} for the autoionization of water: 2H2O(l)⇌H3O+(aq)+OH−(aq)2H_2O(l) \rightleftharpoons H_3O^+(aq) + OH^-(aq). What is the correct equilibrium constant expression for this reaction?

  1. Kc=[H3O+][OH−][H2O]2K_c = \frac{[H_3O^+][OH^-]}{[H_2O]^2}
  2. Kc=[H3O+][OH−]K_c = [H_3O^+][OH^-] (correct answer)
  3. Kc=[H3O+][OH−][H2O]K_c = \frac{[H_3O^+][OH^-]}{[H_2O]}
  4. Kc=[H2O]2[H3O+][OH−]K_c = \frac{[H_2O]^2}{[H_3O^+][OH^-]}
  5. Kc=[H3O+]2[OH−][H2O]2K_c = \frac{[H_3O^+]^2[OH^-]}{[H_2O]^2}
Explanation: When writing equilibrium constant expressions, you need to understand how the physical state of substances affects their inclusion. The general rule is that only gases and aqueous species appear in the expression, while pure liquids and solids are omitted because their concentrations remain essentially constant. For the autoionization of water, you might initially think to write Kc=[H3O+][OH−][H2O]2K_c = \frac{[H_3O^+][OH^-]}{[H_2O]^2} following the standard format of products over reactants. However, since H2OH_2O is a pure liquid, its concentration is constant and gets incorporated into the equilibrium constant itself. This gives us the familiar water ionization constant: Kw=[H3O+][OH−]K_w = [H_3O^+][OH^-]. The correct answer is B because it properly excludes the pure liquid water from the expression, leaving only the aqueous ion concentrations. Answer A incorrectly includes [H2O]2[H_2O]^2 in the denominator, which violates the rule about pure liquids. Answer C makes the same mistake but with the wrong stoichiometry, using [H2O][H_2O] instead of [H2O]2[H_2O]^2. Answer D inverts the entire expression, putting products in the denominator and reactants in the numerator, which is backwards from the standard convention. Remember that equilibrium expressions for reactions involving pure liquids or solids will look "incomplete" compared to the balanced equation. The key is recognizing that KwK_w is actually KcK_c with the water concentration already factored in. This same principle applies to any heterogeneous equilibrium involving pure phases.

Question 13

The equilibrium constant for reaction (1) A⇌BA \rightleftharpoons B is K1=2.0K_1 = 2.0, and for reaction (2) B⇌CB \rightleftharpoons C is K2=0.50K_2 = 0.50. What is the equilibrium constant for the overall reaction A⇌CA \rightleftharpoons C?

  1. 0.25
  2. 1.0 (correct answer)
  3. 1.5
  4. 2.5
  5. 4.0
Explanation: When you encounter sequential equilibrium reactions, you need to understand how equilibrium constants combine when reactions are coupled together. For sequential reactions where the product of one reaction becomes the reactant of the next, you multiply the equilibrium constants. Here's why: when you add the two given reactions, A⇌BA \rightleftharpoons B (with K1=2.0K_1 = 2.0) and B⇌CB \rightleftharpoons C (with K2=0.50K_2 = 0.50), the intermediate species B cancels out, giving you the overall reaction A⇌CA \rightleftharpoons C. The equilibrium constant for the overall reaction is Koverall=K1×K2=2.0×0.50=1.0K_{overall} = K_1 \times K_2 = 2.0 \times 0.50 = 1.0. This works because equilibrium expressions are ratios of concentrations, and when you multiply them, the intermediate B terms algebraically cancel out. Looking at the incorrect choices: (A) 0.25 results from dividing K2K_2 by K1K_1 instead of multiplying—this reverses the logic. (C) 1.5 comes from averaging the two constants, which has no basis in equilibrium theory. (D) 2.5 results from adding the constants, but equilibrium constants multiply, not add, when reactions are combined sequentially. The correct answer is B) 1.0. Study tip: Remember the rule "multiply for sequential, raise to powers for coefficients." When reactions occur in sequence (product of first becomes reactant of second), always multiply their equilibrium constants. This is one of the most tested concepts in equilibrium chemistry.

