1. CHE1102, Chapter 14 Learn, 1 Chapter 15 Chemical Equilibrium

2. CHE1102, Chapter 14 Learn, 2 – 2 opposing reactions occur simultaneously at the same rate ⇌ D E E D When the rate D E is equal to rate E D, the condition of equilibrium has been established D E ⇌ Chemical Equilibrium

3. CHE1102, Chapter 14 Learn, 3

4. CHE1102, Chapter 14 Learn, 4 2 NO (g) + O 2 (g) 2 NO 2 (g) 2 NO 2 (g) 2 NO (g) + O 2 (g) 2 NO (g) + O 2 (g) 2 NO 2 (g) ⇌ Dynamic Equilibrium

5. CHE1102, Chapter 14 Learn, 5 The same equilibrium is reached whether we start with only reactants (N 2 and H 2 ) or with product, NH 3. Dynamic Equilibrium N 2 (g) + 3 H 2 (g) ⇌ 2 NH 3 (g)

6. CHE1102, Chapter 14 Learn, 6 When a process is at equilibrium, it has been found experimentally, for all processes, that the ratio of products to reactants is constant ! K = [Products] [Reactants] Equilibrium constant, K K c = for [ ] in molar, M K p = for [ ] in partial pressure in atm for gases The Equilibrium Constant

7. CHE1102, Chapter 14 Learn, 7 2 NO (g) + O 2 (g) 2 NO 2 (g) ⇌ K = [NO 2 ] 2 [NO] 2 [O 2 ] Mass Action Expression

8. CHE1102, Chapter 14 Learn, 8 Consider the generalized reaction The equilibrium expression for this reaction would be K c = [C] c [D] d [A] a [B] b aA + bBcC + dD The Equilibrium Constant

9. CHE1102, Chapter 14 Learn, 9 Derivation of Mass Action Expression Consider a system at equilibrium in which both the forward and reverse reactions are being carried out; we write its equation with a double arrow: N 2 O 4 (g) ⇌ 2 NO 2 (g)

10. CHE1102, Chapter 14 Learn, 10 Comparing Rates For the forward reaction N 2 O 4 (g) → 2 NO 2 (g) The rate law is Rate = k f [N 2 O 4 ] For the reverse reaction 2 NO 2 (g) → N 2 O 4 (g) The rate law is Rate = k r [NO 2 ] 2

11. CHE1102, Chapter 14 Learn, 11 The Meaning of Equilibrium At equilibrium Rate f = Rate r k f [N 2 O 4 ] = k r [NO 2 ] 2 Rewriting this, it becomes the mass action expression for the equilibrium constant, K eq. K eq = kfkrkfkr [NO 2 ] 2 [N 2 O 4 ] =

12. CHE1102, Chapter 14 Learn, 12 N 2 O 4 (g) 2 NO 2 (g) ⇌ When equilibrium is established, the concentration of reactants and products vary….but the RATIO of P/R remains CONSTANT ! K c = [NO 2 ] 2 [N 2 O 4 ]

13. CHE1102, Chapter 14 Learn, 13 2 NO (g) + O 2 (g) 2 NO 2 (g) ⇌ K c = [NO 2 ] 2 [NO] 2 [O 2 ] At equilibrium, the following [ ]’s are found: [NO 2 ] = 0.896 M [NO] = 0.0126 M [O 2 ] = 0.00413 M Determine the equilibrium constant, K c

14. CHE1102, Chapter 14 Learn, 14 By convention, equilibrium constants, K c are always UNITLESS !!

15. CHE1102, Chapter 14 Learn, 15 2 NO 2 (g) 2 NO (g) + O 2 (g) ⇌ This equilibrium expression is the reciprocal of the previous example Determine the equilibrium constant, K c for the reverse process K c = [NO] 2 [O 2 ] [NO 2 ] 2 K c = 1 1.22 x 10 6 = 8.17 x 10 - 7 The value of the equilibrium constant, K, for a reaction in one direction is the reciprocal of the equilibrium reaction written in the reverse direction

16. CHE1102, Chapter 14 Learn, 16 2 NO (g) + O 2 (g) 2 NO 2 (g) ⇌ 2 NO 2 (g) 2 NO (g) + O 2 (g) ⇌ K c = 8.17 x 10 - 7 K c = 1.22 x 10 6 reverse reactionreciprocal Manipulating Equilibrium Constants

17. CHE1102, Chapter 14 Learn, 17 Manipulating Equilibrium Constants The equilibrium constant of a reaction that has been multiplied by a number is the equilibrium constant raised to a power that is equal to that number: K c = = 0.212 at 100  C [NO 2 ] 2 [N 2 O 4 ] K c = = (0.212) 2 at 100  C [NO 2 ] 4 [N 2 O 4 ] 2 4NO 2 (g)2N 2 O 4 (g) N2O4(g)N2O4(g)2NO 2 (g)

