Download presentation

Presentation is loading. Please wait.

Published bySavion Leggitt Modified over 2 years ago

1
Jeffrey Mack California State University, Sacramento Chapter 16 Principles of Chemical Reactivity: Equilibria

2
All chemical reactions are reversible, at least in principle. The concept of equilibrium is fundamental to chemistry. The general concept of equilibrium was introduced in Chapter 3 to explain the limited dissociation of weak acids. The goals of this and the following chapter will be to consider chemical equilibria in quantitative terms. The extent to which that equilibrium lies (product favored verses reactant favored) will be discussed. Chemical Equilibrium: A Review

3
At some point in time during the progress of a reaction, if the concentration of the reactants and products remains constant, equilibrium is said to be achieved. The concentrations are NOT equal. Equilibrium

4
Moving towards equilibrium Equilibrium established Equilibrium

5
Equilibrium systems are said to be: Dynamic (in constant motion) Reversible Equilibrium can be approached from either direction. Pink to blue Co(H 2 O) 6 Cl 2 Co(H 2 O) 4 Cl 2 + 2 H 2 O Blue to pink Co(H 2 O) 4 Cl 2 + 2 H 2 O Co(H 2 O) 6 Cl 2 Properties of Chemical Equilibria

6
After a period of time, the concentrations of reactants and products are constant. The forward and reverse reactions continue after equilibrium is attained. They are equal and opposite in rate. Chemical Equilibrium Fe 3+ (aq) + SCN (aq) Fe(SCN) 2+ (aq)

7
Phase changes such as H 2 O(s) vs. H 2 O(liq) Examples of Chemical Equilibria

8
When a general chemical reaction is at equilibrium, the equilibrium constant is given by: If K > 0 then the reaction is said to be product favored. If K < 0 then the reaction is said to be reactant favored. The Equilibrium Constant

9
Product-favored K > 1 Reactant-favored K < 1 The Equilibrium Constant

10
For the formation of HI(g) the equilibrium constant is given by: The Equilibrium Constant

11
11 N 2 O 4 (g) 2NO 2 (g) Start with NO 2 Start with N 2 O 4 Start with NO 2 & N 2 O 4 equilibrium

12
The equilibrium between reactants and products may be disturbed in three ways: (1) by changing the temperature (2) by changing the concentration of a reactant (3) by changing the volume (for systems involving gases) A change in any of these factors will cause a system at equilibrium to shift back towards a state of equilibrium. Le Chatelier’s principle. This statement is often referred to as Le Chatelier’s principle. Disturbing a Chemical Equilibrium

13
Effect of the Addition or Removal of a Reactant or Product If the concentration of a reactant or product is changed from its equilibrium value at a given temperature, equilibrium will be reestablished eventually. The new equilibrium concentrations of reactants and products will be different, but the value of the equilibrium constant expression will still equal K Disturbing a Chemical Equilibrium

14
14 If an external stress is applied to a system at equilibrium, the system adjusts in such a way that the stress is partially offset as the system reaches a new equilibrium position. Le Châtelier’s Principle Changes in Concentration N 2 (g) + 3H 2 (g) 2NH 3 (g) Add NH 3 Equilibrium shifts left to offset stress

15
15 Le Châtelier’s Principle Changes in Concentration continued ChangeShifts the Equilibrium Increase concentration of product(s)left Decrease concentration of product(s)right Decrease concentration of reactant(s) Increase concentration of reactant(s)right left aA + bB cC + dD Add Remove

16
Effect of Volume Changes on Gas-Phase Equilibria For a reaction that involves gases, what happens to equilibrium concentrations or pressures if the size of the container is changed? (Such a change occurs, for example, when fuel and air are compressed in an automobile engine.) To answer this question, recall that concentrations are in moles per liter. If the volume of a gas changes, its concentration therefore must also change, and the equilibrium composition can change. Disturbing a Chemical Equilibrium

17
17 Le Châtelier’s Principle Changes in Volume and Pressure A (g) + B (g) C (g) ChangeShifts the Equilibrium Increase pressureSide with fewest moles of gas Decrease pressureSide with most moles of gas Decrease volume Increase volumeSide with most moles of gas Side with fewest moles of gas When the number of gas moles on either side is the same, there is no effect.

