2. THE STATE OF CHEMICAL EQUILIBRIUM Chemical Equilibrium: The state reached when the concentrations of reactants and products remain constant over time. 2 NO 2 (g)N2O4(g)N2O4(g) BrownColorless

3. THE STATE OF CHEMICAL EQUILIBRIUM 2 NO 2 (g)N2O4(g)N2O4(g)

4. THE STATE OF CHEMICAL EQUILIBRIUM

5. THE EQUILIBRIUM CONSTANT K C c C + d Da A + b B For a general reversible reaction: K c = [A] a [B] b [C] c [D] d Equilibrium constant expression Equilibrium constant Reactants Products For the following reaction:2 NO 2 (g)N2O4(g)N2O4(g) [N 2 O 4 ] [NO 2 ] 2 = 4.64 x 10 –3 (at 25 °C)K c = Equilibrium equation: Concentration is Molarity [ ]: units of mol/L

6. THE EQUILIBRIUM CONSTANT K C = 4.63 x 10 –3 0.0429 (0.0141) 2 Experiment 5 [N 2 O 4 ] [NO 2 ] 2 K c == 4.64 x 10 –3 0.0337 (0.0125) 2 Experiment 1

7. THE EQUILIBRIUM CONSTANT K P What is the relationship of mol/L to P? P = nRT/V well isn’t n/V = M? K P = (P A ) a (P B ) b (P C ) c (P D ) d Equilibrium constant expression Equilibrium constant Reactants Products Equilibrium equation: Pressure is like concentration of a gas: units are atm c C + d Da A + b B ∆n K p = K c (RT) ∆n 0.082 06 K mol L atm R is the gas constant, T is the absolute temperature (Kelvin). is the number of moles of gaseous products minus the number of moles of gaseous reactants. *If the substance is a pure solid or a liquid, put a 1 in its place.

8. HETEROGENEOUS EQUILIBRIA CaO(s) + CO 2 (g)CaCO 3 (s) LimeLimestone (1) (1)[CO 2 ] = [CO 2 ] [CaCO 3 ] [CaO][CO 2 ] = Pure solids and pure liquids are not included. K c = K c = [CO 2 ]K p = P CO 2

9. HETEROGENEOUS EQUILIBRIA

10. USING THE EQUILIBRIUM CONSTANT 2 H 2 (g) + O 2 (g)2 H 2 O(g) K c = 4.2 x 10 –48 (at 500 K) 2 H 2 O(g)2 H 2 (g) + O 2 (g) K c = 2.4 x 10 47 (at 500 K) 2 H I (g)H 2 (g) + I 2 (g) K c = 57.0(at 700 K)

11. EXAMPLE #1 Write the equilibrium expressions for Kc for the following: a.2H 2 O (l)  2H 2 (g) + O 2 (g) b.2HCl (aq)  H 2 (g) + Cl 2 (g)

12. EXAMPLE PROBLEM #2 The following equilibrium concentrations were observed for the Haber process at 127 o C N 2(g) + 3H 2(g)  2NH 3 (g) [NH 3 ] = 0.031M [N 2 ] = 0.85M [H 2 ] = 0.0031M Calculate Kc and Kp.

13. EXAMPLE PROBLEM #3 The reaction below occurs at 25 o C. 2NO (g) + Cl 2(g)  2NOCl (g) The equilibrium partial pressures are: P NOCl = 1.2atm P NO = 0.05atm P Cl 2 = 0.30atm Calculate Kp and Kc.

14. REACTION QUOTIENT Predicting shifts in equilibrium mathematically Q c = [C] c [D] d Q p = (P C ) c (P D ) d [A] a [B] b (P A ) a (P B ) b Q<K: shifts right Q>K: shifts left Q=K: at equilibrium

15. USING THE EQUILIBRIUM CONSTANT If Q c = K c no net reaction occurs. If Q c < K c net reaction goes from left to right (reactants to products). If Q c > K c net reaction goes from right to left (products to reactants).

16. USING THE EQUILIBRIUM CONSTANT

17. EXAMPLE #1 The equilibrium constant, Kp, is 2.33 for the reaction C (s) + CO 2 (g)  2CO (g) at 50 o C. If 0.75g of C, 3.5atm of CO 2 and 4.5 atm CO are present in a container, is the reaction at equilibrium? If not, which way will it shift?

18. EXAMPLE #2 The equilibrium constant, Kp, for the reaction 2H 2 O (l)  2H 2 (g) + O 2 (g) is 5 x 10 -6. If 50g of hydrogen, 25g of oxygen and 100g of water are placed in a 5L flask at 100 o C, will the reaction be at equilibrium? If not, which way will it shift?

19. EXAMPLE #3 Consider the reaction: N 2(g) + 3H 2(g)  2NH 3 (g) The equilibrium constant, Kc, at 25 o C is 0.082. If 0.4mol of nitrogen, 0.098mol of hydrogen and 2.32mol of ammonia are present in a 4L container, is the system at equilibrium? If not, which way will it shift?

20. DETERMINING EQUILIBRIUM CONCENTRATIONS The ICE method aA + bB  cC + dD I # # 0 0 C-ax -bx +cx +dx E #-ax #-bx cx dx I= initial concentrations, C= change during reaction process E = equilibrium concentrations.

