1. Le Chatelier’s Principle
When a chemical system at equilibrium is disturbed by a stress, the system adjusts (shifts) to oppose the change
Stresses include:
Change in concentration
Change in pressure (or volume)
Change in temperature

2. Change in Concentration
A(g) + 3B(g)  2C(g) + heat
Increasing the concentration of the reactants OR
Decreasing the concentration of the products
Will favour the forward reaction, causing the equilibrium to shift to the RIGHT
Decreasing the concentration of the reactants OR
Increasing the concentration of the products
Will favour the reverse reaction, causing the equilibrium to shift to the LEFT
RECALL: Addition or removal of solid or liquids does not change the concentration. Therefore does not cause a shift. I.e. only applies to gases and aqueous solutions.

3. Change in Concentration

4. N2(g) + 3H2(g)  2NH3

5. Change in Pressure
 volume  pressure
 volume  pressure
A(g) + 3B(g)  2C(g) + heat
Increasing the volume of the container OR Decreasing the pressure
Will cause a shift to the side with MORE gas molecules
In our example, it will shift left (4 molreactants > 2 molproducts)
Decreasing the volume of the container OR Increasing the pressure
Will cause a shift to the side with LESS gas molecules
In our example, it will shift right (4 molreactants > 2 molproducts)

7. Change in Temperature In an exothermic reaction:
Increasing the temperature will cause a shift to the LEFT
Decreasing the temperature will cause a shift to the RIGHT

8. Change in Temperature In an endothermic reaction:
Increasing the temperature will cause a shift to the RIGHT
Decreasing the temperature will cause a shift to the LEFT

9. Change in Temperature
Recall: Keq is temperature dependent. Therefore, changes in temperature will also affect Keq
Shift right =  products,  Keq
Shift left =  reactants, Keq

10. DEMONSTATION

11. Variables that do NOT Affect Equilibrium
Catalysts
Increases reaction rate by lowering activation energy (of BOTH the forward and the reverse reactions equally)
Decreases the time required to reach equilibrium but does not affect the final position of equilibrium
Inert Gases
Increases the pressure, which will increase reaction rate
Increases the probability of successful collisions for BOTH products and reactants equally

12. Practice

13. The Reaction Quotient (Q)
If a chemical system begins with reactants only, it is obvious that the reaction will shift right (to form products).
However, if BOTH reactants and products are present initially, how can we tell which direction the reaction will proceed?
Use a trial value called the reaction quotient, Q
When a reaction is NOT at equilibrium
Q=Keq  the system is at equilibrium
Q > Keq  the system shifts towards reactants to reach equilibrium
Q < Keq the system shifts towards products to reach equilibrium

14. Practice #1
(p. 464) In a container at 450°C, N2 and H2 react to produce NH3. K = When the system is analysed, [N2] = 4.0 mol/L, [H2] = 2.0 X 10-2 mol/L, and [NH3] = 2.2 X 10-4 mol/L. Is the system at equilibrium, if not, predict the direction in which the reaction will proceed.

15. Practice #2
In a container, carbon monoxide and water vapour are producing carbon dioxide and hydrogen at 900oC.
CO(g) + H2O(g)  H2(g) + CO2(g) Keq = 4.00 at 900oC
If the concentrations at one point in the reaction are: [CO(g)] = 4.00 mol/L, [H2O(g)] = 2.00 mol/L, [CO2(g)] = 4.00 mol/L, and [H2(g)] = 2.00 mol/L. Determine whether the reaction has reached equilibrium, and, if not, in which direction it will proceed to establish equilibrium.

16. Practice #2 Answer

17. Practice #3 Calculating Equilibrium Concentrations from Initial Concentrations
Carbon monoxide reacts with water vapour to produce carbon dioxide and hydrogen. At 900oC, Keq is Calculate the concentrations of all entities at equilibrium if mol of each entity are initially place in a 1.00L closed container.

18. Practice #4 Calculating Equilibrium Concentrations Involving a Quadratic Equation
If 0.50 mol of N2O4 is placed in a 1.0L closed container at 150oC, what will be the concentrations of N2O4 and NO2 at equilibrium? (Keq = 4.50)

19. Practice #5
Simplifying Assumption: 100 rule (for small K values) If: [reactant] > 100, you can simplify the Keq expression K Ex: 2CO2(g)   2CO(g) + O2 (g) If K = 6.40 x 10-7, determine the concentrations of all substances at equilibrium if it starts with [CO2] = mol/L

20. Solubility Equilibrium
Not all ionic compounds are equally soluble
Ionic compounds dissolve into individual ions
This can be a reversible system
Example: CaCl2(s) Ca2+(aq) Cl-(aq)
Equilibrium can be reached between the solid substance and its dissolved ions (saturation point)
The solution is saturated at equilibrium

21. Solubility Product Constant (Ksp)
An equilibrium equation can be written for solubility reactions
Ex: AgCl (s) Ag+ (aq) + Cl- (aq)
Since AgCl is a solid, the concentration is not changing, so it is “built in” to the K value:
The new constant is the solubility product constant (Ksp)

22. Example
Eg: Lead (II) chloride has a molar solubility of 1.62x10-2mol/L at 25oC. What is the Ksp of this salt?
PbCl2   Pb Cl-
Ksp = [Pb2+][Cl-]2
[Pb2+] = [PbCl2] = 1.62x10-2mol/L
[Cl-] = 2[PbCl2] = 2(1.62x10-2mol/L) = 3.24 x 10-2mol/L
Ksp = [1.62x10-2mol/L][3.24 x 10-2mol/L]2
= 1.7x10-5

23. Example 2 Ksp = [Ag+][Cl-] 1.8x10-10 = [X][X] 1.8x10-10 = X2
The Ksp of silver chloride at 25oC is 1.8x What is the molar solubility of AgCl?
AgCl   Ag Cl-
AgCl  
Ag+
Cl-
Initial -
Change
+ X
Equilibrium X
Ksp = [Ag+][Cl-]
1.8x10-10 = [X][X]
1.8x10-10 = X2
X = 1.34x10-5M

24. The size of Ksp depends on the solubility of the salt.
Large Ksp: [ions] at equilibrium is high, salt is very soluble
Small Ksp: [ions] at equilibrium is low, salt has low solubility
To determine whether a precipitate will form during a reaction, a trial solubility product constant can be determine which is denoted by the symbol Qsp.
Qsp < Ksp : Shifts right to equilibrium – all solid dissolving
Qsp > Ksp : Shifts left to equilibrium – precipitate forms
Qsp = Ksp : Equilibrium (saturated) – no precipitate