18
Effect of Temperatue Changes on Gas-Phase Equilibria Consider the reaction of nitrogen and oxygen to form nitric oxide: As the temperature of the reaction is increased, the equilibrium constant increases. KTemperature (K) 4.5 10 31 298 6.7 10 10 900 1.7 10 3 2300 Why? Disturbing a Chemical Equilibrium

19
Effect of Temperature Changes on Gas-Phase Equilibria Lets write the reaction in this manner: Notice that energy is included as a reactant! Disturbing a Chemical Equilibrium

20
Effect of Temperature Changes on Gas-Phase Equilibria Lets write the reaction in this manner: As temperature (Energy) is increased, equilibrium shifts to the right, favoring products. Disturbing a Chemical Equilibrium

21
Effect of Temperatue Changes on Gas-Phase Equilibria Lets write the reaction in this manner: As the concentration of products increases so does the value of the equilibrium constant. Disturbing a Chemical Equilibrium

22
Effect of Temperatue Changes on Gas-Phase Equilibria This explains the increase in the equilibrium constant with increasing temperature. KTemperature (K) 4.5 10 31 298 6.7 10 10 900 1.7 10 3 2300 Disturbing a Chemical Equilibrium

23
Effect of Temperature Changes on Gas-Phase Equilibria Conclusion Conclusion: Increasing rightIncreasing the temperature of an endothermic reaction favors the products, equilibrium shifts to the right. Increasing leftIncreasing the temperature of an exothermic reaction favors the reactants, equilibrium shifts to the left. Lowering temperature results in the reverse effects. Disturbing a Chemical Equilibrium

24
K c (273 K) = 0.00077 K c (298 K) = 0.0059 Temperature Effects on Equilibrium

25
25 Catalyst lowers E a for both forward and reverse reactions. Catalyst does not change equilibrium constant or shift equilibrium. Adding a Catalyst does not change K does not shift the position of an equilibrium system system will reach equilibrium sooner Le Châtelier’s Principle

26
ChangeReaction Shift Adding more N 2 (g)Right Removing H 2 (g)Left Decreasing the container volumeRight Increasing the container temperatureLeft Increasing the container volumeLeft Decreasing the container temperatureRight adding a catalystno effect adding argon to the containerno effect Le Chatelier’s Principle Practice

27
27 Le Châtelier’s Principle - Summary ChangeShift Equilibrium Change Equilibrium Constant Concentrationyesno Pressureyes*no Volumeyes*no Temperatureyes Catalystno *Dependent on relative moles of gaseous reactants and products

28
In an equilibrium constant expression, all concentrations are reported as equilibrium values. Product concentrations appear in the numerator, and reactant concentrations appear in the denominator. Each concentration is raised to the power of its stoichiometric balancing coefficient. Values of K are dimensionless. The value of the constant K is particular to the given reaction at a specific temperature. Writing Equilibrium Constant Expressions

29
Reactions Involving Solids So long as a solid is present in the course of a reaction, its concentration is not included in the equilibrium constant expression. Equilibrium constant: Writing Equilibrium Constant Expressions

30
Reactions in Solution If water is a participant in the chemical reaction, its concentration based on magnitude is considered to remain constant throughout. Equilibrium constant: Writing Equilibrium Constant Expressions

31
Reactions Involving Gases: K c and K p Concentration data can be used to calculate equilibrium constants for both aqueous and gaseous systems. In these cases, the symbol K is sometimes given the subscript “c” for “concentration,” as in K c. For gases, however, equilibrium constant expressions can be written in another way: in terms of partial pressures of reactants and products. Writing Equilibrium Constant Expressions

32
Reactions Involving Gases: K p Notice that the basic form of the equilibrium constant expression is the same as for K c. In some cases, the numerical values of K c and K p are the same. They are different when the numbers of moles of gaseous reactants and products are different. Writing Equilibrium Constant Expressions

33
Reactions Involving Gases: K p & K c The general relationship between K p and K c is derived in chapter 26, pa726. When the number of gas mole is equivalent on either side of the chemical equation, the two equilibrium constants are the same value. R = the gas constant T = the absolute temperature n = (mols gas product) mols gas reactant) Writing Equilibrium Constant Expressions

34
REACTION QUOTIENT, Q In general, ALL reacting chemical systems are characterized by their REACTION QUOTIENT, Q. If Q = K, then system is at equilibrium. The Reaction Quotient, Q

