1. Slide 2: corrected spelling of LeChatelier Slide 32: reformatted Slide 33: reformatted Slide 40: what is this??

2. Intersection 9: Equilibrium 10/31/06 Reading: 14.1-14.2 (p671-679); 14.3-14.5 (p682- 694); 14.6 Changing Concentrations of Reactants or Product (p694-696); Changing Temperature (p698-701.)

3. Outline Equilibrium defined Equilibrium constant ICE Shifting equilibrium –Equilibrium Law –LeChâtelier’s principle Practice problems

4. Reactions that Don’t Go to Completion Generally, we assume that reactions “go to completion”…as much of the reactants are used up as possible; there may be a limiting reagent and thus reactants may be left over We assume that the reaction can only go in the forward direction.

5. Question 1 2L of a 0.1M solution of magnesium chloride and 1 L of a 0.5 M solution of silver nitrate are mixed together. –Write out the balanced net ionic equation –What is your limiting reagent? –How many grams of silver chloride will you make?

6. Equilibrium What does it mean? That a reaction is going forward and backward at the same rate. (All reactants and products are present and actively interchanging)

7. What's a rate of reaction? For a simple reactions A B, rate = k[A]. Most reactions slow down as they proceed and as the concentration/s of the reactant/s decreases (the rate approaches zero.)

8. Forward and Reverse Rates Reactions in equilibrium have both a significant forward and reverse rate of reaction. A  B When equilibrium is reached, the rates of the forward and reverse reaction are equal AND are NOT equal to 0. Forward reaction: A → B Reverse reaction: B → A

9. An Equilibrium Model http://www.chm.davidson.edu/ronutt/che115/ EquKin/EquKin.htm 2A  B B  2A When has the reaction reached equilibrium? (How can you tell?) Is the forward reaction (2A  B) still taking place? Is the reverse reaction (B  2A) still taking place?

11. Equilibrium Constant A  B Forward Rate = k fwd [A] B  A Reverse Rate = k rev [B] At equilibrium we have the following equality: k fwd [A] = k rev [B] forward rate = reverse rate Rearranging this equation yields: K eq = k fwd /k rev = [B]/[A] K eq is the equilibrium constant

12. What does K eq tell you? A  B K eq = For high values of K eq, do you expect there to be a higher concentration of products or reactants at equilibrium?

13. K eq for more complex Reactions

14. Writing Equilibrium Constants 1) NO(g) + Cl 2 (g) ↔ NOCl(g) First, balance the equation. In any equilibrium expression, the concentration of a pure liquid (e.g water) or pure solid is considered a constant. 2) H 2 (g) + I 2 (g) ↔ HI(g) 3) CaCO 3 (s) ↔ CaO(s) + CO 2 (g)

16. Determining K eq 2NO 2 (g) ↔ N 2 O 4 (g) ΔH = -24.02 KJ/mol (red-brown) (colorless) Suppose that 0.55 moles of NO 2 are placed in an empty 5.00 L flask which is subsequently heated to 407 K. By measuring the intensity of the color of red-brown NO 2, it can be determined that its concentration at equilibrium is 0.10 mol/L. What is K eq at 407 K? How would you determine the equilibrium constant? Is this enough information to solve the problem?

17. Put the Reaction on ICE 2NO 2 (g) ↔ N 2 O 4 (g) Initial Concentration (M) Change in Concentration (M) Equilibrium Concentration (M) K eq = [N 2 O 4 ]/[NO 2 ] 2 = 0.5 0.55 moles/5 L 0.10 mol/L - 2x + x 0 moles/L.11 mol/L – 2x = 0.10 mol/L x = 0.005 mol/L 0.005 mol/L

18. Temperature Does temperature matter? 2NO 2 (g) ↔ N 2 O 4 (g) ΔH = -24.02 KJ/mol (red-brown) (colorless) K eq (407K) = 0.5

19. ICE in action What if you want to know equilibrium concentrations? H 2 (g) + I 2 (g) ↔ 2 HI (g) K eq = 2.5 x 10 1 Two moles of hydrogen and 2 moles of iodine are added to a 4 L container; what are the concentrations of all reactants and products at equilibrium? You know how to find K eq. (What do you need?) In the future, you can look up the equilibrium constants in Table 14.1 p685 as well as the Appendices) *

20. H 2(g) I 2(g) ↔ 2 HI (g) Initial2mol / 4L Change Equilibrium K eq = 25

21. Equilibrium Constant Family K eq -a generic equilibrium constant K c -an equilibrium constant calculated using equilibrium concentrations in M (mol/L) K p - associated with gaseous equilibria; found using equilibrium pressures (atm) Pressure is directly proportional to concentration (PV = nRT or P = (n/V)RT).

22. Making Ammonia Desired for fertilizing (belief that without ammonia, wouldn’t be able to feed world.) N 2(g) + 3 H 2(g) ↔ 2 NH 3(g) K c = 3.5 x10 8 (25 o C) –What is the K c if the reaction were written for the production of 1 mole of ammonia? –1/2 N 2(g) + 3/2 H 2(g) ↔ NH 3(g) –If 10 moles of nitrogen and 10 moles of hydrogen are placed in a 1 L flask, how many moles of ammonia can you make? How many moles of starting material would be left over?

