1. 14-2-1 CHEM 102 Spring 15, LA TECH Instructor: Dr. Upali Siriwardane e-mail: upali@latech.edu Office: CTH 311 Phone 257-4941 Office Hours: M,W 8:00-9:30 & 11:00-12:30 am; Tu,Th, F 8:00 - 10:00 am. or by appointment.; Test Dates : Chemistry 102(001) Spring 2015 March 30, 2015 (Test 1): Chapter 13 April 22, 2015 (Test 2): Chapter 14 &15 May 18, 2015 (Test 3): Chapter 16 &7 May 20, 2015 (Make-up test) comprehensive: Chapters 13-17

2. 14-2-2 CHEM 102 Spring 15, LA TECH Chapter 14. Chemical Equilibrium 14.1 Fetal Hemoglobin and Equilibrium 61 3 14.2 The Concept of Dynamic Equilibrium 61 5 14.3 The Equilibrium Constant (K) 61 8 14.4 Expressing the Equilibrium Constant in Terms of Pressure 622 14.5 Heterogeneous Equilibria: Reactions Involving Solids and Liquids 625 14.6 Calculating the Equilibrium Constant from Measured Equilibrium Concentrations 626 14.7 The Reaction Quotient: Predicting the Direction of Change 629 14.8 Finding Equilibrium Concentrations 631 14.9 Le Châtelier’s Principle: How a System at Equilibrium Responds to Disturbances 641

3. 14-2-3 CHEM 102 Spring 15, LA TECH Law of mass Action Defines an equilibrium constant (K) for the process j A + k B l C + m D j A + k B l C + m D [C] l [D] m [C] l [D] m K = ----------------- ; [A], [B] etc are K = ----------------- ; [A], [B] etc are [A] j [B] k Equilibrium concentrations [A] j [B] k Equilibrium concentrations Pure liquid or solid concentrations are not written in the expression.

4. 14-2-4 CHEM 102 Spring 15, LA TECH 7) What is the difference between initial [A] i and equilibrium [A] eq concentrations? Initial@ Equilibrium N 2 O4NO 2 N 2 O4NO 2 K eq 0.000.020.00140.0170.21 0.000.030.00280.0240.21 0.000.040.00450.0310.21 0.020.000.00450.0310.21 N 2 O 4 (g) colorless 2NO 2 (g) Dark brown

5. 14-2-5 CHEM 102 Spring 15, LA TECH We can predict the direction of a reaction by calculating the reaction quotient. Reaction quotient, Q For the reaction: aA + bB eE + fF Q has the same form as Kc Kc with one important difference. Q can be for any set of concentrations, not just at equilibrium. Q = [E] e [F] f [A] a [B] b Equilibrium calculations

6. 14-2-6 CHEM 102 Spring 15, LA TECH Equilibrium constant calculations 1) Consider the reaction: COCl 2 (g) CO(g) + Cl 2 (g) At equilibrium, [CO] = 4.14 × 10 -6 M; [Cl 2 ] = 4.14 × 10 -6 M; and [COCl 2 ] = 0.0627 M. Calculate the value of the equilibrium constant.

7. 14-2-7 CHEM 102 Spring 15, LA TECH Terminology Initial concentration : concentration (M) of reactants and products before the equilibrium is reached. Equilibrium Concentration Concentration (M) of reactants and products After the equilibrium is reached.

8. 14-2-8 CHEM 102 Spring 15, LA TECH Any set of concentrations can be given and a Q calculated. By comparing Q to the Kc Kc value, we can predict the direction for the reaction. Q < KcNet forward reaction will occur. Q = KcNo change, at equilibrium. Q > KcNet reverse reaction will occur. Reaction quotient

9. 14-2-9 CHEM 102 Spring 15, LA TECH Reaction quotient (Q) calculations

10. 14-2-10 CHEM 102 Spring 15, LA TECH Reaction quotient (Q) calculations

11. 14-2-11 CHEM 102 Spring 15, LA TECH Determining Equilibrium Constants ICE Method 1. Derive the equilibrium constant expression for the balanced chemical equation 2. Construct a Reaction Table with information (ICE) about reactants and products 3. Include the amounts reacted, x, in the Reaction Table 4. Calculate the equilibrium constant in terms of x

