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AP Chemistry

AP Chemistry Help: Calculating The Equilibrium Constant

Review real example questions for Calculating The Equilibrium Constant in AP Chemistry.

Question 1

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

  1. 25
  2. 2.5
  3. 0.25
  4. 0.10
  5. 0.040
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)H2​(g)+I2​(g)⇌2HI(g) is at equilibrium. The equilibrium concentrations are [H2]=0.20 M[\text{H}_2]=0.20\,\text{M}[H2​]=0.20M, [I2]=0.20 M[\text{I}_2]=0.20\,\text{M}[I2​]=0.20M, and [HI]=0.80 M[\text{HI}]=0.80\,\text{M}[HI]=0.80M. What is KcK_cKc​?

  1. 0.063
  2. 4.0
  3. 16
  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)}CH3​COOH(aq)⇌H+(aq)+CH3​COO−(aq) The system is at equilibrium with [CH3COOH]=0.10 M[\mathrm{CH_3COOH}]=0.10\,\text{M}[CH3​COOH]=0.10M, [H+]=0.010 M[\mathrm{H^+}]=0.010\,\text{M}[H+]=0.010M, and [CH3COO−]=0.010 M[\mathrm{CH_3COO^-}]=0.010\,\text{M}[CH3​COO−]=0.010M. What is the value of KcK_cKc​ for this reaction?

  1. 1.0×10−11.0 \times 10^{-1}1.0×10−1
  2. 1.0×10−31.0 \times 10^{-3}1.0×10−3
  3. 1.0×1021.0 \times 10^{2}1.0×102
  4. 1.0×10−21.0 \times 10^{-2}1.0×10−2
  5. 1.0×10−41.0 \times 10^{-4}1.0×10−4
Explanation: This question tests the skill of calculating the equilibrium constant. To calculate KcK_cKc​, first write the equilibrium expression for the reaction CH3COOH(aq)⇌H+(aq)+CH3COO−(aq)\mathrm{CH_3COOH(aq) \rightleftharpoons H^+(aq) + CH_3COO^-(aq)}CH3​COOH(aq)⇌H+(aq)+CH3​COO−(aq), which is Kc=[H+][CH3COO−][CH3COOH]K_c = \frac{[\mathrm{H^+}] [\mathrm{CH_3COO^-}]}{[\mathrm{CH_3COOH}]}Kc​=[CH3​COOH][H+][CH3​COO−]​. 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}[CH3​COOH]=0.10M, [H+]=0.010 M[\mathrm{H^+}] = 0.010\, \text{M}[H+]=0.010M, and [CH3COO−]=0.010 M[\mathrm{CH_3COO^-}] = 0.010\, \text{M}[CH3​COO−]=0.010M, 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}Kc​=(0.010×0.010)/0.10=0.0001/0.10=0.001=1.0×10−3. A tempting distractor is 1.0×10−41.0 \times 10^{-4}1.0×10−4, 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)}A(g)+2B(g)⇌C(g). The equilibrium concentrations are [A]=0.20 M[\mathrm{A}]=0.20\,\mathrm{M}[A]=0.20M, [B]=0.10 M[\mathrm{B}]=0.10\,\mathrm{M}[B]=0.10M, and [C]=0.20 M[\mathrm{C}]=0.20\,\mathrm{M}[C]=0.20M. What is the value of KcK_cKc​?

  1. 100
  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)}2NOCl(g)⇌2NO(g)+Cl2​(g) is at equilibrium. The equilibrium concentrations are [NOCl]=0.20 M[\mathrm{NOCl}]=0.20\,\mathrm{M}[NOCl]=0.20M, [NO]=0.20 M[\mathrm{NO}]=0.20\,\mathrm{M}[NO]=0.20M, and [Cl2]=0.05 M[\mathrm{Cl_2}]=0.05\,\mathrm{M}[Cl2​]=0.05M. What is KcK_cKc​?

  1. 0.25
  2. 1.0
  3. 0.50
  4. 2.0
  5. 0.050
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)}2SO2​(g)+O2​(g)⇌2SO3​(g). The equilibrium concentrations are [SO2]=0.20 M[\mathrm{SO_2}]=0.20\,\mathrm{M}[SO2​]=0.20M, [O2]=0.10 M[\mathrm{O_2}]=0.10\,\mathrm{M}[O2​]=0.10M, and [SO3]=0.30 M[\mathrm{SO_3}]=0.30\,\mathrm{M}[SO3​]=0.30M. What is the value of KcK_cKc​?

  1. 2.25
  2. 0.225
  3. 22.5
  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)}H2​(g)+Br2​(g)⇌2HBr(g) is at equilibrium. The equilibrium concentrations are [H2]=0.30 M[\mathrm{H_2}]=0.30\,\mathrm{M}[H2​]=0.30M, [Br2]=0.10 M[\mathrm{Br_2}]=0.10\,\mathrm{M}[Br2​]=0.10M, and [HBr]=0.30 M[\mathrm{HBr}]=0.30\,\mathrm{M}[HBr]=0.30M. What is the value of KcK_cKc​?

  1. 0.33
  2. 0.030
  3. 3.0
  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)SO2​(g)+NO2​(g)⇌SO3​(g)+NO(g). The equilibrium concentrations are [SO2]=0.50 M[\text{SO}_2]=0.50\,\text{M}[SO2​]=0.50M, [NO2]=0.20 M[\text{NO}_2]=0.20\,\text{M}[NO2​]=0.20M, [SO3]=0.10 M[\text{SO}_3]=0.10\,\text{M}[SO3​]=0.10M, and [NO]=0.40 M[\text{NO}]=0.40\,\text{M}[NO]=0.40M. What is the value of KcK_cKc​?

  1. 0.10
  2. 0.20
  3. 0.40
  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)CO2​(g)+H2​(g)⇌CO(g)+H2​O(g). The equilibrium concentrations are [CO2]=0.25 M[\text{CO}_2]=0.25\,\text{M}[CO2​]=0.25M, [H2]=0.50 M[\text{H}_2]=0.50\,\text{M}[H2​]=0.50M, [CO]=0.10 M[\text{CO}]=0.10\,\text{M}[CO]=0.10M, and [H2O]=0.20 M[\text{H}_2\text{O}]=0.20\,\text{M}[H2​O]=0.20M. What is the value of KcK_cKc​?

  1. 0.080
  2. 0.16
  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)}A(g)+2B(g)⇌AB2​(g) has equilibrium concentrations [A]=0.20 M[\mathrm{A}]=0.20\,\text{M}[A]=0.20M, [B]=0.10 M[\mathrm{B}]=0.10\,\text{M}[B]=0.10M, and [AB2]=0.20 M[\mathrm{AB_2}]=0.20\,\text{M}[AB2​]=0.20M. What is the value of KcK_cKc​ for the reaction?

  1. 10
  2. 100
  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}Kc​=[A][B]2[AB2​]​. 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}[A]=0.20M, [B]=0.10 M[\mathrm{B}] = 0.10\,\text{M}[B]=0.10M, and [AB2]=0.20 M[\mathrm{AB_2}] = 0.20\,\text{M}[AB2​]=0.20M, 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 = 100Kc​=0.20/(0.20×(0.10)2)=0.20/(0.20×0.01)=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.