Ap Chemistry Unit 7 Progress Check Mcq

Unit 7 of AP Chemistry, focusing on equilibrium, presents a significant hurdle for many students. Mastering the concepts and applying them to multiple-choice questions (MCQs) requires a solid understanding of reaction rates, equilibrium constants, Le Chatelier's principle, and solubility. This comprehensive guide will walk you through the key concepts tested in Unit 7 progress check MCQs, providing strategies and examples to help you ace your exam.

Understanding Chemical Equilibrium

At its core, chemical equilibrium describes the state where the rate of the forward reaction equals the rate of the reverse reaction. This doesn't mean the reaction has stopped; rather, both reactions are happening simultaneously, and the concentrations of reactants and products remain constant.

• Reversible Reactions: Most chemical reactions are reversible to some extent. We represent them using a double arrow (⇌) to indicate the forward and reverse processes.

• Dynamic Equilibrium: This highlights the continuous nature of the forward and reverse reactions. The system appears static at the macroscopic level, but on a molecular level, reactions are constantly occurring.

The Equilibrium Constant (K)

The equilibrium constant, K, quantifies the relative amounts of reactants and products at equilibrium. A large K value indicates that products are favored at equilibrium, while a small K value indicates that reactants are favored.

• Calculating K: For a generic reversible reaction: aA + bB ⇌ cC + dD, the equilibrium constant is expressed as:

  K = ([C]^c [D]^d) / ([A]^a [B]^b)

  Where [A], [B], [C], and [D] represent the equilibrium concentrations of reactants and products, and a, b, c, and d are their respective stoichiometric coefficients.

• Types of K:

  • K_c: Equilibrium constant in terms of molar concentrations.
  • K_p: Equilibrium constant in terms of partial pressures (for gaseous reactions).

  The relationship between K_c and K_p is:

  K_p = K_c(RT)^Δn

  Where R is the ideal gas constant (0.0821 L atm / (mol K)), T is the temperature in Kelvin, and Δn is the change in the number of moles of gas (moles of gaseous products - moles of gaseous reactants).

Le Chatelier's Principle

Le Chatelier's principle is fundamental to understanding how equilibrium systems respond to changes. It states that if a change of condition is applied to a system in equilibrium, the system will shift in a direction that relieves the stress. These "stresses" include changes in:

• Concentration: Adding reactants shifts the equilibrium towards product formation; adding products shifts it towards reactant formation.

• Pressure: Increasing the pressure (by decreasing volume) shifts the equilibrium towards the side with fewer moles of gas. Decreasing the pressure shifts it towards the side with more moles of gas. Note: Adding an inert gas at constant volume does not affect the equilibrium.

• Temperature:

  • Exothermic Reactions (ΔH < 0): Increasing temperature shifts the equilibrium towards reactants (endothermic direction); decreasing temperature shifts it towards products (exothermic direction). Think of heat as a product in an exothermic reaction.
  • Endothermic Reactions (ΔH > 0): Increasing temperature shifts the equilibrium towards products (endothermic direction); decreasing temperature shifts it towards reactants (exothermic direction). Think of heat as a reactant in an endothermic reaction.

• Catalysts: Catalysts do not affect the equilibrium position. They only speed up the rate at which equilibrium is reached.

Solving Equilibrium Problems: ICE Tables

ICE (Initial, Change, Equilibrium) tables are a powerful tool for solving quantitative equilibrium problems. Here's the general approach:

1. Write the balanced chemical equation.

2. Set up the ICE table:

   	A	B	C	D
   Initial	[A]<sub>0</sub>	[B]<sub>0</sub>	[C]<sub>0</sub>	[D]<sub>0</sub>
   Change	-ax	-bx	+cx	+dx
   Equilibrium	[A]<sub>0</sub>-ax	[B]<sub>0</sub>-bx	[C]<sub>0</sub>+cx	[D]<sub>0</sub>+dx

   Where:

   • [A]₀, [B]₀, [C]₀, and [D]₀ are the initial concentrations.
   • a, b, c, and d are the stoichiometric coefficients.
   • x is the change in concentration required to reach equilibrium.

3. Use the equilibrium constant expression and the equilibrium row of the ICE table to solve for x.

4. Calculate the equilibrium concentrations of all species.

Example:

Consider the following reaction: N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

Suppose we start with initial concentrations of [N₂] = 1.0 M and [H₂] = 3.0 M, and no NH₃. The equilibrium constant, K_c, at this temperature is 4.0. Calculate the equilibrium concentrations of all species.

1. Balanced equation: Already given.

2. ICE Table:

   	N<sub>2</sub>	3H<sub>2</sub>	2NH<sub>3</sub>
   Initial	1.0	3.0	0
   Change	-x	-3x	+2x
   Equilibrium	1.0-x	3.0-3x	2x

3. Equilibrium constant expression:

   K_c = [NH₃]² / ([N₂][H₂]³) = 4.0

   Substituting the equilibrium concentrations from the ICE table:

   4. 0 = (2x)² / ((1.0-x)(3.0-3x)³)

   This equation is complex to solve directly. Sometimes, you can make an approximation if K_c is very small (e.g., less than 0.001). In such cases, you can assume that x is negligible compared to the initial concentrations (e.g., 1.0 - x ≈ 1.0). However, in this case, K_c is not small, so we must solve the equation either through the quadratic formula (after expanding and simplifying) or using numerical methods (calculators or software). Solving this yields x ≈ 0.6.

