1. 1 Chemical Equilibrium Chapter 17 SAVE PAPER AND INK!!! When you print out the notes on PowerPoint, print "Handouts" instead of "Slides" in the print setup. Also, turn off the backgrounds (Tools>Options>Print>UNcheck "Background Printing")!

2. 2 Equilibrium We’ve already used the phrase “equilibrium” when talking about reactions. In principle, every chemical reaction is reversible... capable of moving in the forward or backward direction. 2 H 2 + O 2 ⇌ 2 H 2 O Some reactions are easily reversible... Some not so easy...

3. 3 Equilibrium: the extent of a reaction In stoichiometry we talk about theoretical yields, and the many reasons actual yields may be lower. Another critical reason actual yields may be lower is the reversibility of chemical reactions: some reactions may produce only 70% of the product you may calculate they ought to produce. Equilibrium looks at the extent of a chemical reaction.

5. Chemical Equilibrium Analogy Let us introduce the idea of chemical equilibrium by an analogy, seemingly far-fetched at first sight, but actually mathematically correct. Imagine that a crabapple tree sits on the dividing line between two homes, one inhabited by a crochety old man, and the other by a father who has told his young son to go out and rid the back yard of crabapples. The boy quickly realizes that the easiest way to dispose of the crabapples is to throw them into the neighbouring yard. He does so, arousing the ire of the old man. The boy and the man start throwing crabapples back and forth across the fence as fast as they can. Who will win? The battle is outlined in five phases, as shown on the following pages.

7. Chemical Equilibrium Analogy Assuming that the boy is more energetic and agile than the old man, you might think at first that the conflict would end with all of the apples on the old man's side (Phases I and II). It is true that with equal numbers of crabapples on either side, the boy will throw apples across the fence faster than the old man can return them. But this only means that apples will become more plentiful on the old man's side, and easier to reach.

8. Chemical Equilibrium Analogy They will become scarcer on the boy's side, and require more running around to locate. Eventually a standoff, or equilibrium, will be reached, in which the number of apples crossing the fence is the same in both directions. The old man will throw less quickly but will have less trouble finding apples (Phase III); the boy will throw more rapidly but will waste time scurrying around hunting for the relatively few crabapples on his side (Phase IV). The ratio of apples on the two sides of the fence ultimately will be determined by the relative agility of the two combatants, but all of the apples will not end up on one side (Phase V).

10. Chemical Equilibrium Analogy We can express the rate at which the old man throws apples by Rate M = k M C M The rate is measured in apples per second across the fence, and C M is the concentration of apples on the man's side of the fence in apples per square foot of ground. The rate constant, k M, has units of square feet per second: The value of k M expresses the agility of the old man, and his speed in covering the territory on his side of the fence. The rate at which the boy throws apples back across the fence is given by Rate B = k B C B in which C B is the concentration of apples in the boy's yard, and k B is the rate constant, or agility constant, which tells how fast the boy gets around on his side of the fence, in square feet per second. Since we have assumed that the boy is livelier than the man, k B is greater than k M.

11. Chemical Equilibrium Analogy If the boy had cleaned up his yard completely before the old man came out, then as the battle began,Rate M would be greater than Rate B, and there would be a net flow of apples to the boy's side. His agility would do him no good if there were no apples on his side to pick up. Conversely, if the battle had begun with equal concentrations of apples on each side, then Rate B would have been greater than Rate M because the agility constant k B is greater than k M. With the same number of apples at their disposal, the boy always can do better than the old man because he gets around faster. In either case, a neutral observer would have found to his surprise that the battle eventually settled down into a stalemate, or equilibrium in which Rate M = Rate B, at a point where the extra apples on the old man's side just compensated for the extra agility of the boy.

13. 13 The Concept of Equilibrium Consider colorless frozen N 2 O 4. At room temperature, it decomposes to brown NO 2 : N 2 O 4 (g)  2NO 2 (g). At some time, the color stops changing and we have a mixture of N 2 O 4 and NO 2. Chemical equilibrium is the point at which the rate of the forward reaction is equal to the rate of the reverse reaction. At that point, the concentrations of all species are constant. Using the collision model: –as the amount of NO 2 builds up, there is a chance that two NO 2 molecules will collide to form N 2 O 4. –At the beginning of the reaction, there is no NO 2 so the reverse reaction (2NO 2 (g)  N 2 O 4 (g)) does not occur.

14. 14 The Concept of Equilibrium As the substance warms it begins to decompose: N 2 O 4 (g)  2NO 2 (g) When enough NO 2 is formed, it can react to form N 2 O 4 : 2NO 2 (g)  N 2 O 4 (g). At equilibrium, as much N 2 O 4 reacts to form NO 2 as NO 2 reacts to re-form N 2 O 4 The double arrow implies the process is dynamic.

15. 15 The Concept of Equilibrium As the reaction progresses –[A] decreases to a constant, –[B] increases from zero to a constant. –When [A] and [B] are constant, equilibrium is achieved.

