1. 1 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Section 14 Chemical Equilibrium

2. 2 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Many chemical reactions do not completely convert reactants to products. Stop somewhere between no rxn and complete rxn. A + B C + D some left some formed reversible (both directions) Chemical Equilibrium Previously we have assumed that chemical reactions results in complete conversion of reactants to products: A + B C + D No A or B remaining or possibly an excess of A or B but not both and eventually reaction stops

3. 3 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. The concentration of all species level off to some individual constant value that persists indefinitely as long as no stress is added to system. The state that is reached when the concentrations of R and P remain a constant in time is called state of chemical equilibrium and the mixture is considered an equilibrium mixture. Individual molecules are continually reacting even though the overall composition of the reaction does not change. There is a exchange going on: A + B forms C+ D but then C + D breaks down into A + B at the same rate; therefore, concentration of each component remains a constant and the rate of exchange is constant at equilibrium. Reaction has not stopped; just the rates of the forward and reverse reactions are equal.

4. 4 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Chemical Equilibrium Therefore, many reactions do not go to completion but rather form a mixture of products and unreacted reactants, in a dynamic equilibrium. –A dynamic equilibrium consists of a forward reaction, in which substances react to give products, and a reverse reaction, in which products react to give the original reactants. –Chemical equilibrium is the state reached by a reaction mixture when the rates of the forward and reverse reactions have become equal.

5. 5 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Graphically we can represent this A + B C + D The concentrations and reaction rate (less collisions, less component) of A and B decreases over time as the concentrations and reaction rate of C and D increases (more collisions, more component) over time until; the rates are equal and the concentrations of each components reaches a constant. This occurs at what we call equilibrium -- R f = R r. If the rates are equal, then there must be a relationship to show this.

6. 6 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Figure : Catalytic methanation reaction approaches equilibrium. CO + 3 H 2 CH 4 + H 2 O

7. 7 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. For the reverse reaction we have, C + D A + B R r = k r [C][D] We know at equil that R f = R r ;therefore, we can set these two expressions as equal k f [A][B] = k r [C][D] Rearrange to put constants on one side we get If we assume these reactions are elementary rxns (based on collisions), we can write the rate laws directly from the reaction: A + B C + D R f = k f [A][B]

8. 8 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Constant divided by constant just call a new constant K. This ratio is given a special name and symbol called equilibrium constant K relating to the equilibrium condition at a certain temperature (temp dependent) for a particular reaction relating conc of each component. This is basically a comparison between forward and reverse reaction rates. At equilibrium, the ratio of conc of species must satisfy K.

9. 9 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. The Equilibrium Constant Every reversible system has its own “position of equilibrium”- K- under any given set of conditions. –The ratio of products produced to unreacted reactants for any given reversible reaction remains constant under constant conditions of pressure and temperature. If the system is disturbed, the system will shift and all the concentrations of the components will change until equilibrium is re-established which occurs when the ratio of the new concentrations equals what -- "K". Different constant conc but ratio same as before. –The numerical value of this ratio is called the equilibrium constant for the given reaction, K.

10. 10 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. The Equilibrium Constant The equilibrium-constant expression for a reaction is obtained by multiplying the equil concentrations ( or partial pressures) of products, dividing by the equil concentrations (or partial pressures) of reactants, and raising each concentration to a power equal to its coefficient in the balanced chemical equation. –The molar concentration of a substance is denoted by writing its formula in square brackets for aq solutions. For gases can put P a - atm. As long as use M or atm, K is unitless. –Temp dependent; any changes, ratio must equal K when equil established

11. 11 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. The Equilibrium Constant The equilibrium constant, K, is the value obtained for the equilibrium-constant expression when equilibrium concentrations (not just any conc but equil conc) are substituted. –A large K indicates large concentrations of products at equilibrium. –A small K indicates large concentrations of unreacted reactants at equilibrium.

12. 12 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. The Equilibrium Constant –Do same set up for all equil equations (future chapters), just different subscript describing the reaction in question: i.e. acid hydrolysis - acid + water - K a, K b, K c, K p, K sp –K c is based on conc (M) and K p is based on pressures (atm). There is a difference between K c and K p and to jump between these two, there is an equation to use. Let's show how to jump between. HW 14

13. 13 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. The Equilibrium Constant, K p In discussing gas-phase equilibria, it is often more convenient to express concentrations in terms of partial pressures rather than molarities. –It can be seen from the ideal gas equation (PV =nRT) that the partial pressure of a gas is proportional to its molarity –(P proportional to M - n/V ; therefore handled the same way.