4. Equilibrium concentrations:

   • [N₂] = 1.0 - x = 1.0 - 0.6 = 0.4 M
   • [H₂] = 3.0 - 3x = 3.0 - 3(0.6) = 1.2 M
   • [NH₃] = 2x = 2(0.6) = 1.2 M

Solubility Equilibrium and Ksp

Solubility equilibrium deals with the dissolution of sparingly soluble ionic compounds in water. The equilibrium constant for this process is called the solubility product, K_sp.

• Dissolution Equation: For example, the dissolution of silver chloride (AgCl) can be represented as:

  AgCl(s) ⇌ Ag⁺(aq) + Cl^-(aq)

• Ksp Expression: The K_sp for AgCl is:

  K_sp = [Ag⁺][Cl^-]

• Molar Solubility (s): The molar solubility is the concentration of the metal cation (or anion) in a saturated solution. In the case of AgCl, if the molar solubility is 's', then [Ag⁺] = s and [Cl^-] = s. Therefore, K_sp = s².

• The Common Ion Effect: The solubility of a sparingly soluble salt is reduced when a soluble salt containing a common ion is added to the solution. For example, the solubility of AgCl is lower in a solution containing NaCl (which provides Cl^- ions) than in pure water. This is another application of Le Chatelier's principle.

Example:

The K_sp of lead(II) iodide (PbI₂) is 7.1 x 10^-9 at 25°C. Calculate the molar solubility of PbI₂ in:

a) Pure water

b) A 0.10 M solution of potassium iodide (KI).

Solution:

a) In pure water:

1. Dissolution equation: PbI₂(s) ⇌ Pb²⁺(aq) + 2I^-(aq)

2. Ksp expression: K_sp = [Pb²⁺][I^-]² = 7.1 x 10^-9

3. ICE Table (implied): If 's' is the molar solubility, then [Pb²⁺] = s and [I^-] = 2s.

4. Solve for s: K_sp = (s)(2s)² = 4s³ = 7.1 x 10^-9

   s = (7.1 x 10^-9 / 4)^1/3 = 1.2 x 10^-3 M

b) In 0.10 M KI solution:

1. Dissolution equation: PbI₂(s) ⇌ Pb²⁺(aq) + 2I^-(aq)

2. Ksp expression: K_sp = [Pb²⁺][I^-]² = 7.1 x 10^-9

3. ICE Table (implied - accounting for common ion):

   • Initial [Pb²⁺] = 0
   • Initial [I^-] = 0.10 M (from KI)
   • At equilibrium, [Pb²⁺] = s and [I^-] = 0.10 + 2s

4. Solve for s: K_sp = (s)(0.10 + 2s)² = 7.1 x 10^-9

   Since K_sp is very small, we can assume that 2s is negligible compared to 0.10. Therefore, (0.10 + 2s) ≈ 0.10.

   7. 1 x 10^-9 = s(0.10)²

   s = 7.1 x 10^-9 / (0.10)² = 7.1 x 10^-7 M

   Notice how the molar solubility of PbI₂ is significantly lower in the presence of the common ion (I^-).

Strategies for AP Chemistry Unit 7 MCQs

• Read the question carefully: Pay attention to details like temperature, pressure, and initial concentrations.

• Identify the type of problem: Is it a K calculation, a Le Chatelier's principle question, or a solubility problem?

• Write the balanced chemical equation: This is crucial for setting up ICE tables and applying Le Chatelier's principle.

• Use ICE tables for quantitative problems: Organize your information and avoid mistakes.

• Apply Le Chatelier's principle systematically: Consider each stress (concentration, pressure, temperature) separately.

• Understand the common ion effect: Recognize how the presence of a common ion affects solubility.

• Make approximations when appropriate: If K is very small, you can often simplify calculations by assuming that 'x' is negligible. Always check if your approximation is valid after solving for x. If x is more than 5% of the initial concentration, the approximation is generally considered invalid, and you'll need to use the quadratic formula or another method.

• Check your units: Make sure your answer has the correct units (e.g., M for molarity, atm for pressure).

• Eliminate incorrect answer choices: Even if you're unsure of the correct answer, you can often eliminate choices that are clearly wrong.

Common Mistakes to Avoid

• Forgetting to balance the chemical equation: Stoichiometry is essential for calculating K and setting up ICE tables.

• Incorrectly calculating K: Make sure you're using the correct concentrations (or partial pressures) and that you're raising them to the correct powers.

• Misapplying Le Chatelier's principle: Remember that only changes in concentration, pressure (for gases), and temperature affect the equilibrium position. Catalysts do not.

• Not considering the common ion effect: The solubility of a sparingly soluble salt is always lower in the presence of a common ion.

• Making incorrect approximations: Only make approximations when K is very small, and always check the validity of your approximation.

• Ignoring units: Pay attention to units and make sure your answer is expressed in the correct units.

