1. If at equilibrium there are significant amounts of both product and reactants, then the reaction can be described in terms of a K value, or equilibrium constant. For any reaction: aA + bB  cC + dD the K eq is defined as: K eq = [C] c [D] d [A] a [B] b Note: The value of the equilibrium constant may be determined from experimental data if the concentrations of both the reactants and the products are known. Equilibrium Constant, K eq

2. where K is a constant for the reaction at constant temperature. (Note: K has no units. ) K eq = [C] c [D] d [A] a [B] b The concentrations at equilibrium always combine in the manner below where the products are in the numerator and the reactants are in the denominator to produce the K value, regardless of the initial concentrations of species.* Equilibrium Constant, K eq *What if the reaction were going in the reverse direction? How would it affect K?

3. An important point about forming the equilibrium constant expression is that: Only aqueous and gaseous substances are included in the expression. Pure solids and liquids are excluded.* Intuitively, this makes sense because the "concentration" of a pure solid or liquid is defined by its density, and is therefore a constant at a given temperature and pressure. Equilibrium Constant, K eq K eq = [C] c [D] d [A] a [B] b *

4. 4 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...

5. Why is Equilibrium Constant Important? Knowing K c and the initial concentrations, we can determine the concentrations of components at equilibrium.

6. Kc= k f /k r, at equilibrium, if K> 1, then more products at equilib. And if k<1, then reactants favored at equilb. K=1 (conc. Of reactants and products nearly same at equilibrium) The magnitude of Kc gives us an indication of how far the reaction has proceeded toward the formation of products, when the equilibrium is achieved. The larger the value of K, the further the reaction will have proceeded towards completion when equilibrium is reached. 6

7. 7 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.

8. 8 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.

9. 9 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.

10. 10 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.

11. 11 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.

12. 12 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.

13. 13 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.

14. 14 The Equilibrium Constant The Magnitude of Equilibrium Constants An equilibrium can be approached from any direction. Example:

15. 15 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.

16. 16 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?

17. 17 The Equilibrium Constant Heterogeneous Equilibria Neither density nor molar mass is a variable, the concentrations of 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 ]

18. 18 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.)

19. 19 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.

20. 20 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)

21. 21 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.

22. 22 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.

23. 23 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

24. 24 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

25. 25 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 [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 Solubility of CaF 2 = 3.68 X 10 -4

26. 26 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

27. 27 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

28. 28 P X or [X] Time → [B] or P B / RT Equilibrium is established Figure 1: Reversible reactions [A] or P A / RT [A] 0 or P A 0 / RT

29. 29 Reversible Reactions and Rate Reaction Rate Time Backward rate Forward rate Equilibrium is established: Forward rate = Backward rate When equilibrium is achieved: [A] ≠ [B] and k f /k r = K eq

30. Fig. 6.1. Progress of a chemical reaction. The rate of the forward reaction diminishes with time, while that of the backward reaction increases, until they are equal. A large K means the reaction lies far to the right at equilibrium. The rate of the forward reaction diminishes with time, while that of the backward reaction increases, until they are equal. A large K means the reaction lies far to the right at equilibrium. ©Gary Christian, Analytical Chemistry, 6th Ed. (Wiley)

31. Equilibrium constants may be written for dissociations, associations, reactions, or distributions. ©Gary Christian, Analytical Chemistry, 6th Ed. (Wiley)

32. Chemical Equilibrium Review of Principles Chemical reactions are never “complete” Chemical reactions proceed to a state where ratio of products to reactants is constant NH 3 + HOH  NH 4 + + OH - [NH 4 + ][OH - ]/[NH 3 ][HOH] = K b o If K b << 1 (little ionization) H 2 SO 4 + HOH  H 3 O + + HSO 4 - [H 3 O + ][HSO 4 - ] / [H 2 SO 4 ][HOH] = K a If K a >> 1 (mostly ionized)

33. Chemical Equilibrium Important Equilibria in Analytical Chemistry Solubility: AgCl(s)  Ag + + Cl - Ag 3 AsO 4 (s)  3 Ag + + AsO 4 3- BaSO 4 (s)  Ba 2+ + SO 4 2- K sp (AgCl) = [Ag + ][Cl - ] = 1.0 x 10 -10 K sp (Ag 3 AsO 4 ) = [Ag + ] 3 [AsO 4 3- ] = 1.0 x 10 -22 K sp (BaSO 4 ) = [Ba 2+ ][SO 4 2- ] = 1.0 x 10 -10

