1. Reaction Rates and Equilibrium
Chapter 17a
Reaction Rates and Equilibrium

2. Dark Energy and Dark Matter-Not in Textbook
How Chemical Reactions Occur
Conditions That Affect Reaction Rates
The Equilibrium Condition
Chemical Equilibrium: A Dynamic Condition

3. Dark Energy
In a 1998 study led by Adam Riess, and a group of scientists from the Mount Stromlo Observatory, which is part of the Australian National University, Harvard University and Johns Hopkins University and Space Telescope Science Institute found, by observing supernovas in distant galaxies that the universe is expanding faster and faster. This violates Newton’s second law of motion that says acceleration is due a force. What force?????

4. Dark Energy
Basically, dark energy is what they attribute to the accelerated expansion of the universe which means it isn’t dark as to light, but dark as to they know nothing about it. Attempts to explain or measure this energy have largely failed.

5. Dark Matter
Dark matter was discovered by observing that the arms of spiral move at the same speed in violation of Kepler’s Law of planetary motion. It was explained by some unknown (dark) matter that distorted known gravitational effects. It was later found that the gravity of super massive black holes at the center of galaxies was not enough to hold the galaxies together. Gravity was not enough to hold the small local clusters of galaxies together as well as the super clusters of galaxies. Long filamental lines of matter, evident of some sort of attractive forces, has also been found between the super clusters of galaxies. Dark Matter has been used to explain all of these

6. New Model of the Cosmos
The new model of the Cosmos puts about 4.6% of the universe being made up of atoms and molecules like what we think we know something about, about 23% is made up by dark matter and the rest (72%) is composed of dark energy.
They say the Universe was different 13.7 billion years ago?????

7. Rates of Chemical Reactions
4 C3H5N3O9  6 N H2O CO2 + O kJ Nitroglycerine Exothermic

8. An Example of Reaction Rates
35/97 people died in 1937
Fast Reaction vs. Slow Reaction

9. Molecules must collide in order for a reaction to occur.
Collision Model
Molecules must collide in order for a reaction to occur.
Rate depends on concentrations of reactants and temperature.

10. Concentration – increases rate because more molecules lead to more collisions.
Temperature – increases rate.
Why?

11. How to Tame Allergic Reactions
How can you slow down a histamine attack?
Histamine attacks are greater when you are hot. Cooling down affected areas can reduce allergy symptoms.

12. Minimum energy required for a reaction to occur.
Activation Energy
Minimum energy required for a reaction to occur.
Wood must have Ea to light and burn!

13. What makes Switzerland unique?

15. Chemical Reactions must go over an energy hill like a mountain (Swiss Alps).

16. A substance that speeds up a reaction without being consumed.
Catalyst
A substance that speeds up a reaction without being consumed.
Enzyme – catalyst in a biological system

17. A substance that speeds up a reaction without being consumed.
Catalyst
A substance that speeds up a reaction without being consumed.
Enzyme – catalyst in a biological system

18. A substance that speeds up a reaction without being consumed.
Catalyst
A substance that speeds up a reaction without being consumed.
Chlorofluoro Carbons (CFC’s) are acting as catalysts to decompose the ozone (O3) layer. The ozone layer is formed from cosmic radiation and protects us from UV light.

19. Depletion is measured by T.O.M.S.
“Total Ozone Mapping Spectrometer”
The below dark shaded are shows the amount of depletion around the Antarctica

20. An Amana refrigerator, one of many appliances that now use HFC-134a
An Amana refrigerator, one of many appliances that now use HFC-134a. This compound is replacing CFC’s, which lead to the destruction of atmospheric ozone.
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21. Use a catalytic converter to convert the polluting exhaust gases of burned lead-free gasoline into harmless gases. Platinum (Pt) is the catalysts used. Only a small amount is needed.

