1. 14-1 CHEM 102, Fall 2011, LA TECH Instructor: Dr. Upali Siriwardane e-mail: upali@chem.latech.edu Office: CTH 311 Phone 257-4941 Office Hours: M,W 8:00-9:00 & 11:00-12:00 am; Tu, Th, F 8:00 - 10:00am. Exams: 11:30-12:45 am, CTH 322. Sept 22, 2011 (Test 1): Chapter 13 Oct 18, 2011 (Test 2): Chapter 14 &15 Nov 15, 2011 (Test 3): Chapter 17 &18 Comprehensive Final Exam: Nov 17, 2011 : Chapters 13, 14, 15, 16, 17, and 18 Chemistry 102(01) Fall 2011

2. 14-2 CHEM 102, Fall 2011, LA TECH Chapter 14. Chemical Equilibrium 14.1Characteristics of Chemical Equilibrium 14.2The Equilibrium Constant 14.3Determining Equilibrium Constants 14.5The Meaning of Equilibrium Constant 14.6Using Equilibrium Constants 14.7Shifting a Chemical Equilibrium: Le Chatelier's Principle 14.8Equilibrium at the Nanoscale 14.9Controlling Chemical Reactions: The Haber-Bosch Process

3. 14-3 CHEM 102, Fall 2011, LA TECH Different types of arrows are used in chemical equations associated with equilibria. Single arrow Assumes that the reaction proceeds to completion as written. Two single-headed arrows Used to indicate a system in equilibrium. Two single-headed arrows of different sizes. May be used to indicate when one side of an equilibrium system is favored. Chemical equilibrium

4. 14-4 CHEM 102, Fall 2011, LA TECH Chemical Equilibrium Branch of chemistry dealing with reactions where reactants and products coexist in a dynamic equilibrium the rates of forward and backward reactions have comparable rates reaction

5. 14-5 CHEM 102, Fall 2011, LA TECH Chemical Equilibrium Equilibrium region. A point is finally reached where the forward and reverse reactions occur at the same rate. H 2 + I 2 2HI There is no net change in the concentration of any of the species.

6. 14-6 CHEM 102, Fall 2011, LA TECH Chemical Equilibrium Partial Pressure Time HI I2I2 H2H2 Equilibrium Region Kinetic Region

7. 14-7 CHEM 102, Fall 2011, LA TECH Complete Reaction Concentration Time KineticNo change Region

8. 14-8 CHEM 102, Fall 2011, LA TECH Equilibrium A state where the forward and reverse conditions occur at the same rate. Dynamic Equilibrium I’m in static equilibrium.

9. 14-9 CHEM 102, Fall 2011, LA TECH This type of plot shows the energy changes during a reaction. This type of plot shows the energy changes during a reaction. Forward and Backward Reactions HH activation energy Potential Energy Reaction coordinate

10. 14-10 CHEM 102, Fall 2011, LA TECH Value of K rate of forward Reaction Reaction k+ K = ------------------------------ = --- rate of backward k- K = (infinity) -> Irreversible reactions K = 0 -> No reaction K = between 0 and 1 -> Equilibrium reactions

11. 14-11 CHEM 102, Fall 2011, 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.

12. 14-12 CHEM 102, Fall 2011, LA TECH Equilibrium Expression An equilibrium expression could be written for any reaction [HI] 2 [HI] 2 K = ----------- = 16 L/mol K = ----------- = 16 L/mol [H 2 ][I 2 ] [H 2 ][I 2 ] K eq >> 1reaction will go mainly to products K eq ~ 1 reaction will produce roughly equal amounts of product and reactant K eq << 1reaction will go mainly to reactants

13. 14-13 CHEM 102, Fall 2011, LA TECH k is constant at a temperature 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 N2O4(g) colorless 2NO2(g) Dark brown

14. 14-14 CHEM 102, Fall 2011, LA TECH Calculating Stepwise Equilibrium Add two equations with K 1 and K 2 to get K eq K eq = K 1 x K 2 Subtract one equations with K 2 from another with K 2 to get K eq K eq = K 1 / K 2 Doubling K 1 to get K eq K eq = (K 1 ) 2 ;tripling K eq = (K 1 ) 3 etc. Reversing a reaction with K 1 get K eq K eq = (K 1 ) ½