34. Chemical Equilibrium Equilibrium Constants 2 A + 3 B  C + 4 D K e = [C][D] 4 /[A] 2 [B] 3 Concentrations [ ] : – molar for solutes – partial pressures (atm) for gases – [1.0] for pure liquid, solid, or solvent

35. Chemical Equilibrium Important Equilibria in Analytical Chemistry Autoprotolysis: HOH + HOH  H 3 O + + OH - K e = [H 3 O + ][OH - ]/[HOH] 2 K e [HOH] 2 = K w = [H 3 O + ][OH - ] = 1.0 x 10 -14 @ 25 o C In pure water @ 25 o C [H 3 O + ] = [OH - ] = 10 -7 Acid Dissociation: H 2 CO 3 + HOH  H 3 O + + HCO 3 - K a = [H 3 O + ][HCO 3 - ]/[H 2 CO 3 ] = 4.3 x 10 -7

36. Chemical Equilibrium Important Equilibria in Analytical Chemistry H 2 CO 3 + HOH  H 3 O + + HCO 3 - acid 1 base 1 Dissociation of Conjugate Base 1 : HCO 3 - + HOH  H 3 O + + CO 3 2- K a (HCO 3 - ) = [H 3 O + ][CO 3 2- ]/[HCO 3 - ] = 4.8 x 10 -11 Hydrolysis of Conjugate Base 1 : HCO 3 - + HOH  H 2 CO 3 + OH - K b (HCO 3 - ) = K w /K a (H 2 CO 3 ) = 10 -14 /4.3x 10 -7 K b (HCO 3 - ) = 2.3 x 10 -8

37. Chemical Equilibria Electrolyte Effects Diverse Ion (Inert Electrolyte) Effect: Add Na 2 SO 4 to saturated solution of AgCl K sp o = a Ag +. a Cl - = 1.75 x 10 -10 At high concentration of diverse (inert) electrolyte: higher ionic strength,  a Ag + < [Ag + ] ; a Cl - < [Cl - ] a Ag +. a Cl - < [Ag + ] [Cl - ] K sp o < [Ag + ] [Cl - ] ;  K sp o < [Ag + ] = solubility Solubility = [Ag + ] >  K sp o

38. Multiple Chemical Equilibria Example Problem #1 What is pH of Mg(OH) 2 solution ? (assume a x = [X]) Equilibria: Mg(OH) 2(s)  Mg 2+ + 2 OH - K sp = [Mg 2+ ][OH - ] = 1.8 x 10 -11 HOH + HOH  H 3 O + + OH - K e = K w = 1.0 x 10 -14 Mass Balance: [OH - ] = [H 3 O + ] + 2 [Mg 2+ ] Charge Balance: [OH - ] = [H 3 O + ] + 2 [Mg 2+ ] Expression for unknown : [H 3 O + ] = K w / [OH - ]

39. Chemical Equilibrium Important Equilibria in Analytical Chemistry Base Dissociation: NH 3 + HOH  NH 4 + + OH - K b (NH 3 ) = [NH 4 + ][OH - ]/[NH 3 ] = 1.75 x 10 -5 Hydrolysis of Salts: NH 4 Cl(s)  NH 4 + + Cl - NH 4 + + HOH  NH 3 + H 3 O + K a (NH 4 + ) = K w /K b (NH 3 ) = 10 -14 /1.75 x 10 -5 K a (NH 4 + ) = 5.7 x 10 -10

40. 40 – For a balanced chemical reaction in equilibrium: a A + b B ↔ c C + d D – Equilibrium constant expression (K eq ): K eq is strictly based on stoichiometry of the reaction (is independent of the mechanism). Units: K eq is considered dimensionless (no units) or

41. 41 Magnitude of K eq Since K eq  [products]/[reactants], the magnitude of K eq predicts which reaction direction is favored: – If K eq > 1 then [products] > [reactants] and equilibrium “lies to the right” – If K eq < 1then [products] < [reactants] and equilibrium “lies to the left”

42. 42 Relationship Between Q and K Reaction Quotient (Q): The particular ratio of concentration terms that we write for a particular reaction is called reaction quotient. For a reaction, A  B, Q= [B]/[A] At equilibrium, Q= K Reaction Direction: Comparing Q and K Q<K, reaction proceeds to right, until equilibrium is achieved (or Q=K) Q>K, reaction proceeds to left, until Q=K

