AP Chemistry Quiz: Calculating Equilibrium Concentrations
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Calculating Equilibrium ConcentrationsQuestion 1 of 20

At 427°C, Kc=55.3K_c = 55.3 for H2(g)+Br2(g)⇌2HBr(g)\text{H}_2(g) + \text{Br}_2(g) \rightleftharpoons 2\text{HBr}(g). If the initial concentrations are [H2]=0.200 M[\text{H}_2] = 0.200 \text{ M}, [Br2]=0.100 M[\text{Br}_2] = 0.100 \text{ M}, and [HBr]=0.300 M[\text{HBr}] = 0.300 \text{ M}, what is the equilibrium concentration of Br2\text{Br}_2?

0.0485 M0.0485 \text{ M}
0.0525 M0.0525 \text{ M}
0.0565 M0.0565 \text{ M}
0.0605 M0.0605 \text{ M}
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AP Chemistry Quiz

AP Chemistry Quiz: Calculating Equilibrium Concentrations

Practice Calculating Equilibrium Concentrations in AP 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 Calculating Equilibrium Concentrations, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Chemistry.

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Question 1

At 427°C, Kc=55.3K_c = 55.3 for H2(g)+Br2(g)⇌2HBr(g)\text{H}_2(g) + \text{Br}_2(g) \rightleftharpoons 2\text{HBr}(g). If the initial concentrations are [H2]=0.200 M[\text{H}_2] = 0.200 \text{ M}, [Br2]=0.100 M[\text{Br}_2] = 0.100 \text{ M}, and [HBr]=0.300 M[\text{HBr}] = 0.300 \text{ M}, what is the equilibrium concentration of Br2\text{Br}_2?

  1. 0.0485 M0.0485 \text{ M}
  2. 0.0525 M0.0525 \text{ M}
  3. 0.0565 M0.0565 \text{ M} (correct answer)
  4. 0.0605 M0.0605 \text{ M}

Explanation: Calculate Q: Q = [HBr]²/([H₂][Br₂]) = (0.300)²/(0.200 × 0.100) = 0.09/0.02 = 4.5. Since Q < Kc (55.3), reaction proceeds forward. Let x = mol/L of H₂ and Br₂ consumed. At equilibrium: [H₂] = 0.200 - x, [Br₂] = 0.100 - x, [HBr] = 0.300 + 2x. Substituting: Kc = (0.300 + 2x)²/[(0.200 - x)(0.100 - x)] = 55.3. Expanding: (0.09 + 1.2x + 4x²)/[(0.02 - 0.3x + x²)] = 55.3. Solving the resulting quadratic gives x = 0.0435 M, so [Br₂] = 0.100 - 0.0435 = 0.0565 M.

Question 2

The synthesis of hydrogen chloride is represented by H2(g)+Cl2(g)⇌2HCl(g)H_2(g) + Cl_2(g) \rightleftharpoons 2HCl(g), which has a large equilibrium constant, Kp=2.5×104K_p = 2.5 \times 10^{4}, at a certain temperature. If 1.0 atm of H2H_2 and 1.0 atm of Cl2Cl_2 are mixed in a container, what will be the approximate partial pressure of H2H_2 at equilibrium?

  1. 0.013 atm0.013~atm (correct answer)
  2. 0.026 atm0.026~atm
  3. 0.50 atm0.50~atm
  4. 0.99 atm0.99~atm

Explanation: Because KpK_p is very large, the reaction proceeds nearly to completion. Assume it goes to completion first: PH2P_{H_2} and PCl2P_{Cl_2} become 0 atm, and PHClP_{HCl} becomes 2.0 atm. Then, let the reaction shift back to equilibrium by an amount x. At equilibrium, PH2=xP_{H_2} = x, PCl2=xP_{Cl_2} = x, and PHCl=2.0−2xP_{HCl} = 2.0 - 2x. The equilibrium expression is Kp=(PHCl)2(PH2)(PCl2)K_p = \frac{(P_{HCl})^2}{(P_{H_2})(P_{Cl_2})}. So, 2.5×104=(2.0−2x)2x22.5 \times 10^{4} = \frac{(2.0 - 2x)^2}{x^2}. We can approximate 2.0−2x≈2.02.0 - 2x \approx 2.0. This gives 2.5×104≈(2.0)2x22.5 \times 10^{4} \approx \frac{(2.0)^2}{x^2}. Solving for x gives x2=4.02.5×104=1.6×10−4x^2 = \frac{4.0}{2.5 \times 10^4} = 1.6 \times 10^{-4}, so x≈0.013x \approx 0.013 atm. This is the equilibrium pressure of H2H_2.

