2. Le Chatelier’s Principle

3. The 4 most commons changes to make for equilibrium reactions are: 1. Concentration changes for reactants 2. Concentration changes for products 3. Temperature changes for reaction 4. Volume/Pressure changes for gaseous reactions Ways to disturb equilibrium:

4.  Concentration changes for Reactants and Products

5. Think of equilibrium as a big seesaw. To the eye, no changes are occurring to the amount of reactants on the left or to the amount of product on the right. At equilibrium, the seesaw is balanced. A + BC + D

6. When you increase the concentration of a reactant A, you are adding weight to the left side of the seesaw. A + B C + D How can you re-balance the seesaw? How can you achieve equilibrium again?

7. By shifting some of the weight toward the right! A + B C + D A + B C + D Equilibrium has been re-established when concentrations stop changing.

8. Before A was added, the system was at equilibrium. A + BC + D

9. At the moment that A was added, the [A] went up. In our example, increased [A] is symbolized as more weight on the left side of the seesaw. A + B C + D

10. In the process of re-establishing equilibrium, the concentration of C and D went up. In our example, increased concentration is symbolized as more weight on the right side of the seesaw. A + B C + D A + B C + D

11. Let’s look at the overall process one more time. A + BC + D System was at equilibrium. A + B C + D [A] increased. System not at equilibrium. A + BC + D System regains equilibrium.

12. Conc. Time Initial Concentrations First Equilibrium [A] up. New Equilibrium [C] & [D] go up

13. What would the seesaw look like if we increased the [D]? A + B C + D In which direction do we need to shift “weight” in order to regain equilibrium? Shift to the LEFT!

14. A + B C + D A + B C + D [C] and [D] increase. [A] and [B] increase. Caused by increasing [A]. Shift RIGHT to regain equilibrium. Result: Caused by increasing [D]. Shift LEFT to regain equilibrium. Result:

15. What if we removed D as it was formed? This would be the same as decreasing [D]. What would the seesaw look like? A + B C + D When you decrease the [D], you are removing “weight” from that side of the seesaw.

16. How would you re-establish equilibrium? A + B C + D Shift “weight” to the right. More products will form. This is a common way to make an equilibrium reaction go to “completion.”

17. Remember this all ties into Keq. Recall if: Keq>1 are favored Keq<1 are favored products reactants

18.  Change in Temperature

19. What happens when you change the temperature of a reaction? It will depend on whether the reaction is exothermic or endothermic. ExothermicA + B C + D + heat reaction: EndothermicA + B + heat C + D reaction:

20. Treat “heat” like a reactant or product. If you increase the heat, you are adding “weight” to the seesaw. If you decrease the heat, you are removing “weight from the seesaw. If the reaction is exothermic, the change in “weight” occurs on the product side. If the reaction is endothermic, the change in “weight” occurs on the reactant side.

21. Exothermic Reaction with increased temperature. ( w ant to produce LESS heat) A + B C + D + Heat Exothermic Reaction with decreased temperature. (want to produce MORE heat) A + B C + D + heat Shift Left! More Reactants Shift Right! More Products

22. Endothermic Reaction with increased temperature. (need less heat) A + B + heat C + D Endothermic Reaction with decreased temperature. (need more heat) A + B + heat C + D Shift Right! More Products Shift Left! More Reactants

23.  Changing Volume/Pressure

24. When you have gaseous reactants or products and you change volume, you are changing concentration: PV = nRT Pressure and volume are inversely related. For gases, we use K p and measure pressure changes.

25. Let’s rearrange the ideal gas law to show concentration: P = n RT V n/V is concentration. Pressure is directly related to concentration and inversely related to volume.

26. Let’s look at a reaction with gaseous components: 2A(g) + B(g) 2C(g) There are THREE moles of gas on the reactant side and TWO moles of gas on the product side. 3 moles2 moles

27. If we cut the volume in half, the pressure will double. This means that the concentration of ALL gases went up. The side of the reaction with the most moles of gas, will be most disturbed by the increased concentration. 3 moles 2 moles

28. If we shift the reaction toward the side with fewer moles of gas, the effect of cutting the volume in half will be minimized. 3 moles 2 moles For this reaction, cutting the volume in half results in MORE product. 3 moles2 moles

29. Is it always true that cutting the volume in half will cause more products to form? NO! You have to examine each reaction with gaseous components to see, first, which side has more moles of gas. Cutting the volume in half, increases pressure. Reaction will shift toward side with FEWER moles of gas. Doubling the volume, decreases pressure. Reaction will shift toward side with MOST moles of gas.

30. If volume is increased, which direction will reaction shift? If volume increases, pressure decreases. Shift to the side with the MOST moles of gas. Reaction will shift to the right. 3A(s) + B(g) 2C(g) 1 mole of gas2 moles of gas 1 mole 2 moles

31. Let’s see if you can put it all together !!!

32. How does a system at equilibrium respond to a stress? What factors are considered to be stresses on an equilibrium system? Changes in concentration, pressure (volume), and temperature. If possible, the equilibrium shifts in the direction that relieves the stress.

33. 2 A(g) + B(g) C(g)  H = + 200 kJ 1. In which direction will reaction shift if [A] is doubled? Increase [A], shift right. More products formed. 2. In which direction will reaction shift if [C] is removed? Decrease [C], shift right. More products formed. 3. In which direction will reaction shift if temp goes up? Endothermic reaction. Heat is a reactant. Shift right. More products formed. 4. In which direction will reaction shift if volume goes up? Increase V, decrease P. Side with most gas moles loses “weight.” Shift left. More reactants formed. First, note that the reaction is endothermic. Second, note that the reactant side has more moles of gas.