22. Equilibrium
The exact balancing of two processes, one of which is the opposite of the other.
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23. Chemical Equilibrium
A dynamic state where the concentrations of all reactants and products remain constant.
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24. Macroscopically static Microscopically dynamic
Chemical Equilibrium
On the molecular level, there is frantic activity. Equilibrium is not static, but is a highly dynamic situation.
Macroscopically static 
Microscopically dynamic

25. The Reaction of H2O and CO to Form CO2 and H2 as Time Passes
Equal numbers of moles of H2O and CO are mixed in a closed container.
The reaction begins to occur, and some products (H2 and CO2) are formed.
The reaction continues as time passes and more reactants are changed to products.
Although time continues to pass, the numbers of reactant and product molecules are the same as in (c). No further changes are seen as time continues to pass. The system has reached equilibrium.
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27. Chemical Equilibrium
Concentrations reach levels where the rate of the forward reaction equals the rate of the reverse reaction.
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28. H2O(g) + CO(g) H2(g) + CO2(g)
Concept Check
Consider an equilibrium mixture in a closed vessel reacting according to the equation:
H2O(g) + CO(g) H2(g) + CO2(g)
You add more H2O(g) to the flask. How does the concentration of each chemical compare to its original concentration after equilibrium is reestablished? Justify your answer.
The concentrations of each product will increase, the concentration of CO will decrease, and the concentration of water will be higher than the original equilibrium concentration, but lower than the initial total amount.
Students may have many different answers (hydrogen goes up, but carbon dioxide in unchanged, etc.) Let them talk about this for a while – do not go over the answer until each group of students has come up with an explanation. This question also sets up LeChâtelier’s principle for later.
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29. H2O(g) + CO(g) H2(g) + CO2(g)
Concept Check
Consider an equilibrium mixture in a closed vessel reacting according to the equation:
H2O(g) + CO(g) H2(g) + CO2(g)
You add more H2 to the flask. How does the concentration of each chemical compare to its original concentration after equilibrium is reestablished? Justify your answer.
This is the opposite scenario of the previous slide. The concentrations of water and CO will increase. The concentration of carbon dioxide decreases and the concentration of hydrogen will be higher than the original equilibrium concentration, but lower than the initial total amount.
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30. Reactions Rates and EquilibriumW
Chapter 17b
Reactions Rates and Equilibrium

31. 17.5 The Equilibrium Constant: An Introduction
17.6 Heterogeneous Equilibria
17.7 Le Châtelier’s Principle
17.8 Applications Involving the Equilibrium Constant
17.9 Solubility Equilibria

32. Consider the following reaction at equilibrium:
jA + kB lC + mD
A, B, C, and D = chemical species.
Square brackets = concentrations of species at equilibrium.
j, k, l, and m = coefficients in the balanced equation.
K = equilibrium constant (given without units). l m
[C]
[D] K =
[A] j
[B] k
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33. N2(g) + 3H2(g) 2NH3(g) Example
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34. Equilibrium position is a set of equilibrium concentrations.
K always has the same value at a given temperature regardless of the amounts of reactants or products that are present initially.
For a reaction, at a given temperature, there are many equilibrium positions but only one value for K.
Equilibrium position is a set of equilibrium concentrations.
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35. Consider the following equilibrium reaction:
Concept Check
Consider the following equilibrium reaction:
HC2H3O2(aq) H+(aq) + C2H3O2–(aq)
Determine the equilibrium constant expression for the dissociation of acetic acid.
a) b)
c) d)
The correct answer is a.
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36. Exercise
For the reaction below, calculate the value of the equilibrium constant, given the equilibrium concentrations.
N2O4(g) NO2(g)
[N2O4] = M [NO2] = M
a) K = 0.050
b) K = 0.92
c) K = 1.1
K = 0.065
K = (0.060)2/0.055 = 0.065
The correct answer is d. Putting concentrations into the equilibrium expression yields:
K = (0.060)2/0.055 = 0.065
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37. What is the equilibrium expression for the following?
CH4(g) + 2 H2S(g) <==> CS2(g) + 4 H2(g)
[CS2][H2]4
Keq =
[CH4][H2S]2
H2(g) + I2(g) <==> 2 HI(g)
[HI]2
Keq =
[H2][I2]
Fe3+(aq) + SCN-(aq) <==> Fe(SCN)+2(aq)
[Fe(SCN)+2]
Keq =
[Fe3+][SCN-]

