1. Chemical Equilibrium
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2. Equilibrium
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3. Not all reactions “go to completion”
Reactions in which limiting reagents are used up are said to “go to completion”
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4. Not all reactions “go to completion”
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
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5. The Concept of Equilibrium
Chemical equilibrium occurs when a reaction and its reverse reaction proceed at the same rate.
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6. Not all reactions “go to completion”
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
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7. Equilibrium-Concentration versus time graph
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8. A System at Equilibrium
Once equilibrium is achieved, the amount of each reactant and product remains constant.
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9. Rate versus time graph
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10. 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.
N2O4 (g)
2 NO2 (g)
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11. The Equilibrium Constant
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12. The Equilibrium Constant
Forward reaction:
N2O4 (g)  2 NO2 (g)
Rate Law:
Rate = kf [N2O4]
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13. The Equilibrium Constant
Reverse reaction:
2 NO2 (g)  N2O4 (g)
Rate Law:
Rate = kr [NO2]2
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14. The Equilibrium Constant
Therefore, at equilibrium
Ratef = Rater
kf [N2O4] = kr [NO2]2
Rewriting this, it becomes kf kr
[NO2]2
[N2O4] =
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15. The Equilibrium Constant
The ratio of the rate constants is a constant at that temperature, and the expression becomes
Keq = kf kr
[NO2]2
[N2O4] =
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16. The Equilibrium Constant
Consider the generalized reaction
aA + bB
cC + dD
The equilibrium expression for this reaction would be
Kc =
[C]c[D]d
[A]a[B]b
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17. 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
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18. The Equilibrium Constant
K values are written without units
Kc equilibrium constant for concentration (aqueous)
Kp- Equilibrium constant for partial pressure (gases)
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19. 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.
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20. Write the equilibrium Expression for 4 NH3 (g)+ 7 O2(g)D4 NO2(g)+ 6 H2 O (g)
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21. Write the equilibrium Expression for 4 NH3 (g)+ 7 O2(g)D4 NO2(g)+ 6 H2 O (g)
K= [NO2 ]4 [H2O]6
[NH3 ]4 [O2 ]7
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22. Write the equilibrium expression for 2KClO3 (s) ↔ 2KCl(s) + 3O2 (g)
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23. Write the equilibrium expression for 2KClO3 (s) ↔ 2KCl(s) + 3O2 (g)
K= [O2]3
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24. Write the equilibrium expression for H2O(l) ↔ H+ (aq) + OH- (aq)
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25. Write the equilibrium expression for H2O(l) ↔ H+ (aq) + OH- (aq)
K= [H+ ] [ OH- ]
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26. The Equilibrium Constant Kp
Since pressure is proportional to concentration for gases in a closed system, the equilibrium expression can also be written
Kp =
(PCc) (PDd)
(PAa) (PBb)
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27. Write the K and Kp of the following:
The decomposition of solid phosphorus pentachloride to liquid phosphorus trichloride and chlorine gas
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28. Write the K and Kp of the following:
The decomposition of solid phosphorus pentachloride to liquid phosphorus trichloride and chlorine gas
K= [Cl2 ]
Kp = P Cl2
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29. Write the K and Kp of the following:
Deep blue solid copper II sulfate pentahydrate is heated to drive off water vapor to form solid copper II sulfate.
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30. Write the K and Kp of the following:
Deep blue solid copper II sulfate pentahydrate is heated to drive off water vapor to form solid copper II sulfate.
K=[H2O]5
Kp= PH2O 5
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31. 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
[NH3 ]= 3.1 x 10-2 M, [N2]= 8.5 x 10-1 M
[H2]= 3.1 x 10-3 M
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32. Calculating the values of K
[NH3 ]= 3.1 x 10-2 M, [N2]= 8.5 x 10-1 M
[H2]= 3.1 x 10-3 M
Calculate the value of K at 127 C for this reaction
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33. Calculating the values of K
[NH3 ]= 3.1 x 102 M, [N2]= 8.5 x 10-1 M
[H2]= 3.1 x 10-3 M
Calculate the value of K at 127 C for this reaction
3.8 x 104
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34. Calculating the values of K
[NH3 ]= 3.1 x 102 M, [N2]= 8.5 x 10-1 M
[H2]= 3.1 x 10-3 M
Calculate the value of K at 127 C for the reverse of this reaction (K’ )
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35. Calculating the values of K
[NH3 ]= 3.1 x 102 M, [N2]= 8.5 x 10-1 M
[H2]= 3.1 x 10-3 M
Calculate the value of K at 127 C for the reverse of this reaction (K’ )
2.6 x 10-5
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36. Calculating the values of K
What is the relationship between K and K’ for a reaction?