14. 14 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. The Equilibrium Constant, K p –Consider the reaction below. –The equilibrium-constant expression in terms of partial pressures becomes (same way but partial pressures instead of M):

15. 15 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. The Equilibrium Constant, K p In general if you need to jump between K's, the numerical value of K p differs from that of K c. where  n is the sum of the moles of gaseous products in a reaction minus the sum of the moles of gaseous reactants.

16. 16 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. A Problem to Consider Consider the reaction –K c for the reaction is 280 at 1000. K. Calculate K p for the reaction at this temperature.

17. 17 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. A Problem to Consider –We know that Consider the reaction at 1000. K and K c = 280

18. 18 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. A Problem to Consider –Since Consider the reaction

19. 19 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. The Equilibrium Constant The law of mass action states that the value of the equilibrium constant expression K is constant for a particular reaction at a given temperature, whenever equilibrium concentrations are substituted. When at equil, conc ratio equals K and concs are constant but if disturbed, values change until equil reached, different constant conc but same ratio of K).

20. 20 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Obtaining Equilibrium Constants for Reactions Equilibrium concentrations for a reaction must be obtained experimentally and then substituted into the equilibrium- constant expression in order to calculate K c.

21. 21 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Obtaining Equilibrium Constants for Reactions Consider the reaction below –Suppose we started with initial concentrations of CO and H 2 of 0.100 M and 0.300 M, respectively. –Obviously shift to right, decrease CO and H 2 and increase CH 4 and water until equil established.

22. 22 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. –When the system finally settled into equilibrium we determined the equilibrium concentrations to be as follows. ReactantsProducts [CO] = 0.0613 M [H 2 ] = 0.1839 M [CH 4 ] = 0.0387 M [H 2 O] = 0.0387 M Obtaining Equilibrium Constants for Reactions

23. 23 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. –If we substitute the equilibrium concentrations, we obtain: Obtaining Equilibrium Constants for Reactions

24. 24 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. –Regardless of the initial concentrations (whether they are reactants or products or mixture), the law of mass action dictates that the reaction will always settle into an equilibrium where the equilibrium-constant expression equals K c. Obtaining Equilibrium Constants for Reactions

25. 25 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. –As an example, let’s repeat the previous experiment, only this time starting with initial concentrations of products (note: if only products to start shift left until equil established): [CH 4 ] initial = 0.2000 M and [H 2 O] initial = 0.2000 M –Obviously shift to left decrease CH 4 and H 2 O and increase CO and H 2 until equil established. Obtaining Equilibrium Constants for Reactions

26. 26 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. –We find that these initial concentrations result in the following equilibrium concentrations. ReactantsProducts [CO] = 0.0990 M [H 2 ] = 0.2970 M [CH 4 ] = 0.1010 M [H 2 O] = 0.1010 M Obtaining Equilibrium Constants for Reactions

27. 27 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. –Substituting these values into the equilibrium- constant expression, we obtain the same result. –Whether we start with reactants or products at any initial conc, the system establishes the same ratio. Obtaining Equilibrium Constants for Reactions

28. 28 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. A Problem to Consider Applying Stoichiometry to an Equilibrium Mixture (basic setup for future problems). –What is the composition of the equilibrium mixture if it contains 0.080 mol NH 3 at equilibrium? –Suppose we place 1.000 mol N 2 and 3.000 mol H 2 in a reaction vessel at 450 o C and 10.0 atmospheres of pressure. The reaction is

29. 29 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. –This procedure for many types of problems, however in this problem given equil quantity of NH 3 ;therefore, figure out rest. –The equilibrium amount of NH 3 was given as 0.080 mol. Therefore, 2x = 0.080 mol NH 3 (x = 0.040 mol). Using the information given, set up the following table. (ratio works for atm, M, mols, etc.) Initial, n o Change,  n Equil, n eq

30. 30 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. A Problem to Consider Using the information given, set up the following table. Equilibrium amount of N 2 = Equilibrium amount of H 2 = Equilibrium amount of NH 3 = Starting 1.0003.0000 Change Equilibrium HW 15

31. 31 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Equilibrium Constant for the Sum of Reactions Similar to the method of combining reactions that we saw using Hess’s law in Chapter 6, we can combine equilibrium reactions whose K values are known to obtain K for the overall reaction. –With Hess’s law, when we reversed reactions (change sign) or multiplied them prior to adding them together (mult by factor). We had to manipulate the  H’s values to reflect what we had done. –The rules are a bit different for manipulating K.

