1. CH 13 Chemical Equilibrium

2. The Concept of Equilibrium Chemical equilibrium occurs when a reaction and its reverse reaction proceed at the same rate. As a system approaches equilibrium, both the forward and reverse reactions are occurring. At equilibrium, the forward and reverse reactions are proceeding at the same rate.

3. Chemical equilibrium Dynamic processDynamic process Rate of forward Rxn = Rate of reverse RxnRate of forward Rxn = Rate of reverse Rxn H 2 O (l) H 2 O (g) H 2 O (l) H 2 O (g) (reactant) (product) (reactant) (product) Dynamic Equilibrium Concentration of reactants and products remain constant over time.

4. 2SO 2(g) + O 2(g) 2SO 3(g) At Equilibium SO 2(g) +O 2(g) Initially SO 3(g) Initially

5. 2SO 2(g) + O 2(g) 2SO 3(g) At Equilibium SO 2(g) +O 2(g) Initially SO 3(g) Initially

6. Depicting Equilibrium Consider the reversible reaction between N 2 O 4 and NO 2 : In a system at equilibrium, both the forward and reverse reactions are being carried out; as a result, we write its equation with a double arrow.

7. A System at Equilibrium Once equilibrium is achieved, the amount of each reactant and product remains constant.

8. The Equilibrium Constant Mathematically, the rate of a chemical reaction is described though its rate law. For the forward reaction: The forward rate law is given as: Where:k f is the rate constant [N 2 O 4 ] is the concentration of N 2 O 4

9. The Equilibrium Constant For the reverse reaction: And the rate law is:

10. The Equilibrium Constant Therefore, at equilibrium Rate f = Rate r k f [N 2 O 4 ] = k r [NO 2 ] 2 Rewriting, it becomes:

11. The Equilibrium Constant The ratio of the rate constants is a constant, at that temperature, thus: Where K eq is the “Equilibrium Constant” for the reversible reaction.

12. The Equilibrium Constant To generalize this expression, consider the general chemical reaction: The equilibrium expression for this reaction would be: This is the generalized equilibrium constant based on concentration of the components in the system.

13. [C]c [D]d[A]a [B]b[C]c [D]d[A]a [B]b K eq = Equilibrium constant (K) Equilibrium expressionEquilibrium expression (for any reaction at constant temperature) aA + bB cC + dD moles per liter coefficients productsreactants

14. [C]c [D]d[A]a [B]b[C]c [D]d[A]a [B]b K eq = aA + bB cC + dD productsreactants Equilibrium constant (K)

15. N 2(g) + 3 H 2(g) 2 NH 3(g) K eq = [ NH 3 ] 2 [ N 2 ] [ H 2 ] 3 Equilibrium constant (K)

16. Equilibrium Expression 4NH 3 (g) + 7O 2 (g)  4NO 2 (g) + 6H 2 O(g)

17. The Equilibrium Constant Because pressure is proportional to concentration for gases in a closed system, the equilibrium expression can also be written as a pressure-dependent constant: Since, from the Ideal Gas Law:

18. Relationship between K c and K p Plugging this into the expression for K p for each substance, the relationship between K c and K p becomes: Where: δn = (moles of product) – (moles of reactant)

19. Relationship between K c and K p Relationship between concentration and pressure obtained from the ideal gas law. Recall PV = nRT or Substitute for P in equilibrium expression. Consider the reaction: aA + bB  cC + dD Use this relationship to relate K P and K c E.g. Determine K p for the synthesis of ammonia at 25°C; N 2 (g) + 3H 2 (g)  2NH 3 (g) K c = 4.1x10 8 E.g. 2 Determine K p for the synthesis of HI at 458°C; ½ H 2 (g) + ½ I 2 (g)  HI(g) K c = 0.142.

20. It does not matter whether we start with N 2 and H 2 or whether we start with NH 3. We will have the same proportions of all three substances at equilibrium.

21. What Does the Value of K Mean? If K >> 1, the reaction is product-favored; product predominates at equilibrium. If K << 1, the reaction is reactant-favored; reactant predominates at equilibrium.

