2. 13.1 EQUILIBRIUM CONDITION

3. CHEMICAL EQUILIBRIUM The state where the concentrations of all reactants and products remain constant with time. On the molecular level, there is frantic activity. Equilibrium is not static, but is a highly dynamic situation.

4. EQUILIBRIUM IS: Macroscopically static Microscopically dynamic

5. CHANGES IN CONCENTRATION N 2 (g) + 3H 2 (g) 2NH 3 (g)

6. CHEMICAL EQUILIBRIUM Concentrations reach levels where the rate of the forward reaction equals the rate of the reverse reaction.

7. THE CHANGES WITH TIME IN THE RATES OF FORWARD AND REVERSE REACTIONS

8. Consider an equilibrium mixture in a closed vessel reacting according to the equation: H 2 O(g) + CO(g) H 2 (g) + CO 2 (g) You add more H 2 O(g) to the flask. How does the concentration of each chemical compare to its original concentration after equilibrium is reestablished? Justify your answer.

9. Consider an equilibrium mixture in a closed vessel reacting according to the equation: H 2 O(g) + CO(g) H 2 (g) + CO 2 (g) You add more H 2 to the flask. How does the concentration of each chemical compare to its original concentration after equilibrium is reestablished? Justify your answer.

10. 13.2THE EQUILIBRIUM CONSTANT

11. CONSIDER THE FOLLOWING REACTION AT EQUILIBRIUM: jA + kB lC + mD A, B, C, and D = chemical species. Square brackets = concentrations of species at equilibrium. j, k, l, and m = coefficients in the balanced equation. K = equilibrium constant (given without units). j l k m [B][A] [D] [C] K =

12. CONCLUSIONS ABOUT THE EQUILIBRIUM EXPRESSION Equilibrium expression for a reaction is the reciprocal of that for the reaction written in reverse. When the balanced equation for a reaction is multiplied by a factor of n, the equilibrium expression for the new reaction is the original expression raised to the nth power; thus K new = (K original ) n. K values are usually written without units.

13. K always has the same value at a given temperature regardless of the amounts of reactants or products that are present initially. For a reaction, at a given temperature, there are many equilibrium positions but only one equilibrium constant, K. – Equilibrium position is a set of equilibrium concentrations.

14. 13.3 EQUILIBRIUM EXPRESSIONS INVOLVING PRESSURES

15. K involves concentrations. K p involves pressures.

16. EXAMPLE N 2 (g) + 3H 2 (g) 2NH 3 (g)

17. EXAMPLE N 2 (g) + 3H 2 (g) 2NH 3 (g) Equilibrium pressures at a certain temperature:

18. EXAMPLE N 2 (g) + 3H 2 (g) 2NH 3 (g)

19. THE RELATIONSHIP BETWEEN K AND K P K p = K(RT) Δn Δn = sum of the coefficients of the gaseous products minus the sum of the coefficients of the gaseous reactants. R = 0.08206 L·atm/mol·K T = temperature (in Kelvin)

20. EXAMPLE N 2 (g) + 3H 2 (g) 2NH 3 (g) Using the value of K p (3.9 × 10 4 ) from the previous example, calculate the value of K at 35°C.

21. 13.4 HETEROGENEOUS EQUILIBRIA

22. HOMOGENEOUS EQUILIBRIA Homogeneous equilibria – involve the same phase: N 2 (g) + 3H 2 (g) 2NH 3 (g) HCN(aq) H + (aq) + CN - (aq)

23. HETEROGENEOUS EQUILIBRIA Heterogeneous equilibria – involve more than one phase: 2KClO 3 (s) 2KCl(s) + 3O 2 (g) 2H 2 O(l) 2H 2 (g) + O 2 (g)

24. The position of a heterogeneous equilibrium does not depend on the amounts of pure solids or liquids present. – The concentrations of pure liquids and solids are constant. 2KClO 3 (s) 2KCl(s) + 3O 2 (g)

25. 13.5 APPLICATIONS OF THE EQUILIBRIUM CONSTANT

26. THE EXTENT OF A REACTION A value of K much larger than 1 means that at equilibrium the reaction system will consist of mostly products – the equilibrium lies to the right. – Reaction goes essentially to completion.

