Reaction Rates and Equilibrium Processes

1
Unit 8

• Reaction Rates and Equilibrium Processes
• Chapter 11

2
11-1 What is a reaction rate?

• Rate is a ratio
• Reaction rate the change in reactant
  concentration per unit of time as the reaction
  proceeds
• The rate of reaction is a positive quantity that
  expresses how the concentration of a reactant or
  product changes with time
• Concentration the amount of moles per liter of
  the reactants
• Also called molarity

3
Reaction rate is an experimental quantity

• A reaction rate describes how rapidly a chemical
  change takes place
• An experimental quantity found by measuring the
  disappearance of a reactant or the appearance of
  a product over a period of time
• Has the units concentration / time

4
Example

• N2O5 (g) ? 2NO2 (g) ½ O2 (g)
• Concentration of reactants will decrease over
  time
• While concentration of products will increase
• Different coefficients show that the
  concentrations do not change at the same rate
• -?N2O5/1 ?NO2/2 ?O2/½

5

• http//www.chem.iastate.edu/group/Greenbowe/sectio
  ns/projectfolder/flashfiles/kinetics2/rxnRate01.ht
  ml

6
Rate of reaction

• Rate -?N2O5/?t ?NO2/2?t ?O2/½?t
• More general for the reaction
• aA bB ? cC dD
• Rate -?A/a ?t - ?B/b?t ?C/c?t
  ?D/d?t

7
Practice

• N2(g) 3 H2(g) ? 2NH3 (g)
• Write the reaction rate expression for this
  reaction
• Rate -?N2/1 ?t - ?H2/3?t ?NH3 /2?t
• Lets take it further
• Nitrogen is disappearing at a rate of 0.10 M per
  minute (rate -?N2/1 ?t -0.10 mol/L min) ,
  at what rate is the hydrogen disappearing and at
  what rate is the ammonia appearing?
• Due to the coefficient of three, hydrogen must be
  decreasing three times as fast (0.30 M per
  minute) and the ammonia must be increasing by 2 x
  the rate of nitrogen

8
Practice

• Example. For the reaction, 2N2O5 ? 4NO2 O2, if
  the rate of decomposition of N2O5 is 4.2x10-7
  mol/(Ls), what is the rate of appearance of (a)
  NO2 (b) O2?

9
Example

• 4 Fe 3 O2 ? 2 Fe2O3
• Rate of rusting of Fe is 1 mole per year. What
  is rate for using oxygen and making rust?
• -Fe/4 -O2/3 Fe2O3/2

10
(No Transcript)
11
11.2 Relative reaction rates

• Why do some reactions go fast and others go slow?
• 2 H2 O2 ? 2 H2O (fast)
• 4 Fe 3 O2 ? 4 Fe2O3 (slow)

12
How can reaction rate be described?

• There are various factors that can affect the
  reaction rate of a reaction
• Increase
• Decrease
• Depends on what happens to the reaction

13
Factors affecting reaction rate

• Time
• Reaction rates decrease with time
• Most reactions are fast at the beginning when all
  of the reactant concentrations are high and slow
  down as the reactants are consumed

14
Factors affecting reaction rate

• Surface area
• Which has more surface area a sugar cube or a
  grain of sugar?
• Solid, liquid, or gas which one has the most
  surface area?
• If the surface area of the solid is increased,
  the number of particles available for reaction
  would increase, and the reaction rate would
  increase
• How can you increase the surface area?
• Crush, grind, break, etc

15
Factors affecting reaction rate

• Temperature
• Increasing the temperature increases the reaction
  rate
• Increase temperature, then we are increasing
  particle movement
• Decreasing the temperature decreases the reaction
  rate

16
Effect of Temperature

• Usually Increasing Temp. leads to Increased Rate
• Rate doubles for each increase of 10 Celsius

17
Factors affecting reaction rate

• Concentration
• Amount
• Increase the concentration of reactants, then we
  increase the reaction rate
• Decrease the amount of reactants, then we
  decrease the reaction rate

18

• http//www.chem.iastate.edu/group/Greenbowe/sectio
  ns/projectfolder/animations/O2NO220kinetics8.htm
  l

