Ap Chem Unit 7 Practice Problems

The world of chemical reactions, equilibrium, and thermodynamics can seem daunting, especially when preparing for the AP Chemistry exam. Unit 7, focusing on equilibrium, is a crucial area where many students face challenges. To conquer this unit and excel in your exam, let's dive into some practice problems that will solidify your understanding of equilibrium concepts and calculations.

Mastering AP Chemistry Unit 7: Equilibrium Through Practice Problems

This article will guide you through various types of equilibrium problems, providing step-by-step solutions and explanations. By working through these examples, you'll gain confidence in your ability to tackle any equilibrium question on the AP exam. We'll cover topics such as:

• Writing equilibrium expressions
• Calculating equilibrium constants (K)
• Using ICE tables
• Le Chatelier's Principle
• Acid-base equilibria
• Solubility equilibria

Understanding Equilibrium Expressions

The foundation of equilibrium lies in understanding how to write equilibrium expressions. These expressions relate the concentrations of reactants and products at equilibrium.

Key Concepts:

• Equilibrium Constant (K): A numerical value that represents the ratio of products to reactants at equilibrium. A large K indicates that the equilibrium favors the products, while a small K indicates that it favors the reactants.
• Law of Mass Action: States that the rate of a chemical reaction is proportional to the product of the concentrations of the reactants, each raised to a power equal to its coefficient in the balanced chemical equation.
• Homogeneous vs. Heterogeneous Equilibria: Homogeneous equilibria involve reactants and products in the same phase, while heterogeneous equilibria involve reactants and products in different phases. Solids and liquids do not appear in the equilibrium expression.

Practice Problem 1:

Write the equilibrium expression (K) for the following reversible reaction:

N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

Solution:

The equilibrium expression is written as follows:

K = [NH₃]² / ([N₂] [H₂]³)

Notice that the concentrations of the products (NH₃) are in the numerator, and the concentrations of the reactants (N₂ and H₂) are in the denominator. Each concentration is raised to the power of its stoichiometric coefficient in the balanced equation.

Practice Problem 2:

Write the equilibrium expression (K) for the following heterogeneous reaction:

CaCO₃(s) ⇌ CaO(s) + CO₂(g)

Solution:

Since CaCO₃(s) and CaO(s) are solids, they do not appear in the equilibrium expression. Therefore, the expression is:

K = [CO₂]

Calculating Equilibrium Constants (K)

Once you know how to write equilibrium expressions, you can calculate the value of the equilibrium constant (K) if you know the equilibrium concentrations of all the reactants and products.

Practice Problem 3:

Consider the following reaction:

H₂(g) + I₂(g) ⇌ 2HI(g)

At a certain temperature, the equilibrium concentrations are found to be:

[H₂] = 0.10 M [I₂] = 0.20 M [HI] = 0.40 M

Calculate the value of the equilibrium constant (K).

Solution:

First, write the equilibrium expression:

K = [HI]² / ([H₂] [I₂])

Now, plug in the equilibrium concentrations:

K = (0.40)² / (0.10 * 0.20) = 1.6 / 0.02 = 80

Therefore, the equilibrium constant (K) for this reaction at this temperature is 80.

Practice Problem 4:

For the reaction:

2SO₂(g) + O₂(g) ⇌ 2SO₃(g)

At a certain temperature, the equilibrium constant K is 4.3 x 10⁵. If the equilibrium concentrations of SO₂ and O₂ are 0.10 M and 0.20 M, respectively, what is the equilibrium concentration of SO₃?

Solution:

Write the equilibrium expression:

K = [SO₃]² / ([SO₂]² [O₂])

Plug in the known values:

4. 3 x 10⁵ = [SO₃]² / (0.10² * 0.20)

Solve for [SO₃]:

[SO₃]² = 4.3 x 10⁵ * (0.10² * 0.20) = 860

[SO₃] = √860 ≈ 29.3 M

Therefore, the equilibrium concentration of SO₃ is approximately 29.3 M.

Using ICE Tables

ICE tables (Initial, Change, Equilibrium) are essential for solving equilibrium problems where you don't know all the equilibrium concentrations. They help you organize your information and systematically calculate the equilibrium concentrations.

Practice Problem 5:

Consider the following reaction:

N₂O₄(g) ⇌ 2NO₂(g)

Initially, a container is filled with N₂O₄(g) at a pressure of 0.50 atm. At equilibrium, the pressure of NO₂(g) is found to be 0.20 atm. Calculate the K_p for this reaction.

Solution:

First, set up an ICE table:

	N<sub>2</sub>O<sub>4</sub>(g)	2NO<sub>2</sub>(g)
Initial (I)	0.50	0
Change (C)	-x	+2x
Equilibrium (E)	0.50 - x	2x

We know that at equilibrium, the pressure of NO₂(g) is 0.20 atm. Therefore, 2x = 0.20 atm, which means x = 0.10 atm.