27. THE EXTENT OF A REACTION A very small value of K means that the system at equilibrium will consist of mostly reactants – the equilibrium position is far to the left. – Reaction does not occur to any significant extent.

28. ON YOUR WHITE BOARDS If the equilibrium lies to the right, the value for K is __________. large (or >1) If the equilibrium lies to the left, the value for K is ___________. small (or <1)

29. REACTION QUOTIENT, Q Used when all of the initial concentrations are nonzero. Apply the law of mass action using initial concentrations instead of equilibrium concentrations.

30. REACTION QUOTIENT, Q Q = K; The system is at equilibrium. No shift will occur. Q > K; The system shifts to the left. – Consuming products and forming reactants, until equilibrium is achieved. Q < K; The system shifts to the right. – Consuming reactants and forming products, to attain equilibrium.

31. TYPES OF EQUILIBRIUM PROBLEMS 1.Manipulation of the equilibrium constant, K 2.Given all equilibrium concentrations, calculate the value of the constant, K 3.Given the value of the equilibrium constant, K, and all but one of the equilibrium concentrations, solve for the missing concentration 4.Given the value of the initial concentrations of the reactants and one of the equilibrium concentrations of either the reactants of products, solve for all equilibrium concentrations and the value of K 5.Given the initial concentrations and the value of K, solve by approximation for the equilibrium concentrations First 4 types can be solved using basic algebra, the 5 th can be solved by approximation using an ICE chart

32. ICE: STANDS FOR INITIAL, CHANGE, EQUILIBRIUM To begin, write the balanced equation for the reaction and the equilibrium expression, omitting pure solids and liquids Under a balanced chemical equation make an ICE chart I= initial concentration in mol/L (units are omitted on the chart) C= the change to reach equilibrium represented by a + or – (these will be the coefficients from the balanced equation) E= equilibrium concentration which are obtained by adding the I and C lines together

33. EXAMPLE PROBLEM USING ICE At a particular temperature, K=1.00 x 10 2 for the reaction: H 2 (g) + I 2 (g)  2HI(g) In an experiment, 1.00 mol H 2, 1.00 mole I 2, and 1.00 mol HI are introduced into a 1.00 L container. Calculate the equilibrium concentrations of all reactions and products H 2 (g) + I 2 (g)  2HI(g)

34. Plug the equilibrium values into the expression and solve for x:

35. Consider the reaction represented by the equation: Fe 3+ (aq) + SCN - (aq) FeSCN 2+ (aq) Trial #1: 6.00 M Fe 3+ (aq) and 10.0 M SCN - (aq) are mixed at a certain temperature and at equilibrium the concentration of FeSCN 2+ (aq) is 4.00 M. What is the value for the equilibrium constant for this reaction? 35

36. SET UP ICE TABLE Fe 3+ (aq) + SCN – (aq) FeSCN 2+ (aq) Initial6.00 10.000.00 Change – 4.00 – 4.00 +4.00 Equilibrium 2.00 6.004.00 K = 0.333

37. Consider the reaction represented by the equation: Fe 3+ (aq) + SCN - (aq) FeSCN 2+ (aq) Trial #2: Initial: 10.0 M Fe 3+ (aq) and 8.00 M SCN − (aq) (same temperature as Trial #1) Equilibrium: ? M FeSCN 2+ (aq) 5.00 M FeSCN 2+

38. Consider the reaction represented by the equation: Fe 3+ (aq) + SCN - (aq) FeSCN 2+ (aq) Trial #3: Initial: 6.00 M Fe 3+ (aq) and 6.00 M SCN − (aq) Equilibrium: ? M FeSCN 2+ (aq) 3.00 M FeSCN 2+

39. 13.6 SOLVING EQUILIBRIUM PROBLEMS

40. SOLVING EQUILIBRIUM PROBLEMS 1)Write the balanced equation for the reaction. 2)Write the equilibrium expression using the law of mass action. 3)List the initial concentrations. 4)Calculate Q, and determine the direction of the shift to equilibrium.