Question 14

The equilibrium constant for the reaction N2(g)+O2(g)⇌2NO(g)N_2(g) + O_2(g) \rightleftharpoons 2NO(g) is Kc=4.2×10−31K_c = 4.2 \times 10^{-31} at 298 K. What does this extremely small equilibrium constant indicate about the reaction?

  1. The reaction is very fast at 298 K and reaches equilibrium quickly
  2. The reaction is endothermic and will not occur at room temperature
  3. At equilibrium, virtually no NONO will be present compared to N2N_2 and O2O_2 (correct answer)
  4. The reaction will go to completion, consuming all reactants
  5. Equal amounts of reactants and products will be present at equilibrium
Explanation: When you encounter equilibrium constant problems, focus on what the magnitude of KcK_c tells you about the position of equilibrium. The equilibrium constant expression for this reaction is Kc=[NO]2[N2][O2]K_c = \frac{[NO]^2}{[N_2][O_2]}. An extremely small KcK_c value like 4.2×10−314.2 \times 10^{-31} means the numerator (products) must be much, much smaller than the denominator (reactants) at equilibrium. Since KcK_c involves [NO]2[NO]^2 in the numerator, this tiny value indicates that virtually no nitric oxide will form compared to the amounts of nitrogen and oxygen present. The equilibrium lies far to the left, favoring reactants. Option A is incorrect because KcK_c tells you nothing about reaction rate or how quickly equilibrium is reached—only about the equilibrium position. Option B makes an unsupported claim about thermodynamics; while the reaction may indeed be endothermic, you cannot determine this from KcK_c alone, and the small constant doesn't mean the reaction "will not occur"—it just means very little product forms. Option D contradicts what the small KcK_c tells you; if the reaction went to completion, KcK_c would be extremely large, not small. Therefore, C correctly identifies that the equilibrium mixture will contain essentially only N2N_2 and O2O_2 with negligible NONO. Study tip: Remember the KcK_c magnitude rule: Kc>>1K_c >> 1 favors products, Kc<<1K_c << 1 favors reactants. Values around 10−3010^{-30} indicate virtually no product formation.

Question 15

For which of the following reactions is Kp=KcK_p = K_c?

  1. H2(g)+Br2(g)⇌2HBr(g)H_2(g) + Br_2(g) \rightleftharpoons 2HBr(g) (correct answer)
  2. N2(g)+3H2(g)⇌2NH3(g)N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)
  3. 2SO2(g)+O2(g)⇌2SO3(g)2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)
  4. PCl5(g)⇌PCl3(g)+Cl2(g)PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)
  5. 2NO2(g)⇌N2O4(g)2NO_2(g) \rightleftharpoons N_2O_4(g)
Explanation: When you encounter equilibrium constant problems, you need to understand the relationship between KpK_p (pressure-based) and KcK_c (concentration-based). The key equation is Kp=Kc(RT)ΔnK_p = K_c(RT)^{\Delta n}, where Δn\Delta n is the change in moles of gas (moles of gaseous products minus moles of gaseous reactants). For Kp=KcK_p = K_c, we need (RT)Δn=1(RT)^{\Delta n} = 1, which only occurs when Δn=0\Delta n = 0. This means the number of gas molecules on both sides of the equation must be equal. Let's check each reaction: A) H2(g)+Br2(g)⇌2HBr(g)H_2(g) + Br_2(g) \rightleftharpoons 2HBr(g): Here, 2 moles of reactants produce 2 moles of products, so Δn=2−2=0\Delta n = 2 - 2 = 0. This makes Kp=KcK_p = K_c. B) N2(g)+3H2(g)⇌2NH3(g)N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g): We have 4 moles of reactants and 2 moles of products, giving Δn=2−4=−2\Delta n = 2 - 4 = -2. Since Δn≠0\Delta n \neq 0, Kp≠KcK_p \neq K_c. C) 2SO2(g)+O2(g)⇌2SO3(g)2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g): This gives us 3 moles of reactants and 2 moles of products, so Δn=2−3=−1\Delta n = 2 - 3 = -1. Again, Kp≠KcK_p \neq K_c. D) PCl5(g)⇌PCl3(g)+Cl2(g)PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g): Here we have 1 mole of reactant and 2 moles of products, making Δn=2−1=1\Delta n = 2 - 1 = 1, so Kp≠KcK_p \neq K_c. Study tip: Always count gas molecules on each side first. When they're equal, Kp=KcK_p = K_c immediately—no calculations needed.

