1. Chapter 15
Chemical Equilibrium

2. Reversible Reactions- most chemical reactions are reversible under the correct conditions

3. Reversible Reactions
In a reversible reaction, there is both a forward and a
reverse reaction.
Suppose SO2 and O2 are present initially. As they collide, the forward reaction begins.
2SO2(g) + O2(g) → 2SO3(g)
As SO3 molecules form, they also collide in the reverse reaction that forms reactants. A reversible reaction is written with a double arrow.
forward
2SO2(g) + O2(g) SO3(g)
reverse

4. Chemical Equilibrium At equilibrium
The rate of the forward reaction becomes equal to the rate of the reverse reaction.
The forward and reverse reactions continue at equal rates in both directions.
Figure Page 454

5. Chemical Equilibrium When equilibrium is Reached:
There is no further change in the amounts (concentrations) of reactant and product.

6. N2(g) + O2(g) ↔ 2NO(g)
At equilibrium the concentrations remain constant however the forward and reverse reactions are still occuring.

7. Writing Equilibrium Constant-Expressions

8. The Equilibrium Constant, Keq
Two methods for describing the equilibrium constant of a reaction.
Kc- describes the equilibrium of a reaction where the concentrations of the materials is known.
Kp- describes the equilibrium of a gaseous reaction using partial pressures instead of concentrations.

9. Kc c = concentration aA + bBcC + dD
Coefficients in the chemical equation become exponents in the equilibrium constant expression.
Include only substances in the gas or aqueous phase. Solid’s and liquid’s concentrations do not change during a chemical reaction.

10. Write the equilibrium expression (Kc) for the reaction below
Write the equilibrium expression (Kc) for the reaction below. Ni (s) + 4CO (g)  Ni(CO)4(g)

11. Write the equilibrium expression (Kc) for the reaction below
Write the equilibrium expression (Kc) for the reaction below. Ni (s) + 4CO (g)  Ni(CO)4(g)
[Ni(CO)4]
Kc =
[CO]4

12. Determine the equilibrium constant for the reaction: A ↔ B

13. Kp p= pressure
Equilibrium is described in terms of the partial pressures of the reactants and products.
2CO2 (g) 2CO (g) + O2 (g)

14. Write the equilibrium expression (Kp) for the following mixture of gases at equilibrium.
2NO (g) + O2 (g)  2NO2 (g)
(PNO2)2
Kp =
(PNO)2
(PO2)

15. Properties of the Equilibrium Constant
Keq is constant for a particular reaction at a specific temperature.
We generally omit the units of the equilibrium constant.
Note that the equilibrium constant expression has products over reactants.
K>>1 implies products are favored, and Keq lies to the right (towards the products).
K<<1 implies reactants are favored, and Keq lies to the left.
(towards the reactants).

16. The Magnitude of Equilibrium Constants

17. More Facts About Keq
The equilibrium constant of a reaction in the reverse direction is the inverse of the equilibrium constant of the reaction in the forward direction… Keq(forward) = 1/Keq(reverse)
For example:
At 100 ºC, Keq(forward) = 6.49
At 100 ºC, Keq(reverse) = 1/6.49 = 0.154

18. More Facts About Keq
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.
For example: A + B ↔ X + C Keq(1) = 2.0
X + B ↔ D Keq(2) = 5.0
A + 2B ↔ C + D Keq = K1 x K2 = 10.0
When this is written in terms of concentrations at equilibrium…
Keq=10.0 = [C][D]/[A][B]2

19. CaCO3(s) ↔ CaO(s) + CO2(g)
More Facts About Keq
The equilibrium expression only contains the concentrations of gases or aqueous substances and NEVER solids or pure liquids. Why?
- Consider the decomposition of calcium carbonate:
CaCO3(s) ↔ CaO(s) + CO2(g)
- The concentrations of solids and pure liquids are constant. Therefore.
Keq = Kp = PCO2 Or
Keq = Kc = [CO2]
This is an example of a heterogeneous equilibrium.

20. Note: Although the concentrations of these species are not included in the equilibrium expression, they do participate in the reaction and must be present for an equilibrium to be established!

