1. CHEMICAL EQUILIBRIUM …….occurs when the forward and reverse reactions progress at the same rate in a reversible reaction, i.e. dynamic equilibrium. ……..is the state where concentrations of all reactants and products remain constant with time. Factors influencing equilibrium: forward and reverse rates partial pressures concentrations temperature

2. Rules of reversible first order reaction

3. EQUILIBRIUM CONSTANTS Use rate constants for forward and reverse reactions to calculate equilibrium constants. Rate f = k f [Butane] Rate r = k r [iButane] At equilibrium Rate f = rate r and k f [Butane] equil = k r [iButane] equil so Where Kc = equilibrium constant in M, mole/L.

4. EQUILIBRIUM CONSTANT CALCULATION OF K c LAW OF MASS ACTION: At a given T, the same Kc, state of equilibrium, will be reached regardless of initial conditions. For the reaction, jA +kB  lC + mD The standard state is 1 Molar.

5. When the reversible reaction 2 A + B  2 C Reached equilibrium, the following concentrations were measured: [A] = 0.40 M, [B] = 0.30 M, and [C] = 0.55 M. What is the value of Kc for this reaction?

6. EQUILIBRIUM CONSTANT CALCULATION, Kp Kp uses partial pressures of each species for gas phase reactions. jA(g) + kB(g)  lC(g) + mD(g) The standard state is 1 atmosphere At a specific temperature, the equilibrium constant, Kp, remains a constant regardless of the initial conditions.

7. At 427 o C, a 1.0-L flask contains 20.0 mol H 2, 18.0 mol CO 2, 12.0 mol H 2 O, and 5.9 mol CO at equilibrium. Calculate K for the reaction CO 2 (g) + H 2 (g)  CO(g) + H 2 O(g)

8. At 127 o C, K = 2.6 x 10 -5 mol 2 /L 2 for the reaction 2NH 3 (g)  N 2 (g) + 2H 2 (g) Calculate Kp at this temperature

9. REVERSE EQUILIBRIA: Kc = 1/Kc And Kp = 1/Kp When reverse reactions are written When a reaction is multiplied through by a value of n, then K new = (K original ) n

10. At a given temperature, K = 278 for the reaction 2SO 2 (g) + O 2 (g)  2SO 3 (g) Calculate values of K for the following reactions at this temperature. a. SO 2 (g) + ½ O 2 (g)  SO 3 (g) b. 2SO 3 (g)  2SO 2 (g) + O 2 (g) c. SO 3 (g)  SO 2 (g) + ½ O 2 (g) d. 4SO 2 (g) + 2O 2 (g)  4SO 3 (g)

11. Write the equilibrium expression (for Kp) for each of the following gas-phase reactions, which occur in the atmosphere: a. NO(g) + O 3 (g)  NO 2 (g) + O 2 (g) b. O 3 (g)  O 2 (g) + O(g) c. Cl(g) + O 3 (g)  ClO(g) + O 2 (g) d. 2O 3 (g)  3O 2 (g)

12. HETEROGENEOUS EQUILIBRIA For equilibria involving more than one state, (g), (l), (s), (aq), the (s) and (l) = 1 and are omitted (drop out) (g), (aq) use M, moles/L for Kc. (g), (s) and (l) use either Kp or Kc

13. Write expressions for K for the following reactions: a. P 4 (s) + 5O 2 (g)  P 4 O 10 (s) b. NH 4 NO 3 (s)  N 2 O(g) + 2H 2 O(g) c. CO 2 (g) + NaOH(s)  NaHCO 3 (s) d. S 8 (s) + 8O 2 (g)  8SO 2 (g)

14. Heterogeneous Equilibria 2NaHCO 3 (s)  Na 2 CO 3 (s) + H 2 O(g) + CO 2 (g) Kp = P CO2 P H2O Kc = [CO 2 ] [H 2 O] 2FeCl 3 (aq) + 3H 2 O(l)  2Fe(OH) 3 (s) + 6HCl(aq) Kc = ? But HCl(aq)  H + (aq) + Cl - (aq) Kc = ?

15. An equilibrium mixture contains 0.60 g solid carbon and the gases carbon dioxide and carbon monoxide at partial pressures of 2.9 atm and 2.6 atm, respectively. Calculate Kp for the reaction. C(s) + CO 2 (g)  2CO(g) A sample of gaseous PCl 5 was introduced into an evacuated flask so that the pressure of pure PCl 5 would be 0.50 atm at 523 K. However PCl 5 decomposes to gaseous PCl 3 and Cl 2, and the actual pressure in the flask was found to be 0.84 atm. Calculate Kp for the decomposition reaction. PCl 5 (g)  PCl 3 (g) + Cl 2 (g) At 523 K. Calculate Kp at this temperature

16. EQUILIBRIUM CONSTANTS AND SPONTANEITY Knowing Kp or Kc for a reaction and the reaction mixture concentrations or pressures, predict: 1. Will there be any net reaction? 2. If so, in which direction will it go? Calculate Q, then compare it to Kp or Kc. Q = the reaction Quotient, calculated with concentrations or pressures just as the Kp or Kc.