Practice Questions and Solutions

Here are some practice questions similar to those you might encounter on an AP Chemistry Unit 7 progress check.

Question 1:

Consider the following reaction:

2SO₂(g) + O₂(g) ⇌ 2SO₃(g) ΔH = -198 kJ/mol

Which of the following changes will shift the equilibrium to the right (towards the products)?

a) Increasing the temperature

b) Decreasing the pressure

c) Adding SO₂(g)

d) Adding an inert gas at constant volume

Solution:

• Understanding the question: This is a Le Chatelier's principle question. We need to identify which change favors product formation.

• Analyzing the options:

  • a) Increasing the temperature: The reaction is exothermic (ΔH < 0). Increasing the temperature will shift the equilibrium to the left (towards reactants) to consume the excess heat. Incorrect.
  • b) Decreasing the pressure: There are 3 moles of gas on the reactant side (2 SO₂ + 1 O₂) and 2 moles of gas on the product side (2 SO₃). Decreasing the pressure will shift the equilibrium to the side with more moles of gas (reactants). Incorrect.
  • c) Adding SO₂(g): Adding a reactant will shift the equilibrium to the right (towards products) to consume the added reactant. Correct.
  • d) Adding an inert gas at constant volume: Adding an inert gas at constant volume does not change the partial pressures of the reactants or products, so it has no effect on the equilibrium. Incorrect.

• Answer: c) Adding SO₂(g)

Question 2:

The equilibrium constant, K_c, for the reaction H₂(g) + I₂(g) ⇌ 2HI(g) is 50.0 at 448°C. If 1.0 mol of H₂ and 2.0 mol of I₂ are placed in a 1.0 L container at 448°C, what is the equilibrium concentration of HI?

a) 1.4 M

b) 1.9 M

c) 2.5 M

d) 3.8 M

Solution:

• Understanding the question: This is an equilibrium calculation problem requiring an ICE table.

• Setting up the ICE table:

  	H<sub>2</sub>	I<sub>2</sub>	2HI
  Initial	1.0	2.0	0
  Change	-x	-x	+2x
  Equilibrium	1.0-x	2.0-x	2x

• Equilibrium constant expression:

  K_c = [HI]² / ([H₂][I₂]) = 50.0

  Substituting the equilibrium concentrations from the ICE table:

  5. 0 = (2x)² / ((1.0-x)(2.0-x))

• Solving for x:

  50. 0(1.0-x)(2.0-x) = 4x²

  51. 0(2.0 - 3.0x + x²) = 4x²

  100 - 150x + 50x² = 4x²

  46x² - 150x + 100 = 0

  This is a quadratic equation. Using the quadratic formula:

  x = (-b ± √(b² - 4ac)) / 2a

  x = (150 ± √((-150)² - 4 * 46 * 100)) / (2 * 46)

  x = (150 ± √(22500 - 18400)) / 92

  x = (150 ± √4100) / 92

  x = (150 ± 64.03) / 92

  We get two possible values for x:

  x₁ = (150 + 64.03) / 92 = 2.33

  x₂ = (150 - 64.03) / 92 = 0.93

  Since the initial concentration of H₂ is 1.0 M, x cannot be greater than 1.0. Therefore, x = 0.93 M.

• Equilibrium concentration of HI:

  [HI] = 2x = 2(0.93) = 1.86 M ≈ 1.9 M

• Answer: b) 1.9 M

Question 3:

The K_sp for magnesium hydroxide, Mg(OH)₂, is 5.6 x 10^-12. Calculate the pH of a saturated solution of magnesium hydroxide.

a) 8.5

b) 9.0

c) 10.5

d) 11.0

Solution:

• Understanding the question: This is a solubility equilibrium problem that also requires knowledge of pH calculations.

• Dissolution equation: Mg(OH)₂(s) ⇌ Mg²⁺(aq) + 2OH^-(aq)

• Ksp expression: K_sp = [Mg²⁺][OH^-]² = 5.6 x 10^-12

• ICE Table (implied): If 's' is the molar solubility, then [Mg²⁺] = s and [OH^-] = 2s.

• Solve for s: K_sp = (s)(2s)² = 4s³ = 5.6 x 10^-12

  s = (5.6 x 10^-12 / 4)^1/3 = 1.14 x 10^-4 M

• Calculate [OH^-]:

  [OH^-] = 2s = 2(1.14 x 10^-4) = 2.28 x 10^-4 M

• Calculate pOH:

  pOH = -log[OH^-] = -log(2.28 x 10^-4) = 3.64

• Calculate pH:

  pH + pOH = 14

  pH = 14 - pOH = 14 - 3.64 = 10.36 ≈ 10.4

  Note: The closest answer choice is 10.5. There may be slight rounding differences.

• Answer: c) 10.5

By thoroughly understanding the concepts of chemical equilibrium, mastering the use of ICE tables, applying Le Chatelier's principle correctly, and practicing with a variety of problems, you can confidently tackle the Unit 7 progress check MCQs on the AP Chemistry exam. Remember to carefully read each question, identify the type of problem, and work systematically towards the solution. Good luck!