Question 3

For the reaction PCl5(g)⇌PCl3(g)+Cl2(g)PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g), the equilibrium constant KcK_c is 0.040. A reaction mixture is prepared with initial concentrations of [PCl5]=0.20M[PCl_5] = 0.20 M, [PCl3]=0.20M[PCl_3] = 0.20 M, and [Cl2]=0.20M[Cl_2] = 0.20 M. What is the concentration of Cl2Cl_2 once the system reaches equilibrium?

  1. 0.092M0.092 M
  2. 0.11M0.11 M (correct answer)
  3. 0.20M0.20 M
  4. 0.29M0.29 M

Explanation: First, calculate the reaction quotient, Qc=[PCl3][Cl2][PCl5]=(0.20)(0.20)(0.20)=0.20Q_c = \frac{[PCl_3][Cl_2]}{[PCl_5]} = \frac{(0.20)(0.20)}{(0.20)} = 0.20. Since Qc(0.20)>Kc(0.040)Q_c (0.20) > K_c (0.040), the reaction will shift to the left to reach equilibrium. Let x be the change in concentration. At equilibrium, [PCl5]=0.20+x[PCl_5] = 0.20 + x, [PCl3]=0.20−x[PCl_3] = 0.20 - x, and [Cl2]=0.20−x[Cl_2] = 0.20 - x. Then, 0.040=(0.20−x)20.20+x0.040 = \frac{(0.20-x)^2}{0.20+x}. This expands to the quadratic equation x2−0.44x+0.032=0x^2 - 0.44x + 0.032 = 0. Solving gives x≈0.092Mx \approx 0.092 M. The equilibrium concentration of Cl2Cl_2 is 0.20−x=0.20−0.092=0.108M0.20 - x = 0.20 - 0.092 = 0.108 M, which is approximately 0.11M0.11 M.

Question 4

A 2.0 M sample of HI is placed in a container and allowed to decompose according to the reaction 2HI(g)⇌H2(g)+I2(g)2HI(g) \rightleftharpoons H_2(g) + I_2(g). At equilibrium, the concentration of H2H_2 is found to be 0.20 M. What is the concentration of HI at equilibrium?

  1. 0.20M0.20 M
  2. 1.6M1.6 M (correct answer)
  3. 1.8M1.8 M
  4. 2.4M2.4 M

Explanation: From the stoichiometry of the reaction, for every 1 mole of H2H_2 formed, 2 moles of HI must have reacted. If [H2]eq=0.20M[H_2]_{eq} = 0.20 M, then the change in HI concentration is 2×0.20M=0.40M2 \times 0.20 M = 0.40 M. The initial concentration of HI was 2.0 M. Therefore, the equilibrium concentration of HI is the initial concentration minus the amount that reacted: [HI]eq=2.0M−0.40M=1.6M[HI]_{eq} = 2.0 M - 0.40 M = 1.6 M.

Question 5

Initially, 4.0 moles of gaseous reactant A are placed in an empty 2.0 L flask and allowed to establish equilibrium according to the reaction 2A(g)⇌B(g)+C(g)2A(g) \rightleftharpoons B(g) + C(g). The equilibrium constant, KcK_c, is 0.25 for this reaction. What is the equilibrium concentration of B?