32. 32 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. 2.If you multiply/divide each of the coefficients in an equation by the same factor (2, 1/2, …), raise K c to the same power (2, 1/2, …). (known as coefficient rule, K n ) 3.When you finally combine (that is, add) the individual equations together, take the product of the equilibrium constants to obtain the overall K.(rule of multiple equilibria, K 1 x K 2 x K 3 … = K T ) 1.If you reverse a reaction, invert the value of K. (reciprocal rule, 1/K) Equilibrium Constant for the Sum of Reactions

33. 33 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. For example, nitrogen and oxygen can combine to form either NO(g) or N 2 O (g) according to the following equilibria. K c = 6.4 x 10 -16 K c = 2.4 x 10 -18 (1) (2) K c = ? –Using these two equations, we can obtain K for the formation of NO(g) from N 2 O(g): overall Equilibrium Constant for the Sum of Reactions

34. 34 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. (1) (2) overall HW 16

35. 35 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Heterogeneous Equilibria A heterogeneous equilibrium is an equilibrium that involves reactants and products in more than one phase. Up to now all our reactions have been homogeneous - all gases or aqueous solutions. –The equilibrium of a heterogeneous system is unaffected by the amounts of pure solids or liquids present, as long as some of each is present. –The concentrations of pure solids and liquids are always considered to be “1 activity” and therefore, do not appear in the equilibrium expression. Solids and pure liquids have no effect on conc or pressure.

36. 36 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Heterogeneous Equilibria Consider the reaction below. HW 17

37. 37 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Predicting the Direction of Reaction How could we predict the direction in which a reaction at non-equilibrium conditions will shift to reestablish equilibrium? Remember did example with only reactants, obviously had to go right and if only products obviously must go left but what if have some of R and P? –To answer this question, substitute the current concentrations into the reaction quotient expression and compare it to K c. –The reaction quotient, Q c, is an expression that has the same form as the equilibrium-constant expression but whose concentrations are not necessarily at equilibrium.

38. 38 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Predicting the Direction of Reaction For the general reaction the Q c expresssion would be (i=initial):

39. 39 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Predicting the Direction of Reaction For the general reaction –If Q c = K c, then the reaction is at equilibrium. –If Q c > K c, the reaction will shift left toward reactants until equil reached. –If Q c < K c, the reaction will shift right toward products until equil reached.

40. 40 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. A Problem to Consider –Consider the following equilibrium. –A 50.0 L vessel contains 1.00 mol N 2, 3.00 mol H 2, and 0.500 mol NH 3. Is the sytem at equil? If not, in which direction (toward reactants or toward products) will the system shift to reestablish equilibrium at 400 o C? –K c for the reaction at 400 o C is 0.500.

41. 41 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. A Problem to Consider –First, calculate concentrations from moles of substances.

42. 42 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. A Problem to Consider –First, calculate concentrations from moles of substances.

43. 43 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. A Problem to Consider –First, calculate concentrations from moles of substances. HW 18

44. 44 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Calculating Equilibrium Concentrations Once you have determined the equilibrium constant, K, for a reaction, you can use it to calculate the concentrations of substances in the equilibrium mixture.

45. 45 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Calculating Equilibrium Concentrations –Suppose a gaseous mixture contained 0.30 mol CO, 0.10 mol H 2, 0.020 mol H 2 O, and an unknown amount of CH 4 per liter at equilibrium. –What is the concentration of CH 4 at equilibrium in this mixture? The equilibrium constant K c equals 3.92. –Note: the amounts given are equil amounts; therefore can plug into K eq. –For example, consider the following equilibrium mixture.

46. 46 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Calculating Equilibrium Concentrations –First, calculate concentrations from moles of substances.

47. 47 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Calculating Equilibrium Concentrations –First, calculate concentrations from moles of substances.

48. 48 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Calculating Equilibrium Concentrations –First, calculate concentrations from moles of substances.

49. 49 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Calculating Equilibrium Concentrations Suppose we begin a reaction with known amounts of starting materials and want to calculate the quantities at equilibrium.

50. 50 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Calculating Equilibrium Concentrations –Consider the following equilibrium. Suppose you start with 1.000 mol each of carbon monoxide and water in a 50.0 L container. Calculate the molarity of each substance in the equilibrium mixture at 1000 o C. K c for the reaction is 0.58 at 1000 o C. Which way will reaction shift?

