AP Chemistry Quiz: Calculating The Equilibrium Constant
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Calculating The Equilibrium ConstantQuestion 1 of 20

For the equilibrium reaction COCl2(g)⇌CO(g)+Cl2(g)\text{COCl}_2(g)\rightleftharpoons \text{CO}(g)+\text{Cl}_2(g), the system is at equilibrium with [COCl2]=0.50 M[\text{COCl}_2]=0.50\,\text{M}, [CO]=0.10 M[\text{CO}]=0.10\,\text{M}, and [Cl2]=0.20 M[\text{Cl}_2]=0.20\,\text{M}. What is KcK_c?

25
2.5
0.25
0.10
0.040
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AP Chemistry Quiz

AP Chemistry Quiz: Calculating The Equilibrium Constant

Practice Calculating The Equilibrium Constant in AP 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 Calculating The Equilibrium Constant, giving you a quick way to practice the rules, question types, and explanations that matter most for AP 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 equilibrium reaction COCl2(g)⇌CO(g)+Cl2(g)\text{COCl}_2(g)\rightleftharpoons \text{CO}(g)+\text{Cl}_2(g), the system is at equilibrium with [COCl2]=0.50 M[\text{COCl}_2]=0.50\,\text{M}, [CO]=0.10 M[\text{CO}]=0.10\,\text{M}, and [Cl2]=0.20 M[\text{Cl}_2]=0.20\,\text{M}. What is KcK_c?

  1. 25
  2. 2.5
  3. 0.25
  4. 0.10
  5. 0.040 (correct answer)

Explanation: This problem involves calculating the equilibrium constant for COCl₂(g) ⇌ CO(g) + Cl₂(g). The equilibrium expression is Kc = [CO][Cl₂]/[COCl₂], with products in the numerator and reactant in the denominator. Substituting the equilibrium values: Kc = (0.10)(0.20)/(0.50) = 0.02/0.50 = 0.040. A student might mistakenly write the inverse expression Kc = [COCl₂]/([CO][Cl₂]) = 0.50/0.02 = 25, which gives the reciprocal of the correct answer. To avoid this error, always write the equilibrium expression for the forward reaction as given in the problem statement.

Question 2

In a sealed vessel, the reaction H2(g)+I2(g)⇌2HI(g)\text{H}_2(g)+\text{I}_2(g)\rightleftharpoons 2\text{HI}(g) is at equilibrium. The equilibrium concentrations are [H2]=0.20 M[\text{H}_2]=0.20\,\text{M}, [I2]=0.20 M[\text{I}_2]=0.20\,\text{M}, and [HI]=0.80 M[\text{HI}]=0.80\,\text{M}. What is KcK_c?

  1. 0.063
  2. 4.0
  3. 16 (correct answer)
  4. 0.25
  5. 8.0

Explanation: This question asks for calculating the equilibrium constant for H₂(g) + I₂(g) ⇌ 2HI(g). The equilibrium expression is Kc = [HI]²/([H₂][I₂]), where the product HI is squared because its coefficient is 2 in the balanced equation. Substituting the given concentrations: Kc = (0.80)²/((0.20)(0.20)) = 0.64/0.04 = 16. A common mistake would be to write Kc = [HI]/([H₂][I₂]) without squaring [HI], which would give 0.80/0.04 = 20, an incorrect result. Remember to always include stoichiometric coefficients as exponents in the equilibrium expression before performing calculations.

Question 3

A reaction vessel at constant temperature contains the equilibrium system CH3COOH(aq)⇌H+(aq)+CH3COO−(aq)\mathrm{CH_3COOH(aq) \rightleftharpoons H^+(aq) + CH_3COO^-(aq)} The system is at equilibrium with [CH3COOH]=0.10 M[\mathrm{CH_3COOH}]=0.10\,\text{M}, [H+]=0.010 M[\mathrm{H^+}]=0.010\,\text{M}, and [CH3COO−]=0.010 M[\mathrm{CH_3COO^-}]=0.010\,\text{M}. What is the value of KcK_c for this reaction?

