1. 1 Chemical Equilibrium You learned when we studied mechanisms that some rxns are reversible or equilibrium rxns The double arrow is used to show this.

2. 2 The Concept of Equilibrium Let’s start with colorless frozen N 2 O 4. At room temperature, it decomposes to brown NO 2 : N 2 O 4 (g) 2NO 2 (g). At some time, the color stops changing and we have a mixture of N 2 O 4 and NO 2. At this point, the concentrations of N 2 O 4 and NO 2 are no longer changing and so are constant with time. Chemical equilibrium is the point at which the concentrations of all species are constant.

3. 3 The Concept of Equilibrium

4. 4 What does this mean? The point at which the rate of decomposition: N 2 O 4 (g) 2NO 2 (g) equals the rate of dimerization: 2NO 2 (g) N 2 O 4 (g). is dynamic equilibrium. The equilibrium is dynamic because the reaction has not stopped: the opposing rates are equal. Consider frozen N 2 O 4 : only white solid is present. On the microscopic level, only N 2 O 4 molecules are present.

5. 5 The Concept of Equilibrium

6. 6 As the substance warms it begins to decompose: A mixture of N 2 O 4 (initially present) and NO 2 (initially formed) appears light brown. When enough NO 2 is formed, it can react to form N 2 O 4 : 2NO 2 (g) N 2 O 4 (g). At equilibrium, as much N 2 O 4 reacts to form NO 2 as NO 2 reacts to re-form N 2 O 4 : The double arrow implies the process is dynamic.

7. 7 Generally, Consider a general rxn A goes to B reversibly. Assume that this is an elementary rxn: Forward reaction: A → B Rate = k f [A] Reverse reaction: B → A Rate = k r [B] Then, at equilibrium k f [A] = k r [B] For an equilibrium we write If we start with just A, as the reaction progresses: –[A] decreases to a constant, and the forward rate slows; –[B] increases from 0 to a constant, & the reverse rate increases; –When [A] and [B] are constant, equilibrium is achieved, and the rates are equal and constant.

8. 8 Or: –k f [A] decreases to a constant, –k r [B] increases from zero to a constant. –When k f [A] = k r [B] equilibrium is achieved. –However, k f is not usually equal to k r ! –And [A] is not necessarily equal to [B]!

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10. 10 Important Comments:

11. 11 The Haber Process Consider If we start with a mixture of nitrogen and hydrogen (in any proportions), the reaction will reach equilibrium with a constant concentration of nitrogen, hydrogen and ammonia. However, if we start with just ammonia and no nitrogen or hydrogen, the reaction will proceed and N 2 and H 2 will be produced until equilibrium is achieved.

12. 12 The Equilibrium Constant

13. 13 The Equilibrium Constant No matter the starting composition of reactants and products, the same ratio of concentrations is achieved at equilibrium. For a general reaction the equilibrium constant expression is where K c is the equilibrium constant and is the ratio k f /k r.

14. 14 The Equilibrium Constant K c is based on the molarities of reactants and products at equilibrium. Let’s go back to smog! the equilibrium constant expression is where K c is the equilibrium constant. Solve for K c !

15. 15 The Equilibrium Constant K c is based on the molarities of reactants and products at equilibrium. It does not depend on initial concentrations but on equilibrium concentrations! Note that the equilibrium constant expression has products over reactants so it does depend on stoichiometry. It doesn’t depend on the rxn mechanism. It does depend on T however (more on this later). We generally omit the units of the equilibrium constant.

16. 16 The Equilibrium Constant Problems:

17. 17 The Equilibrium Constant The Equilibrium Constant in Terms of Pressure If K P is the equilibrium constant for reactions involving gases, we can write: K P is based on partial pressures measured in atmospheres. Because PV = nRT or P = MRT, it is true that P A = [A](RT)

18. 18 The Equilibrium Constant The Equilibrium Constant in Terms of Pressure P A = [A](RT) This means that we can relate K c and K P : where Δ n is the change in number of moles of gas. It is important to use: Δ n = n gas (products) - n gas (reactants)

19. 19 The Equilibrium Constant Problem: Find K p

20. 20 The Equilibrium Constant The Magnitude of Equilibrium Constants The equilibrium constant, K, is the ratio of products to reactants. Therefore, the larger K the more products are present at equilibrium. Conversely, the smaller K the more reactants are present at equilibrium. If K >> 1, then products dominate at equilibrium and equilibrium lies to the right. If K << 1, then reactants dominate at equilibrium and the equilibrium lies to the left.