35
If Q < K then the system is heading towards equilibrium: There are more reactants than products as expected at equilibrium. The reaction is said to be headed to the “right”. If Q > K the system has gone past equilibrium. There are more products than reactants as expected at equilibrium. The reaction is said to be headed to the left. If Q = K then the system is at equilibrium. The Reaction Quotient, Q

36
36 K >> 1 K << 1 Lie to the rightFavor products Lie to the leftFavor reactants Equilibrium Will K = [C] c [D] d [A] a [B] b aA + bB cC + dD

37
37 The equilibrium concentrations for the reaction between carbon monoxide and molecular chlorine to form COCl 2 (g) at 74 0 C are [CO] = 0.012 M, [Cl 2 ] = 0.054 M, and [COCl 2 ] = 0.14 M. Calculate the equilibrium constants K c and K p. CO (g) + Cl 2 (g) COCl 2 (g) Kc =Kc = [COCl 2 ] [CO][Cl 2 ] = 0.14 0.012 x 0.054 = 220 K p = K c (RT) n n = 1 – 2 = -1 R = 0.0821T = 273 + 74 = 347 K K p = 220 x (0.0821 x 347) -1 = 7.7

38
38 Heterogenous equilibrium applies to reactions in which reactants and products are in different phases. CaCO 3 (s) CaO (s) + CO 2 (g) [CaCO 3 ] = constant [CaO] = constant K c = [CO 2 ] = K p = P CO 2 The concentration of solids and pure liquids are not included in the expression for the equilibrium constant. [CaO][CO 2 ] [CaCO 3 ] K c = ′ [CaCO 3 ] [CaO] K c x ′

39
The value of a reaction’s equilibrium constant is determined by measuring the concentrations of the reactants and products when a system is at equilibrium. The equilibrium constant can also determined by looking at the changes in concentrations as a system achieves equilibrium. This is know as an “ICE” table. Determining the Equilibrium Constant

40
ICE tables: I = Initial concentration C = change in concentration E = concentrations at equilibrium ReactantsProducts I C E Determining the Equilibrium Constant

41
2.00 mol of NOCl is added to 1.00 L flask. At equilibrium the concentration of NO(g) is found to be 0.66 mol/L. What is the value of K? Determining K

42
[NOCl][NO][Cl 2 ] Initial Change Equilibrium Determining K 2.00 mol of NOCl is added to 1.00 L flask. At equilibrium the concentration of NO(g) is found to be 0.66 mol/L. What is the value of K?

43
[NOCl][NO][Cl 2 ] Initial2.0000 Change Equilibrium Determining K

44
2.00 mol of NOCl is added to 1.00 L flask. At equilibrium the concentration of NO(g) is found to be 0.66 mol/L. What is the value of K? [NOCl][NO][Cl 2 ] Initial2.0000 Change-0.66+0.660.33 Equilibrium note the reaction stoichiometry Determining K

45
2.00 mol of NOCl is added to 1.00 L flask. At equilibrium the concentration of NO(g) is found to be 0.66 mol/L. What is the value of K? [NOCl][NO][Cl 2 ] Initial2.0000 Change-0.66+0.66+0.33 Equilibrium1.340.660.33 Determining K

46
[NOCl][NO][Cl 2 ] Initial2.0000 Change-0.66+0.66+0.33 Equilibrium1.340.660.33 Determining K

47
1.00 mol each of H 2 and I 2 in a 1.00 L flask. What are the equilibrium concentrations of all species. K c = 55.3? Equilibrium Concentrations from K

48
1.00 mol each of H 2 and I 2 in a 1.00 L flask. What are the equilibrium concentrations of all species. K c = 55.3? [H 2 ][I 2 ][HI] Initial1.00 0 Change Equilibrium Equilibrium Concentrations from K

49
1.00 mol each of H 2 and I 2 in a 1.00 L flask. What are the equilibrium concentrations of all species. K c = 55.3? Equilibrium Concentrations from K [H 2 ][I 2 ][HI] Initial1.00 0 Change- x + 2x Equilibrium

50
1.00 mol each of H 2 and I 2 in a 1.00 L flask. What are the equilibrium concentrations of all species. K c = 55.3? Equilibrium Concentrations from K [H 2 ][I 2 ][HI] Initial1.00 0 Change- x + 2x Equilibrium1.00 - x + 2 x Where x is defined as amount of H 2 and I 2 consumed on approaching equilibrium in moles.

