Chapter 12: Chemical Equilibrium The Extent of Chemical Reactions

Size: px

Start display at page:

Download "Chapter 12: Chemical Equilibrium The Extent of Chemical Reactions"

Herbert Riley
5 years ago
Views:

1 Chapter 12: Chemical Equilibrium The Extent of Chemical Reactions

2 When a system reaches equilibrium, the [products] and [reactants] remain constant. A + B C + D [5M] [2M] [3M] [1.5M] Rate fwd = Rate rev

3 Let s look at the decomposition of N 2 O 4 to form NO 2. N 2 O 4(g) (colorless) 2 NO 2(g) (brown)

4 A

5 A A. When the experiment begins, the reaction mixture consists mostly of colorless N 2 O 4.

6 A B A. When the experiment begins, the reaction mixture consists mostly of colorless N 2 O 4.

7 A B A. When the experiment begins, the reaction mixture consists mostly of colorless N 2 O 4. B. As N 2 O 4 decomposes to reddish brown NO 2, the color of the mixture becomes pale brown.

8 A B C A. When the experiment begins, the reaction mixture consists mostly of colorless N 2 O 4. B. As N 2 O 4 decomposes to reddish brown NO 2, the color of the mixture becomes pale brown.

9 A B C A. When the experiment begins, the reaction mixture consists mostly of colorless N 2 O 4. B. As N 2 O 4 decomposes to reddish brown NO 2, the color of the mixture becomes pale brown. C. When equilibrium is reached, the concentrations of NO 2 and N 2 O 4 are constant, and the color reaches its final color.

10 A B C D A. When the experiment begins, the reaction mixture consists mostly of colorless N 2 O 4. B. As N 2 O 4 decomposes to reddish brown NO 2, the color of the mixture becomes pale brown. C. When equilibrium is reached, the concentrations of NO 2 and N 2 O 4 are constant, and the color reaches its final color.

11 A B C D A. When the experiment begins, the reaction mixture consists mostly of colorless N 2 O 4. B. As N 2 O 4 decomposes to reddish brown NO 2, the color of the mixture becomes pale brown. C. When equilibrium is reached, the concentrations of NO 2 and N 2 O 4 are constant, and the color reaches its final color. D. Because the reaction continues in the forward and reverse directions at equal rates, the concentrations (and color) remain constant.

13 Example: Ethyne reacting with hydrogen catalyst gas to form ethene: HCCH + H 2 H 2 CCH 2 R 1 = k 1 [C 2 H 2 ][H 2 ] and R -1 = k -1 [C 2 H 4 ] At equilbrium: R 1 = R -1 k 1 [C 2 H 2 ][H 2 ] = k -1 [C 2 H 4 ] or k 1 = [C 2 H 4 ] k -1 [C 2 H 2 ][H 2 ] Where k 1 /k -1 = K eq

14 K eq is the equilibrium constant. K eq is the ratio of [products] over [reactants] when the reaction is at equilibrium. K eq is a constant for a particular chemical reaction and is temperature dependent.

15 Comparing K eq If K eq is small, <<1, then little product forms If K eq is large, >> 1, then little reactant remains If K eq is ~ 1, significant amounts of reactants and products are present.

16 Comparison of K Values

17 Disturbing Equilibrium: Le Chatelier s Principle

18 Le Chatelier s Principle When a chemical system at equilibrium is disturbed by a change in P, T, or [ ], the system shifts in equilibrium composition in a way that reduces the effect of the disturbance.

20 Example Consider the endothermic reaction below: 2 N 2(g) + 4 H 2 O (g) 4H 2(g) + 4NO (g) Predict the shifts, if any, in equilibrium for the following conditions: Remove N 2 Add H 2 O Remove H 2 Reduce the volume of the container Reduce the temperature Add oxygen gas Add a catalyst

21 Another Example Consider the following system at equilibrium: PCl 3 (g) + Cl 2 (g) PCl 5 (g) What will happen if more Cl 2 (g) is added to the system?

22 General Equation: aa + bb cc + dd K = [C] c [D] d [A] a [B] b Example: Methane gas reacts with water vapor to produce carbon monoxide and hydrogen gas. (a) Write a balanced equation (b) Write a reaction quotient expression for this reaction.

23 Another Example Write the equilibrium expression for the following reaction. N 2(g) + 3 H 2(g) 2 NH 3(g)

24 If a chemical reaction is obtained by the summation of other chemical reactions, then K for the overall net reaction equals the product of the equilibrium constants of each of the individual reactions. Example: CO + 3 H 2 CH 4 + H 2 O K 1 CH H 2 S CS H 2 K 2 CO + 2 H 2 S CS 2 + H 2 + H 2 O Overall K eq = K 1 K 2

25 Equilibrium Expressions Cont. Reactants and/or products in the liquid or solid state do not appear in the equilibrium expression. Why? Consider the reaction of calcium carbonate forming calcium oxide and carbon dioxide.

