1. Chapter 15: CHEMICAL EQUILIBRIUM

2. TOPICS 15.1 The Concept of Equilibrium 15.2 The Equilibrium Constant
15.3
Equilibrium Expressions
15.4
Using Equilibrium Expressions to Solve Problems
15.5
Factors that Affect Chemical Equilibrium

3. Introduction: Concept of Equilibrium
Do all chemical reactions go to completion?
When you start with only reactants, do you end up with only products?
Most of chemical reactions are reversible. They do not go to completion (i.e. just products present).
In fact, they will be left, at some instance, with a mixture of reactants and products, which are in equilibrium (their concentrations are not changing).

4. At this instance, does the reaction stop?
Chemical Equilibrium kf kr
N2O4 (g) NO2 (g)
Reversible reaction
As time passes, more NO2 is produced and the brown color intensifies.
After some time, the brown color stops changing. The concentration of NO2 becomes constant. Same does the concentration of N2O4.
At this instance, does the reaction stop?
Dr. Al-Saadi

5. Calculating Equilibrium Constantkf kr

6. LAW OF MASS ACTION For the hypothetical chemical reaction:
a A + b B c C + d D
Qc =
[C]c [D]d
[A]a [B]b
Qc = Reaction Quotient
At equilibrium:
Qc =
[C]c [D]d
[A]a [B]b
= Kc eq eq
Molar concentration eq eq
Equilibrium

7. The Equilibrium Constant
Exercise
In an experiment conducted at 74°C, the equilibrium concentrations of the species involved in the reaction were as follows: [CO] = 1.2 × 10–2 M, [Cl2] = M, and [COCl2] = 0.14 M.
Write the equilibrium expression, and
determine the value of the equilibrium constant for this reaction at 74°C.

8. Setup
The equilibrium expression has the form of concentrations of products over concentrations of reactants, each raised to the appropriate power— in the case of this reaction, all the coefficients are 1, so all the powers will be 1.
Solution

9. Magnitude of Equilibrium Constant
Kc =
[C]c [D]d
[A]a [B]b ( ) eq
a A + b B c C + d D
For the above reaction, three outcomes are possible:
The reaction goes to completion. The equilibrium mixture will consist of predominantly products.
The reaction doesn't occur to any significant degree. The equilibrium mixture will consist of predominantly reactants.
The reaction occurs to a significant degree, but not to completion. The equilibrium mixture will have both reactants and products in comparable quantities.
The value of Kc is very large
The value of Kc is very small
The value of Kc is neither large nor small

10. Magnitude of Equilibrium Constant
Kc =
[C]c [D]d
[A]a [B]b ( ) eq
a A + b B c C + d D Kc
10-2
102
The magnitude of Kc is very small.
The reaction will almost not occur
The reaction will have an equilibrium mixture of both reactants and products.
The magnitude of Kc is very large.
The reaction will go almost to completion

11. When the magnitude of Kc is very large, we expect the first outcome.
Product favored
When the magnitude of Kc is very small, we expect the second outcome.
Reactant favored

12. Heterogeneous Equilibria
CaCO3(s) CaO(s) + CO2(g) Think about it. If you increase the quantity of CaCO3 or CaO solids, would their concentrations change? The concentration of a solid (or a pure liquid) is constant and can be incorporated into Kc*.
K* =
[CaO]eq [CO2]eq
[CaCO3]eq c K*
= [CO2]eq c
[CaO]eq
[CaCO3]eq
= Kc
Real equilibrium constant

13. Heterogeneous Equilibria
Exercise: write Qc and Kc

14. Solution

15. Equilibrium Expressions Involving Pressures
N2O4 (g) 2NO2 (g) The equilibrium expression is given by: In terms of equilibrium partial pressures of NO2 and N2O4 , the equilibrium expression becomes: Are Kc and KP equal?
Kc =
[NO2]2eq
[N2O4]eq
KP =
(PNO2)2eq
(PN2O4)eq
Equilibrium partial pressures of NO2 and N2O4 gases
Generally, they are NOT. Because the partial pressures of reactants and products expressed in atm are not equal to their concentrations expressed in mol/L.