21. Kc = [Pb2+] [Cl−]2 What is the equilibrium-constant expression for
this reaction?
Kc = [Pb2+] [Cl−]2

22. The equilibrium constant (K) is related to the rate constant (k)

23. We can show for a reversible reaction having equal rates in both directions that….

24. The Reaction Quotient- Q
There are times when a chemist may be given information about a chemical reaction that may or may not be at equilibrium.
A “Q” calculation can be used to determine the direction the reaction is heading to reach equilibrium.

25. Predicting Direction of Reaction
We define Q, the reaction quotient, for a general reaction

26. The Reaction Quotient- Q
If Q > K, then the reaction has too many products and is heading towards the reactant side.
If Q < K, then the reaction has too many reactants and is heading towards the product side.
If Q = K, then the reaction is at equilibrium.

27. Reaction Shifts
At 448 oC the equilibrium constant, Keq, for the reaction: H2(g) + I2(g) HI(g) is Predict how the reaction will proceed to reach equilibrium at 448 oC if we start with 2.0 x 10-2 mol of HI, 1.0 x 10-2 mol of H2, and 3.0 x 10-2 mol of I2 in a 2.0-L container.

28. At 2000°C, the equilibrium constant for the reaction
N2(g) + O2(g) ↔ 2NO(g)
is 4.1 x Find the concentration of NO(g) in a mixture of NO(g), N2(g), and O2(g) in which, at equilibrium, [N2] = mol L-1 and [O2] = mol L-1.
[NO] = 3.6 x 10-4M

29. The equilibrium constant for the reaction of nitrogen and hydrogen to produce ammonia at a certain temperature is 6.00 x Calculate the equilibrium concentration of ammonia if the equilibrium concentrations of nitrogen and hydrogen are 4.26M and 2.09M, respectively.
[NH3] = 1.53M

30. Iodine molecules react reversibly with iodide ions to produce triiodide ions.
I2(aq) + I-(aq) ↔ I3-(aq)
If a solution is prepared with the concentrations of both I2 and I- equal to x 10-3M before reaction and if the concentration of I2 changes to 6.61 x 10-4M at equilibrium, what is the equilibrium constant for the reaction?
The “ICE” Table

31. Iodine molecules react reversibly with iodide ions to produce triiodide ions. I2(aq) + I-(aq) ↔ I3-(aq)
If a solution is prepared with the concentrations of both I2 and I- equal to x 10-3M before reaction and if the concentration of I2 changes to 6.61 x 10-4M at equilibrium, what is the equilibrium constant for the reaction?

32. Find the concentration of NO(g) at equilibrium when a mixture O2(g) with a concentration of 0.50M and N2(g) with a concentration of 0.75M is heated at 700°C. THE 5% RULE
N2(g) + O2(g) ↔ 2NO(g) K = 4.1 x 10-9.

34. Example 15.8 Page 462 THE 5% RULE
Under certain conditions the equilibrium constant for the decomposition of PCl5(g) into PCl3(g) and Cl2(g) is What are the equilibrium concentrations of all of three of these gases when the initial concentration of PCl5 was 1.00M?

36. Example 15.8. The Quadratic Equation(See Appendix A.4)

38. See the graphing calculator solutions method in notebook.

39. CH3CO2H + C2H5OH ↔ CH3CO2C2H5 + H2O
Acetic acid, CH3CO2H, reacts with ethanol to form water and ethyl acetate.
CH3CO2H + C2H5OH ↔ CH3CO2C2H5 + H2O
The equilibrium constant for this reaction (when done with dioxane as a solvent) is What are the equilibrium concentrations of all chemical species when 0.15 mol of CH3CO2H, 0.15 mol of C2H5OH, 0.40 mol of CH3CO2C2H5, and 0.40 mol of H2O are mixed in enough dioxane to make 1.0L of solution.
[CH3CO2H] = [C2H5OH ] = 0.18M
[CH3CO2C2H5] = [H2O] = 0.37M
Make An ICE Table

49. Effect of a Catalyst on Equilibrium
A catalyst speeds up the forward and reverse reactions equally and therefore equilibrium is reached faster.
However a catalyst has no effect on the value of the equilibrium constant or on the equilibrium concentrations.

