1. SSS 3 1 st Class General Equilibrium

2. Copyright © Cengage Learning. All rights reserved 2 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.

3. Copyright © Cengage Learning. All rights reserved 3 Equilibrium Is: Macroscopically static Microscopically dynamic

4. Copyright © Cengage Learning. All rights reserved 4 Changes in Concentration N 2 (g) + 3H 2 (g) 2NH 3 (g)

5. How is a graph of an reaction going to equilibrium different than one going to completion

6. Copyright © Cengage Learning. All rights reserved 6 Chemical Equilibrium Concentrations reach levels where the rate of the forward reaction equals the rate of the reverse reaction.

7. Copyright © Cengage Learning. All rights reserved 7 The Changes with Time in the Rates of Forward and Reverse Reactions

8. Write the Equilibrium Expression 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). Copyright © Cengage Learning. All rights reserved 8 j l k m [B] [A] [D] [C] K = 

9. 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) Copyright © Cengage Learning. All rights reserved 9

10. 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. Copyright © Cengage Learning. All rights reserved 10

11. Mechanism A +B C A +B C Kc C A + B 1/Kc ½ A + ½ B ½ C Kc -1 2 A + 2B 2 C Kc 2

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. Copyright © Cengage Learning. All rights reserved 12

13. Solving Equilibrium Problems 1)Write the balanced equation for the reaction. 2)Write the equilibrium expression using the law of mass action. 3)Solve using ICE (I= initial, C= change, E= equilibrium) List the given information These Three Steps will solve any Equilibrium Problem Copyright © Cengage Learning. All rights reserved 13

14. K involves concentrations. K p involves pressures. N 2 (g) + 3H 2 (g) 2NH 3 (g) Copyright © Cengage Learning. All rights reserved 14

15. Problem 1 N 2 (g) + 3H 2 (g) 2NH 3 (g) Equilibrium pressures at a certain temperature: Copyright © Cengage Learning. All rights reserved 15 Solve for Kp

16. 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) Copyright © Cengage Learning. All rights reserved 16

17. Problem 2 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 Kc at 35°C. Copyright © Cengage Learning. All rights reserved 17 K p = K(RT) Δn

18. 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 Copyright © Cengage Learning. All rights reserved 18 Problem 3

19. 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) Copyright © Cengage Learning. All rights reserved 19 Problem 4

20. Consider the reaction represented by the equation: Fe 3+ (aq) + SCN - (aq) FeSCN 2+ (aq) Kc = 0.333 Initial: 10.0 M Fe 3+ (aq) and 8.00 M SCN − (aq) Equilibrium: ? M FeSCN 2+ (aq) 5.00 M FeSCN 2+ Copyright © Cengage Learning. All rights reserved 20 Problem 5

21. Homogeneous Equilibria Homogeneous equilibria – involve the same phase: N 2 (g) + 3H 2 (g) 2NH 3 (g) HCN(aq) H + (aq) + CN - (aq) Copyright © Cengage Learning. All rights reserved 21

22. 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) Copyright © Cengage Learning. All rights reserved 22

23. 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. 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 Copyright © Cengage Learning. All rights reserved 23

24. Copyright © Cengage Learning. All rights reserved 24 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. CONCEPT CHECK!

25. Reaction Quotient, Q K =Q ; The system is at equilibrium. No shift will occur. K < Q; The system shifts to the left. – Consuming products and forming reactants, until equilibrium is achieved. K > Q; The system shifts to the right. – Consuming reactants and forming products, to attain equilibrium. Copyright © Cengage Learning. All rights reserved 25

26. Copyright © Cengage Learning. All rights reserved 26 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. CONCEPT CHECK!