16. Equilibrium: Q and K If you mix A and B, A will react to become B, and B will react to become A –This equation also states that one molecule of A reacts to give one molecule of B. This will be an important part of the concept. A is just turning into B, and vice versa. No other substances are involved. Associated with this system are two quantities, Q, the reaction quotient, and K, the equilibrium constant. It is important to understand the distinction between Q and K.Q is a quantity that changes as a reaction system approaches equilibrium. K is the numerical value of Q at the "end" of the reaction, when equilibrium is reached. 16

17. 17 If Q < K eq, shift to right (toward product) If Q > K eq, shift to left (toward reactant)

18. 18 The Equilibrium Constant No matter the starting composition of reactants and products, the same ratio of concentrations is achieved at equilibrium. For a general reaction the equilibrium constant expression is where K c is the equilibrium constant.

19. 19 The Equilibrium Constant K c is based on the molarities of reactants and products at equilibrium. We generally omit the units of the equilibrium constant. Note that the equilibrium constant expression has products over reactants.

20. 20 The Equilibrium Expression Write the equilibrium expression for the following reaction:

21. 21 The Equilibrium Constant The Magnitude of Equilibrium Constants The equilibrium constant, K, is the ratio of products to reactants. Therefore, the larger K the more products are present at equilibrium. Conversely, the smaller K the more reactants are present at equilibrium. If K >> 1, then products dominate at equilibrium and equilibrium lies to the right. If K << 1, then reactants dominate at equilibrium and the equilibrium lies to the left.

22. 22 The Equilibrium Constant The Magnitude of Equilibrium Constants An equilibrium can be approached from any direction. Example:

23. 23 The Equilibrium Constant The Magnitude of Equilibrium Constants However, The equilibrium constant for a reaction in one direction is the reciprocal of the equilibrium constant of the reaction in the opposite direction.

24. 24 The Equilibrium Constant Heterogeneous Equilibria When all reactants and products are in one phase, the equilibrium is homogeneous. If one or more reactants or products are in a different phase, the equilibrium is heterogeneous. Consider: –experimentally, the amount of CO 2 does not seem to depend on the amounts of CaO and CaCO 3. Why?

25. 25 The Equilibrium Constant Heterogeneous Equilibria

26. 26 The Equilibrium Constant Heterogeneous Equilibria Neither density nor molar mass is a variable, the concentrations of pure solids and pure liquids are constant. (You can’t find the concentration of something that isn’t a solution!) We ignore the concentrations of pure liquids and pure solids in equilibrium constant expressions. The amount of CO 2 formed will not depend greatly on the amounts of CaO and CaCO 3 present. K c = [CO 2 ]

27. 27 Calculating Equilibrium Constants Steps to Solving Problems: 1.Write an equilibrium expression for the balanced reaction. 2.Write an ICE table. Fill in the given amounts. 3.Use stoichiometry (mole ratios) on the change in concentration line. 4.Deduce the equilibrium concentrations of all species. Usually, the initial concentration of products is zero. (This is not always the case.)

28. 28 Applications of Equilibrium Constants Predicting the Direction of Reaction We define Q, the reaction quotient, for a reaction at conditions NOT at equilibrium as where [A], [B], [P], and [Q] are molarities at any time. Q = K only at equilibrium.

29. 29 Applications of Equilibrium Constants Predicting the Direction of Reaction If Q > K then the reverse reaction must occur to reach equilibrium (go left) If Q < K then the forward reaction must occur to reach equilibrium (go right)

30. 30 Example Problem: Calculate Concentration Note the moles into a 10.32 L vessel stuff... calculate molarity. Starting concentration of HI: 2.5 mol/10.32 L = 0.242 M 2 HI H 2 + I 2 Initial: Change: Equil: 0.242 M00 -2x+x+x 0.242-2xxx What we are asked for here is the equilibrium concentration of H 2...... otherwise known as x. So, we need to solve this beast for x.

31. 31 Example Problem: Calculate Concentration And yes, it’s a quadratic equation. Doing a bit of rearranging: x = 0.00802 or –0.00925 Since we are using this to model a real, physical system, we reject the negative root. The [H 2 ] at equil. is 0.00802 M.

32. 32 Example Problem: Calculate Keq This type of problem is typically tackled using the “three line” approach: 2 NO + O 2 2 NO 2 Initial: Change: Equilibrium:

33. 33 Approximating If Keq is really small the reaction will not proceed to the right very far, meaning the equilibrium concentrations will be nearly the same as the initial concentrations of your reactants. 0.20 – x is just about 0.20 if x is really dinky. If the difference between Keq and initial concentrations is around 3 orders of magnitude or more, go for it. Otherwise, you have to use the quadratic formula.

34. 34Example Initial Concentration of I 2 : 0.50 mol/2.5L = 0.20 M I 2 2 I Initial change equil: 0.20 0 -x +2x 0.20-x 2x With an equilibrium constant that small, whatever x is, it’s near zero, and 0.20 minus almost zero is 0.20 (like a million dollars minus a nickel is still a million dollars). 0.20 – x is the same as 0.20 x = 3.83 x 10 -6 M More than 3 orders of mag. between these numbers. The simplification will work here.

