1. Introduction to Chemical Equilibrium Chapter 15 CHEM 160

2. Consider 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. Chemical equilibrium is the point at which the concentrations of all species are constant. Using the collision model: –as the amount of NO 2 builds up, there is a chance that two NO 2 molecules will collide to form N 2 O 4. –At the beginning of the reaction, there is no NO 2 so the reverse reaction (2NO 2 (g)  N 2 O 4 (g)) does not occur. The Concept of Equilibrium

4. 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. The Concept of Equilibrium

6. As the substance warms it begins to decompose: N 2 O 4 (g)  2NO 2 (g) 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. The Concept of Equilibrium

7. Consider Forward reaction: A  B Rate = k f [A] Reverse reaction: B  A Rate = k r [B] At equilibrium k f [A] = k r [B]. For an equilibrium we write: As the reaction progresses –[A] decreases to a constant, –[B] increases from zero to a constant. –When [A] and [B] are constant, equilibrium is achieved. The Concept of Equilibrium

8. Alternatively: –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. The Concept of Equilibrium

9. 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. The Equilibrium Constant

11. 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. The Equilibrium Constant

12. K c is based on the molarities of reactants and products at equilibrium. We generally omit the units of the equilibrium constant. Note that the equilibrium constant expression has products over reactants. The Equilibrium Constant

13. 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. We can show that P A = [A](RT) The Equilibrium Constant

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

15. Phosphorous Problem 1(a) Phosphorous pentachloride dissociates on heating: PCl 5 (g)  PCl 3 (g) + Cl 2 (g) If K c = 3.26 x 10 -2 at 191  C, what is K p at this temperature? Ans.: K p = 1.24

16. The Equilibrium Constant Phosphorous Problem 1(b) The value of K c for the following reaction at 900  C is 0.28. S 2 (g) + 4H 2 (g)  CH 4 (g) + 2H 2 S(g) What is K p at this temperature? Ans.: K p = 3.0 x 10 -5

17. The Equilibrium Constant Phosphorous Problem 2(a) Consider the following reaction at 1000  C: CO(g) + 3H 2 (g)  CH 4 (g) + H 2 O(g) At equilibrium, the following concentrations are measured: [CO] = 0.0613 M, [H 2 ] = 0.1839 M, [CH 4 ] = 0.0387, [H 2 O] = 0.0387 M. Calculate the value of K c for this reaction. Calculate the value of K p. ? Ans.: K c = 3.93, K p = 3.60 x 10 -4

18. The Equilibrium Constant Phosphorous Problem 2(b) A 5.00 L vessel contained 0.0185 mol of phosphorus trichloride, 0.0158 mol of phosphorus pentachloride, and 0.0870 mol chlorine at 503 K in an equilibrium mixture. Calculate the value of K c at this temperature. PCl 3 (g) + Cl 2 (g)  PCl 5 (g) Ans.: K c = 49.1

19. 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. The Equilibrium Constant

20. The Magnitude of Equilibrium Constants The Equilibrium Constant

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

22. 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. The Equilibrium Constant

23. 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. The Equilibrium Constant

24. Heterogeneous Equilibria The Equilibrium Constant

25. 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. The Equilibrium Constant

26. 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. Calculating Equilibrium Constants

27. Phosphorous Problem 3(a) Carbon dioxide decomposes at elevated temperatures to carbon monoxide and oxygen: 2CO 2 (g)  2CO(g) + O 2 (g) At 3000 K, 2.00 mol of CO 2 is placed into a 1.00 L container and allowed to come to equilibrium. At equilibrium, 0.90 mol CO 2 remains. What is the value for K c at this temperature? Ans.: K c = 0.82

28. Calculating Equilibrium Constants Phosphorous Problem 3(b) Consider the following reaction at 1000  C: CO(g) + 3H 2 (g)  CH 4 (g) + H 2 O(g) The original concentrations of CO and H 2 were 0.2000 M and 0.3000 M, respectively. At equilibrium, the concentration of CH 4 was measured to be 0.0478 M. Calculate the values of K c and K p. Ans.: K c = 3.91, K p = 3.60 x 10 -4