12. 14-2-12 CHEM 102 Spring 15, LA TECH Example: An equilibrium is established by placing 2.00 moles of N 2 O 4(g) in a 5.00 L and heating the flask to 407 K. It was determined that at equilibrium the concentration of the NO 2(g) is 0.525 mol/L. What is the value of the equilibrium constant? N 2 O 4(g) 2 NO 2(g) [NO 2 ] 2 K c = [N 2 O 4 ] N 2 O 4 (g) 2 NO 2 (g) [Initial] (mol/L)0.400 [Change] -x 2x [Equilibrium] 0.40- 0.263= 0.138 0.525 x -1/2 x 0.40 - 1/2x = 0 + x

13. 14-2-13 CHEM 102 Spring 15, LA TECH What is the value of the equilibrium constant? 0.525 = 0 + x [NO 2 ] 2 K c = [N 2 O 4 ] 0.40 - 1/2x x = 0.525 [NO 2 ] = 0.40 - 1/2x = 0.40 - 1/2(0525) = 0.138 [NO2] 2 (0.525) 2 Kc = = [N2O4] 0.138 = 2.00 NO 2 (g ) N 2 O 4 (g)

14. 14-2-14 CHEM 102 Spring 15, LA TECH Equilibrium Calculations Hydrogen iodide, HI, decomposes according to the equation 2 HI(g)  H 2 (g) H 2 (g) + I 2 (g) When 4.00 mol of HI placed in a 5.00-L 5.00-L vessel at 458ºC, the equilibrium mixture was found to contain 0.442 mol I 2. I 2. What is the value of Kc Kc Kc Kc for the reaction?

15. 14-2-15 CHEM 102 Spring 15, LA TECH I nitial I nitial 4.00/5=.80 0 0 C hange C hange -2x x x Equilibrium Equilibrium 0.80-2x 0.80-2x x x=0.442/5 x = 0.0884 Equilibrium concentrations [HI] = 0.80 - 2x = 0.8 - 2 x 0.0884 = 0.62 [H 2 ] [H 2 ] = x = 0.0884 [I 2 ] [I 2 ] = x = 0.0884 [H 2 ] [H 2 ] [I 2 ] [I 2 ] 0.0884 x 0.0884 Kc Kc Kc Kc = ---------------- = ------------------------- = 0.0201 [HI] 2 [HI] 2 (0.62) 2 2 HI(g)  H 2 (g) + I 2 (g)

16. 14-2-16 CHEM 102 Spring 15, LA TECH Equilibrium Calculation Example A sample of COCl 2 COCl 2 is allowed to decompose. The value of Kc Kc Kc Kc for the equilibrium (g) (g) CO CO (g) (g) + Cl 2 Cl 2 (g) is 2.2 x 10 -10 10 -10 at 100 o C. If the initial concentration of COCl 2 COCl 2 is 0.095M, what will be the equilibrium concentrations for each of the species involved?

17. 14-2-17 CHEM 102 Spring 15, LA TECH Equilibrium Calculation Example COCl 2 COCl 2 (g) (g)  CO CO (g) Cl 2 Cl 2 (g) Initial conc., M0.0950.0000.000 Change - X + X + X in conc. due to reaction Equilibrium M(0.095 -X) X X Concentration, K c K c == [ CO ] [ Cl 2 ] [ COCl 2 ] X 2 (0.095 - X)

18. 14-2-18 CHEM 102 Spring 15, LA TECH Equilibrium calculation example X 2 (0.095 - X ) K eq = 2.2 x 10 -10 = Rearrangement gives X 2 + 2.2 x 10 -10 X - 2.09 x 10 -11 = 0 This is a quadratic equation. Fortunately, there is a straightforward equation for their solution a X 2 + b X - c = 0