19
Factors affecting reaction rate

• Reactions occur by collisions
• Not every collision leads to a reaction.
• Colliding molecules must have enough energy to
  break bonds and make an Activated Complex!
• Activation energy
• Remember the minimum amount of energy needed for
  a reaction to proceed
• A small hill means low energy is needed
• A tall hill lots of energy is needed
• As you increase the activation energy, you
  decrease the reaction rate

20

• http//www.chem.iastate.edu/group/Greenbowe/sectio
  ns/projectfolder/animations/NOO3singlerxn.html

21
Activation Energy
Energy
22

• http//www.chem.iastate.edu/group/Greenbowe/sectio
  ns/projectfolder/flashfiles/kinetics2/iodine_clock
  .html

23
Factors that Affect Reaction Rate

• A. Concentration of reactants - higher reactant
  concentrations increase the rate of reaction.
• B. Catalyst substance that accelerates the
  reaction rate without being transformed.
• C. Temperature - higher temperatures usually
  increase the rate of reaction.
• D. Surface area of solid - smaller particles have
  more surface area so the rate increases.

24
Factors affecting reaction rate

• Catalyst
• Definition a chemical that is added to a
  reaction that speeds up the reaction rate, but is
  not consumed during the reaction
• Provide a shortcut to the reaction pathway
• Adding a catalyst increases the reaction rate
• Help the molecules line up in the specific way so
  reactions can occur
• Increases reaction rate while not being consumed
• Provides alternative path or activated complex
  with lower activation energy

25

• http//www.chem.iastate.edu/group/Greenbowe/sectio
  ns/projectfolder/animations/ethylene8.html

26
Catalyst

• In homogeneous catalysis, the catalyst exists in
  the same phase as the reaction system (e.g.
  solution)
• In heterogeneous catalysis, the catalyst is a
  different phase than the reaction system. (eg.
  Solid catalyst, gas phase reaction

27
Heterogeneous catalysis

• Catalytic converters (cars)
• Heterogonous catalysis
• Pt or other metal catalyst
• Gas phase reaction
• Purpose reduce pollution
• m. cat.
• 2 CO (g) O2 (g) ? 2CO2 (g)
• CO byproduct of combustion of gasoline
• The reaction occurs before these gas can leave
  the car

28
Catalysis Example

• 2H2O2 ? 2H2O O2
• Direct reaction requires the correct collision of
  two molecules
• Adding I- is two steps with lower Ea
• H2O2 I- ? H2O IO-
• H2O2 IO- ? H2O O2 I-
• Adding 2H2O2 ? 2H2O O2

29
Factors affecting reaction rate

• Inhibitor
• Slow down the rate of a reaction
• Used in chain reactions to stop one of the steps,
  thus stopping the rest of the reaction
• Adding an inhibitor, decreases the reaction rate

30
Collision theory

• Explains how these factors affect reaction rates
• Reactant molecules must collide
• Colliding particles need the right orientation to
  react
• The molecules must collide at just the right
  orientation and the right speed to react
• The kinetic energy of the reactants must have a
  minimum energy for an effective collision

31
Effective collision
32
Ineffective collision
33
Collision theory continued

• The chances for an effective collision increase
  when the concentration of the reactants increases
• Temperature affects collision theory
• Increase temperature, increase speed, increase
  effective collisions

34
(No Transcript)
35
Bell work

• Which factor do you think has the greatest affect
  on reaction rate?
• Justify your answer.

36
11.2 Reaction rate and concentration

• Reaction rate is directly related to reactant
  concentration
• The higher the concentration of starting
  materials, the more rapidly a reaction takes
  place
• Reactions occurs as the result of collisions
  between reactant molecules
• The higher the concentration of molecules, the
  greater the number of collusions in unit time and
  the faster the reaction
• As reactants are consumed, their concentrations
  drop and collisions occur less frequently and the
  reaction rate decreases
• Reaction rate drops off with time, eventually
  going to zero when the limiting reactant is
  consumed

37
Rate expression

• Tells how the rate of reaction depends on the
  concentration of reactant
• Graph of rate versus concentration is a straight
  line through the origin, which means that the
  rate must be directly proportional to the
  concentration

38
Rate law/expression

• Rate law gives relationship of the reaction rate
  to the rate constant and the concentrations of
  the reactants.
• Rate expression shows how rate depends on
  concentration of reactant
• Rate kreactant
• K Rate Constant k is a numerical constant for
  a reaction at a given temperature.
• k is not affected by Reactants, but reaction
  rate is affected by Reactants.
• E.g. 2N2O5 ? 4NO2 O2
• Rate kN2O5

