# Chapter 14 Lecture Problems

## Introduction to Equlibrium

M&M's Spreadsheet

## Heterogenous Equlibrium

- 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

## 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:
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 |

- Solutions

## K_{p} and K_{c}

- Determining K
_{p}, given K_{c} = 0.06025 liter^{2} mole^{-2} for the reaction

N_{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

## 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_{3} (*g*) + Cl_{2} (*g*) <-> PCl_{5} (*g*)
At 230 C, K_{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

## 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

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

- solutions

## HOCl Equlibrium

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

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

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

- Solutions

## Acid Base Equlibrium

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

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

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

- Solutions

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Widener University

Chester, PA 19013
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Last Updated Friday, May 25, 2001 2:10:36 PM