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Le Chatelier’s Principle

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. Change in Concentration.

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Le Chatelier’s Principle

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  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)

  6. 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

  7. 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

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

  9. DEMONSTATION

  10. 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 • Decreases the time required to reach equilibrium but does not affect the final position of equilibrium

  11. Practice

  12. 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 < Keqthe system shifts towards products to reach equilibrium

  13. Practice #1 (p. 464) In a container at 450°C, N2 and H2 react to produce NH3. K = 0.064. When the system is analysed, [N2] = 4.0 mol/L, [H2] = 2.0 X 10-2mol/L, and [NH3] = 2.2 X 10-4mol/L. Is the system at equilibrium, if not, predict the direction in which the reaction will proceed.

  14. 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.

  15. Practice #2 Answer

  16. Practice #3Calculating Equilibrium Concentrations from Initial Concentrations • Carbon monoxide reacts with water vapour to produce carbon dioxide and hydrogen. At 900oC, Keq is 4.200. Calculate the concentrations of all entities at equilibrium if 4.000 mol of each entity are initially place in a 1.00L closed container.

  17. Practice #4Calculating Equilibrium Concentrations Involving a Quadratic Equation • If 0.50 mol of N2O4is placed in a 1.0L closed container at 150oC, what will be the concentrations of N2O4and NO2 at equilibrium? (Keq = 4.50)

  18. 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] = 0.250 mol/L

  19. 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) + 2Cl-(aq) • Equilibrium can be reached between the solid substance and its dissolved ions (saturation point) • The solution is saturated at equilibrium

  20. 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)

  21. Example • Eg: Lead (II) chloride has a molar solubility of 1.62x10-2mol/L at 25oC. What is the Ksp of this salt? PbCl2 Pb2+ + 2Cl- 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

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

  23. 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

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