38. Homogeneous Equilibria
Homogeneous equilibria – involve the same phase:
N2(g) + 3H2(g) NH3(g)
HCN(aq) H+(aq) + CN-(aq)
C2H5OH(l) + CH3COOH(l)  CH3COOC2H5(l) + H2O(l)
2SO2(g) + O2(g)    2SO3(g)

39. Heterogeneous Equilibria
Heterogeneous equilibria – involve more than one phase:
2KClO3(s) KCl(s) + 3O2(g)
2H2O(l) H2(g) + O2(g)

40. The position of a heterogeneous equilibrium does not depend on the amounts of pure solids or liquids present.
The concentrations of pure liquids and solids are constant.
2KClO3(s) KCl(s) + 3O2(g)
H2O(l) H+(aq) + OH-(aq) K=[H+][OH-]

41. Determine the equilibrium expression for the reaction:
Concept Check
Determine the equilibrium expression for the reaction:
CaF2(s) Ca2+(aq) + 2F–(aq)
a) b)
c) d)
The correct answer is d. CaF2 does not appear in the expression because it is a solid.
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42. Equilibrium shifts to counter a disturbance. Hills and Valleys!
If a change is imposed on a system at equilibrium, the position of the equilibrium will shift in a direction that tends to reduce that change.
Equilibrium shifts to counter a disturbance. Hills and Valleys!
O3 (g) + Cl (g) O2 (g) + OCl(g)

43. Effect of a Change in Concentration
When a reactant or product is added the system shifts away from that added component.
If a reactant or product is removed, the system shifts toward the removed component.
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44. Effect of a Change in Volume
The system is initially at equilibrium.
The piston is pushed in, decreasing the volume and increasing the pressure. The system shifts in the direction that consumes CO2 molecules, lowering the pressure again.

45. Effect of a Change in Volume
Decreasing the volume
The system shifts in the direction that gives the fewest number of gas molecules.
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46. Effect of a Change in Volume
Increasing the volume
The system shifts in the direction that increases its pressure or the greatest number of gas molecules.

47. Effect of a Change in Temperature
The value of K changes with temperature. We can use this to predict the direction of this change.
Exothermic reaction – produces heat (heat is a product)
Adding energy shifts the equilibrium to the left (away from the heat term).
Endothermic reaction – absorbs energy (heat is a reactant)
Adding energy shifts the equilibrium to the right (away from the heat term).
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48. Effect of Temperature on Equilibrium
2NO2(g) N2O4(g) + Heat

49. Industrial Application-The Manufacture of Ammonia
N2(g) H2(g) NH3(g) ΔH = kJ mol-1
To increase production how would you manipulate the equilibrium?
Lower Volume
2. Lower Temperature
3. Remove Product

50. Concept Check Consider the reaction: 2CO2(g) 2CO(g) + O2(g)
How many of the following changes would lead to a shift in the equilibrium position towards the reactant?
I. The removal of CO gas.
II. The addition of O2 gas.
III. The removal of CO2 gas.
IV. Increasing the pressure in the reaction by decreasing the volume of the container.
a) 1
b) 2
c) 3
d) 4
The correct answer is c. In case I, removing the CO gas (product) will shift the equilibrium position to the right. In case II, adding O2 (product) will shift the equilibrium position to the left. In case III, removing CO2 (reactant) will shift the equilibrium position to the left. In case IV, decreasing the volume of the container will shift the equilibrium position to the left in order to produce less moles (2 moles on left versus 3 moles on right).
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51. Concept Check
One method for the production of hydrogen gas can be described by the following endothermic reaction:
CH4(g) + H2O(g) CO(g) + 3H2(g)
How many of the following changes would decrease the amount of hydrogen gas (H2) produced?
I. H2O(g) is added to the reaction vessel. II. The volume of the container is doubled. III. CH4(g) is removed from the reaction vessel. IV. The temperature is increased in the reaction vessel.
a) 1
b) 2
c) 3
d) 4
The correct answer is a. To decrease the amount of hydrogen gas produced, the equilibrium needs to shift left. In case I, adding H2O (reactant) will shift the equilibrium to the right. In case II, increasing the volume of the container will shift the equilibrium position to the right in order to produce more moles (2 moles on left versus 4 moles on right). In case III, removing CH4 (reactant) will shift the equilibrium position to the left (and thus decrease the amount of hydrogen produced). In case IV, increasing the temperature will shift the equilibrium position to the right since the reaction is endothermic (and heat is treated as a reactant).
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52. The Extent of a Reaction
A value of K much larger than 1 means that at equilibrium the reaction system will consist of mostly products – the equilibrium lies to the right.
Reaction goes essentially to completion.