17. Kp, Kc and SPONTANEITY with the “Reaction Quotient, Q” K = QEquilibrium K > QSpontaneous as written. K < QSpontaneous in reverse.

18. At a particular temperature, Kp = 0.133 atm for the reaction N 2 O 4 (g)  2NO 2 (g) Which of the following conditions correspond to equilibrium positions? a. P NO2 = 0.144 atm, P N2O4 = 0.156 atm b. P NO2 = 0.175 atm, P N2O4 = 0.102 atm c. P NO2 = 0.056 atm, P N2O4 = 0.048 atm d. P NO2 = 0.064 atm,P N2O4 = 0.0308 atm

19. EQUILIBRIUM CONSTANT CALCULATIONS - a table of concentrations and changes - substitution into the Kc or Kp - algebraic manipulation including * perfect squares * dropping small components * quadratic equations

20. 1.Write a balanced chemical equation. 2.Select one of the concentration changes and call it x.* 3.Use the stoichiometry to determine all the concentration changes in terms of x. 4.Make a table containing the substances, their initial concentrations, their changes in concentration, and their equilibrium concentrations (calculated from the initial concentrations and the concentration changes). 5.Write the equilibrium-constant expression. 6.Insert the equilibrium concentrations from the table into the equilibrium-constant expression. 7.Solve the equation for x. 8.Check any simplifications for validity. 9.Substitute the value of x into the expressions for the equilibrium concentrations and determine their values. 10.Use the equilibrium concentrations to calculate the reaction quotient, and compare it with the equilibrium constant to verify the accuracy of the answers. *The same product works for partial pressures; simply substitute pressure for concentration in this list of steps.

21. The equilibrium constant, Kc, for the water-gas shift reaction has a value of 0.227 at 2000 K. CO(g) + H 2 O(g)  CO 2 (g) + H 2 (g) Suppose 0.0500 moles of CO and 0.0500 mole of H 2 O(g) are placed in a 2.00 liter flask at 2000 K. What are the equilibrium concentrations of all species?

22. Solving with the Quadratic Equation The equilibrium constant, Kc, at 55 degrees C for the dissociation of dinitrogen tetroxide is 0.0245. N 2 O 4 (g)  2NO 2 (g) Suppose 1.50 mole of N 2 O 4 (g) is place in a 10.0 liter container. What are the equilibrium concentrations of the species involved?

23. Solving with a substance added The equilibrium constant, Kc, at 55 o C for the dissociation of dinitrogen tetroxide is 0.0245. N 2 O 4 (g)  2NO 2 (g) Suppose 1.50 mole of N 2 O 4 (g) is placed in a 10.0 liter container. The equilibrium concentrations of the species are [N 2 O 4 (g)] = 0.123 M and [NO 2 (g)] = 0.0548 M. Now 0.0200 mole/L NO 2 (g) is added to the mixture. What are the new equilibrium concentrations of the species involved?

24. At a particular temperature, assume that K = 1.0 x 10 2 for the reaction H 2 (g) + F 2 (g)  2HF(g) a. In an experiment, 2.0 mol H 2 and 2.0 mol F 2 are introduced into a 1.0-L flask. Calculate the concentrations of all species when equilibrium is reached. b. To the equilibrium mixture in part a, an additional 0.40 mol H 2 is added. Calculate the new equilibrium concentrations of H 2, F 2, and HF.

25. H) A general system: A (g) + B (g)  C (g) + D (g) 1) At time t o, molecules of A and molecules of B are introduced into a reaction flask. 2) Molecules of A react with molecules of B to form molecules of C and D. 3) As time goes on molecules of C react with molecules of D to form molecules of A and B. 4) At time equilibrium the [ ]'s of all the molecules remain constant, not equal.

26. 5) Equilibrium is reached when the ________ ______________________________________. Unless conditions change the concentrations of all species will _______________________. II. THE APPROACH TO EQUILIBRIUM A) Equilibrium can usually be established in a number of different ways.

