- Introduce Idea of equilibrium and balance in a chemical reaction.
- Demonstrate establishing equilibrium with sharing M&M's.
- 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 - Work problems in groups. Assign by row.
- Introduce equilibrium constant (K
_{eq}) - Above reaction as Team A <--> Team B
- K=products over reactants
- Calculate K for example
- Use Spreadsheet to get more data
- Change starting numbers
- Change rate (up and down still same)
- Change ratio of up/down

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 |

- Equlibrium data from textbook for the reaction: N
_{2}O_{4}--> 2 NO_{2}N _{2}O_{4}2 NO _{2}Initial 0.10 M 0.00 M Change -0.06 M +0.12 M Final 0.04 M 0.12 M - Repeat with different starting conditions
Final N _{2}O_{4}Final NO _{2}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 - Calculate Ratio (Not identical)
- Discuss idea of how two NO
_{2}react together so depends upon [NO_{2}]^{2}. - Calculate K

- Introduce general equation for equilibrium constant (K)
For a reaction aA + bB <--> cC + dD

So for N_{2}O_{4} reaction K = .....

- For the reaction PCl
_{5}(s) <-> PCl_{3}(l) + Cl_{2}(g) - Draw a picture
- Determine expression for K
- Concept of solids or pure liquid
- Equilibrium expression for heterogeneous systems
- For the reaction CH
_{3}COOH + H_{2}O <-> CH_{3}COO^{1-}+ H_{3}O^{1+} - Draw a picture
- Equilibrium expression
- [H
_{2}O] - Modify equilibrium expression
- Work a problem:
- For the reaction CaCO
_{3}(s) (limestone) <-> CaO(s) (lime) + CO_{2}(g) - At 1173 K, K
_{p}= 1.04 atm - Start with 250 g of CaCO
_{3}(s) and a 20 liter container - Start with 250 g of CaCO
_{3}(s) and a 400 liter container - Solutions

- Multiple Equlibria. Reactions add, K's multiply.
- Reaction #1: N
_{2}(g) + O_{2}(g) <-> 2 NO(g) - Reaction #2: 2 NO(g) + O
_{2}(g) <-> 2 NO_{2}(g) - Total Rxn: N
_{2}(g) + 2 O_{2}(g) <-> 2 NO_{2}(g) - Write expression for K for each reaction.
- Show K
_{total}= K_{1}* K_{2} - Stoichiometry (Write K expression for each):
- NO(g) + 1/2 O
_{2}(g) <-> NO_{2}(g) - 2 NO(g) + O
_{2}(g) <-> 2 NO_{2}(g) - K
_{2}=K_{1}^{2} - Reaction Quotient Q
- Q expression (just like K)
- If Q<K
- If Q>K

- Reversible Reactions (internet © Saunders, 1997)
- Product Favored Reaction (internet © Saunders, 1997)
- Reactant Favored Reaction (internet © Saunders, 1997)

- Ammonia Equlibrium. For the reaction N
_{2}(*g*) + 3 H_{2}(*g*) <--> 2 NH_{3}(*g*) - Calculate K
_{c} - N
_{2}= 0.921 M - H
_{2}= 0.763 M - NH
_{3}= 0.157 M - Calculate Q if:
- Solutions

N_{2} | H_{2} | NH_{3} |

1.831 | 0.763 | 0.157 |

10.321 | 0.763 | 0.157 |

0.921 | 0.0363 | 0.157 |

0.0259 | 0.0277 | 0.0102 |

- K
_{p}and K_{c}- Determining K
_{p}, given K_{c}= 0.06025 liter^{2}mole^{-2}for the reactionN _{2}(*g*) + 3 H_{2}(*g*) <-> 2 NH_{3}(*g*) - Solve for the equlibrium pressure of NH
_{3}, given equlibrium conditions at 25 °C where P_{N2}= 12 atm and P_{H2}= 35 atm for the reaction:N _{2}(*g*) + 3 H_{2}(*g*) <-> 2 NH_{3}(*g*) - Solutions

- Determining K
- PCl
_{3}- 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:
PCl At 230 C, K_{3}(*g*) + Cl_{2}(*g*) <-> PCl_{5}(*g*)_{c}= 49 liter mole^{-1}- Initial conditions
- mass PCl
_{3}= 50 gm

mass Cl_{2}= 50 gm

mass PCl_{5}= 0 gm

volume = 5 liter

- Solutions

- 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:
- NO Equlibrium. Calculate the equlibrium pressure of NO at 300, 600, and 1500 K. For the reaction:
N _{2}+ O_{2}<-> 2 NO- Given starting pressures (typical atmospheric concentrations) of:
- P
_{N2}= 0.80 atm - P
_{O2}= 0.20 atm - P
_{NO}= 0.00 atm

- P
- And equlibrium constants
- K = 10
^{-30.5}at 300 K - K = 10
^{-14.6}at 600 K - K = 10
^{-5.1}at 1500 K

- K = 10
- solutions

- Given starting pressures (typical atmospheric concentrations) of:
- For the system: H
_{2}O (*g*) + Cl_{2}O (*g*) <-> 2 HOCl (*g*)- Given the equlibrium partial pressures solve for K
_{p}- P
_{H2O}= 200.0 torr - P
_{Cl2O}= 48.7 torr - P
_{HOCl}= 29.6 torr

- P
- Given equlibrium partial pressures of reactants, find product
- P
_{H2O}= 23.4 torr - P
_{Cl2O}= 123.5 torr

- P
- Given initial (non-equlibrium) pressures, find equlibrium conditions
- P
_{H2O}= 100.0 torr - P
_{Cl2O}= 50.0 torr - P
_{HOCl}= 0.0 torr

- P
- Solutions

- Given the equlibrium partial pressures solve for K

- Adding reactant
- Adding product
- Changing temperature (Exothermic Reaction)
- Changing temperature (Endothermic Reaction)
- Animations

- For the following system:
- The base reaction: NH
_{3}+ H_{2}O <-> NH_{4}^{+}+ OH^{-} - Conjugate acid reaction: NH
_{4}^{+}+ H_{2}O <-> NH_{3}+ H_{3}O^{+} - Add the two reactions together 2 H
_{2}O <-> OH^{-}+ H_{3}O^{+}

- The base reaction: NH
- At equlibrium
- [NH
_{3}] = 0.897 M - [NH
_{4}^{+}= 0.345 M

- [NH
- Equlibrium Constants
- K
_{a}= 5.6 x 10^{-10} - K
_{w}= 1.0 x 10^{-14} - K
_{b}= ?

- K
- Find the equlibrium concentrations of:
- H
_{3}O^{+} - OH
^{-}

- H
- Solutions

This page is maintained by

Scott Van Bramer

Department of Chemistry

Widener University

Chester, PA 19013

Please send any comments, corrections, or suggestions to svanbram@science.widener.edu.

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Last Updated Friday, May 25, 2001 1:59:43 PM