51
1.00 mol each of H 2 and I 2 in a 1.00 L flask. What are the equilibrium concentrations of all species. K c = 55.3? Equilibrium Concentrations from K [H 2 ][I 2 ][HI] Initial1.00 0 Change- x + 2x Equilibrium1.00 - x + 2 x

52
1.00 mol each of H 2 and I 2 in a 1.00 L flask. What are the equilibrium concentrations of all species. K c = 55.3? Equilibrium Concentrations from K [H 2 ][I 2 ][HI] Initial1.00 0 Change- x + 2x Equilibrium1.00 - x + 2 x

53
1.00 mol each of H 2 and I 2 in a 1.00 L flask. What are the equilibrium concentrations of all species. K c = 55.3? Equilibrium Concentrations from K

54
Consider the following reaction: If initial concentration of N 2 O 4 is 0.50 M, what are the equilibrium concentrations? Equilibrium Concentrations from K

55
Consider the following reaction: If initial concentration of N 2 O 4 is 0.50 M, what are the equilibrium concentrations? Equilibrium Concentrations from K [N 2 O 4 ][NO 2 ] Initial0.500 Change Equilibrium

56
Consider the following reaction: If initial concentration of N 2 O 4 is 0.50 M, what are the equilibrium concentrations? Equilibrium Concentrations from K [N 2 O 4 ][NO 2 ] Initial0.500 Change-x+2x Equilibrium

57
Consider the following reaction: If initial concentration of N 2 O 4 is 0.50 M, what are the equilibrium concentrations? Equilibrium Concentrations from K [N 2 O 4 ][NO 2 ] Initial0.500 Change-x+2x Equilibrium0.50 - x2x

58
Consider the following reaction: If initial concentration of N 2 O 4 is 0.50 M, what are the equilibrium concentrations? Equilibrium Concentrations from K [N 2 O 4 ][NO 2 ] Initial0.500 Change-x+2x Equilibrium0.50 - x2x

59
Consider the following reaction: If initial concentration of N 2 O 4 is 0.50 M, what are the equilibrium concentrations? Equilibrium Concentrations from K

60
Consider the following reaction: If initial concentration of N 2 O 4 is 0.50 M, what are the equilibrium concentrations? Equilibrium Concentrations from K Solving this requires the Quadratic Equation:

61
Consider the following reaction: If initial concentration of N 2 O 4 is 0.50 M, what are the equilibrium concentrations? Equilibrium Concentrations from K

62
Consider the following reaction: If initial concentration of N 2 O 4 is 0.50 M, what are the equilibrium concentrations? Equilibrium Concentrations from K

63
x = 0.027 or 0.028 The negative value is not reasonable which gives the equilibrium values of concentration to be: Consider the following reaction: If initial concentration of N 2 O 4 is 0.50 M, what are the equilibrium concentrations? Equilibrium Concentrations from K

64
x = 0.027 or 0.028 The negative value is not reasonable which gives the equilibrium values of concentration to be: x = 0.027 M [N 2 O 4 ] = 0.50 M x = 0.47 M [NO 2 ] = 2x = 0.054 M Consider the following reaction: If initial concentration of N 2 O 4 is 0.50 M, what are the equilibrium concentrations? Equilibrium Concentrations from K

65
x = 0.027 or 0.028 The negative value is not reasonable which gives the equilibrium values of concentration to be: x = 0.027 M [N 2 O 4 ] = 0.50 M x = 0.47 M [NO 2 ] = 2x = 0.054 M The results are in agreement with the magnitude of K: The reactant concentration is favored due to the small value of K Consider the following reaction: If initial concentration of N 2 O 4 is 0.50 M, what are the equilibrium concentrations? Equilibrium Concentrations from K

66
What happens when the change in concentration for a reaction is much less than the initial concentration of the reactants? It turns out that when K 100 < [A] 0, then the quadratic equation is not required. since [A] 0 x [A] 0 Approximations in Equilibrium Concentrations

67
Let’s look at what happens to the previous problem if the initial concentration of N 2 O 4 (g) is doubled to 1.00 M. Approximations in Equilibrium Concentrations