  1. 0.25M0.25 M
  2. 0.50M0.50 M (correct answer)
  3. 0.67M0.67 M
  4. 1.0M1.0 M

Explanation: First, calculate the initial concentration of A: [A]initial=4.0 mol2.0 L=2.0M[A]_{initial} = \frac{4.0 \text{ mol}}{2.0 \text{ L}} = 2.0 M. Let x be the equilibrium concentration of B. Then [C]=x[C] = x and [A]=2.0−2x[A] = 2.0 - 2x. The equilibrium expression is Kc=[B][C][A]2=(x)(x)(2.0−2x)2=0.25K_c = \frac{[B][C]}{[A]^2} = \frac{(x)(x)}{(2.0 - 2x)^2} = 0.25. Taking the square root of both sides gives x2.0−2x=0.25=0.50\frac{x}{2.0 - 2x} = \sqrt{0.25} = 0.50. Solving for x: x=0.50(2.0−2x)=1.0−xx = 0.50(2.0 - 2x) = 1.0 - x. This gives 2x=1.02x = 1.0, so x=0.50Mx = 0.50 M. The equilibrium concentration of B is 0.50M0.50 M.

Question 6

The reaction N2O4(g)⇌2NO2(g)N_2O_4(g) \rightleftharpoons 2NO_2(g) has an equilibrium constant Kp=0.66K_p = 0.66. If a container is initially filled with only NO2NO_2 at a pressure of 1.0 atm, what is the partial pressure of N2O4N_2O_4 at equilibrium?

  1. 0.28 atm0.28~atm (correct answer)
  2. 0.33 atm0.33~atm
  3. 0.44 atm0.44~atm
  4. 0.72 atm0.72~atm

Explanation: Since only product is present initially, the reaction will proceed in reverse. Let x be the equilibrium partial pressure of N2O4N_2O_4. The change in NO2NO_2 pressure will be -2x. At equilibrium, PN2O4=xP_{N_2O_4} = x and PNO2=1.0−2xP_{NO_2} = 1.0 - 2x. The equilibrium expression is Kp=(PNO2)2PN2O4K_p = \frac{(P_{NO_2})^2}{P_{N_2O_4}}. So, 0.66=(1.0−2x)2x0.66 = \frac{(1.0 - 2x)^2}{x}. This gives the quadratic equation 4x2−4.66x+1.0=04x^2 - 4.66x + 1.0 = 0. Solving for x gives two possible values, but only x≈0.28x \approx 0.28 atm results in a positive pressure for NO2NO_2. Thus, the equilibrium pressure of N2O4N_2O_4 is 0.280.28 atm.

Question 7

For the gas-phase reaction PCl5(g)⇌PCl3(g)+Cl2(g)PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g), Kp=1.0K_p = 1.0 at a certain temperature. If 2.0 atm of PCl5PCl_5 is initially placed in a container, what is the total pressure at equilibrium?

  1. 2.0 atm2.0~atm
  2. 2.6 atm2.6~atm
  3. 3.0 atm3.0~atm (correct answer)
  4. 4.0 atm4.0~atm

Explanation: Let x be the change in pressure of PCl5PCl_5. At equilibrium, PPCl5=2.0−xP_{PCl_5} = 2.0 - x, PPCl3=xP_{PCl_3} = x, and PCl2=xP_{Cl_2} = x. The equilibrium expression is Kp=(PPCl3)(PCl2)PPCl5K_p = \frac{(P_{PCl_3})(P_{Cl_2})}{P_{PCl_5}}. So, 1.0=x22.0−x1.0 = \frac{x^2}{2.0 - x}. This rearranges to the quadratic equation x2+x−2.0=0x^2 + x - 2.0 = 0, which factors to (x+2)(x−1)=0(x+2)(x-1)=0. The only positive root is x=1.0x=1.0 atm. The equilibrium partial pressures are PPCl5=1.0P_{PCl_5} = 1.0 atm, PPCl3=1.0P_{PCl_3} = 1.0 atm, and PCl2=1.0P_{Cl_2} = 1.0 atm. The total pressure is the sum: 1.0+1.0+1.0=3.01.0 + 1.0 + 1.0 = 3.0 atm.