  1. 1.0×10−11.0 \times 10^{-1}
  2. 1.0×10−31.0 \times 10^{-3} (correct answer)
  3. 1.0×1021.0 \times 10^{2}
  4. 1.0×10−21.0 \times 10^{-2}
  5. 1.0×10−41.0 \times 10^{-4}

Explanation: This question tests the skill of calculating the equilibrium constant. To calculate KcK_c, first write the equilibrium expression for the reaction CH3COOH(aq)⇌H+(aq)+CH3COO−(aq)\mathrm{CH_3COOH(aq) \rightleftharpoons H^+(aq) + CH_3COO^-(aq)}, which is Kc=[H+][CH3COO−][CH3COOH]K_c = \frac{[\mathrm{H^+}] [\mathrm{CH_3COO^-}]}{[\mathrm{CH_3COOH}]}. The stoichiometric coefficients become exponents in the expression, with products in the numerator and reactants in the denominator. Substitute the given equilibrium concentrations: [CH3COOH]=0.10 M[\mathrm{CH_3COOH}] = 0.10\, \text{M}, [H+]=0.010 M[\mathrm{H^+}] = 0.010\, \text{M}, and [CH3COO−]=0.010 M[\mathrm{CH_3COO^-}] = 0.010\, \text{M}, so Kc=(0.010×0.010)/0.10=0.0001/0.10=0.001=1.0×10−3K_c = (0.010 \times 0.010) / 0.10 = 0.0001 / 0.10 = 0.001 = 1.0 \times 10^{-3}. A tempting distractor is 1.0×10−41.0 \times 10^{-4}, which results from omitting the reactant concentration in the denominator. Always write the equilibrium expression based on the balanced equation first, then substitute the given equilibrium concentrations, excluding any pure solids or liquids.

Question 4

A system is at equilibrium for the reaction A(g)+2B(g)⇌C(g)\mathrm{A(g) + 2B(g) \rightleftharpoons C(g)}. The equilibrium concentrations are [A]=0.20 M[\mathrm{A}]=0.20\,\mathrm{M}, [B]=0.10 M[\mathrm{B}]=0.10\,\mathrm{M}, and [C]=0.20 M[\mathrm{C}]=0.20\,\mathrm{M}. What is the value of KcK_c?

  1. 100 (correct answer)
  2. 10
  3. 0.10
  4. 0.010
  5. 1.0

Explanation: This problem requires calculating the equilibrium constant for A(g) + 2B(g) ⇌ C(g). The equilibrium expression is K_c = [C]/([A][B]²), where [B] is squared because its stoichiometric coefficient is 2. Substituting the given concentrations: K_c = 0.20/((0.20)(0.10)²) = 0.20/(0.20 × 0.01) = 0.20/0.002 = 100. Choice C (0.10) represents the error of forgetting to square the B concentration, using [C]/([A][B]) instead of the correct expression. When multiple molecules of a species appear in the balanced equation, always raise that species' concentration to the corresponding power in the equilibrium expression.

Question 5

The reaction 2NOCl(g)⇌2NO(g)+Cl2(g)\mathrm{2NOCl(g) \rightleftharpoons 2NO(g) + Cl_2(g)} is at equilibrium. The equilibrium concentrations are [NOCl]=0.20 M[\mathrm{NOCl}]=0.20\,\mathrm{M}, [NO]=0.20 M[\mathrm{NO}]=0.20\,\mathrm{M}, and [Cl2]=0.05 M[\mathrm{Cl_2}]=0.05\,\mathrm{M}. What is KcK_c?

  1. 0.25
  2. 1.0
  3. 0.50
  4. 2.0
  5. 0.050 (correct answer)

Explanation: This question involves calculating the equilibrium constant. For the balanced equation 2NOCl(g) ⇌ 2NO(g) + Cl₂(g), the equilibrium expression is written as products over reactants, with stoichiometric coefficients as exponents. Thus, Kc = ([NO]² [Cl₂]) / [NOCl]². Substituting the equilibrium concentrations [NOCl] = 0.20 M, [NO] = 0.20 M, and [Cl₂] = 0.05 M gives Kc = ((0.20)² × 0.05) / (0.20)² = 0.002 / 0.04 = 0.050. A tempting distractor is 0.25, which results from forgetting to square [NO], but this is incorrect because exponents are required for the coefficients. Always write the expression first, then substitute equilibrium values only—exclude pure solids and liquids.

