Presentation is loading. Please wait.

Presentation is loading. Please wait.

Equilibrium © 2009, Prentice-Hall, Inc. Chemical Equilibrium.

Similar presentations


Presentation on theme: "Equilibrium © 2009, Prentice-Hall, Inc. Chemical Equilibrium."— Presentation transcript:

1 Equilibrium © 2009, Prentice-Hall, Inc. Chemical Equilibrium

2 Equilibrium © 2009, Prentice-Hall, Inc.

3 Equilibrium Not all reactions “go to completion” © 2009, Prentice-Hall, Inc. Reactions in which limiting reagents are used up are said to “go to completion”

4 Equilibrium Not all reactions “go to completion” © 2009, Prentice-Hall, Inc. Reactions in which limiting reagents are used up are said to “go to completion” Many reactions do not “go to completion”, but approach an equilibrium state in which both reactants are forming products, and products are reforming into reactants

5 Equilibrium © 2009, Prentice-Hall, Inc. The Concept of Equilibrium Chemical equilibrium occurs when a reaction and its reverse reaction proceed at the same rate.

6 Equilibrium Not all reactions “go to completion” © 2009, Prentice-Hall, Inc. Chemical equilibrium- a dynamic state of a chemical reaction where the rate of the forward reaction equals the rate of the reverse reaction in a closed system, but no net effect is observable Quantites of reactants and products are measurably unchanged over time

7 Equilibrium Equilibrium-Concentration versus time graph © 2009, Prentice-Hall, Inc.

8 Equilibrium © 2009, Prentice-Hall, Inc. A System at Equilibrium Once equilibrium is achieved, the amount of each reactant and product remains constant.

9 Equilibrium Rate versus time graph © 2009, Prentice-Hall, Inc.

10 Equilibrium © 2009, Prentice-Hall, Inc. Depicting Equilibrium Since, in a system at equilibrium, both the forward and reverse reactions are being carried out, we write its equation with a double arrow. N 2 O 4 (g) 2 NO 2 (g)

11 Equilibrium © 2009, Prentice-Hall, Inc. The Equilibrium Constant

12 Equilibrium © 2009, Prentice-Hall, Inc. The Equilibrium Constant Forward reaction: N 2 O 4 (g)  2 NO 2 (g) Rate Law: Rate = k f [N 2 O 4 ]

13 Equilibrium © 2009, Prentice-Hall, Inc. The Equilibrium Constant Reverse reaction: 2 NO 2 (g)  N 2 O 4 (g) Rate Law: Rate = k r [NO 2 ] 2

14 Equilibrium © 2009, Prentice-Hall, Inc. The Equilibrium Constant Therefore, at equilibrium Rate f = Rate r k f [N 2 O 4 ] = k r [NO 2 ] 2 Rewriting this, it becomes kfkrkfkr [NO 2 ] 2 [N 2 O 4 ] =

15 Equilibrium © 2009, Prentice-Hall, Inc. The Equilibrium Constant The ratio of the rate constants is a constant at that temperature, and the expression becomes K eq = kfkrkfkr [NO 2 ] 2 [N 2 O 4 ] =

16 Equilibrium © 2009, Prentice-Hall, Inc. The Equilibrium Constant Consider the generalized reaction The equilibrium expression for this reaction would be K c = [C] c [D] d [A] a [B] b aA + bBcC + dD

17 Equilibrium The Equilibrium Constant The products appear in the numerator and the reactants in the denominator Each concentration is raised to the power of its stoichiometric coefficient Is constant for a particular reaction at a as long as the temperature is constant © 2009, Prentice-Hall, Inc.

18 Equilibrium The Equilibrium Constant K values are written without units K c- equilibrium constant for concentration (aqueous) K p- Equilibrium constant for partial pressure (gases) © 2009, Prentice-Hall, Inc.

19 Equilibrium Not part of the equilibrium constant calculation: Pure solids (s) Pure liquids (l) Their concentration is considered to be 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. © 2009, Prentice-Hall, Inc.

20 Equilibrium Write the equilibrium Expression for 4 NH 3 (g)+ 7 O 2(g)  4 NO 2 (g)+ 6 H 2 O (g) © 2009, Prentice-Hall, Inc.