26 (1) Write a balanced equation for this reaction. (2) Write the equilibrium expression for this reaction.

27 Problem Solving Example: Write the equilibrium expression for the below rxns. 4 Fe (s) + 3 O 2(g) 2 Fe 2 O 3(s) 2 H 2 O (l) 2 H 2(g) + O 2(g)

28 Another Form of the Equilibrium Constant Equilibrium Constants can be expressed in terms of pressure for gaseous reactions, where... K p = K c (RT) n n refers to the change in moles of the gas. R = L-atm/mol-K

29 Example General Equation: aa + bb K p = cc + dd c d P P C a P A D b P B

30 The Reaction Quotient The ratio of [products] over [reactants] at some time (not necessarily when equilibrium has been established) is the reaction quotient. When Q = K eq, the reaction is at equilibrium. When Q > K eq, the reaction will proceed in the reverse direction (toward the reactants). When Q < K eq, the reaction will proceed in the forward direction (toward the products).

32 Handout for Problem Solving Do problems 1-6. Place your work and answers on the board. In each of these problems you will either (a) calculate an equilibrium constant or (b) determine a value for the reaction quotient and compare it with the equilibrium constant.

33 Another Type of Problem Solving In this type of chemistry problem solving you will be given initial amounts and need to determine the new concentrations when equilibrium is reached. Pay close attention to how to build an equilibrium table for problem solving.

34 Consider the reaction below. The reaction takes place in a 1.00 L container, and starts out with 2.00 moles of ammonia and 3.00 moles of copper (II) oxide. At equilibrium there are 1.89 moles of copper present. What number of moles of each of the reaction species are present at equilibrium? 2 NH 3(g) + 3 CuO (s) 3 H 2 O (l) + N 2(g) + 3 Cu (s)

35 Another Problem When 1.00 moles of PCl 5 in a 1.00 L container comes to equilibrium, the mixture is found to contain moles of PCl 3. How many moles of each reaction substance are present at equilibrium? PCl 5(g) PCl 3(g) + Cl 2(g)

36 More Problem Solving The following reaction initially contains 1.00 moles of A and 2.00 moles of B in a 1.00 L container. How many moles of each reaction species are present at equilibrium? K c = 5.2 X 10-3 A 2 + B 2 2 AB

37 More Example Problems Let s work problems 7 & 8 from your worksheet. Group Activity: Work problems 9 & 10.

38 Solubility Equilibria of Relatively Insoluble Salts Solubility Product Constant (K sp )

40 Equilibrium Expressions for Solids Write the equilibrium expression for the previous slide. Write the solubility product (K sp ) expression for calcium phosphate. Write the solubility product expression for aluminum hydroxide.

42 The larger the K sp the more soluble the compound. Remember, K values are temperature dependent.

43 Calculating K sp If mg of silver bromide dissolves in 1.00 L of solution. Determine the solubility product constant for silver bromide. An experiment finds the solubility of barium fluoride to be 0.55 grams in 500. ml of solution. Determine the K sp for barium fluoride.

44 Determining Molar Solubility How can we determine how much of a slightly soluble salt dissolves? What is the molar solubility of Al(OH) 3? K sp = 3.0 X What is the molar solubility of Al(OH) 3 in M barium hydroxide?

45 More Molar Solubility Problem Solving What is the molar solubility of aluminum hydroxide in M aluminum nitrate? Mercury (I) chloride (once used as a laxative) called Calomel has a K sp = 1.3 X What is the solubility in grams/l?

46 Predicting the Formation of a Precipitate Determining whether or not a ppt will form you will need to compare Q with K sp. If Q > K sp the rxn will proceed in reverse If Q < K sp the rxn will proceed forward If Q = K sp the rxn is at equilibrium

47 Problem Ca 3 (PO 4 ) 2 forms kidney stones. If a urine sample contains 1.0 X 10-3 M Ca 2+ and 1.0 X 10-8 M PO 4 3-, will a stone (ppt) form? K sp = 1.0 X If 400. ml of 0.50 M lead (II) nitrate are mixed with ml of 2.50 X 10-2 M sodium chloride will a ppt form?

48 Acid Base Equilibria

49 Properties of Acids and Bases Acids Bases Release H + Release OH - Neutralize OH - Neutralize H + Proton Donors Proton Acceptors Electron Pair Acceptors Electron Pair Donors ph < 7 ph > 7 Taste Sour Taste Bitter, Feel Slippery Litmus Paper (turns pink) Litmus paper (turns blue) Reacts with most metals to form H 2 (g)

51 Neutralization Reaction Acid + Base Salt + H 2 O (l) Strong acids and bases dissociate approximately 100% Strong Acids: HCl, HBr, HI, HNO 3, HClO 4, HClO 3, H 2 SO 4 Strong Bases: Soluble metal hydroxides Weak acids and bases dissociate << 100%

52 STRONG ACIDS

53 WEAK ACIDS

55 Weak Acid Equilibrium Expression HA H + + A - K a = [H + ][A - ] or [HA] HA + H 2 O (l) H 3 O + + A - K a = [H 3 O + ][A - ] [HA]

56 Example Write the equilibrium expression for the following dissociation of benzoic acid: O O + H O H O Note: A 2 M solution only dissociates about 0.6%

58 Weak Base Characteristics Weak bases, such as insoluble metal hydroxides, dissociate << 100% in water--- releasing few OH - ions. Weak bases, such as ammonia, amines and amides, slightly react with water to release OH - ions. This process is also called dissociation.