35. 35Example Initial Concentration of I 2 : 0.50 mol/2.5L = 0.20 M I 2 2 I Initial change equil: 0.20 0 -x +2x 0.20-x 2x These are too close to each other... 0.20-x will not be trivially close to 0.20 here. Looks like this one has to proceed through the quadratic...

36. 36 Le Chatelier’s Principle: if you disturb an equilibrium, it will shift to undo the disturbance. Remember, in a system at equilibrium, come what may, the concentrations will always arrange themselves to multiply and divide in the Keq equation to give the same number (at constant temperature). Le Châtelier’s Principle

37. 37 Le Châtelier’s Principle Change in Reactant or Product Concentrations Adding a reactant or product shifts the equilibrium away from the increase. Removing a reactant or product shifts the equilibrium towards the decrease. To optimize the amount of product at equilibrium, we need to flood the reaction vessel with reactant and continuously remove product (Le Châtelier). We illustrate the concept with the industrial preparation of ammonia

38. 38 Le Châtelier’s Principle Change in Reactant or Product Concentrations Consider the Haber process If H 2 is added while the system is at equilibrium, the system must respond to counteract the added H 2 (by Le Châtelier). That is, the system must consume the H 2 and produce products until a new equilibrium is established. Therefore, [H 2 ] and [N 2 ] will decrease and [NH 3 ] increases.

39. 39 Le Châtelier’s Principle Change in Reactant or Product Concentrations The unreacted nitrogen and hydrogen are recycled with the new N 2 and H 2 feed gas. The equilibrium amount of ammonia is optimized because the product (NH 3 ) is continually removed and the reactants (N 2 and H 2 ) are continually being added. Effects of Volume and Pressure As volume is decreased pressure increases. Le Châtelier’s Principle: if pressure is increased the system will shift to counteract the increase.

40. 40 Le Châtelier’s Principle Consider the production of ammonia As the pressure increases, the amount of ammonia present at equilibrium increases. As the temperature decreases, the amount of ammonia at equilibrium increases. Le Châtelier’s Principle: if a system at equilibrium is disturbed, the system will move in such a way as to counteract the disturbance.

41. 41 Le Châtelier’s Principle Change in Reactant or Product Concentrations

42. 42 Example

43. 43 Le Châtelier’s Principle Effects of Volume and Pressure The system shifts to remove gases and decrease pressure. An increase in pressure favors the direction that has fewer moles of gas. In a reaction with the same number of product and reactant moles of gas, pressure has no effect. Consider

44. 44 Le Châtelier’s Principle Effects of Volume and Pressure An increase in pressure (by decreasing the volume) favors the formation of colorless N 2 O 4. The instant the pressure increases, the system is not at equilibrium and the concentration of both gases has increased. The system moves to reduce the number moles of gas (i.e. the forward reaction is favored). A new equilibrium is established in which the mixture is lighter because colorless N 2 O 4 is favored.

45. 45 The Equilibrium Constant The Equilibrium Constant in Terms of Pressure If K P is the equilibrium constant for reactions involving gases, we can write: K P is based on partial pressures measured in atmospheres.

46. 46 Le Châtelier’s Principle Effect of Temperature Changes The equilibrium constant is temperature dependent. For an endothermic reaction,  H > 0 and heat can be considered as a reactant. For an exothermic reaction,  H < 0 and heat can be considered as a product. Adding heat (i.e. heating the vessel) favors away from the increase: –if  H > 0, adding heat favors the forward reaction, –if  H < 0, adding heat favors the reverse reaction.

47. 47 Le Châtelier’s Principle Effect of Temperature Changes Removing heat (i.e. cooling the vessel), favors towards the decrease: –if  H > 0, cooling favors the reverse reaction, –if  H < 0, cooling favors the forward reaction. Consider for which  H > 0. –Co(H 2 O) 6 2+ is pale pink and CoCl 4 2- is blue.

48. 48 Solubility Product Principle Another equilibrium situation is slightly soluble products K sp is the solubility product constant K sp can be found on a chart at a specific temperature Since the product is solid on the left side, only the products (ions) are involved in the K sp expression

49. 49 Solubility Product Principle

50. 50 Solubility Product Principle Example: Find the concentration of ions present in calcium fluoride (in water) and the molar solubility. CaF 2 (s) --> Ca +2 + 2 F - K sp = [Ca +2 ] [F - ] 2 = 2 X 10 -10 If x = [Ca +2 ], then [F - ] = 2x (the x and 2x come from coefficients) [x] [2x] 2 = 2 X 10 -10 4x 3 = 2 X 10 -10 x 3 = 5 X 10 -11 x = 3.68 X 10 -4 [Ca +2 ] = x = 3.68 X 10 -4 [F - ] = 2x = 7.37 X 10 -4 Molar solubility = the lowest of the ion concentrations Solubility of CaF 2 = 3.68 X 10 -4 (same as Ca + )

51. 51 Solubility Product Principle Example: Find the molar solubility of silver chloride (in water). AgCl (s) --> Ag + + Cl -

52. 52 Solubility Product Principle Example: The K sp for PbS(s) is 1x10 -27 at 25°C. Find the solubility of PbS(s) in water at 25°C.