Question 8

A 0.10 mol sample of SO2Cl2(g)SO_2Cl_2(g) is introduced into an evacuated 1.0 L container at 375 K. The sample decomposes according to SO2Cl2(g)⇌SO2(g)+Cl2(g)SO_2Cl_2(g) \rightleftharpoons SO_2(g) + Cl_2(g), for which Kp=2.9K_p = 2.9. What is the partial pressure of SO2SO_2 at equilibrium? (The gas constant R = 0.08206 L atm/mol K)

  1. 1.2 atm1.2~atm
  2. 1.9 atm1.9~atm (correct answer)
  3. 2.9 atm2.9~atm
  4. 3.1 atm3.1~atm

Explanation: First, calculate the initial pressure of SO2Cl2SO_2Cl_2 using the ideal gas law: P=nRTV=(0.10 mol)(0.08206 L atm/mol K)(375 K)1.0 L≈3.08P = \frac{nRT}{V} = \frac{(0.10 \text{ mol})(0.08206 \text{ L atm/mol K})(375 \text{ K})}{1.0 \text{ L}} \approx 3.08 atm. Let x be the change in pressure. At equilibrium, PSO2Cl2=3.08−xP_{SO_2Cl_2} = 3.08 - x, and PSO2=PCl2=xP_{SO_2} = P_{Cl_2} = x. Kp=(PSO2)(PCl2)PSO2Cl2=x23.08−x=2.9K_p = \frac{(P_{SO_2})(P_{Cl_2})}{P_{SO_2Cl_2}} = \frac{x^2}{3.08 - x} = 2.9. Rearranging yields the quadratic equation x2+2.9x−8.93=0x^2 + 2.9x - 8.93 = 0. The positive root is x≈1.87x \approx 1.87 atm. Therefore, the equilibrium partial pressure of SO2SO_2 is approximately 1.9 atm.

Question 9

A solution is made by mixing equal volumes of 0.20 M HClHCl and 0.20 M NaC2H3O2NaC_2H_3O_2. The resulting reaction is H+(aq)+C2H3O2−(aq)⇌HC2H3O2(aq)H^+(aq) + C_2H_3O_2^-(aq) \rightleftharpoons HC_2H_3O_2(aq). The equilibrium constant K for this reaction is 5.6×1045.6 \times 10^4. What is the approximate equilibrium concentration of H+H^+?

  1. 1.3×10−3M1.3 \times 10^{-3} M (correct answer)
  2. 7.5×10−4M7.5 \times 10^{-4} M
  3. 4.2×10−4M4.2 \times 10^{-4} M
  4. 1.8×10−5M1.8 \times 10^{-5} M

Explanation: Mixing equal volumes halves the initial concentrations to 0.10 M for both H+H^+ and C2H3O2−C_2H_3O_2^-. Since K is very large, the reaction proceeds almost to completion, forming 0.10 M HC2H3O2HC_2H_3O_2 and leaving negligible amounts of reactants. To find the small amount of H+H^+ left, assume the reaction goes to completion and then shifts back. Let [H+]eq=x[H^+]_{eq} = x. At equilibrium, [H+]=[C2H3O2−]=x[H^+] = [C_2H_3O_2^-] = x and [HC2H3O2]=0.10−x≈0.10[HC_2H_3O_2] = 0.10 - x \approx 0.10. Then K=[HC2H3O2][H+][C2H3O2−]≈0.10x2=5.6×104K = \frac{[HC_2H_3O_2]}{[H^+][C_2H_3O_2^-]} \approx \frac{0.10}{x^2} = 5.6 \times 10^4. Solving for x gives x2=0.105.6×104≈1.78×10−6x^2 = \frac{0.10}{5.6 \times 10^4} \approx 1.78 \times 10^{-6}, so x=[H+]≈1.3×10−3Mx = [H^+] \approx 1.3 \times 10^{-3} M.

Question 10

For the reaction I2(g)⇌2I(g)I_2(g) \rightleftharpoons 2I(g), the equilibrium constant KcK_c is 3.8×10−53.8 \times 10^{-5}. If the initial concentration of I2I_2 is 0.050 M, what is the approximate equilibrium concentration of I?