16. The Relationship between Kc and KP
Consider the reaction:
aA(g) bB(g)
Assuming ideal behavior:
Similarly:
PB = [B]RT
Substitution in KP gives =>
Kc =
[B]b
[A]a
KP =
(PB)b
(PA)a
PA =
nART V
= RT nA V
= [A]RT Kc
KP =
([B]RT)b
([A]RT)a
= (RT)b-a
[B]b
[A]a

17. The Relationship between Kc and KP
Since: b – a = Δn where: Δn = moles of gaseous products – moles of gaseous reactants
KP = (RT)Δn = Kc (RT)Δn
[B]b
[A]a
R = L. atm/K. mol, T = Kelvin

18. The Relationship between Kc and KP
Exercise
Write KP expressions for

19. The Relationship between Kc and KP
Solution

20. The Relationship between Kc and KP
Exercise
The equilibrium constant, Kc, for the reaction
is 4.63 × 10–3 at 25°C. What is the value of KP at this temperature?

21. The Relationship between Kc and KP
Solution

22. The Relationship between Kc and KP
Exercise
If the value of Kc of the reaction above is 2.3 × 10-2 at 25 0C, calculate Kp?
Kp = Kc

23. Manipulating Equilibrium Expressions
For any chemical reaction, if something gets changed about how its equilibrium is being expressed, the equilibrium expression must also change accordingly.
2NO(g) + O2(g) NO2(g)
At 500 K, Kc = 6.9 × 105.
Kc =
[NO2]2eq
[NO]2eq[O2]eq

24. Manipulating Equilibrium Expressions
By reversing the reaction, we get:
2NO2(g) NO(g) + O2(g)
The new equilibrium constant (K’c) is basically the reciprocal of Kc .
K’c = 1 /(6.9 × 105) = 1.5 × 10-6
K’c =
[NO2]2eq
[NO]2eq [O2]eq
= 1.5 × 10-6

25. Manipulating Equilibrium Expressions
2NO(g) + O2(g) NO2(g)
By multiplying the original reaction by 2:
4NO(g) + 2O2(g) NO2(g)
The new equilibrium constant (K”c) is basically (Kc)2
K”c = (6.9 × 105)2 = 4.8 × 1011
Kc = 6.9 × 105
K”c =
[NO2]4eq
[NO]4eq [O2]2eq
= 4.8 × 1011

26. Manipulating Equilibrium Expressions
2NO2(g) 2NO(g) + O2(g) 2H2(g) + O2(g) 2H2O(g)
Kc1 = 1.5 × 10-6
At 500 K
Kc2 = 2.4 × 1047
2NO2(g) + 2H2(g) + O2(g) NO(g) + 2H2O(g) + O2(g)
K’c =
[NO2]2eq [H2]2eq [O2]eq
[NO]2eq [O2]eq [H2O]2eq =
[NO2]2eq [H2]2eq
[NO]2eq [H2O]2eq
= 3.6 × 1041
In fact when two reactions (one is with an equilibrium constant Kc1 , and the other is with an equilibrium constant Kc2) are added:
K’c = Kc1 × Kc2
= Kc1 × Kc2
= 3.6 × 1041

27. Manipulating Equilibrium Expressions

28. Manipulating Equilibrium Expressions
Exercise:
At 500C, KP = 2.5  for,
2SO2(g) + O2(g) SO3(g)
Compute KP for each of the following reactions at 500C :
(a) SO2(g) + 1/2O2(g) SO3(g)
(b) 3SO3(g) SO2(g) + 3/2 O2(g)
Reaction is divided by 2, so Kp = square root of Kp
Reaction is multiplied by 3/2 and reversed, Kp = 1/(Kp)3/2

29. Exercise
Solution

30. Predicting the Direction of a Reaction
Using Equilibrium Expressions to Solve Problems
Predicting the Direction of a Reaction
Calculating Equilibrium Concentrations

31. The Reaction Quotient and Equilibrium Constant
Qc = Kc ; equilibrium is present.
Qc > Kc ; product conc. > reactant conc. ; some products must convert to reactants in order to reach equilibrium; a net shift to the left should occur ; the reverse reaction is favored.
Qc < Kc ; product conc. < reactant conc. ; some reactants must convert to products in order to reach equilibrium ; a net shift to the right should occur ; the forward reaction is favored.