59 Weak Base Equilibrium Expression B + H 2 O (l) HB + + OH - K b = [HB + ][OH - ] [B]

60 Example Write the equilibrium expression for the following dissociation: NH 3 + H 2 O (l) NH OH - K b = NH 4 + OH - NH 3

62 Autoionization of Water Water self-ionizes. Pure water dissociates, but very little. It is a very weak electrolyte. H 2 O (l) H + + OH - H 2 O (l) + H 2 O (l) H 3 O + + OH -

63 Autoionization of Water Cont. Using your equilibrium table/problem solving technique learned previously, determine the [H + ] and [OH - ] in water. K w = 1.0 X H 2 O (l) H + + OH - Initial 55 M 0 0 change -x +x +x 55M-x x x K w = H+ OH X = x x X = [H + ] = [OH - ] = 1.0 X 10-7 M

65 Strong Acid & Base Problem Solving 1. If HNO 3 is added to H 2 O, the [H 3 O + ] is 2.0 X 10-5 M, what is the [OH - ]? K w = H+ OH X = 2.0 X 10-5 M OH - [OH - ] = 5.0 X M 2. If LiOH is added to water, the [OH - ] is 2.0 X 10-5 M, what is the [H + ]? K w = H+ OH X = H X 10-5 M [ H + ] = 5.0 X M

66 ph The following relationships are often used to determine the ph and poh of a solution. ph = - log [H + ] poh = - log [OH - ] pk w = -log K w 14 = ph + poh If the ph = 7, the solution is neutral; [H + ] = [OH - ] If the ph < 7, the solution is acidic; [H + ] > [OH - ] If the ph > 7, the solution is basic; [H + ] < [OH - ]

67 Problem Solving Determine the ph, poh, [H + ], [OH - ] as appropriate for the following problems. a. A sample of blood has a ph of poh = 6.6 [H + ] = M [OH - ] = M b. A household ammonia solution has a [OH - ]=7.94X10-3 M. poh = 2.1 ph = [H + ] = 1.26 X10-12 M c. A sample of rain water has a ph of poh = 8.6 [H + ] = M [OH - ] = M

68 More Problem Solving What is the ph, poh, [H + ], [OH - ] of a 0.20 M Ba(OH) 2 solution? poh = 0.40 ph = [H + ] = 2.5 X M

69 Bronsted-Lowry Acids and Bases Recall: Acids are proton donors and bases are proton acceptors.

71 Application Time! Look at the following acid-base reactions and determine the acid-base conjugate pairs in each.

72 CO H 2 O (l) HCO OH - C 2 H 3 O HNO 2 HC 2 H 3 O 2 + NO 2 - HCO H 2 O (l) H 3 O + + CO 3 2- HCO H 2 O (l) OH - + H 2 CO 3 Note: A substance that can act as an acid or a base is said to be amphiprotic.

73 Strength relationship between acid-base conjugate pairs.

74 Weak Monoprotic Acid Problem Solving 1. Determine the equilibrium constant for a 0.10 M phenol (HC 6 H 5 O) solution that has a ph of 5.43 at 25 0 C. HC 6 H 5 O C 6 H 5 O - + H + x = [H + ]= M Initial 0.10 M 0 0 change - x + x 0.10 M - x K a = x C 6 H 5 O - H + HC 6 H 5 O + x x K a = M M 0.10 M M K a = 1.4 X 10-10

75 2. What are the concentrations of nicotinic acid, hydrogen ion, and nicotinate ion in a solution of 0.15 M nicotinic acid (HC 6 H 4 O 2 ) at 25 o C? What is the ph? What is the percent ionization? K a = 1.4 x 10-5 Let s say nicotinate is A - AND nicotinic acid is HA HA H + + A - Initial change 0.15 M -x 0 +x 0 +x 0.15 M-x x x 1.4 X 10-5 = x x 0.15 M-x Assumption is True: K/C < 0.01 You can negate x term.

Chapter 12: Chemical Equilibrium The Extent of Chemical Reactions

Chapter 12: Chemical Equilibrium The Extent of Chemical Reactions When a system reaches equilibrium, the [products] and [reactants] remain constant. A + B C + D [5M] [2M] [3M] [1.5M] Rate fwd = Rate rev