Chemical Kinetics

1
Chemical Kinetics Equilibrium

• Chapter 16

2
Collision Model

• Key Idea Molecules must collide to react.
• However, only a small fraction of collisions
  produces a reaction. Why?
• For a reaction to occur, molecules must have
• 1. sufficient energy to break old bonds.
• 2. proper orientation for collision to be
  effective.

3
Three possible collision orientations-- a) b)
produce reactions, while c) does not.
4
Factors Affecting Rate of Reaction

• Concentration The higher the concentration of
  the reactants, the more likely an effective
  collision will occur.
• Temperature An increase in temperature
  increases
• 1. the energy of a collision.
• 2. the number of collisions.

5
Plot showing the number of collisions with a
particular energy at T1 T2, where T2 gt T1 --
Boltzman Distribution.
6
Activation Energy, Ea

• Activation energy for a given reaction is a
  constant and not temperature dependent.
• Activation energy represents the minimum energy
  for a reaction to occur.

7
a) The change in potential energy as a function
of reaction progress. Ea is the activation energy
and ?E is the net energy change -- exothermic.
b) Molecular representation of the reaction.
8
Enthalpy -- ?H

• Enthalpy -- at constant pressure, the change in
  enthalpy equals the energy flow as heat.
• Exothermic -- ?H is negative (-).
• Endothermic -- ?H is positive ().

9
Catalysis

• Catalyst A substance that speeds up a reaction
  without being consumed
• Enzyme A large molecule (usually a protein)
  that catalyzes biological reactions.
• Homogeneous catalyst Present in the same phase
  as the reacting molecules.
• Heterogeneous catalyst Present in a different
  phase than the reacting molecules.

10
Energy plots for a catalyzed and an uncatalyzed
pathway for an endothermic reaction.
11
Effect of a catalyst on the number of
reaction-producing collisions. A greater
fraction of collisions are effective for the
catalyzed reaction.
12
Catalysis

• The breakdown of the ozone layer is illustrated
  by the following equation
• Cl O3 ---gt ClO O2
• O ClO ---gt Cl O2
• Cl O3 O ClO ---gt ClO O2 Cl O2
• Cl
• Net Reaction O O3 ----gt 2O2
• What is the catalyst? The intermediate?

13
Chemical Equilibrium

• The state where the concentrations of all
  reactants and products remain constant with time.
• On the molecular level, there is frantic
  activity. Equilibrium is not static, but is a
  highly dynamic situation.

14
Reactions That Appear to Run to Completion

• 1. Formation of a precipitate.
• 2. Formation of a gas.
• 3. Formation of a molecular substance such as
  water.
• These reactions appear to run to completion, but
  actually the equilibrium lies very far to the
  right. All reactions in closed vessels reach
  equilibrium.

15
Molecular representation of the reaction 2NO2(g)
----gt N2O4(g). c) d) represent equilibrium.
16
Figure 16.9 (a) The initial equilibrium mixture
of N2, H2 and NH3. (b) Addition of N2. (c) The
new equilibrium.
17
Chemical Equilibrium

• 2NO2(g) lt----gt N2O4(g)
• The forward reaction goes to the right.
• The reverse reaction goes to the left.
• At equilibrium the rate of the reverse reaction
  equals the rate of the forward reaction.

18
Concentration profile for the Haber Process which
begins with only H2(g) N2(g).
19
The Law of Mass Action

• For
• jA kB ? lC mD
• The law of mass action (Cato Guldberg Peter
  Waage) is represented by the equilibrium
  expression

20
Equilibrium Expression

• K is the equilibrium constant.
• C is the concentration expressed in mol/L.
• K is temperature dependent.

21
Equilibrium Constant, K

• For an exothermic reaction, if the temperature
  increases, K decreases.
• For an endothermic reaction, if the temperature
  increases, K increases.

22
Writing Equilibrium Expressions

• 2O3(g) lt---gt 3O2(g)

23
Writing Equilibrium Expressions

• Write the equilibrium expression for the
  following
• H2(g) F2(g) lt---gt 2HF(g)
• N2(g) 3H2(g) lt---gt 2NH3(g)

24
Equilibrium Expression

• Write the equilibrium expression for the
  following reaction
• 4NH3(g) 7O2(g) ? 4NO2(g) 6H2O(g)

25
Table 16.1 Results of Three Experiments for the
Reaction N2(g) 1 3H2(g) 2NH3(g) at 500 ºC
26
Equilibrium Position

• For a given reaction at a given temperature,
  there is only one equilibrium constant (K), but
  there are an infinite number of equilibrium
  positions.
• Where the equilibrium position lies is determined
  by the initial concentrations of the reactants
  and products. The initial concentrations do not
  affect the equilibrium constant.