  1. 1.4×10−3M1.4 \times 10^{-3} M (correct answer)
  2. 1.9×10−6M1.9 \times 10^{-6} M
  3. 4.4×10−4M4.4 \times 10^{-4} M
  4. 3.8×10−5M3.8 \times 10^{-5} M

Explanation: Let x be the change in concentration of I2I_2. At equilibrium, [I2]=0.050−x[I_2] = 0.050 - x and [I]=2x[I] = 2x. The equilibrium expression is Kc=[I]2[I2]K_c = \frac{[I]^2}{[I_2]}. So, 3.8×10−5=(2x)20.050−x3.8 \times 10^{-5} = \frac{(2x)^2}{0.050 - x}. Since KcK_c is small, we can approximate 0.050−x≈0.0500.050 - x \approx 0.050. The equation becomes 3.8×10−5≈4x20.0503.8 \times 10^{-5} \approx \frac{4x^2}{0.050}. Solving for x2x^2 gives x2=(3.8×10−5)(0.050)4=4.75×10−7x^2 = \frac{(3.8 \times 10^{-5})(0.050)}{4} = 4.75 \times 10^{-7}. So, x≈6.9×10−4Mx \approx 6.9 \times 10^{-4} M. The equilibrium concentration of I is 2x2x, which is 2×(6.9×10−4)≈1.38×10−3M2 \times (6.9 \times 10^{-4}) \approx 1.38 \times 10^{-3} M, or approximately 1.4×10−3M1.4 \times 10^{-3} M.

Question 11

For the reaction N2(g)+3H2(g)⇌2NH3(g)\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g), the initial concentrations are [N2]=0.50 M[\text{N}_2] = 0.50 \text{ M}, [H2]=1.50 M[\text{H}_2] = 1.50 \text{ M}, and [NH3]=0 M[\text{NH}_3] = 0 \text{ M}. At equilibrium, [NH3]=0.20 M[\text{NH}_3] = 0.20 \text{ M}. What is the equilibrium concentration of N2\text{N}_2?

  1. 0.30 M0.30 \text{ M}
  2. 0.40 M0.40 \text{ M} (correct answer)
  3. 0.50 M0.50 \text{ M}
  4. 0.60 M0.60 \text{ M}

Explanation: Using an ICE table: Initial [N₂] = 0.50 M, change = -x, equilibrium = 0.50 - x. Since 2 mol NH₃ are formed from 1 mol N₂, and [NH₃] at equilibrium = 0.20 M, then x = 0.20/2 = 0.10 M. Therefore, [N₂] at equilibrium = 0.50 - 0.10 = 0.40 M.

Question 12

The reaction H2(g)+I2(g)⇌2HI(g)\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g) has Kc=50.0K_c = 50.0 at 448°C. If equal molar amounts of H2\text{H}_2 and I2\text{I}_2 are mixed, each at an initial concentration of 0.100 M, what is the equilibrium concentration of HI\text{HI}?

  1. 0.124 M0.124 \text{ M}
  2. 0.148 M0.148 \text{ M} (correct answer)
  3. 0.162 M0.162 \text{ M}
  4. 0.186 M0.186 \text{ M}

Explanation: Let x = amount of H₂ and I₂ that react. At equilibrium: [H₂] = [I₂] = 0.100 - x, [HI] = 2x. Kc = (2x)²/((0.100 - x)²) = 50.0. Taking the square root: 2x/(0.100 - x) = √50 = 7.07. Solving: x = 0.074 M, so [HI] = 2(0.074) = 0.148 M.

Question 13

For the equilibrium COCl2(g)⇌CO(g)+Cl2(g)\text{COCl}_2(g) \rightleftharpoons \text{CO}(g) + \text{Cl}_2(g), Kc=8.0×10−4K_c = 8.0 \times 10^{-4} at 400°C. If the initial concentration of COCl2\text{COCl}_2 is 0.500 M and no products are initially present, what is the equilibrium concentration of CO\text{CO}?

  1. 0.0175 M0.0175 \text{ M}
  2. 0.0200 M0.0200 \text{ M} (correct answer)
  3. 0.0225 M0.0225 \text{ M}
  4. 0.0250 M0.0250 \text{ M}

Explanation: Let x = amount of COCl₂ that dissociates. At equilibrium: [COCl₂] = 0.500 - x, [CO] = [Cl₂] = x. Since Kc is small, assume x << 0.500. Then Kc = x²/(0.500) = 8.0 × 10⁻⁴. Solving: x² = 4.0 × 10⁻⁴, so x = 0.0200 M = [CO].

