Lesson Plan | Traditional Methodology | Equilibrium: Partial Pressures

Objectives

Duration: 10 - 15 minutes

The purpose of this stage is to establish a clear and solid foundation for students to understand the concept of equilibrium constant based on partial pressures and its relationship with molar concentrations. This will allow students to better follow subsequent explanations and actively and consciously participate during problem-solving guided by the teacher.

Main Objectives

1. Explain the concept of equilibrium constant in terms of partial pressures (Kp).

2. Demonstrate the relationship between Kp and the equilibrium constant in terms of molar concentrations (Kc).

3. Provide practical and detailed examples to consolidate students' understanding.

Introduction

Duration: 10 - 15 minutes

The purpose of this stage is to establish a clear and solid foundation for students to understand the concept of equilibrium constant based on partial pressures and its relationship with molar concentrations. This will allow students to better follow subsequent explanations and actively and consciously participate during problem-solving guided by the teacher.

Context

To start the lesson on Equilibrium: Partial Pressures, it is essential to contextualize students about the importance of chemical equilibria in everyday life and in industry. Explain that many natural and industrial processes depend on chemical equilibrium. For example, the production of ammonia through the Haber-Bosch process, crucial for fertilizer manufacturing, is a classic example of an equilibrium reaction involving gases. Highlighting that understanding how the partial pressures of gases influence equilibrium is fundamental to optimizing these processes.

Curiosities

An interesting fact about the equilibrium of partial pressures is that it is essential for human respiration. In the lungs, the exchange of gases between the blood and alveolar air occurs due to differences in the partial pressures of oxygen and carbon dioxide. This knowledge is crucial for fields such as medicine and physiology, demonstrating the practical and vital importance of understanding gas equilibria.

Development

Duration: 50 - 60 minutes

The purpose of this stage is to deepen students' understanding of equilibrium constants both in terms of partial pressures and molar concentrations. Through detailed explanations and practical examples, students will be able to apply the concepts learned in specific problems, facilitating the assimilation of content and preparation for future assessments.

Covered Topics

1. Concept of Equilibrium Constant in Terms of Partial Pressures (Kp): Explain that the equilibrium constant for gaseous reactions can be expressed in terms of the partial pressures of the reactants and products. Use the general formula for Kp and provide clear examples of chemical reactions that illustrate this concept. 2. Relationship between Kp and Kc: Detail the formula that relates the equilibrium constant Kp (partial pressures) to Kc (molar concentrations). The formula is Kp = Kc(RT)^(Δn), where Δn is the difference in the number of moles of gases between products and reactants, R is the gas constant, and T is the temperature in Kelvin. Explain each term of the equation and how they affect the relationship between Kp and Kc. 3. Practical Examples and Problem Solving: Present practical examples of calculations of Kp and Kc for various chemical reactions. Conduct step-by-step problem-solving, guiding students to apply the discussed concepts. Ensure that students note the critical steps and formulas used.

Classroom Questions

1. For the reaction N2(g) + 3H2(g) ⇌ 2NH3(g), at 500K, the equilibrium constant Kc is 0.040. Calculate Kp for this reaction. 2. Given the reaction 2NO2(g) ⇌ N2O4(g), where Kp is 0.113 at 298K. Determine Kc for this reaction. 3. A generic reaction has Kc equal to 5.00 at 400K. If the reaction equation is 2A(g) ⇌ B(g) + C(g), calculate Kp.

Questions Discussion

Duration: 20 - 25 minutes

The purpose of this stage is to ensure that students have understood the explanations and calculations carried out during the lesson. The detailed discussion of the questions allows for reviewing concepts, correcting any misunderstandings, and clarifying doubts. Engaging students with reflective questions promotes a deeper and more practical understanding of the content, better preparing them for future assessments.

Discussion

• ▶️ Question 1: For the reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g), at 500K, the equilibrium constant Kc is 0.040. Calculate Kp for this reaction.