22. The Concentrations of Solids and Liquids Are Essentially Constant Both can be obtained by dividing the density of the substance by its molar mass — and both of these are constants at constant temperature. e.g., the concentration of water molecules can be calculated as:

23. The Concentrations of Solids and Liquids Are Essentially Constant Therefore, the concentrations of solids and liquids do not appear in the equilibrium expression – they are taken to have invariant concentrations or “activities” equal to unity (1): And:

24. Equilibrium Calculations A closed system initially containing 0.001M H 2 and 0.002 M I 2 is allowed to reach equilibrium at 448 o C. Analysis of the equilibrium mixture shows that the concentration of HI is 1.87 x 10 −3 M. Calculate K c at 448 o C for the reaction taking place, which is:

25. 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 equilibrium 1.87 x 10 -3

26. [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 equilibrium 1.87 x 10 -3

27. 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 equilibrium 1.87 x 10 -3

28. 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 equilibrium 6.5 x 10 -5 1.065 x 10 -3 1.87 x 10 -3

29. …and, therefore, the equilibrium constant is calculated as:

30. Heterogeneous Equilibria... are equilibria that involve more than one phase. CaCO 3 (s)  CaO(s) + CO 2 (g) K = [CO 2 ] The position of a heterogeneous equilibrium does not depend on the amounts of pure solids or liquids present, so only gases and/or aqueous species are included in the expression.

31. Reaction Quotient If the law of mass action is applied with initial concentrations, the expression becomes a value called the reaction quotient (Q). This helps to determine the direction of the move toward equilibrium. If Q > K then the reaction will favor reactants If Q < K then the reaction will favor products If Q=K the reaction is at equilibrium already

32. Reaction Quotient (continued) H 2 (g) + F 2 (g)  2HF(g) Compare Q with K to determine direction : Q > K to left Q < K to right Q = K at Equilibrium

33. The Reaction Quotient (Q) To calculate Q, one substitutes the initial concentrations on reactants and products into the equilibrium expression. Q gives the same ratio the equilibrium expression gives, but for a system that is not at equilibrium.

34. If Q = K, then the system is at equilibrium. If Q > K, there is too much product and the equilibrium will shift to the left. If Q < K, there is too much reactant and the equilibrium will shift to the right.

35. Le Châtelier’s Principle “If a system at equilibrium is disturbed by a change in temperature, pressure, or the concentration of one of the components, the system will shift its equilibrium position so as to counteract the effect of the disturbance.”

36. Le Chatelier’s principle Stress causes shift in equilibriumStress causes shift in equilibrium Adding or removing reagent N 2(g) + 3 H 2(g) 2 NH 3(g) Stress causes shift in equilibriumStress causes shift in equilibrium Adding or removing reagent N 2(g) + 3 H 2(g) 2 NH 3(g) Add more N 2 ? N2N2N2N2 Reaction shifts to the right [ NH 3 ] inc, [H 2 ] dec

37. Le Chatelier’s principle Adding or removing reagent N 2(g) + 3 H 2(g) 2 NH 3(g) Adding or removing reagent N 2(g) + 3 H 2(g) 2 NH 3(g) Add more NH 3 ? NH 3 Reaction shifts to the left N 2 [ N 2 ] and [H 2 ] inc

38. Le Chatelier’s principle Adding Pressure affects an equilibrium with gases N 2(g) + 3 H 2(g) 2 NH 3(g) P Add P? Increasing pressure causes the equilibrium to shift to the side with the least moles of gas. 4 mol of reactants 2 mol of products

39. Solving Equilibrium Problems From Initial Conditions 1.Balance the equation. 2.Write the equilibrium expression. 3.List the I nitial concentrations. 4.Calculate Q and determine the shift ( C hange) to equilibrium.

40. Solving Equilibrium Problems 5.Define E quilibrium concentrations. 6.Substitute equilibrium concentrations into equilibrium expression and solve. 7.Check calculated concentrations by calculating K. Take note of ICE calculations

41. Example 13.9

42. Example 13.10 Carbon monoxide reacts with steam to produce carbon dioxide and hydrogen. At 700K the value of K is 5.10. Find equilibrium concentrations if 1.00 mol of each component is mixed in a 1.00 L flask.

43. Example 13.12

44. Example 13.13—For Small Values of K