Question 16

At equilibrium for the reaction 2NOBr(g)⇌2NO(g)+Br2(g)2NOBr(g) \rightleftharpoons 2NO(g) + Br_2(g), the concentrations are [NOBr]=0.40 M[NOBr] = 0.40 \, M, [NO]=0.20 M[NO] = 0.20 \, M, and [Br2]=0.10 M[Br_2] = 0.10 \, M. What is the value of KcK_c?

  1. 0.025 (correct answer)
  2. 0.050
  3. 0.10
  4. 0.20
  5. 0.40
Explanation: When you encounter an equilibrium expression problem, you need to write the equilibrium constant expression using the balanced chemical equation, then substitute the given concentrations. For the reaction 2NOBr(g)⇌2NO(g)+Br2(g)2NOBr(g) \rightleftharpoons 2NO(g) + Br_2(g), the equilibrium constant expression is: Kc=[NO]2[Br2][NOBr]2K_c = \frac{[NO]^2[Br_2]}{[NOBr]^2} Notice that the exponents match the stoichiometric coefficients from the balanced equation. Products go in the numerator, reactants in the denominator. Substituting the given equilibrium concentrations: Kc=(0.20)2(0.10)(0.40)2=(0.040)(0.10)0.16=0.0040.16=0.025K_c = \frac{(0.20)^2(0.10)}{(0.40)^2} = \frac{(0.040)(0.10)}{0.16} = \frac{0.004}{0.16} = 0.025 This confirms answer A is correct. Looking at the wrong answers: B (0.050) might result from forgetting to square the NO concentration or making an arithmetic error in the calculation. C (0.10) could come from incorrectly using just the Br₂ concentration or setting up the expression incorrectly. D (0.20) might result from using the NO concentration directly without proper calculation or inverting part of the expression. The key strategy here is to always write out the KcK_c expression first based on the balanced equation, double-check that exponents match stoichiometric coefficients, then carefully substitute and calculate. Remember: products over reactants, and small KcK_c values (like 0.025) indicate the equilibrium lies toward reactants, which makes sense since NOBr has the highest concentration.

Question 17

At 700 K, Kp=0.76K_p = 0.76 for the reaction H2(g)+I2(g)⇌2HI(g)H_2(g) + I_2(g) \rightleftharpoons 2HI(g). What is the value of KcK_c for this reaction at the same temperature? (R=0.0821 L⋅atm⋅mol−1⋅K−1R = 0.0821 \, L \cdot atm \cdot mol^{-1} \cdot K^{-1})