29. Calculating Equilibrium Constants Phosphorous Problem 3(c) Consider the following reaction at 1000  C: 2HI(g)  H 2 (g) + I 2 (g) When 4.00 mol of HI was placed into the 5.0 L reaction vessel at 458  C, the equilibrium mixture was found to contain 0.422 mol I 2. Calculate the value of K c for the decomposition of HI. Ans.: K c = 1.79 x 10 -2

30. Calculating Equilibrium Constants Phosphorous Problem 3(d) Hydrogen sulfide, a colorless gas with a foul odor, dissociates on heating: 2H 2 S(g)  2H 2 (g) + S 2 (g) When 0.100 mol H 2 S was put into a 10.0 L vessel and heated to 1132  C, it gave an equilibrium mixture containing 0.0285 mol H 2. Calculate the value of K c at this temperature. Ans.: K c = 1.35 x 10 -6 M

31. 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. Applications of Equilibrium Constants

32. 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. Applications of Equilibrium Constants

33. 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. Applications of Equilibrium Constants

34. Phosphorous Problem 4 The following reaction has an equilibrium constant K c of 3.07 x 10 -4 at 24  C: 2NOBr(g)  2NO(g) + Br 2 (g) For each of the following compositions, decide whether the reaction is at equilibrium. If not, decide which direction the reaction should go. (a)[NOBr] = 0.0610 M,[NO] = 0.0151 M, [Br 2 ] = 0.0108 M (b) [NOBr] = 0.115 M,[NO] = 0.0169 M, [Br 2 ] = 0.0142 M (c)NOBr] = 0.181 M,[NO] = 0.0123 M, [Br 2 ] = 0.0201 M Ans.: a.) goes left; b.) equilibrium; c.) goes right

35. Applications of Equilibrium Constants Phosphorous Problem 5(a) Nitrogen and oxygen form nitric oxide: N 2 (g) + O 2 (g)  2NO(g) If an equilibrium mixture at 25  C contains 0.040 mol/L N 2 and 0.010 mol/L O 2, what is the concentration of NO in this mixture? K c = 1 x 10 -30 at this temperature. Ans.: 2 x 10 -17 M

36. Applications of Equilibrium Constants Phosphorous Problem 5(b) An equilibrium mixture at 1200 K contains 0.30 mol CO, 0.10 mol H 2, and 0.020 mol H 2 O, plus an unknown amount of CH 4 in each liter. What is the concentration of CH 4 in this mixture if K c = 3.92? The reaction is: CO(g) + 3H 2 (g)  CH 4 (g) + H 2 O(g) Ans.: 0.059 M

37. Applications of Equilibrium Constants Phosphorous Problem 6(a) The reaction: CO(g) + H 2 O(g)  CO 2 (g) + H 2 (g) is used to increase the ratio of hydrogen in synthesis gas (mixtures of CO and H 2 ). Suppose you start with 1.00 mol each of carbon monoxide and water in a 50.0 L vessel. What is the concentration of each substance in the equilibrium mixture at 1000  C given that K c = 0.58 at this temperature? Ans.: [CO] = [H 2 O] = 0.0114 M, [CO 2 ] = [H 2 ] = 0.0086 M

38. Applications of Equilibrium Constants Phosphorous Problem 6(b) Hydrogen iodide decomposes to hydrogen gas and iodine gas: 2HI(g)  H 2 (g) + I 2 (g) At 800 K, K c for this reaction is 0.016. If 0.50 mol HI is placed into a 5.0 L flask, what will be the composition of the mixture at equilibrium? Ans.: [HI] = 0.080 M, [H 2 ] = [I 2 ] = 0.010 M