43. 43 Value of K For the reference rxn, A  >B, For the reverse rxn, B  >A, For the reaction, 2A  > 2B For the rxn, A  > C C  > B K(ref)= [B]/[A] K= 1/K(ref)K= K(ref) 2 K (overall)= K 1 X K 2

44. 44 15.3: Types of Equilibria Homogeneous: all components in same phase (usually g or aq) N 2 (g) + H 2 (g) ↔ NH 3 (g) 321 Fritz Haber (1868 – 1934)

45. 45 NH 3 (aq) + H 2 O (l) ↔ NH 4 1+ (aq) + OH 1- (aq) Initial Change Equilibrium 0.0124 M - x 0.0119 M 0 M + x 4.64 x 10 -4 M NH 3 (aq) H 2 O (l) NH 4 1+ (aq)OH 1- (aq) X X X x = 4.64 x 10 -4 M

46. 46 Ex: Enough ammonia is dissolved in 5.00 L of water at 25ºC to produce a solution that is 0.0124 M ammonia. The solution is then allowed to come to equilibrium. Analysis of the equilibrium mixture shows that [OH 1- ] is 4.64 x 10 -4 M. Calculate K eq at 25ºC for the reaction: NH 3 (aq) + H 2 O (l) ↔ NH 4 1+ (aq) + OH 1- (aq)

47. 47 : Calculating Equilibrium Constants Steps to use “ICE” table: 1.“I” = Tabulate known initial and equilibrium concentrations of all species in equilibrium expression 2.“C” = Determine the concentration change for the species where initial and equilibrium are known Use stoichiometry to calculate concentration changes for all other species involved in equilibrium 3.“E” = Calculate the equilibrium concentrations

48. 48 Heterogeneous: different phases CaCO 3 (s) ↔ CaO (s) + CO 2 (g) Definition:What we use: Even though the concentrations of the solids or liquids do not appear in the equilibrium expression, the substances must be present to achieve equilibrium. Concentrations of pure solids and pure liquids are not included in K eq expression because their concentrations do not vary, and are “already included” in K eq (see p. 548).

49. Calculation of Equilibrium concentration Initial moles11 0 Moles at eqma – xb – x 2x

50. Calculation of Equilibrium concentration Considering the equilibrium Let the reaction starts with ‘a’ moles of H 2 and ‘b’ moles of I 2 taken in a container of volume V litres. ‘x’ moles of H 2 have reacted when the equilibrium is attained, then ‘x’ moles of I 2 will also react and 2x moles of HI are produced at T K temperature.

51. Characteristics of Equilibrium constant (K) 1.Its value is independent of original concentration of reactants, pressure or presence of a catalyst. 2.It is independent of the direction from which equilibrium is attained. 3.Its value is constant at certain temperature and would change with temperature. 4.The larger the value of ‘K’, greater will be the value of products as compared to reactants. 5.Its value, however, depends on the coefficients of balanced reactants and products in the chemical equation.

52. Prediction of direction of reaction – reaction quotient When, Q c > K c — (Backward reaction will occur, the reaction will proceed in direction of reactants) When, Q c < K c — (Forward reaction will occur, the reaction will proceed in direction of products) When, Q c = K c — The reaction will be in equilibrium

53. Types: Equilibrium Heterogeneous equilibria: and Where is partial pressure of carbon dioxide.

54. Types: Equilibrium All the reactants and products are present in more than one phases The concentration of pure solids and pure liquids is taken as 1. Heterogeneous equilibria: For example

55. Types: Equilibrium All the reactants and products of an equilibrium reaction are present in the same phase Homogeneous equilibria For example,

56. LeChatelier’s Principle and Chemical Equilibrium LeChatelier’s Principle how equilibrium systems change when equilibrium conditions are changed. The changes imposed on an equilibrium are called stresses. The response to these changes (stresses) are called shifts. Stresses consist of changes in concentrations, pressures and temperatures. Shifts that occur are characterized as “shift towards products” (shift right) which means increased rate of forward reaction occurs or “shift towards reactants” (shift left) which means increased rate of reverse reaction occurs. LeChatelier’s Principle defines the relationship between applied stresses and equilibrium shifts which result.