  1. 0.76 (correct answer)
  2. 0.013
  3. 1.3
  4. 44
  5. 57
Explanation: When you encounter equilibrium constant problems involving both KpK_p and KcK_c, you need to understand the relationship between these two constants. KpK_p uses partial pressures while KcK_c uses molar concentrations, and they're connected by the equation: Kp=Kc(RT)ΔnK_p = K_c(RT)^{\Delta n}, where Δn\Delta n is the change in moles of gas. For the reaction H2(g)+I2(g)⇌2HI(g)H_2(g) + I_2(g) \rightleftharpoons 2HI(g), calculate Δn\Delta n by subtracting moles of gaseous reactants from moles of gaseous products: Δn=2−(1+1)=0\Delta n = 2 - (1 + 1) = 0. When Δn=0\Delta n = 0, the equation becomes Kp=Kc(RT)0=Kc×1=KcK_p = K_c(RT)^0 = K_c \times 1 = K_c. Therefore, Kp=Kc=0.76K_p = K_c = 0.76. Looking at the wrong answers: B) 0.013 likely results from incorrectly calculating Kc=KpRTK_c = \frac{K_p}{RT} by assuming Δn=−1\Delta n = -1. C) 1.3 might come from using Kc=Kp×RTK_c = K_p \times RT with Δn=+1\Delta n = +1. D) 44 could result from using Kc=Kp(RT)K_c = K_p(RT) with the full value of RT=57.5RT = 57.5 at 700 K, showing a fundamental misunderstanding of the relationship. The correct answer is A) 0.76. Study tip: Always calculate Δn\Delta n first in Kp/KcK_p/K_c problems. When Δn=0\Delta n = 0 (equal moles of gas on both sides), Kp=KcK_p = K_c regardless of temperature. This is a common scenario that saves you from complex calculations.

Question 18

For the gas-phase reaction 2A⇌B+C2A \rightleftharpoons B + C, the equilibrium constant Kp=0.85K_p = 0.85 at 600 K. If the equilibrium partial pressures are PB=PC=1.2 atmP_B = P_C = 1.2 \, atm, what is the partial pressure of AA at equilibrium?

  1. 0.59 atm
  2. 1.2 atm
  3. 1.3 atm (correct answer)
  4. 1.7 atm
  5. 2.0 atm
Explanation: When you encounter gas-phase equilibrium problems, you need to set up the equilibrium expression using partial pressures and solve for the unknown quantity. For the reaction 2A⇌B+C2A \rightleftharpoons B + C, the equilibrium expression is: Kp=PB×PCPA2K_p = \frac{P_B \times P_C}{P_A^2} Notice that PAP_A is squared because the stoichiometric coefficient of A is 2. Given that Kp=0.85K_p = 0.85 and PB=PC=1.2 atmP_B = P_C = 1.2 \, \text{atm}, you can substitute and solve: 0.85=(1.2)(1.2)PA2=1.44PA20.85 = \frac{(1.2)(1.2)}{P_A^2} = \frac{1.44}{P_A^2} Rearranging: PA2=1.440.85=1.694P_A^2 = \frac{1.44}{0.85} = 1.694 Therefore: PA=1.694=1.3 atmP_A = \sqrt{1.694} = 1.3 \, \text{atm} This confirms answer choice C is correct. Looking at the wrong answers: A (0.59 atm) results from incorrectly taking the square root of 0.85/2.44 instead of 1.44/0.85. B (1.2 atm) assumes all partial pressures are equal, ignoring the stoichiometry. D (1.7 atm) comes from failing to take the square root of 1.694. Remember that stoichiometric coefficients become exponents in equilibrium expressions. Always double-check that you've applied the correct powers and performed the algebra carefully. Gas-phase equilibrium problems frequently test whether you properly account for stoichiometry in your KpK_p expression.

Question 19

The equilibrium constant for the reaction A(g)⇌B(g)+C(g)A(g) \rightleftharpoons B(g) + C(g) is K1=0.50K_1 = 0.50 at 298 K. The equilibrium constant for the reaction 2B(g)⇌D(g)2B(g) \rightleftharpoons D(g) is K2=4.0K_2 = 4.0 at the same temperature. What is the equilibrium constant for the overall reaction 2A(g)⇌2C(g)+D(g)2A(g) \rightleftharpoons 2C(g) + D(g)?