27. III. THE EQUILIBRIUM CONSTANT A) At 100 o C, the reaction of: N 2 O 4  2NO 2 has been investigated.

28. What do you notice about the change in the number of moles of NO 2 compared to the change in the number of moles of N 2 O 4 ? Example from Experiment 1 N 2 O 4  2NO 2

29. C) A quantitative relationship exists between the equilibrium concentration of NO 2 and N 2 O 4 which is valid for all three experiments:

30. D) This relationship holds for any equilibrium system containing NO 2 and N 2 O 4 at 100 o C. The equilibrium constant for the reaction changes with temperature as follows:

31. This constant is dependent on the temperature, BUT independent of the amounts of N 2 O 4 and NO 2 with which you start, the volume of the container or the total pressure.

32. E) The general expression for the equilibrium constant, K c, for the general equation: aA (g) + bB (g)  cC (g) + dD (g) IS

33. If the reaction is written as follows: cC (g) + dD (g)  aA (g) + bB (g) What will be the expression for the equilibrium constant? How does this K relate to the K c from the original equation? If K c is 10, then K c ' is equal to _______________.

34. F) Heterogeneous equilibrium involves equations containing ___________________, and ____________________________ as well as gases in the balanced equation. 1)For example the following equation: CaCO 3(s)  CaO (s) + CO 2(g) It has been experimentally shown that the position of equilibrium is independent of the amount of CaCO 3(s) and CaO (s) in the flask and the concentrations of these species do not appear in the K c expression.

35. K c =[CO 2 ] This can be understood in terms of the following expression for the equation: CaCO 3(s)  CaO (s) + CO 2(g)

36. 2) Write the equilibrium expression for the following reactions with respect to Kc. N 2(g) + 3 H 2(g)  2 NH 3(g) BaSO 4(s )  BaO (s) + SO 3(g) 4 NH 3(g) + 3 O 2(g) 2 N 2(g) + 6 H 2 O (g)

37. IV. Calculating the Value of the Equilibrium Constant A) The following reaction has be studied at 2000 o C: N 2(g) + O 2(g)  2 NO (g) If in one experiment, the equilibrium [N 2 ] = 0.036 M, the eq. [O 2 ] = 0.0089 and the eq. [NO] = 3.6 X 10-4, what is the value of Kc?

38. B) A more difficult problem follows with respect to the following equation: 2 NO (g) + O 2(g)  2 NO 2(g) Calculate Kc from the following data: After mixing 0.50 mol NO and 0.50 mol of O 2 in a 1.0 Liter flask at 1.0 atm and 10 o C, it is found that the equilibrium concentration of NO is 0.30 M (mol/L). Why is this problem more difficult?

39. We set up the Kc expression as follows: How do we find the concentrations of NO 2 and O 2 ?

40. We make a chart like the following: How do we find the value of x? What is the value of K c ?

41. V. APPLICATIONS OF K c A) K c can be used to decide the 1) extent to which a reaction will occur. 2) direction a chemical reaction will move to achieve equilibrium. 3) effect a change of conditions will have on a chemical system.

42. B) Predicting the extent of a reaction. 1)A large value of Kc indicates that the reaction proceeds far to the right as the reaction is written. Most of the molecules in the flask are HCl molecules, very few molecules are H 2(g) and Cl 2(g) molecules. H 2(g) + Cl 2(g)  2 HCl (g) K c = 2.18 x 10 33 at 25 o C

43. 2) A small value of K c indicates that the reaction proceeds only slightly to the right. N 2(g) + O 2(g)  2 NO (g) K c = 4.26 x 10 -31 at 25 o C There are very few molecules of ____in the flask. There are lots of _____and _______ molecules in the flask. That is great, since we do not want lots of _____ molecules in the air since _________________________.

44. C) Prediction of direction of the reaction. 1) If a reaction is not at equilibrium, or we wish to know whether or not it is, we will use the concept of Q c, known as the reaction quotient. 2) For the general reaction: aA (g) + bB (g)  cC (g) + dD (g)

45. If Q c is less than (<) K c then [C] and the [D] must _____________and the [A] and the [B] must _____________. There must be a shift to the ____________. If Q c is greater than (>) K c then [C] and the [D] must __________ and ________________ ________________ and the reaction shifts __. If Qc is equal to (=) Kc then the ___________ __________________________.

46. D) Calculation of the equilibrium concentrations of the reactants and products given a variety of information. 1)For the reaction: 2 NO(g)  N 2(g) + O 2(g) K c = 8.36 X 10 3 What concentration of NO is in equilibrium with N 2 at 0.0500 M and O 2 at 0.0500 M?

47. THINK BEFORE PERFORMING ANY OPERATION! What do I want you to think about? Is the answer going to be big or small? Since K c is large, the product concentration will be large, so the reactant concentration will be small.