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37. Calculating the values of K
What is the relationship between K and K’ for a reaction?
K’ = 1/K
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38. Calculating Kp
The reaction for the formation of nitrosyl chloride is 2NO(g) + Cl2 g) D2 NOCl(g)
Was studied at 25 C. The pressures at equilibrium were found to be
PNOCl = atm
PNO = 5.0 x atm
P Cl2 = 3.0 x atm
Calculate the Kp for this reaction at 25 C
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39. Calculating Kp
The reaction for the formation of nitrosyl chloride is 2NO(g) + Cl2 g) D2 NOCl(g)
Was studied at 25 C. The pressures at equilibrium were found to be
PNOCl = atm
PNO = 5.0 x atm
P Cl2 = 3.0 x atm
Calculate the Kp for this reaction at 25 C
1.9 x 103
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40. Relationship Between Kc and Kp
From the Ideal Gas Law we know that
PV = nRT
R= L atm/mol K
Rearranging it, we get
P = RT n V
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41. Relationship Between Kc and Kp
ARE NOT INTERCHANGABLE
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42. Relationship Between Kc and Kp
Plugging this into the expression for Kp for each substance, the relationship between Kc and Kp becomes
Kp = Kc (RT)n
where T is temp in kelvin , R=.0821 L atm/mol K
n = (moles of gaseous product) - (moles of gaseous reactant)
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43. Relationship Between Kc and Kp
Plugging this into the expression for Kp for each substance, the relationship between Kc and Kp becomes
Kp = Kc (RT)n
When would Kp = Kc ??
n = (moles of gaseous product) - (moles of gaseous reactant)
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44. Calculating K from Kp
Using the value for Kp in our previous question, calculate the value of K at 25 C for the reaction
2NO(g) + Cl2 g) D2 NOCl(g)
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45. Calculating K from Kp
Using the value for Kp in our previous question, calculate the value of K at 25 C for the reaction
2NO(g) + Cl2 g) D2 NOCl(g)
4.6 x 104
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46. What Does the Value of K Mean?
If K>>1, the reaction is product-favored; product predominates at equilibrium.
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47. 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.)
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48. If K<<1, the reaction is
There are substantial amounts of reactant and product present at equilibrium
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49. Manipulating Equilibrium Constants
The equilibrium constant of a reaction in the reverse reaction is the reciprocal of the equilibrium constant of the forward reaction.
Kc = = at 100 C
[NO2]2
[N2O4]
N2O4 (g) D
2 NO2 (g)
Kc = = 4.72 at 100 C
[N2O4]
[NO2]2
N2O4 (g)
2 NO2 (g) D
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50. 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.
Kc = = at 100 C
[NO2]2
[N2O4]
N2O4(g) D
2 NO2(g)
Kc = = (0.212)2 at 100 C
[NO2]4
[N2O4]2
2 N2O4(g) D
4 NO2(g)
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51. 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)
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52. Multiple equilibria rule
I) 2A (aq) + B(aq)D4C(aq) K1= [C]4 [A]2 [B] II) 4 C(aq) + E(aq)D2F(aq) KII [F]2 [C]4 [E] III) 2A(aq) + B(aq) +E(aq) D2F(aq) KIII= KI x KII
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53. 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 Kc to Kp ?
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54. Equilibrium Calculations
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55. THE REACTION QUOTIENT Q
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56. 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
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57. The Reaction Quotient Qc = [C]c[D]d [A]a[B]b
Consider the generalized reaction
aA + bB
cC + dD
The reaction quotient Q for this reaction would be
Qc =
[C]c[D]d
[A]a[B]b
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58. 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
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59. 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
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60. If Q equals K, the reaction is at equilibrium!!!
What is Q good for???
If Q equals K, the reaction is at equilibrium!!!
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61. there is too much reactant, and the equilibrium shifts to the right.
If Q < K,
there is too much reactant, and the equilibrium shifts to the right.
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62. the system is at equilibrium.
If Q = K,
the system is at equilibrium.
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63. there is too much product, and the equilibrium shifts to the left.
If Q > K,
there is too much product, and the equilibrium shifts to the left.