18. CHE1102, Chapter 14 Learn, 18 The equilibrium constant for a net reaction made up of two or more steps is the product of the equilibrium constants for the individual steps. 2 NO 2 (g) NO 3 (g) + NO(g) NO 3 (g) + CO(g) NO 2 (g) + CO 2 (g) NO 2 (g) + CO(g) NO(g) + CO 2 (g) Therefore Manipulating Equilibrium Constants

19. CHE1102, Chapter 14 Learn, 19 when K >> 1 (bigger than 1000) when K << 1 (smaller than 0.001) [Products] > [Reactants] [reactants] > [products] “Equilibrium lies to the right” “Equilibrium lies to the left” K = P R K = P R

20. CHE1102, Chapter 14 Learn, 20 K is constant and does NOT vary with [ ] K does vary with temperature If K  1 (0.001 – 1000) The equilibrium mixture will have similar (or comparable) amounts of reactants and products

21. CHE1102, Chapter 14 Learn, 21 K c and K p are Related K p = K c (RT) Δn  n = (sum of coefficients of gaseous products) − (sum of coefficients of gaseous reactants) R = 0.08206 L·atm/mole·K T = temperature in Kelvin

22. CHE1102, Chapter 14 Learn, 22 The production of ammonia from hydrogen and nitrogen has a K p of 41. If the reaction is run at 400 K what would K c be? A. 41 B. 2.3 x 10 -5 C. 24 D. 4.42 x 10 4 E. None of these Δn = (2 – 4) = -2 K c = K p/ (RT) Δn K c = 41/(0.082057 × 400) -2 K c = 4.42 x 10 4 N 2 (g) + 3 H 2 (g) 2 NH 3 (g)

23. CHE1102, Chapter 14 Learn, 23 An Equilibrium Problem A closed system initially containing 1.000 x 10  3 M H 2 and 2.000 x 10  3 M I 2 at 448  C is allowed to reach equilibrium. Analysis of the equilibrium mixture shows that the concentration of HI is 1.87 x 10  3 M. Calculate K c at 448  C for the reaction taking place, which is H 2 (g) + I 2 (g)2HI(g)

24. CHE1102, Chapter 14 Learn, 24 What Do We Know? [H 2 ], M[I 2 ], M[HI], M Initially1.000 x 10  3 2.000 x 10  3 0 Change At equilibrium1.87 x 10  3

25. CHE1102, Chapter 14 Learn, 25 [HI] Increases by 1.87 x 10 −3 M [H 2 ], M[I 2 ], M[HI], M Initially1.000 x 10  3 2.000 x 10  3 0 Change+1.87 x 10 −3 At equilibrium1.87 x 10  3

26. CHE1102, Chapter 14 Learn, 26 Stoichiometry tells us [H 2 ] and [I 2 ] decrease by half as much. [H 2 ], M[I 2 ], M[HI], M Initially1.000 x 10  3 2.000 x 10  3 0 Change−9.35 x 10 −4 +1.87 x 10  3 At equilibrium1.87 x 10  3 H 2 (g) + I 2 (g)2HI(g)

27. CHE1102, Chapter 14 Learn, 27 We can now calculate the equilibrium concentrations of all three compounds [H 2 ], M[I 2 ], M[HI], M Initially1.000 x 10  3 2.000 x 10  3 0 Change  9.35 x 10  4 +1.87 x 10  3 At equilibrium6.5 x 10 −5 1.065 x 10 −3 1.87 x 10  3

28. CHE1102, Chapter 14 Learn, 28 Therefore, the equilibrium constant is: K c = [HI] 2 [H 2 ] [I 2 ] = 51 = (1.87 x 10  3 ) 2 (6.5 x 10  5 )(1.065 x 10  3 )

29. CHE1102, Chapter 14 Learn, 29 A container is initially charged with 2.00 M phosgene, COCl 2 (g) at 395 °C. An equilibrium with carbon monoxide and chlorine gas is established. The equilibrium concentration of chlorine gas was found to be 0.0398 M. What is K c for this reaction ? 1. Write the balanced chemical reaction 2. Build a chart under the reaction 3. Fill in the appropriate information (this is hard !!!) 4. Solve the problem that is presented

30. CHE1102, Chapter 14 Learn, 30 A container is initially charged with 0.260 atm Cl 2 (g) and 0.520 atm Br 2 (g) at 75 °C. The reactants combine to produce BrCl (g). K p = 56.9 at this temperature. What are the partial pressures of all species at equilibrium ?

31. CHE1102, Chapter 14 Learn, 31 A container is initially charged with 0.18 M CH 4 (g) and 0.18 M CCl 4 (g) at 455 °C. The reactants combine to produce CH 2 Cl 2 (g). Kc = 0.559 at this temperature. What are the molarities of all species at equilibrium ?

32. CHE1102, Chapter 14 Learn, 32 Homogeneous vs. Heterogeneous Homogeneous equilibria occur when all reactants and products are in the same phase. Heterogeneous equilibria occur when something in the equilibrium is in a different phase. The value used for the concentration of a pure substance is always 1.