Question 6

A reaction mixture is at equilibrium for the reaction 2SO2(g)+O2(g)⇌2SO3(g)\mathrm{2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)}. The equilibrium concentrations are [SO2]=0.20 M[\mathrm{SO_2}]=0.20\,\mathrm{M}, [O2]=0.10 M[\mathrm{O_2}]=0.10\,\mathrm{M}, and [SO3]=0.30 M[\mathrm{SO_3}]=0.30\,\mathrm{M}. What is the value of KcK_c?

  1. 2.25
  2. 0.225
  3. 22.5 (correct answer)
  4. 0.444
  5. 4.50

Explanation: This question asks for calculating the equilibrium constant for 2SO₂(g) + O₂(g) ⇌ 2SO₃(g). The equilibrium expression is K_c = [SO₃]²/([SO₂]²[O₂]), with both SO₃ and SO₂ concentrations squared due to their coefficients of 2. Substituting the equilibrium concentrations: K_c = (0.30)²/((0.20)²(0.10)) = 0.09/(0.04 × 0.10) = 0.09/0.004 = 22.5. Choice B (0.225) likely results from forgetting to square the concentrations, using [SO₃]/([SO₂][O₂]) instead of the correct expression with squared terms. Remember to always include stoichiometric coefficients as exponents in the equilibrium expression before performing calculations.

Question 7

At a fixed temperature, the reaction H2(g)+Br2(g)⇌2HBr(g)\mathrm{H_2(g) + Br_2(g) \rightleftharpoons 2HBr(g)} is at equilibrium. The equilibrium concentrations are [H2]=0.30 M[\mathrm{H_2}]=0.30\,\mathrm{M}, [Br2]=0.10 M[\mathrm{Br_2}]=0.10\,\mathrm{M}, and [HBr]=0.30 M[\mathrm{HBr}]=0.30\,\mathrm{M}. What is the value of KcK_c?

  1. 0.33
  2. 0.030
  3. 3.0 (correct answer)
  4. 30
  5. 0.10

Explanation: This problem involves calculating the equilibrium constant for H₂(g) + Br₂(g) ⇌ 2HBr(g). The equilibrium expression is K_c = [HBr]²/([H₂][Br₂]), where [HBr] is squared because its stoichiometric coefficient is 2. Substituting the given concentrations: K_c = (0.30)²/((0.30)(0.10)) = 0.09/0.03 = 3.0. Choice B (0.030) likely results from forgetting to square the HBr concentration, using [HBr]/([H₂][Br₂]) instead of [HBr]²/([H₂][Br₂]). Remember that stoichiometric coefficients become exponents in the equilibrium expression, so always check the balanced equation carefully.

Question 8

A system is at equilibrium for the reaction SO2(g)+NO2(g)⇌SO3(g)+NO(g)\text{SO}_2(g)+\text{NO}_2(g)\rightleftharpoons \text{SO}_3(g)+\text{NO}(g). The equilibrium concentrations are [SO2]=0.50 M[\text{SO}_2]=0.50\,\text{M}, [NO2]=0.20 M[\text{NO}_2]=0.20\,\text{M}, [SO3]=0.10 M[\text{SO}_3]=0.10\,\text{M}, and [NO]=0.40 M[\text{NO}]=0.40\,\text{M}. What is the value of KcK_c?

  1. 0.10
  2. 0.20
  3. 0.40 (correct answer)
  4. 2.5
  5. 5.0

Explanation: This problem asks for calculating the equilibrium constant for SO₂(g) + NO₂(g) ⇌ SO₃(g) + NO(g). The equilibrium expression is Kc = [SO₃][NO]/([SO₂][NO₂]), with products in the numerator and reactants in the denominator, all with exponents of 1. Substituting the concentrations: Kc = (0.10)(0.40)/((0.50)(0.20)) = 0.04/0.10 = 0.40. A typical error would be mixing up which species are products versus reactants, or perhaps calculating (0.50)(0.20)/(0.10)(0.40) = 2.5 by inverting the expression. When calculating equilibrium constants, carefully identify products and reactants from the reaction arrow, then construct your expression methodically before substituting values.