27. 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 Copyright © Cengage Learning. All rights reserved 27 Problem 6

28. 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. Change or stress – Concentration – Temperature – Pressure (gas) – Volume (gas Copyright © Cengage Learning. All rights reserved 28 Le Chatelier’s Principle

29. 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. If reactant is added, the forward reaction is favored. If a product is added the reverse reaction is favored. 2.Temperature: K will change depending upon the temperature (endothermic – energy is a reactant; Increasing temperature the forward is favored. Decreasing temperature the reverse reaction is favored. exothermic – energy is a product). Copyright © Cengage Learning. All rights reserved 29

30. 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 or increasing pressure shifts the equilibrium toward the side with fewer moles of gas. d)Increasing the volume or decreasing pressure shifts the equilibrium toward the side with the most moles of gas. Copyright © Cengage Learning. All rights reserved 30

31. Multiple Choice 1 PCl 3 (g) + Cl 2 (g) PCl 5 (g) + energy Some PCl 3 and Cl 2 are mixed in a container at 200 °C and the system reaches equilibrium according to the equation above. Which of the following causes an increase in the number of moles of PCl 5 present at equilibrium? I. Decreasing the volume of the container II. Raising the temperature III. Adding a mole of He gas at constant volume (A) I only (B) II only (C) I and III only (D) II and III only (E) I, II, and III

32. Multiple Choice 2 2NO(g) + O 2 (g) 2 NO 2 (g) DH < 0 Which of the following changes alone would cause a decrease in the value of K eq for the reaction represented above? A) Decreasing the temperature B) Increasing the temperature C) Decreasing the volume of the reaction vessel D) Increasing the volume of the reaction vessel E) Adding a catalyst

33. Multiple Choice 3 HgO(s) + 4 I¯ + H 2 O HgI 4 2 ¯ + 2 OH¯ DH < 0 Consider the equilibrium above. Which of the following changes will increase the concentration of HgI 4 2 ¯? (A) Increasing the concentration of OH¯ (B) Adding 6 M HNO 3 (C) Increasing the mass of HgO present (D) Increasing the temperature (E) Adding a catalyst

34. Free Response 1 2 H2S(g)  2 H2(g) + S2(g) When heated, hydrogen sulfide gas decomposes ac­cording to the equation above. A 3.40 g sample of H2S(g) is introduced into an evacuated rigid 1.25 L container. The sealed container is heated to 483 K, and 3.72  10–2 mol of S2(g) is present at equilibrium. (a) Write the expression for the equilibrium constant, Kc, for the decomposition reaction represented above. (b)Calculate the equilibrium concentration, in mol  L-1, of the following gases in the container at 483 K. (i)H2(g) (ii)H2S(g) (c)Calculate the value of the equilibrium constant, Kc, for the decomposition reaction at 483 K. (d)Calculate the partial pressure of S2(g) in the con­tainer at equilibrium at 483 K. (e)For the reaction H2(g) + S2(g)  H2S(g) at 483 K, calculate the value of the equilibrium constant, Kc.

35. Answer: (a)Kc = (b)(i) ´ = 5.95´10–2 M H2 (ii) = 2.05´10–2 M H2S (c)Kc = = 0.251 (d)PV=nRT = 1.18 (e)K’c = = 2.00

36. Free Response 2 1998 D C(s) + H 2 O(g)  CO(g) + H 2 (g)  Hº = +131kJ A rigid container holds a mixture of graphite pellets (C(s)), H 2 O(g), CO(g), and H 2 (g) at equilibrium. State whether the number of moles of CO(g) in the container will increase, decrease, or remain the same after each of the following disturbances is applied to the original mixture. For each case, assume that all other variables remain constant except for the given disturbance. Explain each answer with a short statement. (a)Additional H 2 (g) is added to the equilibrium mixture at constant volume. (b)The temperature of the equilibrium mixture is increased at constant volume. (c)The volume of the container is decreased at constant temperature. (d)The graphite pellets are pulverized.

37. Answer (a)CO will decrease. An increase of hydrogen gas molecule will increase the rate of the reverse reaction which consumes CO. A LeChatelier Principle shift to the left. (b)CO will increase. Since the forward reaction is endothermic (a  H > 0) an increase in temperature will cause the forward reaction to increase its rate and produce more CO. A LeChatelier Principle shift to the right. (c)CO will decrease. A decrease in volume will result in an increase in pressure, the equilibrium will shift to the side with fewer gas molecules to decrease the pressure, , a shift to the left. (d)CO will remain the same. Once at equilibrium, the size of the solid will affect neither the reaction rates nor the equilibrium nor the concentrations of reactants or products.