39. Applications of Equilibrium Constants Phosphorous Problem 7(a) N 2 O 4 decomposes to NO 2 according to the reaction: N 2 O 4 (g)  2NO 2 (g) At 100  C, K c = 0.36. If a 1.00 L flask initially contains 0.100 mol N 2 O 4, what will be the concentrations of N 2 O 4 and NO 2 at equilibrium? Ans.: [N 2 O 4 ] =0.040 M, [NO 2 ] = 0.12 M

40. Applications of Equilibrium Constants Phosphorous Problem 7(b) Hydrogen and iodine react according to the equation: H 2 (g) + I 2 (g)  2HI(g) Suppose 1.00 mol H 2 and 2.00 mol I 2 are placed into a 1.00 L vessel. How many moles of each substance are in the equilibrium mixture at 458  C if K c = 49.7 at this temperature? Ans.: 1.86 mol HI, 0.07 mol H 2, 1.07 mol I 2

41. Applications of Equilibrium Constants Phosphorous Problem 7(c) Iodine and bromine react to give iodine monobromide: I 2 (g) + Br 2 (g)  2IBr(g) What is the equilibrium composition of a mixture at 150  C that initially contained 0.0015 mol each of iodine and bromine in a 5.0 L vessel if K c = 1.2 x 10 2 at this temperature? Ans.: [IBr] = 5.1 x 10 -4 M, [Br 2 ] = [I 2 ] = 4.7 x 10 -5 M

42. Consider the production of ammonia As the pressure increases, the amount of ammonia present at equilibrium increases. As the temperature decreases, the amount of ammonia at equilibrium increases. Can this be predicted? Le Châtelier’s Principle: if a system at equilibrium is disturbed, the system will move in such a way as to counteract the disturbance. Le Châtelier’s Principle

44. 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. Le Châtelier’s Principle

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

46. 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 Le Châtelier’s Principle

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

48. 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. Le Châtelier’s Principle

49. 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. Le Châtelier’s Principle

50. Phosphorous Problem 8 Predict the effect of the following concentration changes on the reaction below. CH 4 (g) + 2S 2 (g)  CS 2 (g) + 2H 2 S(g) a)(a) Some CH 4 (g) is removed. (b)Some S 2 (g) is added. (c)Some CS 2 (g) is added. (d)Some H 2 S(g) is removed. (e)Some argon gas is added. Ans.: a.) goes left; b.) goes right; c.) goes left; d.) goes right ; e.) no effect

51. 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 Le Châtelier’s Principle

52. 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. Le Châtelier’s Principle

53. Phosphorous Problem 9 Predict the effect of increasing pressure (decreasing volume) on each of the following reactions. a)(a) CH 4 (g) + 2S 2 (g)  CS 2 (g) + 2H 2 S(g) (b)H 2 (g) + Br 2 (g)  2HBr(g) (c)CO 2 (g) + C(s)  2CO(g) (d)PCl 5 (g)  PCl 3 (g) + Cl 2 (g) (e)N 2 O 4 (g)  2NO 2 (g) Ans.: a.) no effect; b.) no effect; c.) goes left; d.) goes right ; e.) goes left

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

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

56. Effect of Temperature Changes Le Châtelier’s Principle

57. 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  H > 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  H > 0, removing heat favors the reverse reaction which is the formation of pink Co(H 2 O) 6 2+. Le Châtelier’s Principle

58. Phosphorous Problem 10 Predict the effect of increasing temperature on each of the following reactions. What effect does this change have on K c ? a)(a) CO(g) + 3H 2 (g)  CH 4 (g) + H 2 O(g)  H < 0 (b)CO 2 (g) + C(s)  2CO(g)  H > 0 (c)4NH 3 (g) + 5O 2 (g)  4NO(g) + 6H 2 O(g)  H < 0 (d)2H 2 O(g)  2H 2 (g) + O 2 (g)  H > 0 Ans.: a.) goes left, K c decreases; b.) goes right, K c increases; c.) goes left, K c decreases; d.) goes right, K c increases

59. 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. Le Châtelier’s Principle