57. Relationships between chemical equations and the expressions of equilibrium constants The expression of equilibrium constant depends on how the equilibrium equation is written. For example, for the following equilibrium: H 2 (g) + I 2 (g) ⇄ 2 HI (g) ; For the reverse reaction: 2HI (g) ⇄ H 2 (g) + I 2 (g) ; And for the reaction: HI (g) ⇄ ½H 2 (g) + ½I 2 (g) ;

58. Relationship between K c and K p In general, for reactions involving gases such that, aA + bB ⇄ cC + dD where A, B, C, and D are all gases, and a, b, c, and d are their respective coefficients, K p = K c (RT)  n and  n = (c + d) – (a + b) (In heterogeneous systems, only the coefficients of the gaseous species are counted.)

59. Relationship between K c and K p For other reactions: 1. 2NO 2 (g) ⇄ N 2 O 4 (g) ;K p = K c (RT) -1 2. H 2 (g) + I 2 (g) ⇄ 2 HI (g) ; K p = K c 3. N 2 (g) + 3H 2 (g) ⇄ 2 NH 3 (g) ;K p = K c (RT) -2

60. Homogeneous & Heterogeneous Equilibria Homogeneous equilibria: CH 4 (g) + H 2 O (g) ⇄ CO (g) + 3H 2 (g) ; CO (g) + H 2 O (g) ⇄ CO 2 (g) + H 2 (g) ; Heterogeneous equilibria: CaCO 3 (s) ⇄ CaO (s) + CO 2 (g) ; HF (aq) + H 2 O (l) ⇄ H 3 O + (aq) + F - (aq) ; PbCl 2 (s) ⇄ Pb 2+ (aq) + 2 Cl - (aq) ;

61. Equilibrium Constant Expressions for Heterogeneous System Examples: CaCO 3 (s) ⇄ CaO (s) + CO 2 (g) ; K c = [CO 2 ] K p = P CO2 ; K p = K c (RT) HF (aq) + H 2 O (l) ⇄ H 3 O + (aq) + F - (aq) ;

62. Solubility Eqilibrium PbCl 2 (s) ⇄ Pb 2+ ( aq) + 2Cl - (aq) ; K sp = [Pb 2+ ][Cl - ] 2 (K sp is called solubility product)

63. Combining Equations and Equilibrium Constants when two or more equations are added to yield a net equation, the equilibrium constant for the net equation, K net, is equal to the product of equilibrium constants of individual equations. For example, Eqn(1): A + B ⇄ C + D; Eqn(2): C + E ⇄ B + F;

64. Combining Equations and Equilibrium Constants Net equation: A + E ⇄ D + F; = K 1 x K 2 If Eqn (1) + Eqn (2) = Net equation, then K 1 x K 2 = K net

65. Equilibrium Exercise #1 A flask is charged with 2.00 atm of nitrogen dioxide and 1.00 atm of dinitrogen tetroxide at 25 o C and allowed to reach equilibrium. When equilibrium is established, the partial pressure of NO 2 has decreased by 1.24 atm. (a) What are the partial pressures of NO 2 and N 2 O 4 at equilibrium? (b) Calculate K p and K c for following reaction at 25 o C. 2 NO 2 (g) ⇄ N 2 O 4 (g) (Answer: K p = 2.80; K c = 68.6)

66. Applications of Equilibrium Constant For any system or reaction: 1.Knowing the equilibrium constant, we can predict whether or not a reaction mixture is at equilibrium, and we can predict the direction of net reaction. Q c = K c  equilibrium (no net reaction) Q c < K c  a net forward reaction; Q c > K c  a net reverse reaction 2.The value of K tells us whether a reaction favors the products or the reactants.

67. Equilibrium constant is used to predict the direction of net reaction For a reaction of known K c value, the direction of net reaction can be predicted by calculating the reaction quotient, Q c. Q c is called the reaction quotient, where for a reaction such as: aA + bB ⇄ cC + dD; Q c has the same expression as K c, but Q c is calculated using concentrations that are not necessarily at equilibrium.

68. Calculating equilibrium concentrations using initial concentrations and value of K c Consider the reaction: H 2 (g) + I 2 (g) ⇄ 2 HI (g), where K c = 55.6 at 425 o C. If [H 2 ] 0 = [I 2 ] 0 = 0.1000 M, and [HI] 0 = 0.0 M, what are their concentrations at equilibrium?

69. Using the ICE table to calculate equilibrium concentrations Equation: H 2 (g) + I 2 (g) ⇄ 2 HI (g),  Initial [ ], M 0.1000 0.1000 0.0000 Change [ ], M -x -x +2x Equilibrium [ ], M (0.1000 - x) (0.1000 - x) 2x  