32. Predicting the Direction of a Reaction
Exercise:
N2(g) + 3H2(g) NH3(g)
375C is 1.2.
At the start of the reaction the concentrations of N2, H2, and NH3 are M, 9.2×10-3 M and 1.83×10-4 M, respectively.
(a) Is the system at equilibrium?
(b) If not, determine to which direction it must proceed in order to establish equilibrium.
Qc = 0.61

33. Calculating Equilibrium Concentrations
Note: Equilibrium pressure can also be calculated following the same method!!
Calculate the equilibrium concentration of A and B, if the equilibrium constant (Kc) is 24 and initial concentration of A is M? A B

34. Calculating Equilibrium Concentrations
15.4
Calculating Equilibrium Concentrations A B
Kc =
[A]eq
[B]eq
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35. Calculating Equilibrium Concentrations
15.4
Calculating Equilibrium Concentrations A B
Check:
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36. Calculating Equilibrium Concentrations
15.8
Kc for the reaction of hydrogen and iodine to produce hydrogen iodide,
is 54.3 at 430°C. What will the concentrations be at equilibrium if we start with M concentrations of both H2 and I2?
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37. Calculating Equilibrium Concentrations
15.8
Setup
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38. Calculating Equilibrium Concentrations
15.8
Solution
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39. Calculating Equilibrium Concentrations
15.8
Solution
Think about it: How about if we start with HI…
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40. Calculating Equilibrium Concentrations
15.9
For the reaction given below, calculate the equilibrium concentrations of all three species if the starting concentrations are as follows: [H2] = M, [I2] = M, and [HI] = M, and KC of the reaction is 54.3.
Setup
H2(g)
2HI(g)
I2(g) +
Initial conc. (M)
Change in conc. (M)
Equilibrium conc. (M)
0.0424
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41. Setup
Qc > Kc, so the system will have to proceed to the left.

42. Calculating Equilibrium Concentrations
15.9
Solution
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43. Calculating Equilibrium Concentrations
15.9
Solution
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44. Calculating Equilibrium Concentrations
15.9
Solution
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45. Calculating Equilibrium Concentrations
15.4
Calculating Equilibrium Concentrations
When the magnitude of K (either Kc or KP) is very small, the solution to an equilibrium problem can be simplified—making it unnecessary to use the quadratic equation.
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46. Calculating Equilibrium Concentrations
15.10
At elevated temperatures, iodine molecules break apart to give iodine atoms according to the equation
Kc for this reaction at 205°C is 3.39 × 10–12. Determine the concentration of atomic iodine when a 1.00-L vessel originally charged with mol of molecular iodine at this temperature is allowed to reach equilibrium.
Setup
The initial concentration of I2(g) is M and the original concentration of I(g) is zero. Kc = 3.39 × 10–12.
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47. Calculating Equilibrium Concentrations
15.10
Solution
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48. Calculating Equilibrium Concentrations
15.10
Solution
According to our ice table, the equilibrium concentration of atomic iodine is 2x; therefore,
[I(g)] = 2 × 3.62 × 10–8 = 7.24 × 10–8 M.
Check
Kc =
[ ]eq
[7.24 × 10–8 ]2eq
= 3.38 × 10–12
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49. Factors that Affect Chemical Equilibrium
The chemical equilibria can be affected by several factors. Affecting the chemical equilibrium of a chemical reaction may result with an increase or decrease of the amount of its products.
Le Châtelier's principle can be used to predict the effect of a change in conditions on a chemical equilibrium.

50. Le Châtelier's Principle
Le Châtelier's principle states that:
If a change is imposed on a system at equilibrium, the system will respond by shifting in the (forward or reverse) direction that minimizes the effect of that change. As a result, a new equilibrium position will be re-established. Changes made on the system can be:
Addition or removal of a reactant or product.
change in the volume and pressure of the system.
change in temperature.

51. Addition of a Substance
N2(g) + 3H2(g) NH3(g)
Consider the following equilibrium concentrations:
[N2]eq = 2.05 M ; [H2]eq = 1.56 M ; [NH3]eq = 1.52 M =
[N2]eq [H2]3eq
[NH3]2eq Kc
= 0.297
Let’s add more N2 by increasing its conc. from 2.05 M to 3.51 M.
How do you think the system will respond to this change? =
(3.51) (1.56)3
(1.52)2 Qc
= 0.173
< Kc
The system responds by shifting to the right in order to reestablish equilibrium.

52. Addition or Removal of a Substance

53. Exercise:
determine whether the equilibrium will shift to the right, shift to the left, or neither:
addition of O2(g),
removal of H2S(g),
removal of H2O(g), and
addition of S(s).

54. Change in Volume and Pressure
Generally,
A decrease in the volume of a reaction vessel will cause a shift in the equilibrium in the direction that has lower number of moles of gaseous species.
An increase in the volume of a reaction vessel will cause a shift in the equilibrium in the direction that higher has number of moles of gaseous species.
If equal moles are present both sides, decrease or increase in the volume will not affect the equilibrium.