68
Consider the following reaction: If initial concentration of N 2 O 4 is 1.00 M, what are the equilibrium concentrations? Approximations in Equilibrium Concentrations

69
Consider the following reaction: If initial concentration of N 2 O 4 is 1.00 M, what are the equilibrium concentrations? Solving for the equilibrium concentration using the quadratic equation yields:

70
Approximations in Equilibrium Concentrations Consider the following reaction: If initial concentration of N 2 O 4 is 1.00 M, what are the equilibrium concentrations? Solving for the equilibrium concentration using the quadratic equation yields: x = 0.038 M, [N 2 O 4 ] = 0.096 & [N 2 O] = 0.076

71
Approximations in Equilibrium Concentrations Consider the following reaction: If initial concentration of N 2 O 4 is 1.00 M, what are the equilibrium concentrations? Solving for the equilibrium concentration using the approximation, ([N 2 O 4 ] 0 x) = [N 2 O 4 ] 0 also yields: x = 0.038 M, [N 2 O 4 ] = 0.096 & [N 2 O] = 0.076 The two results are in agreement.

72
Multiplying the coefficients of a reaction: The second reaction is 2 the first. The general relationship is: When “n” is the multiplication factor. More About Balanced Equations & Equilibrium Constants

73
Reversing a reaction The second reaction is the reverse the first. The general relationship is: The equilibrium constant is the inverse of the original. More About Balanced Equations & Equilibrium Constants

74
Adding reactions at equilibrium: Net. The new equilibrium constant is the product of the individual. More About Balanced Equations & Equilibrium Constants

75
butane isobutane Butane-Isobutane Equilibrium

76
At equilibrium [isobutane] = 1.25 M and [butane] = 0.50 M. K = 2.50 If 1.50 M butane is added to the system, what is the mew concentration of each when the system returns to equilibrium? Butane-Isobutane Equilibrium

77
Let’s first calculate the value of Q to see which direction the reaction will take to reestablish equilibrium. At equilibrium [isobutane] = 1.25 M and [butane] = 0.50 M. If 1.50 M butane is added to the system, what is the new concentration of each when the system returns to equilibrium? Butane-Isobutane Equilibrium

78
Let’s first calculate the value of Q to see which direction the reaction will take to reestablish equilibrium. At equilibrium [isobutane] = 1.25 M and [butane] = 0.50 M. If 1.50 M butane is added to the system, what is the new concentration of each when the system returns to equilibrium? Butane-Isobutane Equilibrium

79
Let’s first calculate the value of Q to see which direction the reaction will take to reestablish equilibrium. Q < K: There are more reactants than products as expected at equilibrium. The reaction shift to the “right”. Reactants must react to form products. At equilibrium [isobutane] = 1.25 M and [butane] = 0.50 M. If 1.50 M butane is added to the system, what is the new concentration of each when the system returns to equilibrium? Butane-Isobutane Equilibrium

80
Let’s now calculate the changes in concentration by setting up an “ICE” table. [butane][isobutane] Initial Change Equilibrium At equilibrium [isobutane] = 1.25 M and [butane] = 0.50 M. If 1.50 M butane is added to the system, what is the new concentration of each when the system returns to equilibrium? Butane-Isobutane Equilibrium

81
Let’s now calculate the changes in concentration by setting up an “ICE” table. At equilibrium [isobutane] = 1.25 M and [butane] = 0.50 M. If 1.50 M butane is added to the system, what is the new concentration of each when the system returns to equilibrium? Butane-Isobutane Equilibrium [butane][isobutane] Initial0.5 +1.501.25 Change Equilibrium

82
Let’s now calculate the changes in concentration by setting up an “ICE” table. At equilibrium [isobutane] = 1.25 M and [butane] = 0.50 M. If 1.50 M butane is added to the system, what is the new concentration of each when the system returns to equilibrium? Butane-Isobutane Equilibrium [butane][isobutane] Initial0.5 +1.501.25 Change x +x Equilibrium

83
Let’s now calculate the changes in concentration by setting up an “ICE” table. At equilibrium [isobutane] = 1.25 M and [butane] = 0.50 M. If 1.50 M butane is added to the system, what is the new concentration of each when the system returns to equilibrium? Butane-Isobutane Equilibrium [butane][isobutane] Initial0.5 +1.501.25 Change x +x Equilibrium 2.00 x 1.25 + x

