**General Equilibrium: A Review.**

**From thermodynamics:**

All chemical process proceed to an equilibrium at which point there exists an equilibrium constant:

K = P a(i)^{m}(i)

(K has no units: activities are defined as ratios without units)

**General Equilibrium: A Review.**

**From thermodynamics:**

All chemical process proceed to an equilibrium at which point there exists an equilibrium constant:

K = P a(i)^{m}(i)

(K has no units: activities are defined as ratios without units)

In practice:

Experimentally, the equilibrium constant is defined only when the activities can be measured. When all components are ideal gases or all components are ideal solutions two other K values can be defined.

**K _{c} is defined as P [i]^{m}(i) M^{Sm}(i)**

(note that K_{c} has units which depend on the process.)

**K _{p} is defined as P P(i)^{m}(i) kPa^{Sm}(i)**

(note that K_{p} has units which depend on the process and on the pressure units chosen.)

**These K values must be converted to the activities K for use in the equation: DGº _{T}
= -RT ln K.**

Units of K values.

For the process

6 H_{2}O(g) + 4 NO_{2}(g) ¾ 7 O_{2}(g) + 4 NH_{3}(g)

define K, K_{p}, and K_{c}.

Interconversion of K_{p} and K_{c} values.

If __all components are ideal gases__, then there is a relationship between K_{p} and K_{c}:

K_{p} = P P(i)^{m}(i) kPa^{Sm}(i)

For an ideal gas: PV = nRT

or

P(i) = {n/V}RT = [i]RT

Substituting:

K_{p} = P {[i]RT}^{Sm}(i) M^{Sm}(i)

= K_{c} {RT}^{Sm}(i) M^{Sm}(i)

Note: when Sm(i) = 0, all K values are identical, and there are no units for any of them.

Example: for the above reaction:

6 H_{2}O(g) + 4 NO_{2}(g) ¾ 7 O_{2}(g) + 4 NH_{3}(g)

K_{p} = K_{c} {RT}^{Sm}(i) kPa^{Sm}(i)

K_{p} = K_{c}*{RT}^{+1} kPa

Manipulating Equilibrium Constants.

In thermodynamics, we have learned that whatever you do to a chemical equation, you do the same to the intensive thermodynamic parameters.

Reverse an equation | - | negate the thermodynamic parameter |

Double an equation | - | double the thermodynamic parameter. |

Add two equations | - | add thermodynamic parameters. |

Thermodynamic parameters for which these rules apply:

DE, DH, DS, DG.

For equilibrium purposes, simply remember that the thermodynamic parameter is ln K which is related to DG.

If an equation is reversed, ln K is negated, K is inverted.

If an equation is doubled, ln K is doubled, K is squared.

If equations are added, ln K values are added, the K values are multiplied.

Examples:

At a certain temperature,

for N_{2}(g) + 2 O_{2}(g) ¾ 2 NO_{2}(g) K = 1 x 10^{-5}

and for 2 NO_{2}(g) ¾ N_{2}O_{4}(g) K = 1 x 10^{4}

Calculate K for the following:

- ½ N
_{2}(g) + O_{2}(g) ¾ NO_{2}(g) - N
_{2}O_{4}(g) ¾ 2 NO_{2}(g) - NO
_{2}(g) ¾ ½ N_{2}(g) + O_{2}(g) - N
_{2}O_{4}(g) ¾ N_{2}(g) + 2 O_{2}(g)

Answers:

a 3 x 10^{-3}

b 1 x 10^{-4}

c 3 x 10^{3}

d 10

Equilibrium Problem Solving - Review.

Problem solving involves the five step process:

Analyze the problem.

Brainstorm for a solution.

Calculate the answer.

Defend and present your answer.

Evaluate the process used.

(Dave has a handout on problem solving if you would like one. Or you could visit my web site:

http://www.sci.ouc.bc.ca/chem/probsol/ps_intro.html.)

In stoichiometry and many other aspects of chemistry, routine analysis of a problem involves writing a balanced equation and analyzing the data given on the basis of the equation.

In equilibrium problems, the same applies: write a balanced equation and analyze the data given in the problem using the I.C.E. method:

1. Balanced equation.

2. Initial conditions.

3. Change occurring to move to equilibrium, use m_{i}x.

4. Equilibrium conditions.

5. Write down a suitable expression for K.

6. Substitute any values from the ICE into the expression for K.

Example:

For the process

H_{2}(g) + I_{2}(g) ¾ 2 HI(g),

K_{c} is 46.0. If the initial concentrations of H_{2} and I_{2} are equal at 0.120 M, calculate the
concentrations of all species at equilibrium.