55. Change in Volume and Pressure
N2O4(g) NO2(g)
Consider the following equilibrium concentrations:
[N2O4]eq = M ; [NO2]eq = M =
[N2O4]eq
[NO2]2eq Kc
= 4.63×10-3
Suppose that a pressure is applied and the volume is decreased by one-half.
As a result, the concentrations of N2O4 and NO2 will be doubled. 

56. Change in Volume and Pressure
15.5
N2O4(g) NO2(g)
Consider the following equilibrium concentrations:
[N2O4]eq = M ; [NO2]eq = M =
1.286
(0.1094)2 Qc
= 9.31×10-3
> Kc = 4.63×10-3
The system responds by shifting to the left (producing more of N2O4 and consuming more of NO2) in order to reestablish equilibrium.

57. When volume is increased:
N2O4(g) NO2(g)
Consider the following equilibrium concentrations:
[N2O4]eq = M ; [NO2]eq = M =
0.3215
(0.0274)2 Qc
= 2.34×10-3
< Kc = 4.63×10-3
The system responds by shifting to the right (producing more of NO2 and consuming more of N2O4) in order to reestablish equilibrium.

58. Change in Volume and Pressure
15.5
Exercise:
Predict the change in direction of the following reactions:
(a) PCl5(g) Cl2(g) + PCl3(g) (Volume decreased)
(b) 2PbS(s) + 3O2(g) PbO(s) + 2SO2(g)
(Volume increased)
(c) H2(g) + F2(g) HF(g)
Shift to the left
Shift to the left
No shift

59. Change in Temperature
15.5
Unlike the case with concentration and pressure changes, the change in temperature of a chemical reaction can change the value of the equilibrium constant.
It makes the reaction faster or slower, depending on the enthalpy change (ΔH) “heat” accompanying the reaction.

60. Change in Temperature N2O4(g) 2NO2(g) ΔH= 58.0 kJ/mol
15.5
N2O4(g) NO2(g) ΔH= 58.0 kJ/mol
Let’s apply here Le Châtelier's principle to the heat absorbed as a reactant.
N2O4(g) NO2(g)
Adding heat means the reaction will be shifted to the right. Also, addition of heat means an increase in temperature.
In general, increasing the temperature of endothermic reactions shifts it to the right.
Heat +

61. 2H2 (g) + O2(g) 2H2O(l) ΔH= -780 kJ/mol
2H2 (g) + O2(g) H2O(l) + Heat
In general, increasing the temperature of exothermic reactions shifts it to the left.
Exercise: which of the following equilibria will shift to the left
when the temperature is increased.
S + H H2O(l) ΔH = -20 kJ/mol
C + H2O CO + H2 ΔH = 131 kJ/mol

62. Does a Catalyst Affect the Equilibrium?
15.5
Does a Catalyst Affect the Equilibrium?
A catalyst
speeds up a reaction by lowering its activation energy,
lowers the activation energy of the forward and backward reactions to the same extent,
neither changes the value of the equilibrium constant nor the equilibrium position, and
causes a reaction mixture that is not at equilibrium to reach equilibrium faster.

63. Adding more C(graphite) Adding more CO2 (g) Removing CO(g)
Exercise: which of the following will cause the equilibrium
to shift to the right. Select all that apply)
C(graphite) + CO2(g) CO(g)
Decreasing the volume
Increasing the volume
Adding more C(graphite)
Adding more CO2 (g)
Removing CO(g)

64. ? Change in Temperature N2O4(g) 2NO2(g) ΔH= 58.0 kJ/mol yellow brown
15.5
Change in Temperature
N2O4(g) NO2(g) ΔH= 58.0 kJ/mol
yellow
brown
yellow
Heating ?
brown

65. CoCl42- and Co(H2O)62+ ions at equilibrium
15.5
Change in Temperature
Consider the following exothermic reaction:
CoCl42‒ + 6H2O Co(H2O) Cl‒ + Heat
blue
pink
Heating
Cooling
CoCl42- and Co(H2O)62+ ions at equilibrium

66. 15.5
Change in Temperature
The increase in temperature favors endothermic reactions.
The decrease in temperature favors exothermic reactions.
The change in temperature not only affects the equilibrium position, but also alters the value of the equilibrium constant.