21. 21 The Equilibrium Constant What if K = 1?

22. 22 The Equilibrium Constant The Magnitude of Equilibrium Constants

23. 23 The Equilibrium Constant The Magnitude of Equilibrium Constants An equilibrium can be approached from any direction. Example: has

24. 24 The Equilibrium Constant The Magnitude of Equilibrium Constants However, has The equilibrium constant for a reaction in one direction is the reciprocal of the equilibrium constant of the reaction in the opposite direction.

25. 25 The Equilibrium Constant Heterogeneous Equilibria When all reactants and products are in one phase, the equilibrium is homogeneous. If one or more reactants or products are in a different phase, the equilibrium is heterogeneous. Consider: –experimentally, the amount of CO 2 does not seem to depend on the amounts of CaO and CaCO 3. Why? The concentration of a solid or pure liquid is its density divided by molar mass.

26. 26 The Equilibrium Constant Heterogeneous Equilibria

27. 27 The Equilibrium Constant Heterogeneous Equilibria Neither density nor molar mass is a variable, the concentrations of solids and pure liquids are constant. For the decomposition of CaCO 3 : We ignore the concentrations of pure liquids and pure solids in equilibrium constant expressions. The amount of CO 2 formed will not depend greatly on the amounts of CaO and CaCO 3 present.

28. 28 Calculating Equilibrium Constants Proceed as follows: –Tabulate initial and equilibrium concentrations (or partial pressures) given. –If an initial and equilibrium concentration is given for a species, calculate the change in concentration. –Use stoichiometry on the change in concentration line only to calculate the changes in concentration of all species. –Deduce the equilibrium concentrations of all species. Usually, the initial concentration of products is zero. (This is not always the case.) When in doubt, assign the change in concentration a variable.

29. 29 Applications of Equilibrium Constants Predicting the Direction of Reaction We define Q, the reaction quotient, for a general reaction as where [A], [B], [P], and [Q] are molarities at any time. Q = K only at equilibrium.

30. 30 Applications of Equilibrium Constants Predicting the Direction of Reaction If Q > K then the reverse reaction must occur to reach equilibrium (i.e., products are consumed, reactants are formed, the numerator in the equilibrium constant expression decreases and Q decreases until it equals K). If Q < K then the forward reaction must occur to reach equilibrium.

31. 31 Applications of Equilibrium Constants Calculation of Equilibrium Concentrations The same steps used to calculate equilibrium constants are used. Generally, we do not have a number for the change in concentration line. Therefore, we need to assume that x mol/L of a species is produced (or used). The equilibrium concentrations are given as algebraic expressions.

32. 32 Le Châtelier’s Principle What can affect the production of ammonia? If the pressure increases, the amount of ammonia present at equilibrium increases. If the temperature decreases, the amount of ammonia at equilibrium increases. Can this be predicted? Or, what happens when we mess with an equilibrium system?

33. 33 Le Châtelier’s Principle Le Châtelier’s Principle: if a system at equilibrium is disturbed, the system will move in such a way as to counteract the disturbance. Or simply: for every action, there is an opposite (but NOT equal) reaction. So if we disturb an equilibrium system, it will shift left or right to minimize the disturbance, until a NEW equilibrium is reached.

34. 34 Le Châtelier’s Principle What can we do to disrupt an equilibrium system? –Change concentration of a reactant or product –Change Volume (changes concentration) –Change Pressure for gaseous systems change volume add more gaseous reactant or product add an inert gas –Change Temperature

35. 35 Le Châtelier’s Principle What do you think about a catalyst? Will adding a catalyst change the equilibrium?