Question 9

A reaction mixture is at equilibrium for CO2(g)+H2(g)⇌CO(g)+H2O(g)\text{CO}_2(g)+\text{H}_2(g)\rightleftharpoons \text{CO}(g)+\text{H}_2\text{O}(g). The equilibrium concentrations are [CO2]=0.25 M[\text{CO}_2]=0.25\,\text{M}, [H2]=0.50 M[\text{H}_2]=0.50\,\text{M}, [CO]=0.10 M[\text{CO}]=0.10\,\text{M}, and [H2O]=0.20 M[\text{H}_2\text{O}]=0.20\,\text{M}. What is the value of KcK_c?

  1. 0.080
  2. 0.16 (correct answer)
  3. 0.25
  4. 4.0
  5. 12

Explanation: This problem involves calculating the equilibrium constant for CO₂(g) + H₂(g) ⇌ CO(g) + H₂O(g). The equilibrium expression is Kc = [CO][H₂O]/([CO₂][H₂]), with all species having exponents of 1 since all coefficients are 1. Substituting the given concentrations: Kc = (0.10)(0.20)/((0.25)(0.50)) = 0.02/0.125 = 0.16. A common mistake would be to invert the expression, calculating [CO₂][H₂]/([CO][H₂O]) = 0.125/0.02 = 6.25, which gives the reciprocal of the correct answer. When calculating equilibrium constants, always ensure products are in the numerator and reactants are in the denominator, then substitute values carefully.

Question 10

At equilibrium in a closed vessel at constant temperature, the reaction A(g)+2B(g)⇌AB2(g)\mathrm{A(g) + 2B(g) \rightleftharpoons AB_2(g)} has equilibrium concentrations [A]=0.20 M[\mathrm{A}]=0.20\,\text{M}, [B]=0.10 M[\mathrm{B}]=0.10\,\text{M}, and [AB2]=0.20 M[\mathrm{AB_2}]=0.20\,\text{M}. What is the value of KcK_c for the reaction?

  1. 10
  2. 100 (correct answer)
  3. 0.10
  4. 1.0
  5. 4.0

Explanation: This question tests the skill of calculating the equilibrium constant. To calculate K_c, first write the equilibrium expression for the reaction \mathrm{A(g) + 2B(g) \rightleftharpoons \mathrm{AB_2(g)}, which is Kc=[AB2][A][B]2K_c = \frac{[\mathrm{AB_2}]}{[\mathrm{A}][\mathrm{B}]^2}. The stoichiometric coefficients become exponents in the expression, with products in the numerator and reactants in the denominator. Substitute the given equilibrium concentrations: [A]=0.20 M[\mathrm{A}] = 0.20\,\text{M}, [B]=0.10 M[\mathrm{B}] = 0.10\,\text{M}, and [AB2]=0.20 M[\mathrm{AB_2}] = 0.20\,\text{M}, so Kc=0.20/(0.20×(0.10)2)=0.20/(0.20×0.01)=0.20/0.002=100K_c = 0.20 / (0.20 \times (0.10)^2) = 0.20 / (0.20 \times 0.01) = 0.20 / 0.002 = 100. A tempting distractor is 10, which results from forgetting to raise the concentration of B to the second power. Always write the equilibrium expression based on the balanced equation first, then substitute the given equilibrium concentrations, excluding any pure solids or liquids.

Question 11

A container holds the equilibrium system PCl5(g)⇌PCl3(g)+Cl2(g)\text{PCl}_5(g)\rightleftharpoons \text{PCl}_3(g)+\text{Cl}_2(g). At equilibrium, [PCl5]=0.40 M[\text{PCl}_5]=0.40\,M, [PCl3]=0.20 M[\text{PCl}_3]=0.20\,M, and [Cl2]=0.20 M[\text{Cl}_2]=0.20\,M. What is the value of KcK_c?

  1. 4.0
  2. 0.40
  3. 0.10 (correct answer)
  4. 0.25
  5. 2.0

Explanation: This question tests the skill of calculating the equilibrium constant. For the balanced equation PCl₅(g) ⇌ PCl₃(g) + Cl₂(g), the equilibrium expression is K_c = ([PCl₃][Cl₂]) / [PCl₅]. To calculate K_c, substitute the given equilibrium concentrations: [PCl₃] = 0.20 M, [Cl₂] = 0.20 M, and [PCl₅] = 0.40 M. This yields K_c = (0.20 × 0.20) / 0.40 = 0.04 / 0.40 = 0.10. A tempting distractor might be 4.0, which results from inverting the expression. Always write the expression first, then substitute equilibrium values only—exclude pure solids and liquids.

