1. Equilibrium Basic Concepts Reversible reactions do not go to completion. –They can occur in either direction Chemical equilibrium exists when two opposing reactions occur simultaneously at the same rate. –A chemical equilibrium is a reversible reaction that the forward reaction rate is equal to the reverse reaction rate. Chemical equilibria are dynamic equilibria. –Molecules are continually reacting, even though the overall composition of the reaction mixture does not change.

2. The Equilibrium Constant K c is the equilibrium constant. K c is defined for a reversible reaction at a given temperature as the product of the equilibrium concentrations (in M) of the products, each raised to a power equal to its stoichiometric coefficient in the balanced equation, divided by the product of the equilibrium concentrations (in M) of the reactants, each raised to a power equal to its stoichiometric coefficient in the balanced equation. Equilibrium constants are dimensionless because they actually involve a thermodynamic quantity called activity. –Activities are directly related to molarity

3. The Equilibrium Constant

4. Basic Concepts One of the fundamental ideas of chemical equilibrium is that equilibrium can be established from either the forward or reverse direction. The rates of the forward and reverse reactions can be represented as: When system is at equilibrium: Rate f = Rate r

5. The Equilibrium Constant Write equilibrium constant expressions for the following reactions at 500 o C. All reactants and products are gases at 500 o C.

6. The Equilibrium Constant

7. One liter of equilibrium mixture from the following system at a high temperature was found to contain 0.172 mole of phosphorus trichloride, 0.086 mole of chlorine, and 0.028 mole of phosphorus pentachloride. Calculate K c for the reaction. Equil []’s 0.028 M 0.172 M 0.086 M

8. The Equilibrium Constant At a given temperature 0.80 mole of N 2 and 0.90 mole of H 2 were introduced in an evacuated 1.00-liter container after equilibrium was established the equilibrium concentrations were determined top be 0.20 mole of NH 3, 0.70 moles N 2, and 0.06 moles H 2. Calculate K c for the reaction.

9. Product-favored K > 1 Reactant-favored K < 1 The Equilibrium Constant Products favored 10 3 > K > 10 -3 Reactants favored

10. PCl 3 + Cl 2 PCl 5 Variation of K c with the Form of the Balanced Equation The value of K c depends upon how the balanced equation is written. From the prior examples we have: PCl 5 PCl 3 + Cl 2 K c = [PCl 3 ][Cl 2 ]/[PCl 5 ] = 0.53 Equil. []’s 0.172 M 0.086 M 0.028 M Equil []’s 0.028 M 0.172 M 0.086 M PCl 3 + Cl 2 PCl 5 K c ’ = [PCl 5 ]/[PCl 3 ][Cl 2 ] = 1.89 Equil. []’s 0.172 M 0.086 M 0.028 M 2PCl 5 2PCl 3 + 2Cl 2 K c “ = [PCl 5 ] 2 /[PCl 3 ] 2 [Cl 2 ] 2 = 3.56 K c ’’’ = [PCl 3 ] 2 [Cl 2 ] 2 /[PCl 5 ] 2 = 0.28 Equil []’s 0.028 M 0.172 M 0.086 M

11. Partial Pressures and the Equilibrium Constant For gas phase reactions the equilibrium constants can be expressed in partial pressures rather than concentrations. For gases, the pressure is proportional to the concentration. We can see this by looking at the ideal gas law. –PV = nRT –P = nRT/V –n/V = M –P= MRT and M = P/RT

12. Partial Pressures and the Equilibrium Constant Consider this system at equilibrium at 500 0 C. 2SO 2 (g) + O 2 (g) 2SO 3 (g)

13. Relationship Between K p and K c K c is 49 for the following reaction at 450 o C. If 1.0 mole of H 2 and 1.0 mole of I 2 are allowed to reach equilibrium in a 3.0-liter vessel, (a) How many moles of I 2 remain unreacted at equilibrium? (b) What are the equilibrium partial pressures of H 2, I 2 and HI? (c) What is the total pressure in the reaction vessel?

14. (a) How many moles of I 2 remain unreacted at equilibrium?

15. (b) What are the equilibrium partial pressures of H 2, I 2 and HI?

16. (c) What is the total pressure in the reaction vessel?

17. Relationship Between K p and K c Nitrosyl bromide, NOBr, is 34% dissociated by the following reaction at 25 o C, in a vessel in which the total pressure is 0.25 atmosphere. What is the value of K p ?

18. Heterogeneous Equlibria Heterogeneous equilibria have more than one phase present. –For example, a gas and a solid or a liquid and a gas. How does the equilibrium constant differ for heterogeneous equilibria? –Pure solids and liquids have activities of unity. –Solvents in very dilute solutions have activities that are essentially unity. –The Kc and Kp for the reaction shown above are:

19. Heterogeneous Equlibria

20. What are K c and K p for this reaction?

21. Heterogeneous Equlibria What are K c and K p for this reaction?

22. An aside: Similar to Hess’ Law we can add several reactions together to get to an overall reaction not listed in a table. To determine K for the sum of the reactions simply multiply all the values for each intermediate step to get the K for the overall reaction.

23. Uses of the Equilibrium Constant, K c The equilibrium constant, K c, is 3.00 for the following reaction at a given temperature. If 1.00 mole of SO 2 and 1.00 mole of NO 2 are put into an evacuated 2.00 L container and allowed to reach equilibrium, what will be the concentration of each compound at equilibrium?