21. USING THE EQUILIBRIUM CONSTANT At 700 K, 0.500 mol of H I is added to a 2.00 L container and allowed to come to equilibrium. Calculate the equilibrium concentrations of H 2, I 2, and H I. K c is 57.0 at 700 K. 2 H I (g)H 2 (g) + I 2 (g)

22. USING THE EQUILIBRIUM CONSTANT 2 H I (g)H 2 (g) + I 2 (g) K c = [H 2 ][ I 2 ] [H I ] 2 I 000.250 C+x –2x Exx0.250 – 2x 57.0 = x2x2 (0.250 – 2x) 2 Substitute values into the equilibrium expression: Set up a table:

23. USING THE EQUILIBRIUM CONSTANT Determine the equilibrium concentrations: x = 0.0262 x2x2 (0.250 – 2x) 2 57.0 = Solve for “x”: H I : 0.250 – 2(0.0262) = 0.198 M H 2 : 0.0262 M I 2 : 0.0262 M

24. EXAMPLE PROBLEM #1 For the reaction N 2 O 4 (g)  2NO 2 (g) the equilibrium constant, Kp, is 0.133. If 2.71 atm of N 2 O 4 are introduced into a container, what will the equilibrium partial pressure be for all the species in the reaction?

25. EXAMPLE PROBLEM #2 For the reaction: CO (g) +H 2 O (g)  CO 2(g) + H 2 (g) the equilibrium constant, Kc, is 5.10 at 700K. If 3 moles of CO and 3 moles of H 2 O are introduced into a 5L flask, what will the equilibrium concentrations be for all the species present in the reaction?

26. EXAMPLE #3 For the reaction 2NOCl (g)  2NO (g) + Cl 2(g) the equilibrium constant, Kc, is 1.6 x 10 -5 at 35 o C. If 3 moles of NOCl are placed in a 6L container, what will the concentrations of all the species reach at equilibrium?

27. EXAMPLE #4 In a 2L flask 2 moles of HF are introduced. The reaction is allowed to come to equilibrium. At equilibrium the concentration of the hydrogen gas reaches 0.064M. Determine the equilibrium constant. 2HF (g)  H 2 (g) + F 2 (g)

28. EXAMPLE #5 Initially the system contains 0.4M A and 1.2M B. After the system reaches equilibrium, 26% of the A has reacted. Determine the equilibrium constant for this reaction. 4A + 2B  3C

29. REVERSE ICE

30. EXAMPLE #1 The equilibrium constant for the reaction below at 448 o C is 50.5. H 2 (g) + I 2 (g)  2HI (g) Determine the equilibrium concentrations if 3 moles of HI are introduced into a 4 L reaction vessel.

31. EXAMPLE #2 For the reaction below at 500 o C the equilibrium constant, K p, is 68956.52atm 2. 2NH 3(g)  N 2(g) + 3H 2 (g) If 2.5 atm of N 2 and 5.5 atm of H 2 are placed in a 10 L reaction vessel, what are the equilibrium partial pressures of all the gases?

32. EXAMPLE #3 Fifteen grams of solid NH 4 HS (s) is placed in an evacuated flask at 24 o C. At equilibrium the total pressure in the flask is 0.614atm. What is the equilibrium constant, K p, for the reaction below? NH 4 HS (s)  NH 3(g) + H 2 S (g)

33. EXAMPLE #4 For the reaction 2SO 2(g) + O 2(g)  2SO 3 (g) The equilibrium constant, K c, is 3 x 10 4. If 240 grams of SO 3 are introduced into a 5L reaction vessel, what will the equilibrium concentrations be for all the species present?

34. EXAMPLE #5 For the reaction: 2H 2 O (l)  2H 2 (g) + O 2 (g) If 5000g of H 2 O are introduced into a 5L flask, the total pressure at equilibrium is 4.5atm. Calculate the K p value for this reaction.

35. LECHATELIER’S PRINCIPLE If a stress is placed on a system at equilibrium, that system will shift to relieve the stress.

36. CHANGE CONCENTRATION Add a substance, the arrow points away from what you added. Remove a substance, the arrow points toward what you removed. You cannot change the concentration of solids or liquids (adding solid or liquid has no effect on equilibrium)

37. THE EFFECT OF CONCENTRATION CHANGES ON AN EQUILIBRIUM MIXTURE 2 NH 3 (g)N 2 (g) + 3 H 2 (g)

38. EXAMPLE OF CONCENTRATION CHANGES 2H 2(g) + O 2(g)  2H 2 O (l) a.Add H 2 b.Remove O 2 c.Add H 2 O d.Remove H 2

39. CHANGING TEMPERATURE  H: positive, energy is a reactant (endothermic)  H: negative, energy is a product (exothermic) Heat adds energy Cooling removes energy

40. EXAMPLE OF TEMPERATURE CHANGE C (s) + O 2(g)  CO 2(g)  H = -394kJ/mol a.Heat b.Cool

41. CHANGING PRESSURE Changing pressure by changing volume: -increase pressure, shifts to the side with less gas particles. -decrease pressure, shifts to the side with more gas particles. *Change pressure by adding a gas not present in the reaction, no effect.

42. THE EFFECT OF PRESSURE AND VOLUME CHANGES ON AN EQUILIBRIUM MIXTURE 2 NH 3 (g)N 2 (g) + 3 H 2 (g)

43. EXAMPLE FOR PRESSURE CHANGES CaCO 3 (s)  CaO (s) + CO 2(g) a.Increase volume b.Decrease volume c.Increase pressure by adding neon.

44. ADD A CATALYST A catalyst will increase the rate of both the forward and reverse reactions at the same time, thus, no effect on equilibrium.

45. THE EFFECT OF A CATALYST ON EQUILIBRIUM

46. EXAMPLE PROBLEM Na 2 CO 3(s)  Na 2 O (s) + CO 2 (g)  H >0 a.Add Na 2 O b.Heat c.Add CO 2 d.Increase volume e.Cool f.Remove Na 2 CO 3