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64. 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:
[NH3]= 1.0 x 10-3 M; [N2]=1.0 x M; [H2]= 2.0 x M
(toward the reactants)
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65. Some Calculations with the Equilibrium Constant
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66. An Equilibrium Problem
A closed system initially containing x 10-3 M H2 and x 10-3 M I2 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 Kc at 448 C for the reaction taking place, which is
H2 (g) + I2 (s) D
2 HI (g)
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67. What Do We Know? [H2], M [I2], M [HI], M Initia 1.000 x 10-3
Change
Equilibrium
1.87 x 10-3
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68. [HI] Increases by 1.87 x 10-3 M [H2], M [I2], M [HI], M Initially
Change
+1.87 x 10-3
Equilibrium
1.87 x 10-3
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69. Stoichiometry tells us [H2] and [I2] decrease by half as much.
[H2], M
[I2], M
[HI], M
Initially
1.000 x 10-3
2.000 x 10-3
Change
-9.35 x 10-4
+1.87 x 10-3
Equilibrium
1.87 x 10-3
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70. We can now calculate the equilibrium concentrations of all three compounds…
[H2], M
[I2], M
[HI], M
Initially
1.000 x 10-3
2.000 x 10-3
Change
-9.35 x 10-4
+1.87 x 10-3
Equilibrium
6.5 x 10-5
1.065 x 10-3
1.87 x 10-3
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71. …and, therefore, the equilibrium constant.
Kc =
[HI]2
[H2] [I2]
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72. …and, therefore, the equilibrium constant.
Kc =
[HI]2
[H2] [I2]
= 51 =
(1.87 x 10-3)2
(6.5 x 10-5)(1.065 x 10-3)
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73. 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.
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74. Le Châtelier’s Principle
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75. 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.”
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76. 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]
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77. 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)
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78. 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).
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79. Le Châtelier’s Principle
Effects 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 N2O4 (g D
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80. An increase in pressure (by decreasing the volume) favors the formation of colorless N2O4.
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 N2O4 is favored.
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81. Le Châtelier’s Principle
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.
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82. Le Châtelier’s Principle
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.
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83. The Haber Process
The transformation of nitrogen and hydrogen into ammonia (NH3) is of tremendous significance in agriculture, where ammonia-based fertilizers are of utmost importance.
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84. The Haber Process
If H2 is added to the system, N2 will be consumed and the two reagents will form more NH3.
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85. The Haber Process
This apparatus helps push the equilibrium to the right by removing the ammonia (NH3) from the system as a liquid.
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86. Consider for which DH > 0.
Co(H2O)62+(aq) + 4 Cl(aq D CoCl4 (aq) + 6 H2O (l)
for which DH > 0.
Co(H2O)62+ is pale pink and CoCl42- is blue.
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87. The Effect of Changes in Temperature
Co(H2O)62+(aq) + 4 Cl(aq)
CoCl4 (aq) + 6 H2O (l)
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88. © 2009, Prentice-Hall, Inc.

89. ex CaCO3 (s) ↔ CaO(s) + CO2 (g)
What effect will adding CaCO3 to the system have on the equilibrium?
What will happen to the concentration of CO and CO2?
If the volume of the reaction vessel is decreased, what will happen to the pressure?
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90. If the pressure is increased, the equilibrium will shift (left/right)____________ . The reverse reaction is favored because(more/less) CO2 is produced.
What effect will this volume change have on the equilibrium constant K?
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91. Gibbs fee energy and equilibrium
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92. Do you remember… DG= DH-TDS
Calculating Gibbs free energy enables us to determine whether a reaction is spontaneous at a particular temperature
When DG is negative, a reaction is spontaneous
DG= DH-TDS
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93. K is an expression of theconcentations (or partial presures) of reactants and products at equ’m
DG indicates whether a reaction will have a higher forward or reverse rate and “move” in the forward or reverse reaction to achieve equilibrium
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94. © 2009, Prentice-Hall, Inc.

95. Relationship between DG and K
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96. DG= Gproducts- Greactants
The relationship between Gibbs free energy and the position of equilibrium in a reaction are brought together by the expression:
DGf = -RT lnK
DG= Gproducts- Greactants
DGf- Gibbs free energy when all reactants and products are in their stadard states
1M concentrtion 1 atm presure ( not 298 K!)
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97. K, lnK, and DG K ln K DG Situation Greater than 1 positive negative
Products favored at eq’m(forward reaction is spontaneous)
Less than 1
Reactants favored at eq’m (reverse reaction is spontaneous) 1
Products and reactants equally favored at eqq’m
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98. Calculating Gibbs free energy from Q
DG = DGf RTlnQ
Can be used to calculate DG in nonstandard conditions
At equilibrium DG=0
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99. Catalysts
Catalysts increase the rate of both the forward and reverse reactions.
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100. Catalysts
When one uses a catalyst, equilibrium is achieved faster, but the equilibrium composition remains unaltered.
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