41. SOLVING EQUILIBRIUM PROBLEMS 5)Define the change needed to reach equilibrium, and define the equilibrium concentrations by applying the change to the initial concentrations. 6)Substitute the equilibrium concentrations into the equilibrium expression, and solve for the unknown. 7)Check your calculated equilibrium concentrations by making sure they give the correct value of K.

42. Consider the reaction represented by the equation: Fe 3+ (aq) + SCN - (aq) FeSCN 2+ (aq) Fe 3+ SCN - FeSCN 2+ Trial #1 9.00 M5.00 M1.00 M Trial #2 3.00 M2.00 M5.00 M Trial #32.00 M9.00 M6.00 M Find the equilibrium concentrations for all species.

43. Trial #1: [Fe 3+ ] = 6.00 M; [SCN - ] = 2.00 M; [FeSCN 2+ ] = 4.00 M Trial #2: [Fe 3+ ] = 4.00 M; [SCN - ] = 3.00 M; [FeSCN 2+ ] = 4.00 M Trial #3: [Fe 3+ ] = 2.00 M; [SCN - ] = 9.00 M; [FeSCN 2+ ] = 6.00 M Answer

44. A 2.0 mol sample of ammonia is introduced into a 1.00 L container. At a certain temperature, the ammonia partially dissociates according to the equation: NH 3 (g) N 2 (g) + H 2 (g) At equilibrium 1.00 mol of ammonia remains. Calculate the value for K. K = 1.69

45. A 1.00 mol sample of N 2 O 4 (g) is placed in a 10.0 L vessel and allowed to reach equilibrium according to the equation: N 2 O 4 (g) 2NO 2 (g) K = 4.00 × 10 -4 Calculate the equilibrium concentrations of: N 2 O 4 (g) and NO 2 (g). Concentration of N 2 O 4 = 0.097 M Concentration of NO 2 = 6.32 × 10 -3 M

46. 13.7 LE CHATELIER’S PRINCIPLE

47. If a change is imposed on a system at equilibrium, the position of the equilibrium will shift in a direction that tends to reduce that change.

48. EFFECTS OF CHANGES ON THE SYSTEM 1.Concentration: The system will shift away from the added component. If a component is removed, the opposite effect occurs. 2.Temperature: K will change depending upon the temperature (endothermic – energy is a reactant; exothermic – energy is a product).

49. EFFECTS OF CHANGES ON THE SYSTEM 3.Pressure: a)The system will shift away from the added gaseous component. If a component is removed, the opposite effect occurs. b)Addition of inert gas does not affect the equilibrium position. c)Decreasing the volume shifts the equilibrium toward the side with fewer moles of gas.

51. EQUILIBRIUM DECOMPOSITION OF N 2 O 4

52. RELATIONSHIP OF K TO FREE ENERGY,  G 

53. The magnitude of the equilibrium constant, K, is directly related to the change in Gibbs free energy associated with the reaction  G . The species that have lower free energy will have larger relative concentrations at equilibrium. When both reactants and products have significant concentrations at equilibrium, the magnitude of  G  is similar to the thermal energy (RT) The equilibrium constant at standard temperature can be determined from the thermodynamic values of a reaction using the equation:

54. The equilibrium constant can be determined quantitatively through calculation or qualitatively through estimation. The thermal energy (RT) at room temperature is 2.4 kJ/mol. If  G  is large compared to RT, then K will be significantly larger than 1.

55. EXAMPLE