24. Uses of the Equilibrium Constant, K c The equilibrium constant is 49 for the following reaction at 450 o C. If 1.00 mole of HI is put into an evacuated 1.00-liter container and allowed to reach equilibrium, what will be the equilibrium concentration of each substance?

25. The Reaction Quotient The mass action expression or reaction quotient has the symbol Q. –Q has the same form as Kc The major difference between Q and Kc is that the concentrations used in Q are not necessarily equilibrium values. Why do we need another “equilibrium constant” that does not use equilibrium concentrations? Q will help us predict how the equilibrium will respond to an applied stress. To make this prediction we compare Q with K c.

26. The Reaction Quotient The equilibrium constant for the following reaction is 49 at 450 o C. If 0.22 mole of I 2, 0.22 mole of H 2, and 0.66 mole of HI were put into an evacuated 1.00-liter container, would the system be at equilibrium? If not, what must occur to establish equilibrium?

27. Disturbing a System at Equlibrium: Predictions LeChatelier’s Principle - If a change of conditions (stress) is applied to a system in equilibrium, the system responds in the way that best tends to reduce the stress in reaching a new state of equilibrium. –We first encountered LeChatelier’s Principle in Chapter 14. Some possible stresses to a system at equilibrium are: 1.Changes in concentration of reactants or products. 2.Changes in pressure or volume (for gaseous reactions) 3.Changes in temperature.

28. Disturbing a System at Equlibrium: Predictions 1Changes in Concentration of Reactants and/or Products Also true for changes in pressure for reactions involving gases.

29. Disturbing a System at Equlibrium: Predictions 2Changes in Volume (and pressure for reactions involving gases) –Predict what will happen if the volume of this system at equilibrium is changed by changing the pressure at constant temperature:

30. Disturbing a System at Equlibrium: Predictions 3Changing the Reaction Temperature

31. Disturbing a System at Equlibrium: Predictions Introduction of a Catalyst –Catalysts decrease the activation energy of both the forward and reverse reaction equally. Catalysts do not affect the position of equilibrium. –The concentrations of the products and reactants will be the same whether a catalyst is introduced or not. –Equilibrium will be established faster with a catalyst.

32. Disturbing a System at Equlibrium: Predictions Given the reaction below at equilibrium in a closed container at 500 o C. How would the equilibrium be influenced by the following?

33. Disturbing a System at Equlibrium: Predictions How will an increase in pressure (caused by decreasing the volume) affect the equilibrium in each of the following reactions?

34. Disturbing a System at Equlibrium: Predictions How will an increase in temperature affect each of the following reactions?

35. Disturbing a System at Equilibrium: Calculations A 2.00 liter vessel in which the following system is in equilibrium contains 1.20 moles of COCl 2, 0.60 moles of CO and 0.20 mole of Cl 2. Calculate the equilibrium constant.

36. Disturbing a System at Equilibrium: Calculations An additional 0.80 mole of Cl 2 is added to the vessel at the same temperature. Calculate the molar concentrations of CO, Cl 2, and COCl 2 when the new equilibrium is established.

37. The Haber Process: An Application of Equilibrium The Haber process is used for the commercial production of ammonia. –This is an enormous industrial process in the US and many other countries. –Ammonia is the starting material for fertilizer production. Fritz Haber 1868-1934 Nobel Prize, 1918 Carl Bosch 1874-1940 Nobel Prize, 1931

38. ∆G, ∆G˚, and K eq ∆G is change in free energy at non-standard conditions. ∆G is related to ∆G˚ ∆G = ∆G˚ + RT ln Q where Q = reaction quotient When Q K, reaction is spontaneous. When Q = K reaction is at equilibrium When ∆G = 0 reaction is at equilibrium Therefore, ∆G˚ = - RT ln K

39. Relationship Between  G o rxn and the Equilibrium Constant The relationships among  G o rxn, K, and the spontaneity of a reaction are:  G o rxn KSpontaneity at unit concentration < 0> 1Forward reaction spontaneous = 0= 1System at equilibrium > 0< 1Reverse reaction spontaneous

40. But systems can reach equilibrium when reactants have NOT converted completely to products. In this case ∆G rxn is < ∆G o rxn, so state with both reactants and products present is MORE STABLE than complete conversion. ∆G, ∆G˚, and K eq Product Favored, ∆G˚ negative, K > 1

41. Product-favored 2 NO 2 ---> N 2 O 4 ∆G o rxn = – 4.8 kJ State with both reactants and products present is more stable than complete conversion. K > 1, more products than reactants. ∆G, ∆G˚, and K eq

42. Reactant-favored N 2 O 4 --->2 NO 2 ∆G o rxn = +4.8 kJ State with both reactants and products present is more stable than complete conversion. K < 1, more reactants than products ∆G, ∆G˚, and K eq

43.  K eq is related to reaction favorability.  When ∆G o rxn < 0, reaction moves energetically “downhill”  ∆G o rxn is the change in free energy when reactants convert COMPLETELY to products. Thermodynamics and K eq

44. Relationship Between  G o rxn and the Equilibrium Constant The relationship for K at conditions other than thermodynamic standard state conditions is derived from this equation.