Now we can calculate the equilibrium pressures:

P(N₂O₄) = 0.50 - 0.10 = 0.40 atm P(NO₂) = 0.20 atm

Write the K_p expression:

K_p = (P(NO₂))² / P(N₂O₄)

Plug in the equilibrium pressures:

K_p = (0.20)² / 0.40 = 0.04 / 0.40 = 0.10

Therefore, the K_p for this reaction is 0.10.

Practice Problem 6:

Consider the following reaction:

CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g)

At a certain temperature, the equilibrium constant K is 4.0. If 0.60 mol of CO and 0.60 mol of H₂O are placed in a 1.0 L container, what are the equilibrium concentrations of all species?

Solution:

First, calculate the initial concentrations:

[CO]_initial = 0.60 mol / 1.0 L = 0.60 M [H₂O]_initial = 0.60 mol / 1.0 L = 0.60 M [CO₂]_initial = 0 M [H₂]_initial = 0 M

Set up an ICE table:

	CO(g)	H<sub>2</sub>O(g)	CO<sub>2</sub>(g)	H<sub>2</sub>(g)
Initial (I)	0.60	0.60	0	0
Change (C)	-x	-x	+x	+x
Equilibrium (E)	0.60 - x	0.60 - x	x	x

Write the equilibrium expression:

K = [CO₂] [H₂] / ([CO] [H₂O])

Plug in the equilibrium concentrations from the ICE table:

4. 0 = (x * x) / ((0.60 - x) * (0.60 - x)) = x² / (0.60 - x)²

Take the square root of both sides:

2. 0 = x / (0.60 - x)

Solve for x:

1. 20 - 2x = x 3x = 1.20 x = 0.40

Now we can calculate the equilibrium concentrations:

[CO] = 0.60 - 0.40 = 0.20 M [H₂O] = 0.60 - 0.40 = 0.20 M [CO₂] = 0.40 M [H₂] = 0.40 M

Therefore, the equilibrium concentrations are: [CO] = 0.20 M, [H₂O] = 0.20 M, [CO₂] = 0.40 M, and [H₂] = 0.40 M.

Le Chatelier's Principle

Le Chatelier's Principle 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 changes can include changes in concentration, pressure, volume, or temperature.

Key Concepts:

• Concentration Changes: Adding reactants shifts the equilibrium towards the products, and adding products shifts the equilibrium towards the reactants.
• Pressure/Volume Changes: Increasing pressure (decreasing volume) shifts the equilibrium towards the side with fewer moles of gas. Decreasing pressure (increasing volume) shifts the equilibrium towards the side with more moles of gas. If the number of moles of gas is the same on both sides, pressure/volume changes have no effect.
• Temperature Changes: For exothermic reactions (ΔH < 0), increasing temperature shifts the equilibrium towards the reactants. For endothermic reactions (ΔH > 0), increasing temperature shifts the equilibrium towards the products.

Practice Problem 7:

Consider the following reaction:

N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = -92 kJ/mol

Predict the effect of each of the following changes on the equilibrium:

a) Adding N₂(g) b) Increasing the pressure c) Increasing the temperature d) Adding a catalyst

Solution:

a) Adding N₂(g): The equilibrium will shift towards the products (right) to consume the added N₂ and produce more NH₃.

b) Increasing the pressure: There are 4 moles of gas on the reactant side (1 N₂ + 3 H₂) and 2 moles of gas on the product side (2 NH₃). Increasing the pressure will shift the equilibrium towards the side with fewer moles of gas, which is the product side (right).

c) Increasing the temperature: The reaction is exothermic (ΔH < 0), so heat is released as a product. Increasing the temperature will shift the equilibrium towards the reactants (left) to consume the added heat.

d) Adding a catalyst: A catalyst speeds up the rate of both the forward and reverse reactions equally. Therefore, adding a catalyst will not shift the equilibrium; it will only cause the system to reach equilibrium faster.

Practice Problem 8:

Consider the following reaction:

CO(g) + 2H₂(g) ⇌ CH₃OH(g)

How will the equilibrium shift if the volume of the container is increased?

Solution:

Increasing the volume decreases the pressure. There are 3 moles of gas on the reactant side (1 CO + 2 H₂) and 1 mole of gas on the product side (1 CH₃OH). Decreasing the pressure will shift the equilibrium towards the side with more moles of gas, which is the reactant side (left).

Acid-Base Equilibria

Acid-base equilibria involve the transfer of protons (H⁺) between acids and bases. Key concepts include acid and base strength, pH, and buffers.

Key Concepts:

• Acid Dissociation Constant (K_a): A measure of the strength of an acid. A larger K_a indicates a stronger acid.
• Base Dissociation Constant (K_b): A measure of the strength of a base. A larger K_b indicates a stronger base.
• pH: A measure of the acidity or basicity of a solution. pH = -log[H⁺].
• pOH: A measure of the hydroxide ion concentration. pOH = -log[OH^-].
• Buffers: Solutions that resist changes in pH upon the addition of small amounts of acid or base.

Practice Problem 9:

Calculate the pH of a 0.10 M solution of acetic acid (CH₃COOH). The K_a for acetic acid is 1.8 x 10^-5.