15. 14-15 CHEM 102, Fall 2011, LA TECH Stepwise Equilibrium (1)N 2 (g ) + O 2 (g)  2NO(g) [NO] 2 K c1 = [N 2 ][O 2 ] (2)2NO(g) + O 2 (g)  2NO 2 (g) [NO 2 ] 2 K c2 = [NO] 2 [O 2 ] Add to Combine (1.) & (2.) N2(g) + 2O 2 (g)  2NO 2 (g) [NO] 2 [NO 2 ] 2 K c =  = K c1  K c2 [N 2 ][O 2 ] [NO] 2 [O 2 ]

16. 14-16 CHEM 102, Fall 2011, LA TECH Stepwise Equilibrium Consider the reactions 2NO + O 2 2 NO 2 K = a 2 NO 2 N 2 O 4 K = b The value of the equilibrium constant for the reaction 2NO + O 2 N 2 O 4 is a.a + b b.ab c.(a/b) 2 d.(ab) 2 e.ab/2

17. 14-17 CHEM 102, Fall 2011, LA TECH Stepwise Equilibrium Consider the reactions 2NO + O 2 2 NO 2 K = a 2 NO 2 N 2 O 4 K = b The value of the equilibrium constant for the reaction 4NO + 2O 2 2 N 2 O 4 is a.a + b b.ab c.(a/b) 2 d.(ab) 2 e.ab/2

18. 14-18 CHEM 102, Fall 2011, LA TECH Homogenous equilibrium: Chemical equilibrium where reactants and products are in same phase. Heterogeneous equilibrium: Chemical Equilibrium where at least one phase of a reactant or product is different from the rest. Types of Equilibria

19. 14-19 CHEM 102, Fall 2011, LA TECH Homogenous equilibrium: Chemical equilibrium where reactants and products are in same phase. Heterogeneous equilibrium: Chemical Equilibrium where at least one phase of a reactant or product is different from the rest. Types of Equilibria

20. 14-20 CHEM 102, Fall 2011, 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)]

21. 14-21 CHEM 102, Fall 2011, 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 ]

22. 14-22 CHEM 102, Fall 2011, 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 ]

23. 14-23 CHEM 102, Fall 2011, 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

24. 14-24 CHEM 102, Fall 2011, 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

25. 14-25 CHEM 102, Fall 2011, 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 = Kp (1/RT )-2 (1/RT)(1/RT) 3

26. 14-26 CHEM 102, Fall 2011, 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(RT)   n K p = K(RT)   n R = universal gas constant R = universal gas constant T = Kelvin Temperature, T = Kelvin Temperature,  n = (sum of stoichiometric coefficients of gaseous products) - (sum of the stoichiometric coefficients of gaseous reactants)

27. 14-27 CHEM 102, Fall 2011, 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

28. 14-28 CHEM 102, Fall 2011, 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

29. 14-29 CHEM 102, Fall 2011, 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

30. 14-30 CHEM 102, Fall 2011, 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.

31. 14-31 CHEM 102, Fall 2011, 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- 0.243= 0.138 0.525 x-1/2 x 0.40 - 1/2x = 0 + x

32. 14-32 CHEM 102, Fall 2011, 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)

33. 14-33 CHEM 102, Fall 2011, 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 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?

34. 14-34 CHEM 102, Fall 2011, LA TECH I nitial I nitial 4.00/5=.80 0 0 C hange C hange -2x x x Equilibrium Equilibrium 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)

35. 14-35 CHEM 102, Fall 2011, LA TECH Selected Equilibrium Constants

36. 14-36 CHEM 102, Fall 2011, 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.

37. 14-37 CHEM 102, Fall 2011, 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

38. 14-38 CHEM 102, Fall 2011, 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

39. 14-39 CHEM 102, Fall 2011, LA TECH Predicting the Direction of a Reaction

40. 14-40 CHEM 102, Fall 2011, 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

41. 14-41 CHEM 102, Fall 2011, 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][SO3] Q = ------------- [SO2][NO2] 0.040 mole 0.500 mole 0.30 mole 0.020 mole [SO2] = -------------; [NO2] = ----------- ; [NO] = ------------; [SO3] = ----------- 5.00 L 5.00L 5.00L 5.00 L [SO2] = 8 x 10 -3 mole/L ; [NO2] =0.1mole/L; [NO] = 0.06 mole/L; [SO3] = 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

42. 14-42 CHEM 102, Fall 2011, 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?