Question 14

At 1000 K, Kc=0.263K_c = 0.263 for C(s)+2H2(g)⇌CH4(g)\text{C}(s) + 2\text{H}_2(g) \rightleftharpoons \text{CH}_4(g). If 2.00 mol of H2\text{H}_2 is placed with excess carbon in a 2.00 L container, what is the equilibrium concentration of CH4\text{CH}_4?

  1. 0.235 M0.235 \text{ M}
  2. 0.265 M0.265 \text{ M} (correct answer)
  3. 0.295 M0.295 \text{ M}
  4. 0.325 M0.325 \text{ M}

Explanation: Initial [H₂] = 2.00 mol/2.00 L = 1.00 M. Let x = [CH₄] formed. At equilibrium: [H₂] = 1.00 - 2x, [CH₄] = x. Kc = x/(1.00 - 2x)² = 0.263. Solving the quadratic equation: 1.052x² - 1.053x + 0.263 = 0 gives x = 0.265 M.

Question 15

The reaction N2O4(g)⇌2NO2(g)\text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g) has Kc=4.6×10−3K_c = 4.6 \times 10^{-3} at 25°C. If the initial concentration of N2O4\text{N}_2\text{O}_4 is 0.0500 M, what is the equilibrium concentration of N2O4\text{N}_2\text{O}_4?

  1. 0.0385 M0.0385 \text{ M}
  2. 0.0415 M0.0415 \text{ M}
  3. 0.0445 M0.0445 \text{ M} (correct answer)
  4. 0.0475 M0.0475 \text{ M}

Explanation: Let x = amount of N₂O₄ that dissociates. At equilibrium: [N₂O₄] = 0.0500 - x, [NO₂] = 2x. Kc = (2x)²/(0.0500 - x) = 4.6 × 10⁻³. Solving: 4x² + 4.6 × 10⁻³x - 2.3 × 10⁻⁴ = 0. Using the quadratic formula: x = 0.0055 M, so [N₂O₄] = 0.0500 - 0.0055 = 0.0445 M.

Question 16

For CaCO3(s)⇌CaO(s)+CO2(g)\text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g), Kc=1.9×10−23K_c = 1.9 \times 10^{-23} at 25°C. If excess CaCO3\text{CaCO}_3 is placed in a sealed container with no initial CO2\text{CO}_2, what is the equilibrium concentration of CO2\text{CO}_2?

  1. 1.4×10−12 M1.4 \times 10^{-12} \text{ M}
  2. 1.9×10−23 M1.9 \times 10^{-23} \text{ M} (correct answer)
  3. 4.4×10−12 M4.4 \times 10^{-12} \text{ M}
  4. 9.5×10−24 M9.5 \times 10^{-24} \text{ M}

Explanation: For this heterogeneous equilibrium, Kc = [CO₂] since solids don't appear in the equilibrium expression. At equilibrium, [CO₂] = Kc = 1.9 × 10⁻²³ M. The extremely small value indicates very little decomposition occurs at 25°C.

Question 17

For the reaction SO2(g)+Cl2(g)⇌SO2Cl2(g)\text{SO}_2(g) + \text{Cl}_2(g) \rightleftharpoons \text{SO}_2\text{Cl}_2(g), Kc=84.7K_c = 84.7 at 100°C. If equal molar amounts of SO2\text{SO}_2 and Cl2\text{Cl}_2 are mixed, each at 0.100 M initially, what is the equilibrium concentration of SO2Cl2\text{SO}_2\text{Cl}_2?

  1. 0.0785 M0.0785 \text{ M}
  2. 0.0825 M0.0825 \text{ M}
  3. 0.0865 M0.0865 \text{ M} (correct answer)
  4. 0.0905 M0.0905 \text{ M}

Explanation: Let x = amount of SO₂Cl₂ formed. At equilibrium: [SO₂] = [Cl₂] = 0.100 - x, [SO₂Cl₂] = x. Kc = x/(0.100 - x)² = 84.7. Taking the square root: √84.7 = 9.20 = √x/(0.100 - x). Solving: x = 0.920(0.100 - x), so x = 0.0865 M.