  1. 1.0 (correct answer)
  2. 2.0
  3. 4.5
  4. 8.0
  5. 9.0
Explanation: When you encounter problems involving multiple equilibrium reactions that combine to form an overall reaction, you need to understand how equilibrium constants relate when reactions are added, multiplied, or reversed. To find the equilibrium constant for the target reaction 2A(g)⇌2C(g)+D(g)2A(g) \rightleftharpoons 2C(g) + D(g), you must combine the given reactions algebraically. Start by manipulating the given equations: From reaction 1: A(g)⇌B(g)+C(g)A(g) \rightleftharpoons B(g) + C(g), K1=0.50K_1 = 0.50 From reaction 2: 2B(g)⇌D(g)2B(g) \rightleftharpoons D(g), K2=4.0K_2 = 4.0 To get the target reaction, multiply reaction 1 by 2: 2A(g)⇌2B(g)+2C(g)2A(g) \rightleftharpoons 2B(g) + 2C(g) When you multiply a reaction by a coefficient, you raise the equilibrium constant to that power: K12=(0.50)2=0.25K_1^2 = (0.50)^2 = 0.25 Now add this doubled reaction 1 to reaction 2: 2A(g)⇌2B(g)+2C(g)2A(g) \rightleftharpoons 2B(g) + 2C(g) 2B(g)⇌D(g)2B(g) \rightleftharpoons D(g) The 2B(g)2B(g) cancels out, giving: 2A(g)⇌2C(g)+D(g)2A(g) \rightleftharpoons 2C(g) + D(g) When adding reactions, you multiply their equilibrium constants: Koverall=K12×K2=0.25×4.0=1.0K_{overall} = K_1^2 \times K_2 = 0.25 \times 4.0 = 1.0 Answer A (1.0) is correct. Answer B (2.0) incorrectly adds the constants. Answer C (4.5) mistakenly adds K1+K2K_1 + K_2. Answer D (8.0) incorrectly uses K1×K2×22K_1 \times K_2 \times 2^2. Remember: when combining equilibrium reactions, multiply constants for added reactions and raise to powers for multiplied reactions.

Question 20

At a certain temperature, Kc=1.5K_c = 1.5 for the reaction 2NO(g)+Br2(g)⇌2NOBr(g)2NO(g) + Br_2(g) \rightleftharpoons 2NOBr(g). If at equilibrium [NO]=0.80 M[NO] = 0.80 \, M, [Br2]=0.60 M[Br_2] = 0.60 \, M, and [NOBr]=0.40 M[NOBr] = 0.40 \, M, which statement is correct?

  1. The given concentrations represent a valid equilibrium state
  2. The reaction quotient Qc>KcQ_c > K_c, so the reaction will shift left
  3. The reaction quotient Qc<KcQ_c < K_c, so the reaction will shift right (correct answer)
  4. The reaction is not at equilibrium because the concentrations are not equal
  5. More information is needed to determine if the system is at equilibrium
Explanation: When you encounter equilibrium problems with given concentrations, you need to determine whether the system is actually at equilibrium by comparing the reaction quotient (QcQ_c) to the equilibrium constant (KcK_c). First, calculate QcQ_c using the same expression as KcK_c, but with the given concentrations: Qc=[NOBr]2[NO]2[Br2]=(0.40)2(0.80)2(0.60)=0.160.64×0.60=0.160.384=0.42Q_c = \frac{[NOBr]^2}{[NO]^2[Br_2]} = \frac{(0.40)^2}{(0.80)^2(0.60)} = \frac{0.16}{0.64 \times 0.60} = \frac{0.16}{0.384} = 0.42 Since Qc=0.42<Kc=1.5Q_c = 0.42 < K_c = 1.5, the reaction will shift right (toward products) to reach equilibrium. This confirms answer C is correct. Now let's examine why the other answers are wrong: A) is incorrect because Qc≠KcQ_c \neq K_c, so these concentrations don't represent equilibrium. B) is wrong because Qc<KcQ_c < K_c, not Qc>KcQ_c > K_c. When Qc>KcQ_c > K_c, the reaction would shift left, but that's not the case here. D) reflects a fundamental misunderstanding—equilibrium concentrations don't need to be equal to each other. They only need to satisfy the equilibrium constant expression. Remember this key strategy: Always calculate QcQ_c first when given concentrations in an equilibrium problem. Compare it to KcK_c to determine the direction of shift: if Qc<KcQ_c < K_c, shift right; if Qc>KcQ_c > K_c, shift left; if Qc=KcQ_c = K_c, you're at equilibrium.