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Equilibrium Chapter 15 Chemical Equilibrium. Equilibrium What is a chemical equilibrium? The reaction of hemoglobin with oxygen is a reversible reaction.

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Presentation on theme: "Equilibrium Chapter 15 Chemical Equilibrium. Equilibrium What is a chemical equilibrium? The reaction of hemoglobin with oxygen is a reversible reaction."— Presentation transcript:

1 Equilibrium Chapter 15 Chemical Equilibrium

2 Equilibrium What is a chemical equilibrium? The reaction of hemoglobin with oxygen is a reversible reaction.

3 Equilibrium If the forward and reverse rxns move at the same speed we combine both rxns and represent it like this:

4 Equilibrium High Altitude Training Increases The Concentration of Hemoglobin An Altitude or Hypoxic Tent

5 Equilibrium

6 An Altitude Training Tent for Dogs

7 Equilibrium What is chemical equilibrium?

8 Equilibrium Assuming the forward and reverse rxns are elementary, then the hemoglobin reaction can be defined in terms of chemical kinetics.

9 Equilibrium We can show for a reversible rxn having equal rates in both directions that….

10 Equilibrium What is a chemical equilibrium? A chemical equilibrium is a reversible reaction where the rate fwd = rate rev

11 Equilibrium

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15 Note: The reactant and product concentrations do not change when a reaction is at equilibrium. What is the equilibrium concentration of the product? What is the equilibrium concentration of the reactant?

16 Equilibrium Note: The reactant and product concentrations are not equal when a reaction is at equilibrium but ….. but … but two or more reactants or two or more products can have equal but constant equilibrium concentrations.

17 Equilibrium Writing Equilibrium Constant-Expressions

18 Equilibrium The equilibrium constant expression is the ratio of product concentrations to reactant concentrations. K c = [C] c [D] d [ A] a [B] b aA + bBcC + dD

19 Equilibrium

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22 The Equilibrium Constant The equilibrium expression can also be written in terms of pressure when the expression contain only gases.

23 Equilibrium Recall how K c is derived

24 Equilibrium Equilibrium Constant Expression Workshop

25 Equilibrium Relationship between K c and K p Using the Ideal Gas Law (PV = nRT) we can show derive and equation relating K p and K c and K p. where K p = K c (RT)  n  n = (moles of gaseous product) − (moles of gaseous reactant )

26 Equilibrium Calculating Equilibrium Concentrations (Using the equilibrium expression to determine [ ] eq

27 Equilibrium Equilibrium Calculation Workshop Calculate the [ ] eqs for this reaction. The K c is 0.110 and 0.500 mol of ICl was placed in a 5.00-L flask.

28 Equilibrium Organize the process from beginning to end. Step I. Write the equilibrium expression Step II. Express the unknown [ ]eqs in terms of x and the respective initial concentrations ( constructing a table is helpful) Step III. Substitute the unknown [ ]eqs and K c in the equilibrium expression and solve for x

29 Equilibrium Practice Problem 15-6 Calculate the [ ] eqs for this reaction. The K c is 5.10 at 700K and 1.000 mol of each compound was placed in a 1.000-L flask.

30 Equilibrium Equilibrium Calculations A system initially contains 1.000 x 10 −3 M H 2 and 2.000 x 10 −3 M I 2 and is allowed to reach equilibrium. The concentration of HI in the equilibrium mixture is 1.87 x 10 −3 M. What is K c at 448  C for the reaction shown below. H 2 (g) + I 2 (g) 2 HI (g)

31 Equilibrium What Do We Know? [H 2 ], M[I 2 ], M[HI], M Initially1.000 x 10 -3 2.000 x 10 -3 0 Change At equilibrium 1.87 x 10 -3

32 Equilibrium [HI] Increases by 1.87 x 10 -3 M [H 2 ], M[I 2 ], M[HI], M Initially1.000 x 10 -3 2.000 x 10 -3 0 Change+1.87 x 10 -3 At equilibrium 1.87 x 10 -3

