In this lesson students will understand the connections between the equilibrium constant (K) and the reaction quotient (Q) as well as how they determine the favorability of a reaction. Additionally students will be able to determine if a reaction is kinetically favored or thermodynamically favored.
High School (AP Chemistry)
This lesson will help prepare your students to meet the performance expectations in the following standards:
- HS-PS1-5: Apply scientific principles and evidence to provide an explanation about the effects of changing the temperature or concentration of the reacting particles on the rate at which a reaction occurs.
- HS-PS1-6: Refine the design of a chemical system by specifying a change in conditions that would produce increased amounts of products at equilibrium.
- Scientific and Engineering Practices:
- Using Mathematics and Computational Thinking
- Analyzing and Interpreting Data
- Engaging in Argument from Evidence
AP Chemistry Curriculum Framework
- Big Idea 5: The laws of thermodynamics describe the essential role of energy and explain and predict the direction of changes in matter.
- 5.18 The student can explain why a thermodynamically favored chemical reaction may not produce large amounts of product (based on consideration of both initial conditions and kinetic effects), or why a thermodynamically unfavored chemical reaction can produce large amounts of product for certain sets of initial conditions.
- Big Idea 6: Any bond or intermolecular attraction
that can be formed can be broken. These two processes are in a dynamic
competition, sensitive to initial conditions and external perturbations.
- 6.25 The student is able to express the equilibrium constant in terms of Δ Go and RT and use this relationship to estimate the magnitude of K and, consequently, the thermodynamic favorability of the process.
By the end of this lesson, students should be able to
- Use K vs. Q to determine the direction a reaction proceeds.
- Understand how equilibrium relates to favorability of a reaction.
- Determine the favorability of a reaction based upon kinetics and thermochemistry.
This lesson supports students’ understanding of
- Equilibrium Constant
- Reaction Quotient
- Activation Energy
- Thermodynamic Favorability
- Energy Diagrams
- Free Energy
Teacher Preparation: 30 minutes (Preparing and practicing the demonstration)
Lesson: Activity: 60-90 minutes
Demo: 30 minutes including discussion
- Demo Materials
- (50 mL per demo)
- MnO2 solid (a few grams per demo–exact amount not required)
- 250 mL volumetric flask
- 50 or 100 mL graduated cylinder (to measure peroxide solution)
- Scoopula or spoon to obtain solid catalyst
- Flinn Disposal #26 is a disposal option. (*If you do not have a Flinn Reference Catalog please order a free copy.)
- Always wear safety goggles when handling chemicals in the lab.
- Students should wash their hands thoroughly before leaving the lab.
- Students should wear proper safety gear during chemistry demonstrations. Safety goggles and lab apron are required.
- 30% Hydrogen Peroxide is a strong oxidizer, please read the safety sheet prior to handling this chemical.
- Wear gloves when using 30% Hydrogen Peroxide.
- This lesson is intended to help students make connections between the following concepts: equilibrium, thermochemistry and kinetics. The students should have knowledge of general equilibrium, kinetics, catalysts, enthalpy and entropy before completing the activity.
- The demonstration should follow the Guided Inquiry Activity as a formative assessment of the students understanding of the lesson.
- Guided Inquiry Activity: Determining Reaction Favorability (student handout provided). Randomly divide students into teams of 2-4. Students should work through this activity together and divide the work among the members as noted in the activity.
- An Answer Key document has been provided for teacher reference.
- After the groups have finished, discuss the answers and be certain all questions have been addressed.
- Demo: Catalyzed decomposition of hydrogen peroxide using manganese dioxide.
- *Note: ALWAYS practice a demonstration prior to attempting in front of students.
- Video of Demonstration for reference.
- Introduce the demo by drawing the potential energy diagram for H 2O2 without a catalyst on the board.
*Graph Created by author
- Ask the students “What is the sign of ∆H? Why?”
- Confirm the student’s choice. Answer: the sign of ∆H is negative, the products are lower in energy than the reactants so energy is released.
- Pour some (50 mL) hydrogen peroxide (30% preferred) in a 250 mL volumetric flask.
- Ask the students “What do you observe?” Answer: students may observe some small bubbles, the solution is clear.
- Ask the students “What is the sign of ∆S? Why?
- Confirm the student’s choice. Answer: the sign of ∆S is positive, the reactants include one aqueous solution and the products contain liquid water and oxygen gas, disorder increases.
- Ask the students “Considering the sign of ∆H and of ∆S, predict the sign of ∆G, why?”
- Confirm the student’s choice. Answer: the sign of ∆H is negative and the sign of ∆S is positive which means the sign of ∆G is negative at all temperatures.
- Ask the students “Considering the sign of ∆G is negative, what does this indicate about the thermodynamic favorability of this reaction?”
- Students should say “thermodynamically favorable”
- Ask the students “If this reaction is thermodynamically favorable, why do we not see a reaction taking place?”
- Students should say things like “it is slow, has a high activation energy, needs a catalyst, etc.
- Introduce the Manganese Dioxide catalyst to the students and ask “what will adding the catalyst do to the graph on the board?
- Confirm the student’s choice by adding the catalyst line on the graph lowering the activation energy.
*Graph Created by Christine Taylor
- Slowly add a small scoop of MnO2 to the H2O2 in the volumetric flask.
- The reaction will quickly occur decomposing the hydrogen peroxide and releasing copious amounts of water vapor, oxygen gas and heat.
- Ask the students “Do you think this reaction is under thermodynamic or kinetic control?”
- Confirm the student’s choice. Answer: Due to the high activation energy and the fact that the reaction is spontaneous but does not visibly occur without a catalyst, the reaction is under kinetic control.