21 Equilibrium Write the equilibrium Expression for 4 NH 3 (g)+ 7 O 2(g)  4 NO 2 (g)+ 6 H 2 O (g) K= [NO 2 ] 4 [H 2 O] 6 [NH 3 ] 4 [O 2 ] 7 © 2009, Prentice-Hall, Inc.

22 Equilibrium Write the equilibrium expression for 2KClO3 (s) ↔ 2KCl(s) + 3O 2 (g) © 2009, Prentice-Hall, Inc.

23 Equilibrium Write the equilibrium expression for 2KClO3 (s) ↔ 2KCl(s) + 3O 2 (g) K= [O 2 ] 3 © 2009, Prentice-Hall, Inc.

24 Equilibrium Write the equilibrium expression for H 2 O(l) ↔ H + (aq) + OH - (aq) © 2009, Prentice-Hall, Inc.

25 Equilibrium Write the equilibrium expression for H 2 O(l) ↔ H + (aq) + OH - (aq) K= [H + ] [ OH - ] © 2009, Prentice-Hall, Inc.

26 Equilibrium © 2009, Prentice-Hall, Inc. The Equilibrium Constant K p Since 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 )

27 Equilibrium Write the K and K p of the following: The decomposition of solid phosphorus pentachloride to liquid phosphorus trichloride and chlorine gas © 2009, Prentice-Hall, Inc.

28 Equilibrium Write the K and K p of the following: The decomposition of solid phosphorus pentachloride to liquid phosphorus trichloride and chlorine gas K= [Cl 2 ] K p = P Cl2 © 2009, Prentice-Hall, Inc.

29 Equilibrium Write the K and K p of the following: Deep blue solid copper II sulfate pentahydrate is heated to drive off water vapor to form solid copper II sulfate. © 2009, Prentice-Hall, Inc.

30 Equilibrium Write the K and K p of the following: Deep blue solid copper II sulfate pentahydrate is heated to drive off water vapor to form solid copper II sulfate. K=[H 2 O] 5 K p = P H2O 5 © 2009, Prentice-Hall, Inc.

31 Equilibrium Calculating the values of K The process in which ammonia is manufactured form nitrogen gas and hydrogen gas at high temperature and pressure is called the Haber process. The following equilibrium concentrations were observed for this process at 127 C [NH 3 ]= 3.1 x M, [N 2 ]= 8.5 x M [H 2 ]= 3.1 x M © 2009, Prentice-Hall, Inc.

32 Equilibrium Calculating the values of K [NH 3 ]= 3.1 x M, [N 2 ]= 8.5 x M [H 2 ]= 3.1 x M Calculate the value of K at 127 C for this reaction © 2009, Prentice-Hall, Inc.

33 Equilibrium Calculating the values of K [NH 3 ]= 3.1 x 10 2 M, [N 2 ]= 8.5 x M [H 2 ]= 3.1 x M Calculate the value of K at 127 C for this reaction 3.8 x 10 4 © 2009, Prentice-Hall, Inc.

34 Equilibrium Calculating the values of K [NH 3 ]= 3.1 x 10 2 M, [N 2 ]= 8.5 x M [H 2 ]= 3.1 x M Calculate the value of K at 127 C for the reverse of this reaction (K ’ ) © 2009, Prentice-Hall, Inc.

35 Equilibrium Calculating the values of K [NH 3 ]= 3.1 x 10 2 M, [N 2 ]= 8.5 x M [H 2 ]= 3.1 x M Calculate the value of K at 127 C for the reverse of this reaction (K ’ ) 2.6 x © 2009, Prentice-Hall, Inc.

36 Equilibrium Calculating the values of K What is the relationship between K and K ’ for a reaction? © 2009, Prentice-Hall, Inc.

37 Equilibrium Calculating the values of K What is the relationship between K and K ’ for a reaction? K ’ = 1/K © 2009, Prentice-Hall, Inc.

38 Equilibrium Calculating K p The reaction for the formation of nitrosyl chloride is 2NO(g) + Cl 2 g)  2 NOCl(g) Was studied at 25 C. The pressures at equilibrium were found to be P NOCl = 1.2 atm P NO = 5.0 x atm P Cl2 = 3.0 x atm Calculate the K p for this reaction at 25 C © 2009, Prentice-Hall, Inc.