Question 18

For the reaction 2NOCl(g)⇌2NO(g)+Cl2(g)2\text{NOCl}(g) \rightleftharpoons 2\text{NO}(g) + \text{Cl}_2(g), Kc=1.6×10−5K_c = 1.6 \times 10^{-5} at 35°C. Starting with [NOCl]=0.040 M[\text{NOCl}] = 0.040 \text{ M}, what is the equilibrium concentration of NO\text{NO}?

  1. 8.0×10−4 M8.0 \times 10^{-4} \text{ M} (correct answer)
  2. 1.0×10−3 M1.0 \times 10^{-3} \text{ M}
  3. 1.2×10−3 M1.2 \times 10^{-3} \text{ M}
  4. 1.4×10−3 M1.4 \times 10^{-3} \text{ M}

Explanation: Let x = amount of Cl₂ formed. At equilibrium: [NOCl] = 0.040 - 2x, [NO] = 2x, [Cl₂] = x. Since Kc is very small, assume 2x << 0.040. Then Kc = (2x)²(x)/(0.040)² = 1.6 × 10⁻⁵. Solving: 4x³ = 2.56 × 10⁻⁸, so x = 4.0 × 10⁻⁴ M and [NO] = 8.0 × 10⁻⁴ M.

Question 19

A 500. mL solution of 0.10 M acetic acid, CH3COOHCH_3COOH, is prepared. Given that the acid-dissociation constant, KaK_a, for acetic acid is 1.8×10−51.8 \times 10^{-5}, what is the approximate number of moles of H+H^+ ions at equilibrium?

  1. 1.3×10−31.3 \times 10^{-3} moles
  2. 6.7×10−46.7 \times 10^{-4} moles (correct answer)
  3. 9.0×10−69.0 \times 10^{-6} moles
  4. 1.8×10−61.8 \times 10^{-6} moles

Explanation: For the dissociation CH3COOH⇌H++CH3COO−CH_3COOH \rightleftharpoons H^+ + CH_3COO^-, let [H+]=[CH3COO−]=x[H^+] = [CH_3COO^-] = x and [CH3COOH]=0.10−x[CH_3COOH] = 0.10 - x. The expression is Ka=x20.10−xK_a = \frac{x^2}{0.10-x}. Since KaK_a is small, we can approximate 0.10−x≈0.100.10-x \approx 0.10. So, 1.8×10−5≈x20.101.8 \times 10^{-5} \approx \frac{x^2}{0.10}. Solving for x gives x2=1.8×10−6x^2 = 1.8 \times 10^{-6}, so x=[H+]≈1.34×10−3Mx = [H^+] \approx 1.34 \times 10^{-3} M. To find the number of moles, multiply the concentration by the volume in liters: moles=(1.34×10−3 mol/L)×(0.500 L)≈6.7×10−4moles = (1.34 \times 10^{-3} \text{ mol/L}) \times (0.500 \text{ L}) \approx 6.7 \times 10^{-4} moles.

Question 20

For the reaction N2O4(g)⇌2NO2(g)N_2O_4(g) \rightleftharpoons 2NO_2(g), Kp=0.66K_p = 0.66. If the initial pressure of N2O4N_2O_4 in a closed container is 1.0 atm and there is no initial NO2NO_2, what is the partial pressure of NO2NO_2 at equilibrium?

  1. 0.33 atm0.33~atm
  2. 0.50 atm0.50~atm
  3. 0.66 atm0.66~atm (correct answer)
  4. 0.81 atm0.81~atm

Explanation: Let x be the change in pressure of N2O4N_2O_4. At equilibrium, PN2O4=1.0−xP_{N_2O_4} = 1.0 - x and PNO2=2xP_{NO_2} = 2x. The equilibrium expression is Kp=(PNO2)2PN2O4K_p = \frac{(P_{NO_2})^2}{P_{N_2O_4}}. Substituting gives 0.66=(2x)21.0−x=4x21.0−x0.66 = \frac{(2x)^2}{1.0 - x} = \frac{4x^2}{1.0 - x}. Rearranging gives the quadratic equation 4x2+0.66x−0.66=04x^2 + 0.66x - 0.66 = 0. Solving for x using the quadratic formula yields x≈0.332x \approx 0.332 atm. The partial pressure of NO2NO_2 is 2x2x, which is 2×0.332=0.6642 \times 0.332 = 0.664 atm, approximately 0.660.66 atm.