36. 36 Le Châtelier’s Principle Change in Reactant or Product Concentrations Consider the Haber process If H 2 is added while the system is at equilibrium, the system must respond to counteract the added H 2 (by Le Châtelier). That is, the system must consume the H 2 and produce products until a new equilibrium is established. Therefore, [H 2 ] and [N 2 ] will decrease and [NH 3 ] increases.

37. 37 Le Châtelier’s Principle Change in Reactant or Product Concentrations

38. 38 Le Châtelier’s Principle Change in Reactant or Product Concentrations Adding a reactant or product shifts the equilibrium away from the increase. Removing a reactant or product shifts the equilibrium towards the decrease. To optimize the amount of product at equilibrium, we need to flood the reaction vessel with reactant and continuously remove product (Le Châtelier). We illustrate the concept with the industrial preparation of ammonia

39. 39 Le Châtelier’s Principle Change in Reactant or Product Concentrations

40. 40 Le Châtelier’s Principle Change in Reactant or Product Concentrations N 2 and H 2 are pumped into a chamber. The pre-heated gases are passed through a heating coil to the catalyst bed. The catalyst bed is kept at 460 – 550°C under high pressure. The product gas stream (containing N 2, H 2 and NH 3 ) is passed over a cooler to a refrigeration unit. In the refrigeration unit, ammonia liquefies but not N 2 or H 2.

41. 41 Le Châtelier’s Principle Change in Reactant or Product Concentrations The unreacted nitrogen and hydrogen are recycled with the new N 2 and H 2 feed gas. The equilibrium amount of ammonia is optimized because the product (NH 3 ) is continually removed and the reactants (N 2 and H 2 ) are continually being added. Effects of Volume and Pressure As volume is decreased pressure increases. Le Châtelier’s Principle: if pressure is increased the system will shift to counteract the increase.

42. 42 Le Châtelier’s Principle Effects of Volume and Pressure That is, the system shifts to remove gases and decrease pressure. An increase in pressure favors the direction that has fewer moles of gas. In a reaction with the same number of product and reactant moles of gas, pressure has no effect. Consider

43. 43 Le Châtelier’s Principle Effects of Volume and Pressure An increase in pressure (by decreasing the volume) favors the formation of colorless N 2 O 4. The instant the pressure increases, the system is not at equilibrium and the concentration of both gases has increased. The system moves to reduce the number moles of gas (i.e. the forward reaction is favored). A new equilibrium is established in which the mixture is lighter because colorless N 2 O 4 is favored.

44. 44 Le Châtelier’s Principle Effects of Volume and Pressure What if you change the pressure of the system by adding an inert gas like Ar or He? What will the reaction system do?

45. 45 Le Châtelier’s Principle Effect of Temperature Changes The equilibrium constant is temperature dependent. For an endothermic reaction, ΔH > 0 and heat can be considered as a reactant. For an exothermic reaction, ΔH < 0 and heat can be considered as a product. Adding heat (i.e. heating the vessel) favors away from the increase: –if ΔH > 0, adding heat favors the forward reaction, –If ΔH < 0, adding heat favors the reverse reaction.

46. 46 Le Châtelier’s Principle Effect of Temperature Changes Removing heat (i.e. cooling the vessel), favors towards the decrease: –if ΔH > 0, cooling favors the reverse reaction, –If ΔH < 0, cooling favors the forward reaction. Consider for which ΔH > 0. –Co(H 2 O) 6 2+ is pale pink and CoCl 4 2- is blue.

47. 47 Le Châtelier’s Principle Effect of Temperature Changes

48. 48 Le Châtelier’s Principle Effect of Temperature Changes –If a light purple room temperature equilibrium mixture is placed in a beaker of warm water, the mixture turns deep blue. –Since ΔH > 0 (endothermic), adding heat favors the forward reaction, i.e. the formation of blue CoCl 4 2-. –If the room temperature equilibrium mixture is placed in a beaker of ice water, the mixture turns bright pink. –Since ΔH > 0, removing heat favors the reverse reaction which is the formation of pink Co(H 2 O) 6 2+.

49. 49 Le Châtelier’s Principle The Effect of Catalysts A catalyst lowers the activation energy barrier for the reaction. Therefore, a catalyst will decrease the time taken to reach equilibrium. A catalyst does not effect the composition of the equilibrium mixture.