43. 14-43 CHEM 102, Fall 2011, 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)

44. 14-44 CHEM 102, Fall 2011, LA TECH Equilibrium calculation example X 2 (0.095 - X ) Kc = 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

45. 14-45 CHEM 102, Fall 2011, 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

46. 14-46 CHEM 102, Fall 2011, 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 = - 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

47. 14-47 CHEM 102, Fall 2011, 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.

48. 14-48 CHEM 102, Fall 2011, 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?

49. 14-49 CHEM 102, Fall 2011, 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.

50. 14-50 CHEM 102, Fall 2011, 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

51. 14-51 CHEM 102, Fall 2011, LA TECH N 2 O 4(g)  2 NO 2(g) ;  H=? (+or -) Shifts with Temperature N2O4(g) colorless 2NO2(g) Dark brown

52. 14-52 CHEM 102, Fall 2011, LA TECH Probability, Entropy and Chemical Equilibrium

53. 14-53 CHEM 102, Fall 2011, 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

54. 14-54 CHEM 102, Fall 2011, 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

55. 14-55 CHEM 102, Fall 2011, 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

56. 14-56 CHEM 102, Fall 2011, LA TECH Shifting of Equilibrium N 2 O 4 (g)  2 NO 2(g)

57. 14-57 CHEM 102, Fall 2011, 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

58. 14-58 CHEM 102, Fall 2011, 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

59. 14-59 CHEM 102, Fall 2011, LA TECH Production of Ammonia N 2(g) + 3 H 2(g)  2 NH 3(g) ;  H = - catalysis high pressure and temperature

60. 14-60 CHEM 102, Fall 2011, 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

61. 14-61 CHEM 102, Fall 2011, LA TECH Haber-Bosch Process

62. 14-62 CHEM 102, Fall 2011, 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

63. 14-63 CHEM 102, Fall 2011, 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.

64. 14-64 CHEM 102, Fall 2011, 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

65. 14-65 CHEM 102, Fall 2011, LA TECH At 100o C the equilibrium constant (K) for the reaction: H 2 (g) + I 2 (g)  2HI(g) is 1.15 x 10 2. If 0.400 moles of H 2 and 0.400 moles of 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? Calculating Concentrations at Equilibrium

66. 14-66 CHEM 102, Fall 2011, LA TECH At a certain temperature the value of the equilibrium constant is 3.24 for the reaction: H 2 (g) + CO 2 (g)  H 2 O(g) + CO(g) If 0.400 mol H 2 and 0.400 mol CO 2 are placed in a 1.00 L vessel, what is the concentration of of CO at equilibrium? Calculating Concentrations at Equilibrium

67. 14-67 CHEM 102, Fall 2011, 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?

68. 14-68 CHEM 102, Fall 2011, 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.

69. 14-69 CHEM 102, Fall 2011, 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

70. 14-70 CHEM 102, Fall 2011, LA TECH N 2 O 4(g)  2 NO 2(g) ;  H=? (+or -) Shifts with Temperature N2O4(g) colorless 2NO2(g) Dark brown

71. 14-71 CHEM 102, Fall 2011, LA TECH Probability, Entropy and Chemical Equilibrium

72. 14-72 CHEM 102, Fall 2011, 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

73. 14-73 CHEM 102, Fall 2011, 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

74. 14-74 CHEM 102, Fall 2011, LA TECH Changes in pressure In general, increasing the pressure by decreasing volume shifts equilibria 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

75. 14-75 CHEM 102, Fall 2011, LA TECH Shifting of Equilibrium N2O4(g)  2 NO 2(g)

76. 14-76 CHEM 102, Fall 2011, 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

77. 14-77 CHEM 102, Fall 2011, 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

78. 14-78 CHEM 102, Fall 2011, LA TECH Production of Ammonia N 2(g) + 3 H 2(g)  2 NH 3(g) ;  H = - catalysis high pressure and temperature

79. 14-79 CHEM 102, Fall 2011, 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

80. 14-80 CHEM 102, Fall 2011, 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

81. 14-81 CHEM 102, Fall 2011, 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.

82. 14-82 CHEM 102, Fall 2011, 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