**Analysis step:**

1 Equation | H_{2}(g) |
+ I_{2}(g) |
¾ | 2 HI(g) |

2 Initial (M) | 0.120 | 0.120 | 0 | |

3 Change (m_{i}x) |
(-1)x | (-1)x | (+2)x | |

4 Equilibrium | 0.12-x | 0.12-x | 2x |

5 K_{c} = [HI]^{2} / [H_{2}]*[I_{2}]

6 46.0 = (2x)^{2} / (0.12-x)(0.12-x) = { 2x / (0.12-x)}^{2}

**Brainstorm step:**

Unknown are equilibrium concentrations: need to find x.

**Calculation step:**

x = 0.0927

**Defend and present answer step:**

At equilibrium, [HI] = 0.185 M, [H2] = [I2] = 0.0273 M

(Check the answer by substituting x back into the expression for K_{c})

Example.

For the reaction

N_{2}O_{4}(g) ¾ 2 NO_{2}(g),

at 416 K the equilibrium constant, K_{p}, is 9850 kPa.

If the initial pressure of N_{2}O_{4}(g) is 177 kPa, and that of NO_{2}(g) is 280 kPa, find the
pressure of N_{2}O_{4}(g) at equilibrium.

**Problem Solving Step A: Analysis:**

1 Equation | N_{2}O_{4} |
¾ | 2 NO_{2} |

2 Initial (kPa) | 177 | 280 | |

3 Change (m_{i}x) |
(-1)x | (+2)x | |

4 Equilibrium | 177 - x | 280 + 2x |

5 K_{p} = P(NO_{2})^{2} / P(N_{2}O_{4})

6 9850 = (280 + 2x)^{2} / (177 - x)

**Problem Solving Step B: Brainstorm for a solution.**

P(N_{2}O_{4}) at equilibrium is required. This is (177 - x) kPa. We need to find x.

This can be done by solving the quadratic.

**Problem Solving Step C: Calculate.**

x = 144.2

**Problem Solving Step D: Defend and present the solution.**

The pressure of N_{2}O_{4}(g) at equilibrium is 32.8 kPa

Example:

At a certain temperature K_{p} = 46.0 kPa for the process

PCl_{5}(g) ¾ PCl_{3}(g) + Cl_{2}(g).

What initial pressure of PCl_{5} must be placed in a reaction vessel if the final total
pressure is to be 100 kPa?

**Analysis:**

1 Equation | PCl_{5}(g) |
¾ | PCl_{3}(g) |
+ Cl_{2}(g). |

2 Initial (kPa ) | x | 0 | 0 | |

3 Change | (-1)y | (+1)y | (+1)y | |

4 Equilibrium | x - y | y | y |

5 Kp = P(PCl_{3}) * P(Cl_{2}) / P(PCl_{5})

6 46.0 = y * y / (x - y)

**Brainstorm:**

Unknown required is x, the initial pressure of PCl_{5}. To find this we need to solve the
quadratic.....but that involves two unknowns (x and y), so we need a second equation.

Total pressure = 100 kPa, so (x - y) + y + y = 100

100 = x + y

This gives us a second equation and the problem is now solvable.

**Calculate:**

substitute y = 100 - x

46 = (100 - x)^{2} / (x - (100 - x))

46 = (100 - x)^{2} / (2x - 100)

Remember the quadratic formula:

Solving x = 228 or x = 64.0

Note that the physical limits of this problem (100 = x + y), preclude x > 100 (since y cannot be negative!)

Thus x = 64.0

**Defend and present the answer:**

Answer:

the initial pressure of PCl_{5} in the container was 64.0 kPa.

The Le Châtellier Principle

When a reaction at equilibrium is disturbed, the reaction will move in the direction that will minimize the disturbance.

A disturbance is defined as a change in the activity of at least one component in the reaction.

(If no activities are changed, then no disturbance occurs.)

Example:

For the reaction:

6 H_{2}O(g) + 4 NO_{2}(g) ¾ 7 O_{2}(g) + 4 NH_{3}(g)

(endothermic)

fill in the following table:

Change | Equilibrium movement | Change in Equilibrium Constant, K |

Addition of O_{2}(g) |
||

Removal of NO_{2}(g) |
||

Increased pressure | ||

Increased temperature | ||

Add He (inert) at constant P | ||

Add catalyst |

Example:

For the process:

H_{2}(g) + I_{2}(g) ¾ 2 HI(g), (exothermic)

fill in the following table:

Change | Equilibrium movement | Change in Equilibrium Constant, K |

Addition of H_{2}(g) |
||

Removal of HI(g) | ||

Increased volume | ||

Increased temperature | ||

Add He (inert) at constant P | ||

Add catalyst |

Example :

For the process:

CaCO_{3}(s) ¾ CaO(s) + CO_{2}(g) (endothermic)

fill in the following table:

Change | Equilibrium movement | Change in Equilibrium Constant, K |

Addition of CaCO_{3}(s) |
||

Removal of CO_{2}(g) |
||

Increased pressure | ||

Increased temperature | ||

Add He (inert) at constant volume | ||

Add catalyst |