39 Equilibrium Calculating K p The reaction for the formation of nitrosyl chloride is 2NO(g) + Cl 2 g)  2 NOCl(g) Was studied at 25 C. The pressures at equilibrium were found to be P NOCl = 1.2 atm P NO = 5.0 x atm P Cl2 = 3.0 x atm Calculate the K p for this reaction at 25 C 1.9 x 10 3 © 2009, Prentice-Hall, Inc.

40 Equilibrium © 2009, Prentice-Hall, Inc. Relationship Between K c and K p From the Ideal Gas Law we know that Rearranging it, we get PV = nRT R=.0821 L atm/mol K P = RT nVnV

41 Equilibrium © 2009, Prentice-Hall, Inc. Relationship Between K c and K p K c and K p ARE NOT INTERCHANGABLE

42 Equilibrium © 2009, Prentice-Hall, Inc. 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 T is temp in kelvin, R=.0821 L atm/mol K K p = K c (RT)  n  n = (moles of gaseous product) - (moles of gaseous reactant)

43 Equilibrium © 2009, Prentice-Hall, Inc. 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 When would K p = K c ?? K p = K c (RT)  n  n = (moles of gaseous product) - (moles of gaseous reactant)

44 Equilibrium Calculating K from K p Using the value for K p in our previous question, calculate the value of K at 25 C for the reaction 2NO(g) + Cl 2 g)  2 NOCl(g) © 2009, Prentice-Hall, Inc.

45 Equilibrium Calculating K from K p Using the value for K p in our previous question, calculate the value of K at 25 C for the reaction 2NO(g) + Cl 2 g)  2 NOCl(g) 4.6 x 104 © 2009, Prentice-Hall, Inc.

46 Equilibrium © 2009, Prentice-Hall, Inc. What Does the Value of K Mean? If K>>1, the reaction is product-favored; product predominates at equilibrium.

47 Equilibrium © 2009, Prentice-Hall, Inc. What Does the Value of K Mean? If K>>1, the reaction is product-favored ; product predominates at equilibrium.( K greater than 10, reaction nearly goes to completion) If K<<1, the reaction is reactant-favored; reactant predominates at equilibrium(K less than.1, the reaction doesn’t go very far to completion at all.)

48 Equilibrium If K<<1, the reaction is There are substantial amounts of reactant and product present at equilibrium © 2009, Prentice-Hall, Inc.

49 Equilibrium © 2009, Prentice-Hall, Inc. Manipulating Equilibrium Constants The equilibrium constant of a reaction in the reverse reaction is the reciprocal of the equilibrium constant of the forward reaction. K c = = at 100  C [NO 2 ] 2 [N 2 O 4 ] N 2 O 4 (g )  2 NO 2 (g) K c = = 4.72 at 100  C [N 2 O 4 ] [NO 2 ] 2 N2O4 (g)N2O4 (g) 2 NO 2 (g) 

50 Equilibrium © 2009, Prentice-Hall, Inc. Manipulating Equilibrium 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. K c = = at 100  C [NO 2 ] 2 [N 2 O 4 ] N 2 O 4(g)  2 NO 2(g) K c = = (0.212) 2 at 100  C [NO 2 ] 4 [N 2 O 4 ] 2 2 N 2 O 4(g)  4 NO 2(g)

51 Equilibrium © 2009, Prentice-Hall, Inc. Manipulating Equilibrium 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.(multiple equilibrium rule)

52 Equilibrium Multiple equilibria rule I) 2A (aq) + B(aq)  4C(aq) K 1= [C] 4 [A] 2 [B] II) 4 C(aq) + E(aq)  2F(aq) K II [F] 2 [C] 4 [E] III) 2A(aq) + B(aq) +E(aq)  2F(aq) K III= K I x K II © 2009, Prentice-Hall, Inc.

53 Equilibrium Can you….. Write an equilibrium constant expression? Calculate K if given equilibrium concentrations of reactants and products Tell how K changes if the stoichiometric coefficients are changed in an equation? Tell how to find K for a summary equation? Explain what the magnitude of K is telling you about a reaction? Convert K c to K p ? © 2009, Prentice-Hall, Inc.

54 Equilibrium © 2009, Prentice-Hall, Inc. Equilibrium Calculations

55 Equilibrium THE REACTION QUOTIENT Q © 2009, Prentice-Hall, Inc.

56 Equilibrium The Reaction Quotient Used when a system is not in equilibrium Calculated in the same way as K, however the concentrations are not necessarily at equilibrium © 2009, Prentice-Hall, Inc.