Question 12

At equilibrium, the reaction 2NO(g)+O2(g)⇌2NO2(g)\mathrm{2NO(g) + O_2(g) \rightleftharpoons 2NO_2(g)} has [NO]=0.20 M[\mathrm{NO}]=0.20\,\mathrm{M}, [O2]=0.10 M[\mathrm{O_2}]=0.10\,\mathrm{M}, and [NO2]=0.20 M[\mathrm{NO_2}]=0.20\,\mathrm{M}. What is KcK_c?

  1. 10 (correct answer)
  2. 0.10
  3. 1.0
  4. 0.25
  5. 2.5

Explanation: This question involves calculating the equilibrium constant. For the balanced equation 2NO(g) + O₂(g) ⇌ 2NO₂(g), the equilibrium expression is written as products over reactants, with stoichiometric coefficients as exponents. Thus, Kc = [NO₂]² / ([NO]² [O₂]). Substituting the equilibrium concentrations [NO] = 0.20 M, [O₂] = 0.10 M, and [NO₂] = 0.20 M gives Kc = (0.20)² / ((0.20)² × 0.10) = 0.04 / (0.04 × 0.10) = 0.04 / 0.004 = 10. A tempting distractor is 0.10, which results from inverting the expression, but this is incorrect because Kc is products over reactants. Always write the expression first, then substitute equilibrium values only—exclude pure solids and liquids.

Question 13

A closed container at constant temperature contains the equilibrium system 2HBr(g)⇌H2(g)+Br2(g)\mathrm{2HBr(g) \rightleftharpoons H_2(g) + Br_2(g)} At equilibrium, [HBr]=0.20 M[\mathrm{HBr}]=0.20\,\text{M}, [H2]=0.10 M[\mathrm{H_2}]=0.10\,\text{M}, and [Br2]=0.10 M[\mathrm{Br_2}]=0.10\,\text{M}. What is the value of KcK_c?

  1. 0.50
  2. 4.0
  3. 0.25 (correct answer)
  4. 2.0
  5. 0.040

Explanation: This question tests the skill of calculating the equilibrium constant. To calculate KcK_c, first write the equilibrium expression for the reaction 2HBr(g)⇌H2(g)+Br2(g)\mathrm{2HBr(g) \rightleftharpoons H_2(g) + Br_2(g)}, which is Kc=[H2][Br2][HBr]2K_c = \frac{[\mathrm{H_2}][\mathrm{Br_2}]}{[\mathrm{HBr}]^2}. The stoichiometric coefficients become exponents in the expression, with products in the numerator and reactants in the denominator. Substitute the given equilibrium concentrations: [HBr]=0.20 M[\mathrm{HBr}] = 0.20\,\text{M}, [H2]=0.10 M[\mathrm{H_2}] = 0.10\,\text{M}, and [Br2]=0.10 M[\mathrm{Br_2}] = 0.10\,\text{M}, so Kc=(0.10×0.10)(0.20)2=0.010.04=0.25K_c = \frac{(0.10 \times 0.10)}{(0.20)^2} = \frac{0.01}{0.04} = 0.25. A tempting distractor is 4.0, which results from inverting the expression by placing reactants over products. Always write the equilibrium expression based on the balanced equation first, then substitute the given equilibrium concentrations, excluding any pure solids or liquids.

Question 14

In a sealed container at constant temperature, the reaction N2O4(g)⇌2NO2(g)\text{N}_2\text{O}_4(g)\rightleftharpoons 2\text{NO}_2(g) is at equilibrium. The equilibrium concentrations are [N2O4]=0.50 M[\text{N}_2\text{O}_4]=0.50\,M and [NO2]=0.20 M[\text{NO}_2]=0.20\,M. What is the value of KcK_c?

  1. 0.080 (correct answer)
  2. 1.25
  3. 0.40
  4. 0.20
  5. 2.5

Explanation: This question tests the skill of calculating the equilibrium constant. For the balanced equation N₂O₄(g) ⇌ 2NO₂(g), the equilibrium expression is K_c = [NO₂]^2 / [N₂O₄]. To calculate K_c, substitute the given equilibrium concentrations: [NO₂] = 0.20 M and [N₂O₄] = 0.50 M. This yields K_c = (0.20)^2 / 0.50 = 0.04 / 0.50 = 0.080. A tempting distractor might be 1.25, which results from inverting the expression. Always write the expression first, then substitute equilibrium values only—exclude pure solids and liquids.

