CHEMICAL EQUILIBRIUM. Cato Maximilian Guldberg and his brother-in-law Peter Waage developed the Law of Mass Action

Transcription

1 CHEMICAL EQUILIBRIUM Cato Maximilian Guldberg and his brother-in-law Peter Waage developed the Law of Mass Action

2 Chemical Equilibrium Reversible Reactions: A chemical reaction in which the products can react to re-form the reactants Chemical Equilibrium: When the rate of the forward reaction equals the rate of the reverse reaction and the concentration of products and reactants remains unchanged HgO(s) Hg(l) + O (g) Arrows going both directions ( ) indicates equilibrium in a chemical equation

3 NO (g) NO(g) + O (g) Remember this from Chapter 1? Why was it so important to measure reaction rate at the start of the reaction (method of initial rates?)

4 NO (g) NO(g) + O (g)

5 Law of Mass Action For the reaction: ja + kb lc + md K [ C] l [ A] [ D] j [ B] m k Where K is the equilibrium constant, and is unitless

6 Product Favored Equilibrium Large values for K signify the reaction is product favored When equilibrium is achieved, most reactant has been converted to product

7 Reactant Favored Equilibrium Small values for K signify the reaction is reactant favored When equilibrium is achieved, very little reactant has been converted to product

8 Writing an Equilibrium Expression Write the equilibrium expression for the reaction: NO (g) NO(g) + O (g) K =??? K [ NO] [ NO [ O ] ]

9 Conclusions about Equilibrium Expressions The equilibrium expression for a reaction is the reciprocal for a reaction written in reverse NO (g) NO(g) + O (g) NO(g) + O (g) NO (g) [ NO] [ O K [ NO ] ] K ' 1 K [ NO [ NO] ] [ O ]

10 Conclusions about Equilibrium Expressions When the balanced equation for a reaction is multiplied by a factor n, the equilibrium expression for the new reaction is the original expression, raised to the nth power. NO (g) NO(g) + O (g) NO (g) NO(g) + ½O (g) [ NO] [ O K [ NO ] K 1 K 1 ] [ NO][ O ] [ NO ] 1

11 LeChatelier s Principle When a system at equilibrium is placed under stress, the system will undergo a change in such a way as to relieve that stress and restore a state of equilibrium. Henry Le Chatelier

12 Le Chatelier Translated: When you take something away from a system at equilibrium, the system shifts in such a way as to replace some what you ve taken away. When you add something to a system at equilibrium, the system shifts in such a way as to use up some of what you ve added.

13 LeChatelier Example #1 A closed container of ice and water is at equilibrium. Then, the temperature is raised. Ice + Energy Water The system temporarily shifts to the right to restore equilibrium.

14 LeChatelier Example # A closed container of N O 4 and NO is at equilibrium. NO is added to the container. N O 4 (g) + Energy NO (g) The system temporarily shifts to the left to restore equilibrium.

15 LeChatelier Example #3 A closed container of water and its vapor is at equilibrium. Vapor is removed from the system. water + Energy vapor The system temporarily shifts to the right to restore equilibrium.

16 LeChatelier Example #4 A closed container of N O 4 and NO is at equilibrium. The pressure is increased. N O 4 (g) + Energy NO (g) The system temporarily shifts to the left to restore equilibrium, because there are fewer moles of gas on that side of the equation.

17 Acid Equilibrium and ph Søren Sørensen

18 Acid/Base Definitions Arrhenius Model Acids produce hydrogen ions in aqueous solutions Bases produce hydroxide ions in aqueous solutions Bronsted-Lowry Model Acids are proton donors Bases are proton acceptors Lewis Acid Model Acids are electron pair acceptors Bases are electron pair donors

19 Acid Dissociation HA H + + A - Acid Proton Conjugate base [ H ][ A ] K a [ HA] Alternately, H + may be written in its hydrated form, H 3 O + (hydronium ion)

20 Dissociation of Strong Acids Strong acids are assumed to dissociate completely in solution. Large K a or small K a? Reactant favored or product favored?

21 Dissociation Constants: Strong Acids Acid Formula Conjugate Base Perchloric HClO 4 ClO 4 - Very large Hydriodic HI I - Very large Hydrobromic HBr Br - Very large Hydrochloric HCl Cl - Very large Nitric HNO 3 NO 3 - Very large Sulfuric H SO 4 HSO 4 - Very large Hydronium ion H 3 O + H O 1.0 K a

22 Dissociation of Weak Acids Weak acids are assumed to dissociate only slightly (less than 5%) in solution. Large K a or small K a? Reactant favored or product favored?