51. 51 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Calculating Equilibrium Concentrations –First, calculate the initial molarities of CO and H 2 O.

52. 52 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Calculating Equilibrium Concentrations The starting concentrations of the products are 0. We must now set up a table of concentrations (starting, change, and equilibrium expressions in x). 0.0200 M 0 M –First, calculate the initial molarities of CO and H 2 O.

53. 53 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Calculating Equilibrium Concentrations –Let x be the moles per liter of product formed. [ ] o 0.0200 00  [ ] [ ] eq –The equilibrium-constant expression is:

54. 54 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. –Solving for x. Starting 0.0200 00 Change Equilibrium

55. 55 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Calculating Equilibrium Concentrations –Solving for x. Starting 0.0200 00 Change -x +x Equilibrium 0.0200-x xx

56. 56 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Calculating Equilibrium Concentrations –Solving for x. Starting 0.0200 00 Change -x +x Equilibrium 0.0200-x xx

57. 57 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Calculating Equilibrium Concentrations –Solving for equilibrium concentrations. Starting 0.0200 00 Change -x +x Equilibrium 0.0200-x xx –If you substitute for x in the last line of the table you obtain the following equilibrium concentrations. If plug into K eq, should equal K or close to it because of sign fig for a check HW 19

58. 58 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Calculating Equilibrium Concentrations The preceding example illustrates the three steps in solving for equilibrium concentrations. 1.Set up a table of concentrations (starting, change, and equilibrium expressions in x). 2.Substitute the expressions in x for the equilibrium concentrations into the equilibrium-constant equation. 3.Solve the equilibrium-constant equation for the values of the equilibrium concentrations.

59. 59 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Another example: If the initial pressure of C is 1.0 atm, what would be the partial pressures of each species at equil. PoPo 001.0 PP P eq HW 20 A + B 2CK p = 9.0

60. 60 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Another example: If the initial pressure of C is 0.10 atm and A and B are 1.00 atm, what would be the partial pressures of each species at equil. PoPo 0.101.00 PP P eq HW 21 2C A + B K p = 0.016

61. 61 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Calculating Equilibrium Concentrations In some cases it is necessary to solve a quadratic equation to obtain equilibrium concentrations - not a perfect square. The next example illustrates how to solve such an equation.

62. 62 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Calculating Equilibrium Concentrations –Consider the following equilibrium. Suppose 1.00 atm H 2 and 2.00 atm I 2 are placed in a 1.00-L vessel. What are the partial pressures of all species when it comes to equilibrium at 458 o C? K p at this temperature is 49.7.

63. 63 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Calculating Equilibrium Concentrations –The concentrations of substances are as follows. PoPo 1.002.000 PP P eq

64. 64 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Calculating Equilibrium Concentrations –The concentrations of substances are as follows. PoPo 1.002.000 PP -x +2x P eq 1.00-x2.00-x2x –Substituting our equilibrium concentration expressions gives:

65. 65 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Calculating Equilibrium Concentrations –Solving for x. –Because the right side of this equation is not a perfect square (sq over sq), you must solve the quadratic equation. Starting1.002.000 Change-x +2x Equilibrium1.00-x2.00-x2x

66. 66 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline.

67. 67 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Calculating Equilibrium Concentrations –However, x = 2.33 gives a negative value to 1.00 - x (the equilibrium concentration of H 2 ), which is not possible. Obviously if you get a negative and positive number, take the positive number and if two positive numbers, take the smaller number. However, if you neglect the shift if all components are present (you don't do Q - not only initial reactants or products) and select it incorrectly, you will end up with either a + and - # or two -#'s. In this case the negative number will be correct or the smaller of the two negative numbers will be correct. Bottom line examine numbers carefully when selecting answer.

68. 68 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Calculating Equilibrium Concentrations Solving for equilibrium concentrations. –If you substitute 0.934 for x in the last line of the table you obtain the following equilibrium concentrations. Starting1.002.000 Change-x +2x Equilibrium1.00-x2.00-x2x HW 22

69. 69 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Le Chatelier’s Principle Obtaining the maximum amount of product from a reaction depends on the proper set of reaction conditions which gets us to: –Le Chatelier’s principle states that when a system in a chemical equilibrium is disturbed by a change of temperature, pressure, or concentration, the equilibrium will shift in a way that tends to counteract this change.

