Chapter 15 Chemical Equilibrium. Equilibrium

Transcription

1 Chapter 15 Chemical

2 The Concept of Chemical equilibrium occurs when a reaction and its reverse reaction proceed at the same rate.

3 The Concept of 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.

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

5 Depicting 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 N 2 O 4 (g) 2 NO 2 (g)

6 The Constant

7 N O (g) 2NO (g) Remember both forward and reverse reactions are elementary reactions

8 The Constant Forward reaction: N 2 O 4 (g) 2 NO 2 (g) Rate law: Rate = k f [N 2 O 4 ]

9 The Constant Reverse reaction: 2 NO 2 (g) N 2 O 4 (g) Rate law: Rate = k r [NO 2 ] 2

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

11 The Constant The ratio of the rate constants is a constant at that temperature, and the expression becomes K eq = k [NO f 2 ] 2 = k r [N 2 O 4 ]

12 We learn following things from these reactions about equilibrium: 1. When a mixture of reactants and products is formed in which the concentration no longer change with time indicates that an equilibrium is reached. 2. For an equilibrium to occur neither reactant or product can escape. 3. At equilibrium the ratio of concentrations remains constant.

13 The Constant To generalize this expression, consider the elementary reaction aa + bb The law of mass action states that if the system is at equilibrium at a given temperature, then the following ratio is a constant K c = [C]c [D] d [A] a [B] b cc + dd The equilibrium constant for this reaction would be Kc i.e. the constant when molarity of the solution is considered. This relationship is called equilibriumconstant expression.

14 The equilibrium constant expression depends only on the stoichiometry of the reaction, not on its mechanism. And remember that the reactants go into the denominator. Kc is independent of the initial concentration of the reactants and products, but on the concentrations at the equilibrium.

15 The equilibrium constant is written without a unit.

16 What Are the Expressions for These Equilibria?

17 Kc is independent of the initial concentration of the reactants: N 2 O 4 (g) 2 NO 2 (g) As you can see, the ratio of [NO 2 ] 2 to [N 2 O 4 ] remains constant at this temperature no matter what the initial concentrations of NO 2 and N 2 O 4 are Calculate the Kc for yourself.

18 The Constant Because pressure is proportional to concentration for gases in a closed system, the equilibrium expression can also be written K p = (P C) c (P D ) d (P A ) a (P B ) b

19 The numerical value of K c is different than the numerical value of K p. We must indicate the subscript c or p

20 Relationship between K c and K p From the ideal gas law we know that PV = nrt Rearranging it, we get P = RT n V

21 n V is nothing but molarity ( moles / liter) So for substance A we can write n A V P= RT P = [A] RT or [A] = P RT

22 Relationship between K c and K p aa + bb K c = [C]c [D] d [A] a [B] b cc + dd PA [A]= RT K p = (P C) c (P D ) d (P A ) a (P B ) b

23 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 K p = K c (RT) n n = (moles of gaseous product) (moles of gaseous reactant)

24 N 2 O 4 (g) 2 NO 2 (g) n = (moles of gaseous product) (moles of gaseous reactant) = 2-1 = 1 Kp = Kc (RT) n So Kp = Kc (RT) If n= 0 i.e. same number of moles of gas appear in the reactant and the product then Kp = Kc As anything raised to 0 is 1

25 In the synthesis of ammonia from nitrogen and hydrogen, N 2 + 3H 2 2NH 3 K c = 9.60 at 300 o C. Calculate K p Δ n = 2-4=-2 Kp = Kc (RT) n = 9.60 (.0821x 573) -2 = 9.60 = 4.34X10-3 (0.0821x 573) 2

26 For the equilibrium 2SO 3 (g) 2SO 2 (g) + O 2 (g) K c = 4.08x 10-3 at 1000K. Calculate K p Kp = Kc (RT) n n = (moles of gaseous product) (moles of gaseous reactant) = 3 2 = 1 Kp = 4.08 x 10-3 ( x 1000) = = 0.3

27 Can Be Reached from Either Direction 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.

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

29 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.

30 SAMPLE EXERCISE 15.3 Interpreting the Magnitude of an Constant The reaction of N 2 with O 2 to form NO might be considered a means of fixing nitrogen: The value for the equilibrium constant for this reaction at 25 C is K c = Describe the feasibility of fixing nitrogen by forming NO at 25 C. Because Kc is so small, very little NO will form at 25 C. The equilibrium lies to the left, favoring the reactants. Consequently, this reaction is an extremely poor choice for nitrogen fixation, at least at 25 C.