19. 14-2-19 CHEM 102 Spring 15, LA TECH Quadratic Equations An equation of the form a X 2 + b X + c = 0 Can be solved by using the following x = Only the positive root is meaningful in equilibrium problems. -b + b 2 - 4ac 2a

20. 14-2-20 CHEM 102 Spring 15, LA TECH Equilibrium Calculation Example -b + b 2 - 4ac 2a 2.2 x 10 -10 2.09 x 10 -11 X 2 + 2.2 x 10 -10 X - 2.09 x 10 -11 = 0 b c a b c X =X = X = - 2.2 x 10 -10 + [(2.2 x 10 -10 ) 2 - (4)(1)(- 2.09 x 10 -11 )] 1/2 2 X = 4.6 x 10 -6 M X = -4.6 x 10 -6 M

21. 14-2-21 CHEM 102 Spring 15, LA TECH Equilibrium Calculation Example Now that we know X, we can solve for the concentration of all of the species. COCl 2 = 0.095 - X = 0.095 M CO= X = 4.6 x 10 -6 M Cl 2 = X = 4.6 x 10 -6 M In this case, the change in the concentration of is COCl 2 negligible.

22. 14-2-22 CHEM 102 Spring 15, LA TECH Equilibrium calculations using ICE 3) Consider the reaction A 2B, where the value of K eq is 1.4 × 10 -12. At equilibrium, the concentration of B is 0.45 M. What is the concentration of A? ICE Calculation [A][B] Initial concentration: Change Equilibrium concentration: K = 1.4 x 10 -12

23. 14-2-23 CHEM 102 Spring 15, LA TECH Equilibrium calculations using ICE 4) Calculation of unknown concentration of reactants or products in an equilibrium mixture At 100 o C the equilibrium constant (K) for the reaction: H 2 (g) + I 2 (g) 2 HI(g) ; K = 1.15 x 10 2. If 0.400 moles of H 2 and 0.400 moles. If I 2 are placed into a 12.0-liter container and allowed to react at this temperature, what is the HI concentration (moles/liter) at equilibrium? ICE Calculation [A][B] Initial concentration: Change Equilibrium concentration:

24. 14-2-24 CHEM 102 Spring 15, LA TECH Equilibrium calculations using ICE 4) H 2 (g) + I 2 (g) 2 HI(g) ; K = 1.15 x 10 2. If 0.400 moles of H 2 and 0.400 moles. If I 2 are placed into a 12.0-liter container and allowed to react at this temperature, what is the HI concentration (moles/liter) at equilibrium? ICE Calculation [H 2 ][I 2 ][HI] Initial concentration: Change Equilibrium concentration: K = 1.15 x 10 2

25. 14-2-25 CHEM 102 Spring 15, LA TECH What is K (K c ) and K p K c (K) - equilibrium constant calculated based on [A]-Concentrations. K p - equilibrium constant calculated based on partial pressure Kp =Kp =Kp =Kp =

26. 14-2-26 CHEM 102 Spring 15, LA TECH Pressure Equilibrium Constants K c & K p N 2 + 3H 2 2NH 3 [NH 3 ] 2 Kc = [N 2 ][H 2 ] 3 = (P NH3 /RT) 2 (P N2 /RT)(P H2 /RT) 3 (P NH3 ) 2 (1/RT) 2 Kc = (P N2 ) (1/RT))(P H2 ) 3 (1/RT) 3 ) P NH3 2 (1/RT) 2 = P N2 P H2 3 (1/RT)(1/RT) 3 (1/RT) 2 = K p (1/RT)(1/RT) 3

27. 14-2-27 CHEM 102 Spring 15, LA TECH K c vs. K p N 2 (g) + 3H 2 (g) 2NH 3 (g) In General K c = K p (1/RT)  n where  n = #moles gaseous products - # moles gaseous reactants (1/RT) 2 K c = K p = K p (1/RT )-2 (1/RT)(1/RT) 3