27
Homogeneous Equilibria

• Homogeneous equilibria are equilibria in which
  all substances are in the same state.
• N2(g) 3H2(g) lt---gt 2NH3(g)
• H2(g) F2(g) lt---gt 2HF(g)

28
Heterogeneous Equilibria

• . . . are equilibria that involve more than one
  phase.
• CaCO3(s) ? CaO(s) CO2(g)
• K CO2
• The position of a heterogeneous equilibrium does
  not depend on the amounts of pure solids or
  liquids present. This does not apply to gases or
  solutions.

29
The position of the equilibrium CaCO3(s) ---gt
CaO(s) CO2(g) does not depend upon the amounts
of solid CaCO3 or CaO.
30
Figure 16.10 The reaction system CaCO3(s) --gt
CaO(s) CO2(g)
31
Heterogeneous Equilibria

• Write equilibrium expressions for the following
• 2HOH(l) lt---gt 2H2(g) O2(g)
• 2HOH(g) lt---gt 2H2(g) O2(g)

K H22O2
32
Heterogeneous Equilibria

• Write equilibrium expressions for the following
• PCl5(s) lt---gt PCl3(l) Cl2(g)
• CuSO4. 5 HOH(s) lt---gt CuSO4(s) 5HOH(g)

K Cl2
K HOH5
33
Le Châteliers Principle

• . . . If a system at equilibrium is subjected to
  a stress, the equilibrium will be displaced in
  such direction as to relieve the stress.

34
Le Chateliers Principle

• If a reactant or product is added to a system at
  equilibrium, the system will shift away from the
  added component.
• If a reactant or product is removed, the system
  will shift toward the removed component.

35
Effect of Changes in Concentration on Equilibrium

• N2(g) 3H2(g) lt---gt 2NH3(g)
• The addition of 1.000 M N2 has the following
  effect
• Equilibrium Position I Equilibrium Position II
• N2 0.399M N2 1.348 M
• H2 1.197 M H2 1.044 M
• NH3 0.203 M NH3 0.304 M

36
Effect of Change in Concentration on Equilibrium

• Position I
• Position II

K 0.0602
K 0.0602
37
Changes in Concentration

• Predict the effect of the changes listed to this
  equilbrium
• As4O6(s) 6C(s) lt---gt As4(g) 6CO(g)
• a) addition of carbon monoxide
• b) addition or removal of C(s) or As4O6(s)
• c) removal of As4(g)

a) left b) none c) right
38
The Effect of Container Volume on Equilibrium

• If the size of a container is changed, the
  concentration of the gases change.
• A smaller container shifts the equilibrium to the
  right -- N2(g) 3H2(g) ---gt 2NH3(g). Four
  gaseous molecules produce two gaseous molecules.
• A larger container shifts to the left -- two
  gaseous molecules produce four gaseous molecules.

39
The system of N2, H2, and NH3 are initially at
equilibrium. When the volume is decreased,
the system shifts to the right -- toward fewer
molecules.
40
The Effect of Container Volume on Equilibrium

• Predict the direction of shift for the following
  equilibrium systems when the volume is reduced
• a) P4(s) 6Cl2(g) lt---gt 4PCl3(l)
• b) PCl3(g) Cl2(g) lt---gt PCl5(g)
• c) PCl3(g) 3NH3(g) lt---gt P(NH2)3(g) 3HCl(g)

a) right b) right c) none
41
Equilibrium Constant, K

• For an exothermic reaction, if the temperature
  increases, K decreases.
• N2(g) 3H2(g) lt---gt 2NH3(g) 92 kJ
• For an endothermic reaction, if the temperature
  increases, K increases.
• CaCO3(s) 556 kJ lt---gt CaO(s) CO2(g)

42
Effect of Temperature on Equilbirum

• Predict how the equilibrium will shift as the
  temperature is increased.
• N2(g) O2(g) lt---gt 2NO(g) (endothermic)
• 2SO2(g) O2(g) lt---gt 2SO3(g) (exothermic)

shift to the right
shift to the left
43
Effects of Changes on the System

• 1. Concentration The system will shift away
  from the added component.
• 2. Temperature K will change depending upon the
  temperature treat heat as a reactant
  (endothermic) and as a product (exothermic).