70. 70 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Removing Products or Adding Reactants If a chemical system at equilibrium is disturbed by adding a gaseous or aqueous species (not solid or liquid R or P), the reaction will proceed in such a direction as to consume part of the added species. Conversely, if a gaseous or aqueous species is removed (complex or escape gas), the system shifts to restore part of that species. This shift will occur until equilibrium is re-established. A + B C + D add - shift to opposite side, remove - shift towards that side.

71. 71 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Effects of Pressure Change A pressure change caused by changing the volume of the reaction vessel can affect the yield of products in a gaseous reaction; only if the reaction involves a change in the total moles of gas present Ex. N 2 O 4 (g) 2NO 2 (g) Suppose system is compressed by pushing down a piston (decrease volume of space), which way would the shift be that would benefit and use the the available space wisely? Shift to smaller number of gas molecules to pack more efficiently and relieve the increase in pressure due to piston coming down.

72. 72 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Effects of Pressure Change Basically the reactants require less volume (that is, fewer moles of gaseous reactant) and by decreasing the volume of the reaction vessel by increasing the pressure, the rxn would shift the equilibrium to the left (toward reactants) until equil is established.

73. 73 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Effects of Pressure Change Literally “squeezing” the reaction (increase P) will cause a shift in the equilibrium toward the fewer moles of gas. Reducing the pressure in the reaction vessel by increasing its volume would have the opposite effect. Decrease P, increase V, shift larger mols of gas (L &S not compressible) Increase P, decrease V, shift to smaller mols of gas In the event that the number of moles of gaseous product equals the number of moles of gaseous reactant, vessel volume/pressure will have no effect on the position of the equilibrium; no advantage to shift one way over the other.

74. 74 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. SO 2 (g) + 1/2 O 2 (g) SO 3 (g) N 2 (g) + 3 H 2 (g) 2NH 3 (g) N 2 (g) + O 2 (g) 2NO (g) C (s) + H 2 O(g) CO (g) + H 2 (g)

75. 75 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Effect of Temperature Change Temperature has a significant effect on most reactions. –Reaction rates generally increase with an increase in temperature. Consequently, equilibrium is established sooner. –However, when you add or remove reactants/products or change pressure, the result is that the system establishes new conc of species but ratio still equal to same K at that temperature. For temp changes, the numerical value of the equilibrium constant K c varies with temperature. K temp dependent.

76. 76 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Effect of Temperature Change Let’s look at “heat” as if it were a product in exothermic reactions and a reactant in endothermic reactions. We see that increasing the temperature is related to adding more product (in the case of exothermic reactions) or adding more reactant (in the case of endothermic reactions). This ultimately has the same effect as if heat were a physical entity.

77. 77 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Effect of Temperature Change For example, consider the following generic exothermic reaction. Increasing temperature would be like adding more product, causing the equilibrium to shift left. Since “heat” does not appear in the equilibrium- constant expression, this change would result in a smaller numerical value for K c (numerator smaller and den larger)

78. 78 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Effect of Temperature Change For an endothermic reaction, the opposite is true. Increasing temperature would be analogous to adding more reactant, causing the equilibrium to shift right. This change results in more product at equilibrium, and a larger numerical value for K c (larger numerator, smaller den)

79. 79 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Effect of Temperature Change In summary: –For an exothermic reaction (  H is negative) the amounts of reactants are increased at equilibrium by an increase in temperature (K c is smaller at higher temperatures). –For an endothermic reaction (  H positive) the amounts of products are increased at equilibrium by an increase in temperature (K c is larger at higher temperatures).

80. 80 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Effect of a Catalyst A catalyst is a substance that increases the rate of a reaction but is not consumed by it. –It is important to understand that a catalyst has no effect on the equilibrium composition of a reaction mixture. –A catalyst merely speeds up the attainment of equilibrium but does not cause it shift one way or the other just get to direction is was going faster.

81. 81 Material was developed by combining Janusa’s material with the lecture outline provided with Ebbing, D. D.; Gammon, S. D. General Chemistry, 8th ed., Houghton Mifflin, New York, NY, 2005. Majority of figures/tables are from the Ebbing lecture outline. Consider the system I 2 (g) 2I (g)  H = 151 kJ Suppose the system is at equilibrium at 1000 o C. In which direction will rxn occur if a.) I atoms are added? b.) the system is compressed? c.)the temp is increased? d.)effect increase temp has on K? e.) add catalyst? HW 23