28. 14-2-28 CHEM 102 Spring 15, LA TECH What is K (K c ) and K p K c (K) - equilibrium constant calculated based on [A]-Concentrations. K p - equilibrium constant calculated based on partial pressure (p) K p = K c (RT)   n K p = K c (RT)   n K c = K p (RT)   n K c = K p (RT)   n R = universal gas constant (0.08206 R = universal gas constant (0.08206 ) T = Kelvin Temperature, T = Kelvin Temperature,  n = (sum of stoichiometric coefficients of gaseous products) - (sum of the stoichiometric coefficients of gaseous reactants) atm L mol K

29. 14-2-29 CHEM 102 Spring 15, LA TECH For the following equilibrium, K c = 1.10 x 10 7 at 700. o C. What is the K p ? 2H 2 (g) + S 2 (g) 2H 2 S (g) K p = K c (RT)  n g T = 700 + 273 = 973 K R= 0.08206  n g = ( 2 ) - ( 2 + 1) = -1 atm L mol K Partial pressure & Equilibrium Constants

30. 14-2-30 CHEM 102 Spring 15, LA TECH K p  = K c (RT)  n g = 1.10 x 10 7 (0.08206 ) (973 K) = 1.378 x10 5 atm L mol K [] Partial pressure & Equilibrium Constants

31. 14-2-31 CHEM 102 Spring 15, LA TECH Types of Equilibrium 1) Heterogeneous Equilibrium 2) Heterogeneous Equilibrium 3) Acid Dissociation Constant- K a 4) Base Dissociation Constant- K b 5) Autoionization Constant- K w 6) Solubility Product Constant-K sp

32. 14-2-32 CHEM 102 Spring 15, LA TECH Heterogeneous Equilibrium CaCO 3(s) CaO (s) + CO 2(g) [CaO(s)][CO 2 (g)] Kc = [CaCO 3 (s)] concentrations of pure solids and liquids are constant are dropped from expression K c = [CO 2 (g)]

33. 14-2-33 CHEM 102 Spring 15, LA TECH Acid Dissociation Constant HC 2 H 3 O 2 (aq) + H 2 O(l) H 3 O + (aq) + C 2 H 3 O 2 - (aq) [H 3 O + ][C 2 H 3 O 2 - ] K = [H 2 O][HC 2 H 3 O 2 ] [H 3 O + ][C 2 H 3 O 2 - ] K a = K  [H 2 O] = [HC 2 H 3 O 2 ]

34. 14-2-34 CHEM 102 Spring 15, LA TECH Base Dissociation Constant NH 3 + H 2 O(l) NH 4 + + OH - [NH 4 + ][OH - ] K = [H 2 O][NH 3 ] [NH 4 + ][OH - ] K b = K  [H 2 O] = [NH 3 ]

35. 14-2-35 CHEM 102 Spring 15, LA TECH Autoionization of Water H 2 O (l) + H 2 O (l) H 3 O + + OH - [H 3 O + ][OH - ] K = [H 2 O] 2 K w = K [H 2 O] 2 = [H 3 O + ][OH - ] = 1.0  10 -14

36. 14-2-36 CHEM 102 Spring 15, LA TECH Solubility Product of Salts in Water AgCl(s) + H 2 O (l) Ag + (aq) + Cl - K sp = [Ag + ] [Cl - ] K sp (AgCl) = 1.77 × 10 -10 K sp (BaSO 4 ) = 1.1 x 10 -10

37. 14-2-37 CHEM 102 Spring 15, LA TECH What is the reaction quotient, Q (Q) is constant in the equilibrium expression when initial concentration of reactants and products are used. SO 2 (g)+ NO 2 (g)  NO(g) +SO 3 (g) [NO][SO 3 ] [NO][SO 3 ] Q = ---------------- Q = ---------------- [SO 2 ][NO 2 ] [SO 2 ][NO 2 ] comparing to K and Q provide the net direction to achieve equilibrium.