Solution:

First, write the equilibrium reaction:

CH₃COOH(aq) ⇌ H⁺(aq) + CH₃COO^-(aq)

Set up an ICE table:

	CH<sub>3</sub>COOH	H<sup>+</sup>	CH<sub>3</sub>COO<sup>-</sup>
Initial (I)	0.10	0	0
Change (C)	-x	+x	+x
Equilibrium (E)	0.10 - x	x	x

Write the K_a expression:

K_a = [H⁺] [CH₃COO^-] / [CH₃COOH]

Plug in the equilibrium concentrations from the ICE table:

1. 8 x 10^-5 = (x * x) / (0.10 - x)

Since K_a is small, we can assume that x is much smaller than 0.10, so 0.10 - x ≈ 0.10:

1. 8 x 10^-5 = x² / 0.10

Solve for x:

x² = 1.8 x 10^-6 x = √(1.8 x 10^-6) ≈ 1.34 x 10^-3

Since x = [H⁺], we can calculate the pH:

pH = -log[H⁺] = -log(1.34 x 10^-3) ≈ 2.87

Therefore, the pH of the 0.10 M acetic acid solution is approximately 2.87.

Practice Problem 10:

A buffer solution is prepared by mixing 25.0 mL of 0.10 M benzoic acid (C₆H₅COOH) and 25.0 mL of 0.10 M sodium benzoate (C₆H₅COONa). The K_a for benzoic acid is 6.3 x 10^-5. Calculate the pH of the buffer solution.

Solution:

This is a buffer solution containing a weak acid (benzoic acid) and its conjugate base (benzoate). We can use the Henderson-Hasselbalch equation to calculate the pH:

pH = pK_a + log([A^-] / [HA])

First, calculate the pK_a:

pK_a = -log(K_a) = -log(6.3 x 10^-5) ≈ 4.20

Since the volumes of the acid and conjugate base solutions are equal, and the concentrations are equal, the ratio of [A^-] / [HA] is 1:

pH = 4.20 + log(1) = 4.20 + 0 = 4.20

Therefore, the pH of the buffer solution is 4.20.

Solubility Equilibria

Solubility equilibria involve the dissolution of sparingly soluble ionic compounds in water. The solubility product (K_sp) is a measure of the extent to which a compound dissolves.

Key Concepts:

• Solubility Product (K_sp): The equilibrium constant for the dissolution of a sparingly soluble ionic compound. A smaller K_sp indicates lower solubility.
• Molar Solubility (s): The number of moles of solute that dissolve in one liter of solution at a given temperature.
• Common Ion Effect: The decrease in the solubility of a sparingly soluble salt when a soluble salt containing a common ion is added to the solution.

Practice Problem 11:

The solubility of silver chloride (AgCl) in water is 1.3 x 10^-5 M at 25°C. Calculate the K_sp for AgCl.

Solution:

First, write the equilibrium reaction:

AgCl(s) ⇌ Ag⁺(aq) + Cl^-(aq)

The K_sp expression is:

K_sp = [Ag⁺] [Cl^-]

Since the molar solubility of AgCl is 1.3 x 10^-5 M, this means that at equilibrium:

[Ag⁺] = 1.3 x 10^-5 M [Cl^-] = 1.3 x 10^-5 M

Plug these values into the K_sp expression:

K_sp = (1.3 x 10^-5) * (1.3 x 10^-5) = 1.69 x 10^-10

Therefore, the K_sp for AgCl is 1.69 x 10^-10.

Practice Problem 12:

Calculate the molar solubility of lead(II) iodide (PbI₂) in a 0.10 M solution of potassium iodide (KI). The K_sp for PbI₂ is 7.1 x 10^-9.

Solution:

First, write the equilibrium reaction:

PbI₂(s) ⇌ Pb²⁺(aq) + 2I^-(aq)

Set up an ICE table, considering the common ion effect from KI:

	Pb<sup>2+</sup>	2I<sup>-</sup>
Initial (I)	0	0.10
Change (C)	+s	+2s
Equilibrium (E)	s	0.10 + 2s

Write the K_sp expression:

K_sp = [Pb²⁺] [I^-]²

Plug in the equilibrium concentrations from the ICE table:

1. 1 x 10^-9 = (s) * (0.10 + 2s)²

Since K_sp is small, we can assume that 2s is much smaller than 0.10, so 0.10 + 2s ≈ 0.10:

1. 1 x 10^-9 = s * (0.10)²

Solve for s:

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

Therefore, the molar solubility of PbI₂ in the 0.10 M KI solution is 7.1 x 10^-7 M.

Conclusion

Equilibrium is a fundamental concept in chemistry, and mastering it is crucial for success in AP Chemistry. By working through these practice problems, you've strengthened your understanding of equilibrium expressions, equilibrium constants, ICE tables, Le Chatelier's Principle, acid-base equilibria, and solubility equilibria. Remember to practice regularly and review these concepts to solidify your knowledge. Good luck with your AP Chemistry exam!