31 PRACTICE EXERCISE For the reaction favored more at the higher or lower temperature? at 298 K and K p = 54 at 700 K. Is the formation of HI

32 PRACTICE EXERCISE For the reaction favored more at the higher or lower temperature? at 298 K and K p = 54 at 700 K. Is the formation of HI Answer: The formation of product, HI, is favored at the lower temperature because Kp is larger at the lower temperature.

33 Manipulating Constants The equilibrium constant of a reaction in the reverse reaction is the reciprocal of the equilibrium constant of the forward reaction. N 2 O 4 (g) 2 NO 2 (g) 2 NO 2 (g) N 2 O 4 (g) [NO K c = 2 ] 2 = at 100 C [N 2 O 4 ] K c = [N 2 O 4 ] [NO 2 ] 2 = = 4.72 at 100 C

34 With this information the direction of the reaction needs to be specified when the equilibrium constant is given. Also the temperature too needs to be given as the constant depends at the temperature.

35 SAMPLE EXERCISE 15.4 Evaluating an Constant When an Equation Is Reversed (a) Write the equilibrium-constant expression for K c for the following reaction: (b) The value for the equilibrium constant for this reaction at 25 C is Kc = , determine the value of this equilibrium constant at 25 C.

36 SAMPLE EXERCISE 15.4 Evaluating an Constant When an Equation Is Reversed (a) Write the equilibrium-constant expression for K c for the following reaction: (b) With the information given in sample exercise 15.3, determine the value of this equilibrium constant at 25 C. SAMPLE EXERCISE 15.3 Interpreting the Magnitude of an Constant The reaction of N 2 with O 2 to form NO might be considered a means of fixing nitrogen: The value for the equilibrium constant for this reaction at 25 C is K c = Describe the feasibility of fixing nitrogen by forming NO at 25 C. a. Writing products over reactants, we have b.

37 PRACTICE EXERCISE For the formation of NH 3 from N 2 and H 2, What is the value of K p for the reverse reaction? at 300 C. Answer:

38 Manipulating Constants The equilibrium constant of a reaction that has been multiplied by a number is the equilibrium constant raised to a power that is equal to that number. N 2 O 4 (g) 2 NO 2 (g) [NO K c = 2 ] 2 = at 100 C [N 2 O 4 ] 2 N 2 O 4 (g) 4 NO 2 (g) [NO K c = 2 ] 4 = (0.212) 2 at 100 C [N 2 O 4 ] 2

39 Manipulating Constants The equilibrium constant for a net reaction made up of two or more steps is the product of the equilibrium constants for the individual steps.

40 [NO] [Br ] 2 2NOBr(g) 2NO+Br (g) Kc = = [NOBr] Br (g) + Cl (g) 2 BrCl (g) Kc = = [BrCl] 2 2 [Br ][Cl ] 2 2 The net sum of these two reactions is 2NOBr(g) + Cl (g) 2NO (g)+ 2 BrCl(g) Kc = [NO] [BrCl] [NO] [Br ] 2 [BrCl] = x 2 2 [NOBr] [Cl [NOBr] 2 ] [Br ][Cl ] 2 2 = x 7.2 =

41 SAMPLE EXERCISE 15.5 Combining Expressions Given the following information, determine the value of K c for the reaction Now we have two equations that sum to give the net equation, and we can multiply the individual K c values to get the desired equilibrium constant.

42 SAMPLE EXERCISE 15.5 continued PRACTICE EXERCISE Given that, at 700 K, K p = 54.0 for the reaction reaction determine the value of K p for the reaction for the

43 Heterogeneous

44 The Concentrations of Solids and Liquids Are Essentially Constant The concentration of a solid and liquids can be derived in terms of moles per unit volume but it is not required in equilibrium constant expressions. Remember we are referring to liquids not dissolved substances.

45 The Concentrations of Solids and Liquids Are Essentially Constant Therefore, the concentrations of solids and liquids do not appear in the equilibrium expression PbCl 2 (s) Pb 2+ (aq) + 2 Cl (aq) K c = [Pb 2+ ] [Cl ] 2

46 CaCO 3 (s) CO 2 (g) + CaO (s) As long as some CaCO 3 or CaO remain in the system, the amount of CO 2 above the solid will remain the same. The equilibrium expression for the reaction is K p =P CO 2

47 SAMPLE EXERCISE 15.6 Writing -Constant Expressions Write the equilibrium-constant expression for Kc for each of the above

48 PRACTICE EXERCISE Write the following equilibrium-constant expressions:

49 Calculations

50 German scientist Fritz Haber developed the method for synthesis of ammonia from N 2 and H 2. Ammonia is used to make to make fertilizes and explosives. He received the noble prize in 1918 for his development of synthetic ammonia.