44
Effects of Changes on the System (continued)

• 3. Pressure
• a. Addition of inert gas does not affect the
  equilibrium position.
• b. Decreasing the volume shifts the
  equilibrium toward the side with fewer gaseous
  molecules.

45
Figure 16.12 Shifting equilibrium by changing
the temperature
46
(No Transcript)
47
Magnitude of K

• A K value much larger than 1 means that the
  equilibrium system contains mostly products --
  equilibrium lies far to the right.
• A very small K value means the system contains
  mostly reactants -- equilibrium lies far to the
  left.
• The size of K and the time required to reach
  equilibrium are not directly related!

48
Calculating Equilibrium Concentrations

• Gaseous phosphorus pentachloride decomposes to
  chlorine gas and gaseous phosphorus trichloride.
  If K 8.96 x 10-2 , and the equilbrium
  concentration of PCl5 is 6.70 x 10-3 M and that
  of PCl3 is 0.300 M, calculate the concentration
  of Cl2 at equilibrium.
• PCl5(g) lt---gt PCl3(g) Cl2(g)

49
Calculating Equilibrium Concentrations

• PCl5(g) lt---gt PCl3(g) Cl2(g)
• K 8.96 x 10-2
• PCl5 6.70 x 10-3 M
• PCl3 0.300 M
• Cl2 ?

Cl2 2.00 x 10-3 M
50
Solubility

• Allows us to flavor foods -- salt sugar.
• Solubility of tooth enamel in acids.
• Allows use of toxic barium sulfate for intestinal
  x-rays.

51
Solubility Product

• For solids dissolving to form aqueous solutions.
• Bi2S3(s) ? 2Bi3(aq) 3S2?(aq)
• Ksp solubility product constant
• and
• Ksp Bi32S2?3
• Why is Bi2S3(s) not included in the solubilty
  product expression?

52
Writing Solubility Product Expressions

• Write the balanced equation for dissolving each
  of the following solids in water. Also write the
  Ksp expression for each solid.
• a) PbCl2(s) b) Bi2S3(s) c)
  Ag2CrO4(s)
• PbCl2(s) lt---gt Pb2(aq) 2Cl-(aq) Ksp
  Pb2Cl-2
• Bi2S3(s) lt---gt 2Bi3(aq) 3S2-(aq) Ksp
  Bi32S2-3
• Ag2CrO4(s) lt---gt 2Ag(aq) CrO42-(aq) Ksp
  Ag2CrO42-

53
Solubility Product Calculations

• Cupric iodate has a measured solubility of 3.3 x
  10-3 mol/L. What is its solubility product?
• Cu(IO3)2(s) lt---gt Cu2(aq) 2 IO3-(aq)
• 3.3 x 10-3 M ---gt 3.3 x 10-3 M 6.6 x 10-3 M
• Ksp Cu2IO3-2
• Ksp 3.3 x 10-36.6 x 10-32
• Ksp 1.4 x 10-7

54
Calculating Solubility from Ksp Values

• The Ksp value for solid AgI(s) is 1.5 x 10-16 at
  25 oC. Calculate the solubility of AgI(s) in
  water at 25 oC.
• AgI(s) lt---gt Ag(aq) I-(aq)
• Ag x Ksp 1.5 x 10-16 AgI-
• I- x Ksp 1.5 x 10-16 x2
• x ?1.5 x 10-16
• x 1.2 x 10-8 M

55
Calculating Solubility from Ksp Values

• The Ksp value for solid lead chromate is 2.0 x
  10-16 at 25 oC. Calculate its solubility in
  water at 25 oC.
• PbCrO4(s) lt---gt Pb2(aq) CrO42-(aq)
• Pb2 x Ksp 2.0 x 10-16 Pb2CrO42-
• CrO42- x Ksp 2.0 x 10-16 x2
• x ?2.0 x 10-16
• x 1.4 x 10-8 M

56
Solubility Product Calculations

• Cu(IO3)2(s) lt---gt Cu2(aq) 2 IO3-(aq)
• Ksp Cu2IO3-2
• If solid cupric iodate is dissolved in HOH
  double square the iodate concentration.
• If mixing two solutions, one containingCu2 and
  and the other IO3-, then use the concentration of
  iodate and only square it.