#Mastering Equilibrium: A Last-Minute Study Guide

Hey there, future AP Chem master! Let's get you prepped and confident for your exam. We're diving into the properties of the equilibrium constant (Keq), building on what you already know about Hess's Law. Think of this as your ultimate cheat sheet for acing those equilibrium problems!

#Review of Hess's Law

Remember Hess's Law? It's all about how enthalpy change (ΔH) is a state function. This means the total ΔH for a reaction is the same, no matter how many steps you take to get from reactants to products. It's like climbing a mountain – you can take different paths, but the total elevation change is always the same.

#Key Rules of Hess's Law:

• Flipping Reactions: When you reverse a reaction, the sign of ΔH flips (e.g., +ΔH becomes -ΔH). 🔄

• Multiplying Reactions: If you multiply a reaction by a factor 'n', you also multiply ΔH by 'n'.

• Adding Reactions: When you add reactions, you add their ΔH values to get the overall ΔH. ➕

#Properties of Keq

Now, let's see how these rules apply to the equilibrium constant, Keq. It's like Hess's Law, but with a twist!

#Flipping Reactions

Flipping a reaction means looking at it from the products-to-reactants perspective. If our original reaction is A ⇌ B, then K = [B]/[A]. For the reverse reaction, B ⇌ A, K₂ = [A]/[B] = 1/K.

#Example of Inverting K

Let's say we have the reaction:

N₂ + 3H₂ ⇌ 2NH₃ with K = 0.118. If we flip it to:

2NH₃ ⇌ N₂ + 3H₂, then K_new = 1/0.118 = 8.47.

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#Multiplying Reactions

What happens when we multiply a reaction by a coefficient? For A ⇌ B, K = [B]/[A]. If we multiply by 'n' to get nA ⇌ nB, then K_new = ([B]/[A])ⁿ = Kⁿ.

#Example of Exponentiating K

Using the same reaction: N₂ + 3H₂ ⇌ 2NH₃ with K = 0.118. If we have (1/2)N₂ + (3/2)H₂ ⇌ NH₃, we've multiplied by 1/2. So, K_new = (0.118)^(1/2) = 0.343.

#Adding Reactions Together

When we add reactions, we multiply their equilibrium constants. Let's see how this works:

A ⇌ B : K₁ = [B]/[A]

C ⇌ D : K₂ = [D]/[C]

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A + C ⇌ B + D : K_total = [B][D] / [A][C] = K₁ * K₂

#Summary of Properties of K

Here's a handy table to summarize everything:

#Practice Problem

Let's put it all together!

Find the equilibrium constant for the reaction N₂ + 2O₂ ⇌ 2NO₂ using the following:

• Reaction 1: (1/2)N₂ + (1/2)O₂ ⇌ NO …. K₁ = 6.55 * 10⁻¹³
• Reaction 2: 2NO + O₂ ⇌ 2NO₂ …. K₂ = 6.9 * 10⁵

Solution:

1. Multiply Reaction 1 by 2: N₂ + O₂ ⇌ 2NO, K_new = (6.55 * 10⁻¹³)² = 4.3 * 10⁻²⁵

2. Add the modified Reaction 1 and Reaction 2: N₂ + 2O₂ ⇌ 2NO₂

3. Multiply the K values: K_total = 4.3 * 10⁻²⁵ * 6.9 * 10⁵ = 3.0 * 10⁻¹⁹

#Final Exam Focus

Alright, let's wrap this up with a focus on what's most important for the exam:

• High-Priority Topics: - Manipulating Keq using the rules we've discussed. - Combining multiple equilibrium steps. - Understanding the relationship between K and ΔG (Gibbs Free Energy).

• Common Question Types: - Multiple-choice questions testing your understanding of how K changes with reaction manipulation. - Free-response questions requiring you to calculate K for a multi-step reaction.

• Last-Minute Tips: - Time Management: Don't get bogged down on one problem. If you're stuck, move on and come back later. - Common Pitfalls: Watch out for sign errors when flipping reactions and remember to multiply K values when adding reactions. - Strategies: Always double-check your work and make sure your final answer makes sense in the context of the problem.

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Practice Question

Practice Questions

Here are some practice questions to test your understanding:

#Multiple Choice Questions

1. The equilibrium constant for the reaction 2A ⇌ B is K. What is the equilibrium constant for the reaction 4A ⇌ 2B? (A) 2K (B) K² (C) K^(1/2) (D) 4K

2. Given the following reactions and their equilibrium constants: X ⇌ Y, K₁ = 2 Y ⇌ Z, K₂ = 3 What is the equilibrium constant for the reaction X ⇌ Z? (A) 1/6 (B) 5 (C) 6 (D) 8

3. For the reaction A + B ⇌ 2C, the equilibrium constant is K. If the reaction is reversed and multiplied by 2, what is the new equilibrium constant? (A) 2K (B) 1/K² (C) 1/(2K) (D) K²

#Free Response Question

Consider the following reactions:

Reaction 1: N₂(g) + O₂(g) ⇌ 2NO(g) K₁ = 4.0 × 10⁻³¹ Reaction 2: 2NO(g) + O₂(g) ⇌ 2NO₂(g) K₂ = 6.0 × 10⁶

(a) Calculate the equilibrium constant for the reaction N₂(g) + 2O₂(g) ⇌ 2NO₂(g)

(b) If the reaction N₂(g) + 2O₂(g) ⇌ 2NO₂(g) was reversed, what would be the new equilibrium constant?

(c) If the reaction N₂(g) + O₂(g) ⇌ 2NO(g) was multiplied by 2, what would be the new equilibrium constant?

Scoring Breakdown:

(a) (2 points) - 1 point for correctly multiplying the equilibrium constants. - 1 point for the correct answer with correct scientific notation.

(b) (1 point) - 1 point for correctly inverting the equilibrium constant from part (a).

(c) (1 point) - 1 point for correctly squaring the equilibrium constant from Reaction 1.

You've got this! Go ace that exam! 🚀