# Chapter 14 Lecture Outline

## The Concept of Dynamic Equilibrium

1. Introduce Idea of equilibrium and balance in a chemical reaction.
2. Demonstrate establishing equilibrium with sharing M&M's.
1. The Setup. Divide up into teams. Each group (2 teams) has 100 M&M's. Group A starts out with all of them. Every 10 seconds you each team gives up half of what they have. Make a table as follows.

 Initial Move Final Time Team A Team B Team A Team B Team A Team B 0:00 100 0 50 0 50 50 0:30 50 50 25 25 50 50

3. Work problems in groups. Assign by row.
4.  Initial Rate Final K Team A Team B Team A Team B Team A Team B 100 0 0.1 0.1 0 100 0.1 0.1 80 20 0.1 0.1 100 0 0.1 0.2 0 100 0.1 0.2 100 0 0.2 0.1

5. Introduce equilibrium constant (Keq)
1. Above reaction as Team A <--> Team B
2. K=products over reactants
3. Calculate K for example

6. Use Spreadsheet to get more data
1. Change starting numbers
2. Change rate (up and down still same)
3. Change ratio of up/down

## The Equilibrium Constant (K)

1. Equlibrium data from textbook for the reaction: N2O4 --> 2 NO2

 N2O4 2 NO2 Initial 0.10 M 0.00 M Change -0.06 M +0.12 M Final 0.04 M 0.12 M

2. Repeat with different starting conditions

 Final N2O4 Final NO2 Ratio K 0.040 0.120 3.0 0.36 0.014 0.072 5.1 0.37 0.160 0.243 1.5 0.37

1. Calculate Ratio (Not identical)
2. Discuss idea of how two NO2 react together so depends upon [NO2]2.
3. Calculate K

3. Introduce general equation for equilibrium constant (K)

For a reaction aA + bB <--> cC + dD

4. So for N2O4 reaction K = .....

## Heterogeneous Equilibrium: Reactions Involving Solids and Liquids

1. For the reaction PCl5(s) <-> PCl3(l) + Cl2 (g)
1. Draw a picture
2. Determine expression for K
3. Concept of solids or pure liquid
4. Equilibrium expression for heterogeneous systems

2. For the reaction CH3COOH + H2O <-> CH3COO1- + H3O1+
1. Draw a picture
2. Equilibrium expression
3. [H2O]
4. Modify equilibrium expression

3. Work a problem:
1. For the reaction CaCO3(s) (limestone) <-> CaO(s) (lime) + CO2(g)
2. At 1173 K, Kp = 1.04 atm
3. Start with 250 g of CaCO3(s) and a 20 liter container
4. Start with 250 g of CaCO3(s) and a 400 liter container
5. Solutions

## Calculating the Equilibrium Constant from Measured Equilibrium Concentrations

1. Multiple Equlibria. Reactions add, K's multiply.
1. Reaction #1: N2(g) + O2(g) <-> 2 NO(g)
2. Reaction #2: 2 NO(g) + O2(g) <-> 2 NO2(g)
3. Total Rxn: N2(g) + 2 O2(g) <-> 2 NO2(g)
4. Write expression for K for each reaction.
5. Show Ktotal = K1 * K2

2. Stoichiometry (Write K expression for each):
1. NO(g) + 1/2 O2(g) <-> NO2(g)
2. 2 NO(g) + O2(g) <-> 2 NO2(g)
3. K2=K12

3. Reaction Quotient Q
1. Q expression (just like K)
2. If Q<K
3. If Q>K

## Equlibrium Video Clips

1. Reversible Reactions (internet © Saunders, 1997)

2. Product Favored Reaction (internet © Saunders, 1997)

3. Reactant Favored Reaction (internet © Saunders, 1997)

## The Reaction Quotient: Predicting the Direction of Change

1. Ammonia Equlibrium. For the reaction N2 (g) + 3 H2 (g) <--> 2 NH3 (g)
1. Calculate Kc
1. N2 = 0.921 M
2. H2 = 0.763 M
3. NH3 = 0.157 M

2. Calculate Q if:
3.  N2 H2 NH3 1.831 0.763 0.157 10.321 0.763 0.157 0.921 0.0363 0.157 0.0259 0.0277 0.0102

4. Solutions

## Finding Equilibrium Concentrations

1. Kp and Kc
1. Determining Kp, given Kc = 0.06025 liter2 mole-2 for the reaction

N2(g) + 3 H2(g) <-> 2 NH3(g)

2. Solve for the equlibrium pressure of NH3, given equlibrium conditions at 25 °C where PN2 = 12 atm and PH2 = 35 atm for the reaction:

N2(g) + 3 H2(g) <-> 2 NH3(g)

3. Solutions

2. PCl3
1. Given INITIAL NON-EQULIBRIUM conditions, calculate the final EQULIBIRUM conditions. In solving this type of reaction the quadratic equation is often required. The example below shows one way to work this type of problem. For the following BALANCED reaction:

PCl3 (g) + Cl2 (g) <-> PCl5 (g)
At 230 C, Kc = 49 liter mole-1

Initial conditions
mass PCl3 = 50 gm
mass Cl2 = 50 gm
mass PCl5 = 0 gm
volume = 5 liter
2. Solutions

3. NO Equlibrium. Calculate the equlibrium pressure of NO at 300, 600, and 1500 K. For the reaction:

N2 + O2 <-> 2 NO

1. Given starting pressures (typical atmospheric concentrations) of:
1. PN2 = 0.80 atm
2. PO2 = 0.20 atm
3. P NO = 0.00 atm
2. And equlibrium constants
1. K = 10-30.5 at 300 K
2. K = 10-14.6 at 600 K
3. K = 10-5.1 at 1500 K
3. solutions

4. For the system: H2O (g) + Cl2O (g) <-> 2 HOCl (g)
1. Given the equlibrium partial pressures solve for Kp
1. PH2O = 200.0 torr
2. PCl2O = 48.7 torr
3. PHOCl = 29.6 torr

2. Given equlibrium partial pressures of reactants, find product
1. PH2O = 23.4 torr
2. PCl2O = 123.5 torr

3. Given initial (non-equlibrium) pressures, find equlibrium conditions
1. PH2O = 100.0 torr
2. PCl2O = 50.0 torr
3. PHOCl = 0.0 torr

4. Solutions

## Le Chatelier's Principle: How a System at Equilibrium Responds to Disturbances

3. Changing temperature (Exothermic Reaction)
4. Changing temperature (Endothermic Reaction)

5. Animations
2. Remove reactant (internet © Saunders, 1997)
4. Remove product (internet © Saunders, 1997)
5. Changing Volume (internet © Saunders, 1997)

## Introduction to Acid Base Equlibrium Problems

1. For the following system:

1. The base reaction: NH3 + H2O <-> NH4+ + OH-
2. Conjugate acid reaction: NH4+ + H2O <-> NH3 + H3O+
3. Add the two reactions together 2 H2O <-> OH- + H3O+

2. At equlibrium
1. [NH3] = 0.897 M
2. [NH4+ = 0.345 M

3. Equlibrium Constants
1. Ka = 5.6 x 10-10
2. Kw = 1.0 x 10-14
3. Kb = ?

4. Find the equlibrium concentrations of:
1. H3O+
2. OH-

5. Solutions