Question 15

A sealed 1.0 L flask contains the reaction H2(g)+I2(g)⇌2HI(g)\text{H}_2(g)+\text{I}_2(g)\rightleftharpoons 2\text{HI}(g). The system is at equilibrium at a certain temperature. The equilibrium concentrations are: [H2]=0.20 M[\text{H}_2]=0.20\,M, [I2]=0.20 M[\text{I}_2]=0.20\,M, and [HI]=0.80 M[\text{HI}]=0.80\,M. What is the value of KcK_c for the reaction at this temperature?

  1. 2.0
  2. 64
  3. 16 (correct answer)
  4. 0.25
  5. 4.0

Explanation: This question tests the skill of calculating the equilibrium constant. For the balanced equation H₂(g) + I₂(g) ⇌ 2HI(g), the equilibrium expression is K_c = [HI]^2 / ([H₂][I₂]). To calculate K_c, substitute the given equilibrium concentrations: [HI] = 0.80 M, [H₂] = 0.20 M, and [I₂] = 0.20 M. This yields K_c = (0.80)^2 / (0.20 × 0.20) = 0.64 / 0.04 = 16. A tempting distractor might be 4.0, which results from forgetting to square [HI] in the numerator. Always write the expression first, then substitute equilibrium values only—exclude pure solids and liquids.

Question 16

A reaction mixture is at equilibrium for the reaction CO(g)+Cl2(g)⇌COCl2(g)\text{CO}(g)+\text{Cl}_2(g)\rightleftharpoons \text{COCl}_2(g). The equilibrium concentrations are [CO]=0.40 M[\text{CO}]=0.40\,M, [Cl2]=0.20 M[\text{Cl}_2]=0.20\,M, and [COCl2]=0.80 M[\text{COCl}_2]=0.80\,M. What is the value of KcK_c?

  1. 0.25
  2. 10 (correct answer)
  3. 2.0
  4. 4.0
  5. 0.10

Explanation: This question tests the skill of calculating the equilibrium constant. For the balanced equation CO(g) + Cl₂(g) ⇌ COCl₂(g), the equilibrium expression is K_c = [COCl₂] / ([CO][Cl₂]). To calculate K_c, substitute the given equilibrium concentrations: [COCl₂] = 0.80 M, [CO] = 0.40 M, and [Cl₂] = 0.20 M. This yields K_c = 0.80 / (0.40 × 0.20) = 0.80 / 0.08 = 10. A tempting distractor might be 0.10, which results from inverting the expression. Always write the expression first, then substitute equilibrium values only—exclude pure solids and liquids.

Question 17

A sealed container is held at constant temperature. The reaction H2(g)+I2(g)⇌2HI(g)\mathrm{H_2(g) + I_2(g) \rightleftharpoons 2HI(g)} is at equilibrium. The equilibrium concentrations are: [H2]=0.20 M[\mathrm{H_2}]=0.20\,\mathrm{M}, [I2]=0.20 M[\mathrm{I_2}]=0.20\,\mathrm{M}, and [HI]=0.80 M[\mathrm{HI}]=0.80\,\mathrm{M}. What is the value of KcK_c for the reaction?

  1. 1.0
  2. 4.0
  3. 0.25
  4. 16 (correct answer)
  5. 0.0625

Explanation: This problem requires calculating the equilibrium constant for the reaction H₂(g) + I₂(g) ⇌ 2HI(g). The equilibrium expression is written as K_c = [products]/[reactants], with each concentration raised to the power of its stoichiometric coefficient, giving K_c = [HI]²/([H₂][I₂]). Substituting the given equilibrium concentrations: K_c = (0.80)²/((0.20)(0.20)) = 0.64/0.04 = 16. Choice E (0.0625) represents the common error of inverting the equilibrium expression, calculating [H₂][I₂]/[HI]² instead of the correct products over reactants. To avoid errors, always write the equilibrium expression first with products in the numerator and reactants in the denominator, then substitute the equilibrium concentrations with proper exponents.