24. Disturbing Equilibrium Sir Isaac Newton claimed that a ball at rest would remain at rest unless disturbed. You might be tempted to apply this logic to equilibrium and get: A reaction at equilibrium will remain at equilibrium unless disturbed; consequently, the reaction will shift so as to come back to equilibrium.

25. Can the Equilibrium Constant be Changed? 2NO 2 (g) ↔ N 2 O 4 (g) ΔH = -24.02 KJ/mol (red-brown) (colorless)

26. Evaluating Changes in Equilibrium Method 1: LeChâtelier's Principle if a system at equilibrium is disturbed or stressed by a change in temperature, pressure or concentration of one of the components, it will shift its equilibrium so as to oppose the stress. Does this method help explain the demonstrations that you just saw?

27. Can Equilibrium be Changed? Fe(NO 3 ) 3 + KSCN ↔ Fe(SCN) +2 + KNO 3  H < 0 red Use LeChâtelier's Principle to predict what you will see:

28. Evaluating Changes in Equiliibriu Method 2: Equilibrium Law (Q) K eq is used at equilibrium to represent the ratio of reactants to products for a give reaction. K eq = Q, the reaction quotient, is used for this ratio under any conditions at any point in time, not just equilibrium. Q = At equilibrium, K eq and Q are EQUAL. According to the equilibrium law, the system will proceed to bring Q and K eq equal to each other. aA + bB ↔ pP + qQ

29. Applying the Equilibrium Law What is the equilibrium expression for this reaction? K eq was determined to be 6.42x10 -5 at 25 o C. At equilibrium is this reaction product favored or reactant favored? H 2 O (l) + C 6 H 5 CO 2 H ↔ C 6 H 5 CO 2 - + H 3 O + (aq)

30. 2.00 moles of C 6 H 5 CO 2 -, 1.00 moles of H 3 O + and 3.00 moles of C 6 H 5 CO 2 H are placed in 1 liter of water. What is Q under these conditions? K eq = 6.42 x10 -5 Will the reaction proceed to form more C 6 H 5 CO 2 - (aq) and H 3 O + (aq) or more C 6 H 5 CO 2 H(aq)? H 2 O (l) + C 6 H 5 CO 2 H ↔ C 6 H 5 CO 2 - + H 3 O + (aq)

31. Q vs. K eq In general, how would the reaction proceed to result in a decreased Q? What if an increased Q were the desired result?

32. Q vs LeChâtelier One instance where Le Châtelier's principle provides us with information that the equilibrium law cannot is in the case of changing temperature. Suppose we have the following reaction, CaCO 3 (s) ↔ CaO(s) + CO 2 (g) ΔH > 0 What happens if you increase the temperature?

33. CaCO 3 (s) ↔ CaO(s) + CO 2 (g) ΔH > 0 Using each method, explain what will happen to the concentration of CO 2 if solid lime (CaCO 3 ) is added to the system? Q vs LeChâtelier

35. Question 2a S 2(g) and C (graphite) when placed together in a closed system form an equilibrium with CS 2(g). C (graphite) + S 2(g) ↔ CS 2(g) Suppose that the equilibrium constant for this reaction is 4.0. Draw a qualitative graph that shows how the concentration of each gas changes with time if the system initially consists of pure S 2(g) and graphite.

36. Question 2b C (graphite) + S 2(g) ↔ CS 2(g) Draw a picture representing the molecules present under initial conditions and when the reaction reaches equilibrium. Will the amount of graphite in the system be the same, more, or less at equilibrium than it was initially? Why?

37. Draw a second graph showing what happens if the system initially contains pure CS 2(g) and graphite. Draw a picture representing the molecules present under initial conditions and when the reaction reaches equilibrium. Will the amount of graphite have changed in this scenario? If so, how? Question 2c C (graphite) + S 2(g) ↔ CS 2(g)

38. Question 3 At room temperature, the equilibrium constant for the reaction: 2NO (g) ↔ N 2(g) + O 2(g) is 1.4 x10 30 Is this reaction product-favored or reactant-favored? In the atmosphere at room temperature, the concentration of N 2 is 0.33 mol/L, and the concentration of O 2 is about 25% of that value. Calculate the equilibrium concentration of NO in the atmosphere.

39. Question 4 CO (g) + H 2 O (g) ↔ CO 2(g) + H 2(g) K c = 4.00 at 500 K. A mixture of 1.00 mol CO and 1.00 mol H 2 O is allowed to come to equilibrium in a flask of volume 0.5 L at 500K, Calculate the final concentrations of all four species: CO, H 2 O, CO 2 and H 2 What would be the equilibrium concentrations if an additional 1.00 mol of each CO and H 2 O were added?

40. ICP- One Minute Paper

42. Equilibrium Representation (Friday 11/10) Your group will create a visual representation of a dynamic equilibrium. The medium is completely up to you (animation, skit, artwork, song, etc.), and creativity is encouraged. –The representative system that is in a stable dynamic equilibrium. –A stress to the system and how it would respond according to Le Chatelier's principle. You will tell the class the system what species (chemical or otherwise) that are present etc., but the class will have to infer the stress placed on the system by its response to that stress.

43. Proposals Proposal Meetings During lab on Friday 11/3 –10 minutes –1 hard copy of proposal for me to keep (in template format) –No more than 5 minutes to present proposal orally –Questions and discussion