50. Le Chatelier’s Principle
Le Chatelier's principle states that if a stress is applied to a system at equilibrium, the system will shift to minimize the stress.
Stresses include changes in:
Concentration
Temperature
Pressure
Reactions can shift:
Right (toward products)
Left (towards reactants)

51. Stress: Change in concentration

52. Stress: Change in concentration

53. Stress: Change in concentration
SCN- is thiocyanate

54. Stress: Change in concentration
Would Adding Water Create a Stress?

55. What can you tell me about the following reaction?

56. It is at equilibrium.

57. Calculate the equilibrium constant.

59. Adjust the graph to illustrate what would happen if some N2 was added.

60. N2(g) + 3H2(g) ↔ 2NH3(g)

61. Stress: Change in Temperature
An increase in temperature will always shift a reaction in the endothermic (heat/energy absorbing direction).
A decrease in temperature will always shift a reaction in the exothermic (heat/energy releasing direction).

62. How can we tell if a reaction is exothermic or endothermic?
Energy term written as a product (exothermic) or reactant (endothermic) within the equation.
-∆H (exothermic) or +∆H (endothermic).
Warms up (exothermic) or cools down (endothermic).

63. Stress: Change in Pressure
An increase in pressure will shift an equilibrium towards fewer moles of gas.
An decrease in pressure will shift an equilibrium towards more moles of gas.

64. CO (g) + 2H2 (g)  CH3OH (g) H° = -21.7 Kcal
Example: Describe how the reaction below adjusts its equilibrium in each of the scenarios.
CO (g) + 2H2 (g)  CH3OH (g) H° = Kcal
Some of the methanol vapor is condensed and removed from the reaction vessel.
The pressure is increased by decreasing the volume of the reaction vessel.
The temperature is increased.

65. CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)
Which way will the reaction shift?
CO2 is removed.
Pressure is increased.
How would your answer to #2 change if H2O were a gas?

66. CaCO3(s) + heat  CaO(g) + CO2(g)
Which way will the reaction shift?
CO2 is added.
CaCO3 is added.
Temperature is increased.
Pressure is decreased.

67. 2NO2(g) ↔N2O4(g) brown colorless
Is this reaction endothermic or exothermic?
Exothermic

68. Photochromatic Sunglasses
AgCl + energy Ago + Clo
(clear) (dark)
Go outside…
Sunlight more intense than inside light;
“energy”
shift to a new equilibrium:
GLASSES DARKEN
Then go inside…
Sensitive Sunglasses
Oxidation-reduction reactions are the basis for many interesting and useful applications of technology. One such application
is photochromic glass, which is used for the lenses in light sensitive glasses. Lenses manufactured by the Corning Glass Company
can change from transmitting 85% of light to only transmitting 22% of light when exposed to bright sunlight.
Photochromic glass is composed of linked tetrahedrons of silicon and oxygen atoms jumbled together in a disorderly array,
with crystals of silver chloride caught in between the silica tetrahedrons. When the glass is clear, the visible light passes right
through the molecules. The glass absorbs ultraviolet light, however, and this energy triggers an oxidation-reduction reaction
between Ag+ and Cl-:
Ag Cl- --> Ag0 + Cl0
To prevent the reaction from reversing itself immediately, a few ions of Cu+ are incorporated into the silver chloride crystal.
These Cu+ ions react with the newly formed chlorine atoms:
Cu Cl0 --> Cu Cl-
The silver atoms move to the surface of the crystal and form small colloidal clusters of silver metal. This metallic silver
absorbs visible light, making the lens appear dark (colored).
As the glass s removed from the light, the Cu2+ ions slowly move to the surface of the crystal where they interact with
the silver metal:
Cu Ag0 --> Cu Ag+
The glass clears as the silver ions rejoin chloride ions in the crystals.
“energy”
shift to a new equilibrium:
GLASSES LIGHTEN