33 Equilibrium Stoichiometry tells us [H 2 ] and [I 2 ] decrease by half as much [H 2 ], M[I 2 ], M[HI], M Initially1.000 x 10 -3 2.000 x 10 -3 0 Change-9.35 x 10 -4 +1.87 x 10 -3 At equilibrium 1.87 x 10 -3

34 Equilibrium We can now calculate the equilibrium concentrations of all three compounds… [H 2 ], M[I 2 ], M[HI], M Initially1.000 x 10 -3 2.000 x 10 -3 0 Change-9.35 x 10 -4 +1.87 x 10 -3 At equilibrium 6.5 x 10 -5 1.065 x 10 -3 1.87 x 10 -3

35 Equilibrium …and, therefore, the equilibrium constant Kc =Kc = [HI] 2 [H 2 ] [I 2 ] = 51 = (1.87 x 10 -3 ) 2 (6.5 x 10 -5 )(1.065 x 10 -3 )

36 Equilibrium Equilibrium Can Be Reached from Either Direction 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.

37 Equilibrium Equilibrium Can Be Reached from Either Direction This is the data from the last two trials from the table on the previous slide.

38 Equilibrium Equilibrium 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.

39 Equilibrium Equilibrium Position

40 Equilibrium The size of K is an indication of the equilibrium position. Equilibrium position is the side of the reaction the equilibrium favors terms of reactants and products. If K >> 1, the reaction is product-favored; or the equilibrium lies to the right.

41 Equilibrium If K << 1, the reaction is reactant-favored; the equilibrium lies to the left.

42 Equilibrium Heterogeneous Equilibria

43 Equilibrium Pure solids and pure liquids are not included in Kc expressions for heterogeneous equilibria What is the equilibrium-constant expression for this reaction? K c = [Pb 2+ ] [Cl − ] 2

44 Equilibrium

45 The Reaction Quotient, Q c

46 Equilibrium The Reaction Quotient (Q) The reaction quotient is a ratio of nonequilibrium or initial reactant and product concentrations. Q is used to predict the direction a reaction will take in order to reach equilibrium.

47 Equilibrium The Reaction Quotient (Q) To do this we must: (1) calculate Q from the initial concentrations and (2) compare Q to K. There are three possibilities: Q = K, Q K

48 Equilibrium If Q = K means the reaction is at equilibrium and Change is required in [product] or [reactant]

49 Equilibrium Q < K means that [products] must be increased for the reaction to attain equilibrium Q < K, equilibrium shifts right

50 Equilibrium Q > K, equilibrium shifts left Q > K means that [products] must be decreased for the reaction to attain equilibrium.

51 Equilibrium Q c = [C] c [D] d [ A] a [B] b Q is calculated the same as K except only initial concentrations can be used in the reaction quotient expression.

52 Equilibrium Le Châtelier’s Principle

53 Equilibrium Le Châtelier’s Principle Le Châtelier’s principle states: a chemical equilibrium will shift left or shift right to counteract a change that disturbs the equilibrium.

54 Equilibrium What Happens When More of a Reactant Is Added to a System?

55 Equilibrium 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.

56 Equilibrium Effect of Pressure

57 Equilibrium The Haber Process

58 Equilibrium Effect of Temperature pinkblue ∆H > 0 Reaction is placed in a hot water bath

59 Equilibrium Effect of Temperature pinkblue ∆H > 0 Reaction is placed in a ice-water bath

60 Equilibrium A catalyst does not affect the equilibrium position Effect of Catalysts

61 Equilibrium

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66 Which reaction does not represent a chemical equilibrium?

67 Equilibrium Manipulating Equilibrium Constants The equilibrium constant of a reaction in the reverse reaction is the reciprocal of the equilibrium constant of the forward reaction. 1 0.212 = K c = = 0.212 at 100  C [NO 2 ] 2 [N 2 O 4 ] N2O4 (g)N2O4 (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)

68 Equilibrium 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 = = 0.212 at 100  C [NO 2 ] 2 [N 2 O 4 ] N2O4 (g)N2O4 (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)

69 Equilibrium 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.


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