Question 18

A system is at equilibrium for the reaction CH3COOH(aq)+H2O(l)⇌H3O+(aq)+CH3COO−(aq)\text{CH}_3\text{COOH}(aq)+\text{H}_2\text{O}(l)\rightleftharpoons \text{H}_3\text{O}^+(aq)+\text{CH}_3\text{COO}^-(aq). The equilibrium concentrations are [CH3COOH]=0.50 M[\text{CH}_3\text{COOH}]=0.50\,\text{M}, [H3O+]=0.10 M[\text{H}_3\text{O}^+]=0.10\,\text{M}, and [CH3COO−]=0.10 M[\text{CH}_3\text{COO}^-]=0.10\,\text{M}. (Liquid water is not included in the equilibrium expression.) What is the value of KcK_c?

  1. 0.020 (correct answer)
  2. 0.050
  3. 0.20
  4. 2.0
  5. 25

Explanation: This problem requires calculating the equilibrium constant for CH₃COOH(aq) + H₂O(l) ⇌ H₃O⁺(aq) + CH₃COO⁻(aq). The equilibrium expression is Kc = [H₃O⁺][CH₃COO⁻]/[CH₃COOH], noting that liquid water is excluded from the expression as stated in the problem. Substituting the concentrations: Kc = (0.10)(0.10)/(0.50) = 0.01/0.50 = 0.020. A common error would be including water in the expression or inverting it to get [CH₃COOH]/([H₃O⁺][CH₃COO⁻]) = 0.50/0.01 = 50. When calculating equilibrium constants, remember to exclude pure solids and liquids from the expression, then carefully place products over reactants.

Question 19

For the reaction H2(g)+I2(g)⇌2HI(g)\text{H}_2(g)+\text{I}_2(g)\rightleftharpoons 2\text{HI}(g) in a sealed container at a constant temperature, the system is at equilibrium with the following concentrations: [H2]=0.20 M[\text{H}_2]=0.20\,\text{M}, [I2]=0.10 M[\text{I}_2]=0.10\,\text{M}, and [HI]=0.80 M[\text{HI}]=0.80\,\text{M}. What is the value of KcK_c for the reaction?

  1. 0.031
  2. 3.2
  3. 0.16
  4. 32 (correct answer)
  5. 160

Explanation: This problem requires calculating the equilibrium constant for the reaction H₂(g) + I₂(g) ⇌ 2HI(g). The equilibrium expression is written as K_c = [HI]²/([H₂][I₂]), where products appear in the numerator and reactants in the denominator, each raised to their stoichiometric coefficients. Substituting the given equilibrium concentrations: K_c = (0.80)²/(0.20 × 0.10) = 0.64/0.02 = 32. A common error would be forgetting to square the HI concentration (choice C gives 0.16, which is 0.80/(0.20 × 0.10) without squaring), failing to account for the coefficient of 2 in the balanced equation. To avoid mistakes, always write the equilibrium expression first with proper exponents matching the balanced equation coefficients, then substitute the equilibrium concentrations.

Question 20

In a rigid flask at constant temperature, the reaction H2(g)+I2(g)⇌2HI(g)\text{H}_2(g)+\text{I}_2(g)\rightleftharpoons 2\text{HI}(g) is at equilibrium. The equilibrium concentrations are [H2]=0.50 M[\text{H}_2]=0.50\,\text{M}, [I2]=0.20 M[\text{I}_2]=0.20\,\text{M}, and [HI]=1.0 M[\text{HI}]=1.0\,\text{M}. What is KcK_c for the reaction?

  1. 1.0 × 10^1 (correct answer)
  2. 2.0 × 10^0
  3. 5.0 × 10^-1
  4. 4.0 × 10^1
  5. 2.5 × 10^-2

Explanation: This problem involves calculating the equilibrium constant for H₂(g) + I₂(g) ⇌ 2HI(g). The equilibrium expression is Kc = [HI]²/([H₂][I₂]), where the product HI is squared because its coefficient is 2 in the balanced equation. Substituting the equilibrium values: Kc = (1.0)²/((0.50)(0.20)) = 1.0/(0.10) = 10 = 1.0 × 10¹. A tempting error would be to forget to square the HI concentration, which would give Kc = 1.0/(0.50 × 0.20) = 10, but without properly accounting for the stoichiometry. When writing equilibrium expressions, always raise each concentration to the power of its stoichiometric coefficient from the balanced equation.