33. CHE1102, Chapter 14 Learn, 33 When Q > K, the reaction will proceed to the left (toward reactants) to establish equilibrium Reaction Quotient, Q When Q = K, the reaction is at equilibrium When Q < K, the reaction will proceed to the right (toward products) to establish equilibrium – an equilibrium expression for a reaction NOT necessarily at equilibrium

34. CHE1102, Chapter 14 Learn, 34

35. CHE1102, Chapter 14 Learn, 35 Homogeneous equilibrium – all species are in the same phase Heterogeneous equilibrium – all species are NOT in the same phase * ALL pure liquids and solids are left out of the equilibrium expression Aqueous solutions, ex. NaCl (aq), are always included in the equilibrium expression

36. CHE1102, Chapter 14 Learn, 36 The Decomposition of CaCO 3 — A Heterogeneous Equilibrium The equation for the reaction is CaCO 3 (s) ⇌ CaO(s) + CO 2 (g) This results in K c = [CO 2 ] and K p = P CO 2

37. CHE1102, Chapter 14 Learn, 37 Heterogeneous Equilibria Write equilibrium laws for the following: Ag + (aq) + Cl – (aq) AgCl(s) H 3 PO 4 (aq) + H 2 O(l) H 3 O + (aq) + H 2 PO 4 – (aq)

38. CHE1102, Chapter 14 Learn, 38 Fact: When a system is at equilibrium, it will remain at equilibrium forever, unless disturbed by some outside force. outside force:- change in concentration - change in temperature - change in pressure (gas phase) Le Châtelier’s Principle – when a “stress” is applied to a system at equilibrium, the reaction “shifts” to relieve the stress, and re-establish the condition of equilibrium Henry Le Chatelier 1850 – 1936

39. CHE1102, Chapter 14 Learn, 39 N 2 (g) + 3 H 2 (g) ⇌ 2 NH 3 (g)

40. CHE1102, Chapter 14 Learn, 40 A Stress Applied to System

41. CHE1102, Chapter 14 Learn, 41 Le Châtelier states, “if a system is at equilibrium, and you… 1. add reactant, the reaction shifts to the Right, to consume the excess reactant and restore equilibrium” 4. remove product, the reaction shifts to the Right, to replace the lost product and restore equilibrium” 3. remove reactant, the reaction shifts to the Left, to replace the lost reactant and restore equilibrium” 2. add product, the reaction shifts to the Left, to consume the excess product and restore equilibrium”

42. CHE1102, Chapter 14 Learn, 42 Changes in applied pressure affects gas phase reactions Increase in pressure results in the reaction shifting toward the side with the fewest number of gas particles Decrease in pressure results in the reaction shifting toward the side with the largest number of gas particles N 2 (g) + 3 H 2 (g) ⇌ 2 NH 3 (g)

43. CHE1102, Chapter 14 Learn, 43 N 2 (g) + 3 H 2 (g) ⇌ 2 NH 3 (g)

44. CHE1102, Chapter 14 Learn, 44 Changes in Temperature Affects Equilibrium heat is a reactant for an endothermic reaction and heat is a product for an exothermic reaction Changes in concentration or pressure result in reactions shifting but do NOT affect the numerical value of the equilibrium constant, K Changes in temperature also result in reactions shifting and DOES affect the numerical value of the equilibrium constant, K predicting how temperature affects equilibria requires thermodynamic knowledge of the equilibrium reaction

45. CHE1102, Chapter 14 Learn, 45 Changes in Temperature

46. CHE1102, Chapter 14 Learn, 46 Catalysts Catalysts increase the rate of both the forward and reverse reactions. Equilibrium is achieved faster, but the equilibrium composition remains unaltered. Activation energy is lowered, allowing equilibrium to be established at lower temperatures.

47. CHE1102, Chapter 14 Learn, 47 An Example A 1.000 L flask is filled with 1.000 mol of H 2 (g) and 2.000 mol of I 2 (g) at 448 °C. Given a K c of 50.5 at 448 °C, what are the equilibrium concentrations of H 2, I 2, and HI? H 2 (g) + I 2 (g) ⇌ 2 HI(g) initial concentration (M) 1.0002.0000 change in concentration (M) –x–x–x–x+2x equilibrium concentration (M) 1.000 – x2.000 – x2x2x

48. CHE1102, Chapter 14 Learn, 48 Example (continued) Set up the equilibrium constant expression, filling in equilibrium concentrations from the table. Solving for x is done using the quadratic formula, resulting in x = 2.323 or 0.935.

49. CHE1102, Chapter 14 Learn, 49 Example (completed) Since x must be subtracted from 1.000 M, 2.323 makes no physical sense. (It results in a negative concentration!) The value must be 0.935. So [H 2 ] eq = 1.000 – 0.935 = 0.065 M [I 2 ] eq = 2.000 – 0.935 = 1.065 M [HI] eq = 2(0.935) = 1.87 M