23 Dissociation Constants: Weak Acids Acid Formula Conjugate Base Iodic HIO 3 IO x 10-1 Oxalic H C O 4 HC O x 10 - Sulfurous H SO 3 HSO x 10 - Phosphoric H 3 PO 4 H PO x 10-3 Citric H 3 C 6 H 5 O 7 H C 6 H 5 O x 10-4 Nitrous HNO NO x 10-4 Hydrofluoric HF F x 10-4 Formic HCOOH HCOO x 10-4 Benzoic C 6 H 5 COOH C 6 H 5 COO x 10-5 Acetic CH 3 COOH CH 3 COO x 10-5 Carbonic H CO 3 HCO x 10-7 Hypochlorous HClO ClO x 10-8 Hydrocyanic HCN CN x K a

24 Self-Ionization of Water H O + H O H 3 O + + OH - At 5, [H 3 O + ] = [OH - ] = 1 x 10-7 K w is a constant at 5 C: K w = [H 3 O + ][OH - ] K w = (1 x 10-7 )(1 x 10-7 ) = 1 x 10-14

25 Calculating ph, poh ph = -log 10 (H 3 O + ) poh = -log 10 (OH - ) Relationship between ph and poh ph + poh = 14 Finding [H 3 O + ], [OH - ] from ph, poh [H 3 O + ] = 10 -ph [OH - ] = 10 -poh

26 [H + ] = 10 -ph ph = -log[h + ] [OH - ] = 10 -poh poh = -log[oh - ] ph and poh Calculations [OH - ] = 1 x [H + ] H + OH - [H + ] = 1 x [OH - ] ph poh = 14 - ph ph = 14 - poh poh

27 The ph Scale Graphic: Wikimedia Commons user Slower

28 A Weak Acid Equilibrium Problem What is the ph of a 0.50 M solution of acetic acid, HC H 3 O, K a = 1.8 x 10-5? Step #1: Write the dissociation equation HC H 3 O C H 3 O - + H +

29 A Weak Acid Equilibrium Problem What is the ph of a 0.50 M solution of acetic acid, HC H 3 O, K a = 1.8 x 10-5? Step #: ICE it! I C E HC H 3 O C H 3 O - + H x +x +x x x x

30 A Weak Acid Equilibrium Problem What is the ph of a 0.50 M solution of acetic acid, HC H 3 O, K a = 1.8 x 10-5? Step #3: Set up the law of mass action HC H 3 O C H 3 O - + H + E x x x 1.8 x10 ( x)( x) x 5 (0.50 x) (0.50)

31 A Weak Acid Equilibrium Problem What is the ph of a 0.50 M solution of acetic acid, HC H 3 O, K a = 1.8 x 10-5? Step #4: Solve for x, which is also [H + ] HC H 3 O C H 3 O - + H + E x x x 1.8 x10 x 5 (0.50) [H + ] = 3.0 x 10-3 M

32 A Weak Acid Equilibrium Problem What is the ph of a 0.50 M solution of acetic acid, HC H 3 O, K a = 1.8 x 10-5? Step #5: Convert [H + ] to ph HC H 3 O C H 3 O - + H + E x x x ph log(3.0 x10 5 ) 4.5

33 Solubility Equilibria Lead (II) iodide precipitates when potassium iodide is mixed with lead (II) nitrate. Graphic: Wikimedia Commons user PRHaney

34 K sp Values for Some Salts at 5 C Name Formula K sp Barium carbonate BaCO 3.6 x 10-9 Barium chromate BaCrO 4 1. x Barium sulfate BaSO x Calcium carbonate CaCO x 10-9 Calcium oxalate CaC O 4.3 x 10-9 Calcium sulfate CaSO x 10-5 Copper(I) iodide CuI 1.3 x 10-1 Copper(II) iodate Cu(IO 3 ) 6.9 x 10-8 Copper(II) sulfide CuS 6.0 x Iron(II) hydroxide Fe(OH) 4.9 x Iron(II) sulfide FeS 6.0 x Iron(III) hydroxide Fe(OH) 3.6 x Lead(II) bromide PbBr 6.6 x 10-6 Lead(II) chloride PbCl 1. x 10-5 Lead(II) iodate Pb(IO 3 ) 3.7 x Lead(II) iodide PbI 8.5 x 10-9 Lead(II) sulfate PbSO x 10-8 Name Formula K sp Lead(II) bromide PbBr 6.6 x 10-6 Lead(II) chloride PbCl 1. x 10-5 Lead(II) iodate Pb(IO 3 ) 3.7 x Lead(II) iodide PbI 8.5 x 10-9 Lead(II) sulfate PbSO x 10-8 Magnesium carbonate MgCO x 10-6 Magnesium hydroxide Mg(OH) 5.6 x 10-1 Silver bromate AgBrO x 10-5 Silver bromide AgBr 5.4 x Silver carbonate Ag CO x 10-1 Silver chloride AgCl 1.8 x Silver chromate Ag CrO x 10-1 Silver iodate AgIO 3 3. x 10-8 Silver iodide AgI 8.5 x Strontium carbonate SrCO x Strontium fluoride SrF 4.3 x 10-9 Strontium sulfate SrSO x 10-7 Zinc sulfide ZnS.0 x 10-5

35 Solving Solubility Problems For the salt AgI at 5 C, K sp = 1.5 x AgI(s) Ag + (aq) + I - (aq) I C E O O +x +x x x 1.5 x = x x = solubility of AgI in mol/l = 1. x 10-8 M

36 Solving Solubility Problems For the salt PbCl at 5 C, K sp = 1.6 x 10-5 PbCl (s) Pb + (aq) + Cl - (aq) I C E O O +x +x x x 1.6 x 10-5 = (x)(x) = 4x 3 x = solubility of PbCl in mol/l = 1.6 x 10 - M

37 Solving Solubility with a Common Ion For the salt AgI at 5 C, K sp = 1.5 x What is its solubility in 0.05 M NaI? AgI(s) Ag + (aq) + I - (aq) I C E O x 0.05+x x 0.05+x 1.5 x = (x)(0.05+x) (x)(0.05) x = solubility of AgI in mol/l = 3.0 x M