51 He evaluated the equilibrium constant at various temperatures. Using the method very similar to what we saw before:

52 We do not need to know the equilibrium concentration of all the species. We can use the stoichiometry of the reaction to deduce the equilibrium concentrations of the unknown ones.

53 1.Tabulate all the known initial and equilibriums of all the species. 2.Calculate the change of concentrations for the species for which the initial and equilibrium concentrations are known. 3. Use the stoichiometry of the reaction to calculate the changes in concentration for all other species. 4. Calculate the K c.

54 Calculations Here is an example A closed system initially containing x 10 3 M H 2 and x 10 3 M I 2 At 448 C is allowed to reach equilibrium. Analysis of the equilibrium mixture shows that the concentration of HI is 1.87 x 10 3 M. Calculate K c at 448 C for the reaction taking place, which is H 2 (g) + I 2 (g) 2 HI (g)

55 What Do We Know? [H 2 ], M [I 2 ], M [HI], M Initially x x Change At equilibrium 1.87 x 10-3 H 2 (g) + I 2 (g) 2 HI (g)

56 [HI] Increases by 1.87 x 10-3 M [H 2 ], M [I 2 ], M [HI], M Initially x x Change x 10-3 At equilibrium 1.87 x 10-3 H 2 (g) + I 2 (g) 2 HI (g)

57 Stoichiometry tells us [H 2 ] and [I 2 ] decrease by half as much [H 2 ], M [I 2 ], M [HI], M Initially x x Change x x x 10-3 At equilibrium 1.87 x 10-3 H 2 (g) + I 2 (g) 2 HI (g)

58 We can now calculate the equilibrium concentrations of all three compounds [H 2 ], M [I 2 ], M [HI], M Initially x x Change x x x 10-3 At equilibrium 6.5 x x x 10-3 H 2 (g) + I 2 (g) 2 HI (g)

59 and, therefore, the equilibrium constant K c = = [HI] 2 [H 2 ] [I 2 ] (1.87 x 10-3 ) 2 (6.5 x 10-5 )(1.065 x 10-3 ) = 51

60 Applications for Constants a. Predict the direction of the reaction b. To calculate equilibrium concentrations.

61 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.

62 If Q = K, the system is at equilibrium.

63 If Q > K, there is too much product and the equilibrium shifts to the left.

64 If Q < K, there is too much reactant, and the equilibrium shifts to the right.

65 SAMPLE EXERCISE Predicting the Direction of Approach to At 448 C the equilibrium constant K c for the reaction is Predict in which direction the reaction will proceed to reach equilibrium at 448 C if we start with mol of HI, mol of H 2, and mol of I 2 in a 2.00-L container. First calculate the concentration or Molarity Then calculate the reaction quotient.

66 PRACTICE EXERCISE At 1000 K the value of K p for the reaction is Calculate the value for Q p, and predict the direction in which the reaction will proceed toward equilibrium if the initial partial pressures are Answer: Qp = 16; Qp > Kp, and so the reaction will proceed from right to left, forming more SO 3.

67 Le Châtelier s Principle

68 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.

69 The Haber Process The transformation of nitrogen and hydrogen into ammonia (NH 3 ) is of tremendous significance in agriculture, where ammonia-based fertilizers are of utmost importance.

70 The Haber Process If H 2 is added to the system, N 2 will be consumed and the two reagents will form more NH 3.

71 The Haber Process This apparatus helps push the equilibrium to the right by removing the ammonia (NH 3 ) from the system as a liquid.

72 Effect of Volume and Pressure on Reducing the volume of the reaction, hence increasing the pressure causes the equilibrium to move in a direction that reduces the number of moles of gas.

73

74 The volume and pressure do not change the value of K as long as the temperature remains constant.

75 The Effect of Changes in Temperature Co(H 2 O) 6 2+ (aq) + 4 Cl (aq) CoCl 4 2- (aq) + 6 H 2 O (l) This is an endothermic reaction and heat is absorbed when the product is formed.

76 We can treat heat as one of the reactants In an endothermic reaction Reactant + heat Product In an Exothermic reaction Reactant Product + heat When the temperature of a reaction is increased the reaction moves in the direction that consumes heat. And vice versa

77 Catalysts increase the rate of both the forward and reverse reactions.

78 is achieved faster, but the equilibrium composition remains unaltered.