53. The Extent of a Reaction
A very small value of K means that the system at equilibrium will consist of mostly reactants – the equilibrium position is far to the left.
Reaction does not occur to any significant extent.

54. The value of K for a system can be calculated from a known set of equilibrium concentrations.
Unknown equilibrium concentrations can be calculated if the value of K and the remaining equilibrium concentrations are known.

55. If the equilibrium lies to the right, the value for K is __________.
Concept Check
If the equilibrium lies to the right, the value for K is __________.
large (or >1)
If the equilibrium lies to the left, the value for K is ___________.
small (or <1)
Large (or >1); small (or < 1)
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56. At a given temperature, K = 50. for the reaction: H2(g) + I2(g) 2HI(g)
Concept Check
At a given temperature, K = 50. for the reaction:
H2(g) + I2(g) HI(g)
Calculate the equilibrium concentration of H2 given:
[I2] = 1.5 × 10–2 M and [HI] = 5.0 × 10–1 M
1.5 × 10–2 M
3.0 × 10–2 M
5.0 × 10–1 M
3.3 × 10–1 M
K = (HI)2/(H2)(I2)
50 = (5.0 × 10–1)2/(H2)(1.5 × 10–2)
(H2) = 3.3 × 10–1 M
The correct answer is d.
K = (HI)2/(H2)(I2)
50 = (5.0 × 10–1)2/(H2)(1.5 × 10–2)
(H2) = 3.3 × 10–1 M

57. The equilibrium conditions also applies to a saturated solution containing excess solid, MX(s).
Ksp = [M+][X] = solubility product constant
The value of the Ksp can be calculated from the measured solubility of MX(s).

58. Solubility Equilibria
Solubility product (Ksp) – equilibrium constant; has only one value for a given solid at a given temperature.
Solubility – an equilibrium position.
Bi2S3(s) Bi3+(aq) + 3S2–(aq)
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59. Concept Check
If a saturated solution of PbCl2 is prepared by dissolving some of the salt in distilled water and the concentration of Pb2+ is determined to be 1.6 × 10–2 M, what is the value of Ksp?
2.6 × 10–4
2.0 × 10– PbCl Pb Cl-
3.2 × 10– x x x
1.6 × 10–5
Ksp = [Pb2+] [Cl–]2 = (x)(2x)2=(1.6 × 10–2) (3.2 × 10–2)2 = 1.6 × 10–5
The correct answer is d.
The equilibrium for the dissolution of the salt is written as: PbCl2(s) <==> Pb2+(aq) + 2Cl–(aq)
We are given a value for the concentration of lead(II) ion. The concentration of chloride ion must be twice that of the lead. Next we put these values into the Ksp expression and solve:
Ksp = [Pb2+] [Cl-]2 = (1.6 × 10-2) (3.2 × 10-2)2 = 1.6 × 10-5

60. Calculate the solubility of silver chloride in water.
Concept Check
Calculate the solubility of silver chloride in water.
Ksp = 1.6 × 10–10
1.3 × 10–5 M
1.6 × 10–10 M AgCl(s) Ag+ + Cl-
3.2 × 10–10 M x x x
8.0 × 10–11 M
Ksp = [Ag+][Cl–]
1.6 × 10–10 = (x)(x) = x2
x = 1.3 × 10–5 M
The correct answer is a.
The equilibrium for the dissolution of silver chloride is written as: AgCl(s) <==> Ag+(aq) + Cl–(aq)
Ksp = [Ag+][Cl-]
We are given the Ksp value so that we can solve for the solubility of silver chloride:
1.6 × = (x)(x) = x2
x = 1.3 × 10-5 M