Explanation: Identify the given data: Kc = 0.040, T = 500K, R = 0.0821 (gas constant). Determine Δn (change in the number of moles of gas): Δn = (2) - (1 + 3) = 2 - 4 = -2. Use the formula Kp = Kc (RT)^(Δn):

Kp = 0.040 * (0.0821 * 500)^(-2)

Calculate (0.0821 * 500)^(-2):

(0.0821 * 500) = 41.05

41.05^(-2) = 1 / (41.05^2) ≈ 0.000594 Substitute into the formula:

Kp = 0.040 * 0.000594 ≈ 0.0000238 Conclusion: Kp ≈ 2.38 x 10⁻⁵.

Note: Ensure that students write down all the steps and understand each stage of the calculation.

• ▶️ Question 2: Given the reaction 2NO₂(g) ⇌ N₂O₄(g), where Kp is 0.113 at 298K. Determine Kc for this reaction.

Explanation: Identify the given data: Kp = 0.113, T = 298K, R = 0.0821. Determine Δn: Δn = (1) - (2) = -1. Use the formula Kp = Kc (RT)^(Δn) and rewrite to find Kc:

Kc = Kp / (RT)^(Δn) Calculate (RT)^(-1):

(0.0821 * 298) = 24.4758

24.4758^(-1) = 1 / 24.4758 ≈ 0.0409 Substitute into the formula:

Kc = 0.113 / 0.0409 ≈ 2.76 Conclusion: Kc ≈ 2.76.

Note: Ensure that students write down and understand the inversion of the formula and the calculation of (RT)^(-1).

• ▶️ Question 3: A generic reaction has Kc equal to 5.00 at 400K. If the reaction equation is 2A(g) ⇌ B(g) + C(g), calculate Kp.

Explanation: Identify the given data: Kc = 5.00, T = 400K, R = 0.0821. Determine Δn: Δn = (1 + 1) - (2) = 2 - 2 = 0. Use the formula Kp = Kc (RT)^(Δn):

Kp = 5.00 * (0.0821 * 400)^(0) Calculate (RT)^(0):

Any number raised to zero is 1. Substitute into the formula:

Kp = 5.00 * 1 = 5.00 Conclusion: Kp = 5.00.

Note: Highlight that Δn being zero simplifies the equation, since (RT)^(0) is equal to 1.

Student Engagement

1. 🎓 Question 1: Why is it important to know the difference between Kp and Kc when studying gas equilibria? 2. 🎓 Question 2: How does temperature affect the relationship between Kp and Kc? Give practical examples. 3. 🎓 Question 3: In what practical situations (industrial or natural) can understanding Kp and Kc be crucial? 4. 🎓 Question 4: Reflect on how changes in the number of moles of gas (Δn) influence the relationship between Kp and Kc. 5. 🎓 Question 5: If the gas constant R were different, how would that affect the calculations of Kp and Kc?

Conclusion

Duration: 10 - 15 minutes

The purpose of this stage is to review the main concepts addressed during the lesson, reinforcing learning and ensuring that students have a clear and consolidated understanding of the content. The conclusion also connects theory to practice, highlighting the relevance of the topic and motivating students to value the knowledge acquired.

Summary

• Concept of equilibrium constant in terms of partial pressures (Kp).
• Relationship between Kp and the equilibrium constant in terms of molar concentrations (Kc).
• Formula Kp = Kc(RT)^(Δn) and detailed explanation of each term.
• Practical examples of calculations of Kp and Kc.
• Step-by-step problem-solving.

The lesson connected theory with practice by using real examples of industrial and biological processes, such as the Haber-Bosch process and human respiration. The step-by-step problem-solving allowed students to apply theoretical concepts in practical calculations, consolidating their understanding of the concepts of Kp and Kc.

Understanding the equilibrium constants Kp and Kc is crucial for various applications in everyday life. For example, in the chemical industry, optimizing reactions such as ammonia production can increase efficiency and reduce costs. In medicine, understanding gas equilibrium in the lungs is vital for respiratory treatments. This knowledge shows the practical importance and relevance of gas equilibria.