38. 14-2-38 CHEM 102 Spring 15, LA TECH Predicting the Direction of a Reaction

39. 14-2-39 CHEM 102 Spring 15, LA TECH Consider the following reaction: SO 2 (g) + NO 2 (g)  NO(g) + SO 3 (g) (Kc = 85.0 at 460oC) Given: 0.040 mole of SO 2 (g), 0.500 mole of NO 2 (g), 0.30 mole of NO(g),and 0.020 mole of SO3(g) are mixed in a 5.00 L flask, Determine: a) The net the reaction quotient, Q. b) Direction to achieve equilibrium at 460 o C. Q Calculation

40. 14-2-40 CHEM 102 Spring 15, LA TECH Q Calculation Q Calculation SO 2 (g) + NO 2 (g)  NO(g) + SO 3 (g) (K c = 85.0 at 460 o C) [NO][SO 3 ] Q = ------------- [SO 2 ][NO 2 ] 0.040 mole 0.500 mole 0.30 mole 0.020 mole [SO 2 ] = -------------; [NO 2 ] = ----------- ; [NO] = ------------; [SO 3 ] = ----------- 5.00 L 5.00L 5.00L 5.00 L [SO 2 ] = 8 x 10 -3 mole/L ; [NO 2 ] =0.1mole/L; [NO] = 0.06 mole/L; [SO 3 ] = 4 x 10 - 3 mole/L 0.06 (4 x 10 -3 ) Q = ---------------------- = 0.3 8.0 x 10 -3 x 0.1 Therefore the equilibrium shift to right

41. 14-2-41 CHEM 102 Spring 15, LA TECH Le Chatelier’s principle Any stress placed on an equilibrium system will cause the system to shift to minimize the effect of the stress. You can put stress on a system by adding or removing something from one side of a reaction. N 2 (g) + 3H 2 (g) 2NH 3 (g) What effect will there be if you added more ammonia? How about more nitrogen?

42. 14-2-42 CHEM 102 Spring 15, LA TECH Predicting Shifts in Equilibria Equilibrium concentrations are based on: The specific equilibrium The starting concentrations Other factors such as: TemperatureTemperature PressurePressure Reaction specific conditionsReaction specific conditions Altering conditions will stress a system, resulting in an equilibrium shift.

43. 14-2-43 CHEM 102 Spring 15, LA TECH Increase in Concentration or Partial Pressure for N 2(g) + 3 H 2(g)  2 NH 3(g) an increase in N 2 and/or H 2 concentration or pressure, will cause the equilibrium to shift towards the production of NH 3

44. 14-2-44 CHEM 102 Spring 15, LA TECH N 2 O 4(g)  2 NO 2(g) ;  H=? (+or -) Shifts with Temperature N2O4(g) colorless 2NO2(g) Dark brown

45. 14-2-45 CHEM 102 Spring 15, LA TECH For the following equilibrium reactions: H 2 (g) + CO 2 (g)  H 2 O(g) + CO(g)  H = 40 kJ Predict the equilibrium shift if: a) The temperature is increased b) The pressure is decreased Predicting Equilibrium Shifts

46. 14-2-46 CHEM 102 Spring 15, LA TECH Shifting of Equilibrium N 2 O 4 (g)  2 NO 2(g)

47. 14-2-47 CHEM 102 Spring 15, LA TECH Changes in pressure In general, increasing the pressure by decreasing volume shifts equilibrium towards the side that has the smaller number of moles of gas. H 2 (g) + I 2 (g) 2HI (g) N 2 O 4 (g) 2NO 2 (g) Unaffected by pressure Increased pressure, shift to left

48. 14-2-48 CHEM 102 Spring 15, LA TECH −41.2 kJ/mol 5) How you would increase the products of following industrially important reactions: a) CO(g) + H 2 O (g) H 2 (g) + CO 2 (g);  H= −41.2 kJ/mol b) N 2 (g) nitrogen + 3H 2 (g) hydrogen 2NH 3 (g) ammonia H = -92.4 kJ mol -1

49. 14-2-49 CHEM 102 Spring 15, LA TECH Equilibrium Systems product-favored if K > 1 exothermic reactions favor products increasing entropy in system favors products at low temperature, product-favored reactions are usually exothermic at high temperatures, product-favored reactions usually have increase in entropy