57 Equilibrium © 2009, Prentice-Hall, Inc. The Reaction Quotient Consider the generalized reaction The reaction quotient Q for this reaction would be Q c = [C] c [D] d [A] a [B] b aA + bBcC + dD

58 Equilibrium What is Q good for??? If Q is less than K, the reaction is not at equilibrium Reactants  products for the reaction to be at equilibrium © 2009, Prentice-Hall, Inc.

59 Equilibrium What is Q good for??? If Q is greater than K, the reaction is not at equilibrium Products  reactants for the reaction to be at equilibrium © 2009, Prentice-Hall, Inc.

60 Equilibrium What is Q good for??? If Q equals K, the reaction is at equilibrium!!! © 2009, Prentice-Hall, Inc.

61 Equilibrium © 2009, Prentice-Hall, Inc. If Q < K, there is too much reactant, and the equilibrium shifts to the right.

62 Equilibrium © 2009, Prentice-Hall, Inc. If Q = K, the system is at equilibrium.

63 Equilibrium © 2009, Prentice-Hall, Inc. If Q > K, there is too much product, and the equilibrium shifts to the left.

64 Equilibrium Using the Reaction Quotient For the synthesis of ammonia at 500 C, the equilibrium constant = 6.0 x Predict the direction in which equilibrium will shift in each of the following cases: [NH 3 ]= 1.0 x M; [N 2 ]=1.0 x M; [H 2 ]= 2.0 x M (toward the reactants) © 2009, Prentice-Hall, Inc.

65 Equilibrium Some Calculations with the Equilibrium Constant © 2009, Prentice-Hall, Inc.

66 Equilibrium © 2009, Prentice-Hall, Inc. An Equilibrium Problem A closed system initially containing x M H 2 and x 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 M. Calculate K c at 448  C for the reaction taking place, which is H 2 (g) + I 2 (s)  2 HI (g)

67 Equilibrium © 2009, Prentice-Hall, Inc. What Do We Know? [H 2 ], M[I 2 ], M[HI], M Initia1.000 x x Change Equilibrium1.87 x 10 -3

68 Equilibrium © 2009, Prentice-Hall, Inc. [HI] Increases by 1.87 x M [H 2 ], M[I 2 ], M[HI], M Initially1.000 x x Change+1.87 x Equilibrium1.87 x 10 -3

69 Equilibrium © 2009, Prentice-Hall, Inc. Stoichiometry tells us [H 2 ] and [I 2 ] decrease by half as much. [H 2 ], M[I 2 ], M[HI], M Initially1.000 x x Change-9.35 x x Equilibrium1.87 x 10 -3

70 Equilibrium © 2009, Prentice-Hall, Inc. We can now calculate the equilibrium concentrations of all three compounds… [H 2 ], M[I 2 ], M[HI], M Initially1.000 x x Change-9.35 x x Equilibrium6.5 x x x 10 -3

71 Equilibrium © 2009, Prentice-Hall, Inc. …and, therefore, the equilibrium constant. K c = [HI] 2 [H 2 ] [I 2 ]

72 Equilibrium © 2009, Prentice-Hall, Inc. …and, therefore, the equilibrium constant. K c = [HI] 2 [H 2 ] [I 2 ] = 51 = (1.87 x ) 2 (6.5 x )(1.065 x )

73 Equilibrium © 2009, Prentice-Hall, Inc. The Reaction Quotient (Q) Q gives the same ratio the equilibrium expression gives, but for a system that is not at equilibrium. To calculate Q, one substitutes the initial concentrations of reactants and products into the equilibrium expression.

74 Equilibrium © 2009, Prentice-Hall, Inc. Le Châtelier’s Principle

75 Equilibrium © 2009, Prentice-Hall, Inc. 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.”

76 Equilibrium © 2009, Prentice-Hall, Inc. Le Châtelier’s Principle When a stress is added, shifts in the reactant and product concentrations occur to reestablish equilibrium position Think about K :[product]/[reactants]

77 Equilibrium Things that “Stress a System” Adding or removing a reactant or product Increasing or decreasing the pressure or volume in systems with gaseous reactants/products (sometimes) Increasing or decreasing the temperature (which will change K) © 2009, Prentice-Hall, Inc.