49. The other way to do the problem is to:

50. 2) For the reaction: H 2(g) + I 2(g)  2 HI (g) K c = 50.5 at 448 o C If initially a 1.00 L container is filled with 0.500 moles of HI at 448 o C, what are the concentrations of H 2, I 2, and HI at equilibrium? We make a table of the [ ]'s of each of the species first.

51. We can then substitute the values obtained from the table into the expression for K c and solve for x.

52. What can we do to make solving for x as easy as possible? You need to be able to recognize that the right hand side is a perfect square, so we can take the square root of both sides.

53. 3) Consider the system N 2 O 4  2 NO 2 K c = 0.36 at 100 o C Suppose we start with pure N 2 O 4 at a concentration of 0.100 mol/L. What are the equilibrium concentrations of NO 2 and N 2 O 4 ?

54. Let's look at the equation of H 2(g) + I 2(g)  2 HI (g) We put H 2 gas and I 2 gas into a container and allow equilibrium to be reached. Then we add more I 2 gas to the flask. We no longer have _____________. Since I 2 is now larger than the equilibrium concentration we have:

55. How do we get Q c = K c ? The [H 2 ] and the new [I 2 ] must decrease and the [HI] must increase. K c STAYS THE SAME.

56. For example: N 2(g) + 3 H 2(g)  2NH 3(g) K c = 5.96 X 10 -2 at some T

57. B) Effect of a catalyst - A catalyst increases the rate of both the forward and the reverse reactions. Equilibrium is reached faster, but there is no change in the equilibrium concentration of the species and no change in K c. C) Effect of a change in pressure - There are three ways to change the pressure:

58. 1) Add or remove a gaseous reactant or product. 2) Add an inert gas (He) that is not involved in the the reaction. 3) Change the volume of the container. D) Adding or removing a gaseous reactant or product is the same as changing the concentration. That has already been examined and we saw that the equilibrium was shifted, BUT there was no change in K c.

59. E) Addition of an inert gas changes the total pressure BUT has no effect on the concentrations or partial pressure of the reactants or products. The system is unaffected and remains at equilibrium. F) The effect of a change in volume - When we affect a change in pressure by a decrease in the volume of the container, the system responds by decreasing its own volume. How is this accomplished?

60. PV = nRT at constant P and T, V is proportional to n. N 2(g) + 3 H 2(g)  2NH 3(g) There are 4 moles of gaseous reactants and on 2 moles of gaseous products. At 200 o C, starting with 1.0 mol of N 2 and 3.0 mol of H 2, how does the percentage of NH 3 produced with changing pressure due to changing the volume?

61. The equilibrium shifts to the side of the equation with ________________________. If you were building an NH 3 factory, what economic factors would you consider with respect to getting more NH 3 with pressure greater than 50 atm?

62. For the reaction: H 2(g) + I 2(g)  2 HI (g) The position of the equilibrium is _________. What about the following reaction? P 4(s) + 6 Cl 2(g)  4 PCl 3(l) If the volume is decreased, pressure increased, the eq. shifts to the __________. Why? If the volume is increased, the pressure decreased the eq. shifts to the _____________.

63. A CHANGE IN TEMPERATURE WILL CHANGE THE POSITION OF EQUILIBRIUM AS WELL AS K c. 1) Changing the temperature has a profound effect on most reactions. K c 's change with temperature.

64. For the reaction: N 2(g) + O 2(g)  2 NO (g)  H = 181 kJ Heat + N 2(g) + O 2(g)  2 NO (g) If we add heat as in the combustion process of "hot" auto engines, it favors _____________. What happens to K c ?

65. As we heat the mixture the concentration of the products _____________, the concentration of reactants _______________, so K c must __________________.

66. For the reaction: N 2(g) + 3 H 2(g)  2NH 3(g) at 200 o C K c = 65 at 400 o C K c = 0.50 Observation of the difference in K c ’s, indicates that the addition of heat _________ the amount of ammonia produced. Therefore we can predict that the production of ammonia is an ____________ reaction. N 2(g) + 3 H 2(g) 2 NH 3(g)  H = ____

67. G) SUMMARY Consider the following reaction: CO (g) + H 2 O (g)  H 2(g) + CO 2(g)  H = - Assume equilibrium has been reached and all species are present in a closed container. In which direction will the position of equilibrium be shifted by each of the following changes? How will K c be affected by the same changes?

68. CO (g) + H 2 O (g)  H 2(g) + CO 2(g)  H = - 1) Add CO (shifts ________) K c ___________ 2) Add H 2 O (g) ___________________________ 3) Add H 2(g) ____________________________ 4) Some CO 2(g) is removed ________________ 5) The temperature is raised ______________ 6) Add a catalyst ________________________ 7) Helium is added __________________ 8) The volume is decreased _______________