50. 14-2-50 CHEM 102 Spring 15, LA TECH Thermodynamics of Equilibrium a) Enthalpy (  H) b) Entropy (  S) c) Free Energy (  G) (  G is a combined term involving  H,  S and T) (  G is a combined term involving  H,  S and T)

51. 14-2-51 CHEM 102 Spring 15, LA TECH Probability, Entropy and Chemical Equilibrium

52. 14-2-52 CHEM 102 Spring 15, LA TECH Entropy measure of the disorder in the system more disorder for gaseous systems than liquid systems, more than solid systems Chapter 18. Thermodynamics  G =  H -T  S   G = Gibbs Free Energy (- for spontaneous)   H = Enthalpy   S = Entropy  T = Kelvin Temperature

53. 14-2-53 CHEM 102 Spring 15, LA TECH Equilibrium Reaction Rates Equilibrium Reaction Rates reactions occur faster in gaseous phase than solids and liquids reactions rates increase as temperature increases reactions rates increase as concentration increases rates increase as particle size decreases rates increase with a catalyst

54. 14-2-54 CHEM 102 Spring 15, LA TECH Industrial Production of Ammonia N 2(g) + 3 H 2(g)  2 NH 3(g) ;  H = - catalysis high pressure and temperature

55. 14-2-55 CHEM 102 Spring 15, LA TECH Ammonia Synthesis reaction is slow at room temperature, raising temperature, increases rate but lowers yield increasing pressure shifts equilibrium to products liquefying ammonia shifts equilibrium to products use of catalyst increases rate

56. 14-2-56 CHEM 102 Spring 15, LA TECH Haber-Bosch Process

57. 14-2-57 CHEM 102 Spring 15, LA TECH Decrease in Concentration or Partial Pressure for N 2(g) + 3 H 2(g)  2 NH 3(g) ;  H = - likewise, a decrease in NH 3 concentration or pressure will cause more NH 3 to be produced

58. 14-2-58 CHEM 102 Spring 15, LA TECH Changes in Temperature for N 2(g) + 3 H 2(g)  2 NH 3(g) ;  H = - for an exothermic reaction, an increase in temperature will cause the reaction to shift back towards reactants and vice versa.

59. 14-2-59 CHEM 102 Spring 15, LA TECH Volume Change for N 2(g) + 3 H 2(g)  2 NH 3(g) ;  H = - an increase in volume, causes the equilibrium to shift to the left where there are more gaseous molecules a decrease in volume, causes the equilibrium to shift to the right where there are fewer gaseous molecules

60. 14-2-60 CHEM 102 Spring 15, LA TECH N 2 O 4(g)  2 NO 2(g) ;  H=? (+or -) Shifts with Temperature N2O4(g) colorless 2NO2(g) Dark brown

61. 14-2-61 CHEM 102 Spring 15, LA TECH Probability, Entropy and Chemical Equilibrium

62. 14-2-62 CHEM 102 Spring 15, LA TECH Entropy measure of the disorder in the system more disorder for gaseous systems than liquid systems, more than solid systems Chapter 17. Thermodynamics  G =  H -T  S   G = Gibbs Free Energy (- for spontaneous)   H = Enthalpy   S = Entropy  T = Kelvin Temperature

63. 14-2-63 CHEM 102 Spring 15, LA TECH For the following equilibrium reactions: H 2 (g) + CO 2 (g)  H 2 O(g) + CO(g);  H = 40 kJ Predict the equilibrium shift if: a) The temperature is increased b) The pressure is decreased Predicting Equilibrium Shifts

64. 14-2-64 CHEM 102 Spring 15, LA TECH Equilibrium Systems product-favored if K > 1 exothermic reactions favor products increasing entropy in system favors products at low temperature, product-favored reactions are usually exothermic at high temperatures, product-favored reactions usually have increase in entropy

65. 14-2-65 CHEM 102 Spring 15, LA TECH Equilibrium Reaction Rates Equilibrium Reaction Rates reactions occur faster in gaseous phase than solids and liquids reactions rates increase as temperature increases reactions rates increase as concentration increases rates increase as particle size decreases rates increase with a catalyst