78 Equilibrium 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). © 2009, Prentice-Hall, Inc.

79 Equilibrium Le Châtelier’s Principle Effects of Volume and PressureEffects of Volume and Pressure 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 N 2 O 4 (g  © 2009, Prentice-Hall, Inc.

80 Equilibrium © 2009, Prentice-Hall, Inc. 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.

81 Equilibrium Le Châtelier’s Principle © 2009, Prentice-Hall, Inc. 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.

82 Equilibrium Le Châtelier’s Principle © 2009, Prentice-Hall, Inc. 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.

83 Equilibrium © 2009, Prentice-Hall, Inc. 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.

84 Equilibrium © 2009, Prentice-Hall, Inc. 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.

85 Equilibrium © 2009, Prentice-Hall, Inc. The Haber Process This apparatus helps push the equilibrium to the right by removing the ammonia (NH 3 ) from the system as a liquid.

86 Equilibrium Consider Co(H 2 O) 6 2+ (aq) + 4 Cl (aq  CoCl 4 (aq) + 6 H 2 O (l) for which  H > 0. –Co(H 2 O) 6 2+ is pale pink and CoCl 4 2- is blue. © 2009, Prentice-Hall, Inc.

87 Equilibrium © 2009, Prentice-Hall, Inc. The Effect of Changes in Temperature Co(H 2 O) 6 2+ (aq) + 4 Cl (aq) CoCl 4 (aq) + 6 H 2 O (l)

88 Equilibrium © 2009, Prentice-Hall, Inc.

89 Equilibrium ex CaCO 3 (s) ↔ CaO(s) + CO 2 (g) What effect will adding CaCO 3 to the system have on the equilibrium? What will happen to the concentration of CO and CO 2 ? If the volume of the reaction vessel is decreased, what will happen to the pressure? © 2009, Prentice-Hall, Inc.

90 Equilibrium © 2009, Prentice-Hall, Inc. If the pressure is increased, the equilibrium will shift (left/right)____________. The reverse reaction is favored because(more/less) CO 2 is produced. What effect will this volume change have on the equilibrium constant K?

91 Equilibrium © 2009, Prentice-Hall, Inc. Gibbs fee energy and equilibrium

92 Equilibrium Do you remember… Calculating Gibbs free energy enables us to determine whether a reaction is spontaneous at a particular temperature When  G is negative, a reaction is spontaneous  G=  H-T  S © 2009, Prentice-Hall, Inc.

93 Equilibrium K is an expression of theconcentations (or partial presures) of reactants and products at equ’m  G indicates whether a reaction will have a higher forward or reverse rate and “move” in the forward or reverse reaction to achieve equilibrium © 2009, Prentice-Hall, Inc.

94 Equilibrium © 2009, Prentice-Hall, Inc.

95 Equilibrium Relationship between  G and K © 2009, Prentice-Hall, Inc.

96 Equilibrium The relationship between Gibbs free energy and the position of equilibrium in a reaction are brought together by the expression:  G   RT lnK  G= G products- G reactants DG  - Gibbs free energy when all reactants and products are in their stadard states 1M concentrtion 1 atm presure ( not 298 K!) © 2009, Prentice-Hall, Inc.

97 Equilibrium K, lnK, and  G Kln K GG Situation Greater than 1positivenegativeProducts favored at eq’m(forward reaction is spontaneous) Less than 1negativepositiveReactants favored at eq’m (reverse reaction is spontaneous) 100Products and reactants equally favored at eqq’m © 2009, Prentice-Hall, Inc.

98 Equilibrium Calculating Gibbs free energy from Q  G =  G  + RTlnQ Can be used to calculate  G in nonstandard conditions At equilibrium  G=0 © 2009, Prentice-Hall, Inc.

99 Equilibrium © 2009, Prentice-Hall, Inc. Catalysts Catalysts increase the rate of both the forward and reverse reactions.

100 Equilibrium © 2009, Prentice-Hall, Inc. Catalysts When one uses a catalyst, equilibrium is achieved faster, but the equilibrium composition remains unaltered.


Download ppt "Equilibrium © 2009, Prentice-Hall, Inc. Chemical Equilibrium."

Similar presentations


Ads by Google