84
Using the value of K, determine the change in concentration, “x”. [butane][isobutane] Initial0.5 +1.501.25 Change x +x Equilibrium 2.00 x 1.25 + x Butane-Isobutane Equilibrium

85
Using the value of K, determine the change in concentration, “x”. [butane][isobutane] Initial0.5 +1.501.25 Change x +x Equilibrium 2.00 x 1.25 + x Butane-Isobutane Equilibrium

86
Using the value of K, determine the change in concentration, “x”. [butane][isobutane] Initial0.5 +1.501.25 Change x +x Equilibrium 2.00 x 1.25 + x Butane-Isobutane Equilibrium X = 1.07

87
The new concentrations for butane and isobutane are: [butane] = 0.93 M[isobutane] = 2.32 M Equilibrium is shifted to the right, forming more products. [butane][isobutane] Initial0.5 +1.501.25 Change x +x Equilibrium 2.00 x 1.25 + x Butane-Isobutane Equilibrium

88
Effect of Volume Changes on Gas-Phase Equilibria Consider the following reaction: At Equilibrium, the concentrations are: [N 2 O 4 ] = 0.0280 M & [NO 2 ] = 0.0128 M What would happen if the total volume of the system was suddenly doubled ? Disturbing a Chemical Equilibrium

89
Effect of Volume Changes on Gas-Phase Equilibria What would happen if the total volume of the system was suddenly doubled? ½ [N 2 O 4 ] = ½ 0.0280M = 0.0140M ½ [NO 2 ] = ½ 0.0128M = 0.00640M Disturbing a Chemical Equilibrium

90
Effect of Volume Changes on Gas-Phase Equilibria What would happen if the total volume of the system was suddenly doubled? ½ [N 2 O 4 ] = ½ 0.0280M = 0.0140M ½ [NO 2 ] = ½ 0.0128M = 0.00640M Calculate Q: Disturbing a Chemical Equilibrium

91
Effect of Volume Changes on Gas-Phase Equilibria What would happen if the total volume of the system was suddenly doubled? Q = 342 > K Q = 342 > K: Therefore some products must shift to reactants (left) to reestablish equilibrium. Disturbing a Chemical Equilibrium

92
Effect of Volume Changes on Gas-Phase Equilibria Conclusion: Increasing the volume of a container favors the side of equilibrium with the greatest number of gas moles. Decreasing the volume favors the side with the least number of moles. When the number of gas moles on either side is the same, there is no effect. Disturbing a Chemical Equilibrium

93

94
94 Chemical Kinetics and Chemical Equilibrium A + 2B AB 2 kfkf krkr rate f = k f [A][B] 2 rate r = k r [AB 2 ] Equilibrium rate f = rate r k f [A][B] 2 = k r [AB 2 ] kfkf krkr [AB 2 ] [A][B] 2 =K c =

95
95 Modifying the Chemical Equation (cont’d) What will be the equilibrium constant K" c for the new reaction? Consider the reaction:2 NO(g) + O 2 (g) 2 NO 2 (g) [NO 2 ] 2 K c = ––––––––– = 4.67 x 10 13 (at 298 K) [NO] 2 [O 2 ] Now consider the reaction: NO 2 (g) NO(g) + ½ O 2 (g) [NO] [O 2 ] 1/2 1 1/2 K" c = ––––––––– = ––– [NO 2 ] 2 K c = 2.14 x 10 –14 = 1.46 x 10 –7

96
96 If the coef in the reaction is:Then K is: DoubledSquared HalvedSquare root Reversed in signInverted Multiplied by a constant nRaised to the n th power Calculate K of reversed reaction, ½ of a reaction or doubled Example:2SO 2(g) + O 2(g) 2SO 3(g) 4SO 2(g) + 2O 2(g) 4SO 3(g) SO 2(g) + ½ O 2(g) SO 3(g) 2SO 3(g) O 2(g) + 2SO 2(g) Example: 2CO (g) + O 2(g) 2CO 2(g) K = 2.75 x 10 20 @1000K CO 2(g) CO (g) + ½ O 2(g) K = 6.03 x 10 -11 @1000K Summarize:

Similar presentations

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google