39
Order of reaction involving a single reactant

• A ? products
• Rate expression has the form
• Rate kAm
• k rate constant
• L/mols
• m order

40
Overall Reaction Order Sum of all exponents in
the rate equation.

• m 0, the reaction is said to be zero-order
• Concentration does not affect the reaction rate
• If m 1, the reaction is first order
• E.g. A reactant is 1st order if doubling
  Reactant causes the rate to double (For a first
  order reaction, rate is directly proportional to
  Reactant).
• What happens to the reactant also happens to the
  rate
• If m 2, the reaction is second-order
• E.g. A reactant is 2nd Order if doubling
  Reactant causes the rate to quadruple.
• What happens to the reactant is doubled when it
  happens to the rate
• Usually a whole number, but fractional orders are
  possible

41
Order of a reaction

• Exponents cannot be obtained by looking at the
  equation they are experimentally determined
  values.
• Only one reaction order for a reaction, but there
  are many ways in which the equation for the
  reaction can be balanced
• Cannot be deduced from the coefficients in the
  balanced equations
• Exponents are usually whole 's (0, 1, 2), but
  can be negative 's or fractions.

42
Determining reaction rate

• Instantaneous rate
• One way to determine the rate of reaction at a
  particular time, the instantaneous rate, is to
  plot the reactant concentration versus time and
  take the slope of the tangent to the curve at
  time t
• If the slopes of the tangents to the curve at two
  different concentrations are calculated, the rate
  law of a reaction can be determined by comparing
  the changes in rate to the changes in
  concentration

43

• What is the instantaneous rate at 1 minute and 3
  minutes, based on the figure below?
• Solution
• To determine the instantaneous rate, draw a line
  tangent to the curve at the relevant time value.
• Determine the slope of the line from n (y1
  y2)/(x1 x2). The slope is the rate of reaction.
  For 1 minute your value should be about 0.1
  M/min. Your tangent line and points chosen to
  determine the slope may be slightly different.
  The process is not very exact. For 3 minutes your
  value should be about 0.03 M/min.

44
Determining reaction order

• Use initial rate (rate at t0)
• Have two different reaction mixtures differing
  only in the concentration of reactant A
• Measure rate at beginning of the experiment, get
  rate1 and rate2
• Rate2 kA2m
• Rate1 kA1m
• Divide the second rate by the first
• Rate2 / rate1 kA2m/ kA1m (A2/ A1)m

45
Practice

• Determining reaction order and explain what that
  means in regards to how the reactant
  concentration and the rate are related
• CH3CHO ? CH4 CO
• Rate 0.185 M/s 0.37 M/s
  0.56 M/s 0.74 M/s
• CH3CHO 0.1 M 0.20 M 0.30 M
  0.40 M
• Choose any two concentrations (lets do 1st and
  2nd)
• Difference in rates
• Rate2/rate 1 0.37 M/s / 0.185 M/s 2.0
• Difference in concentration
• CH3CHO2/ CH3CHO1 0.20 M / 0.1 M 2.0
• General relation is
• 2.0 (2.0)m
• m 1, the reaction is first order

46
example

• Determining reaction order
• CH3CHO ? CH4 CO
• Rate 0.34 M/s 1.36 M/s
  3.06 M/s
• CH3CHO 0.1 M 0.20 M 0.30 M
• Choose any two concentrations (lets do 1st and
  2nd)
• Difference in rates
• Rate2/rate 1 1.36 M/s / .034 M/s 4.0
• Difference in concentration
• CH3CHO2/ CH3CHO1 0.20 M / 0.1 M 2.0
• General relation is
• 4.0 (2.0)m
• m 2, the reaction is second order

47
Why do we care about order of reaction?

• Once the order of reaction is known, the rate
  constant is readily calculated
• Rate k CH3CHO 2
• k rate / CH3CHO 2
• k 0.085 M/s / (0.10 M) 2
• k 8.5 L/mols

48
Once we know k and the reaction order

• We can calculate the rate at any concentration
• k rate / CH3CHO 2
• Rate k CH3CHO 2
• Rate 8.5 L/mols CH3CHO 2
• Pick a concentration (example 0.50 M)
• Rate 8.5 L/mols (0.50 M) 2
• 2.1 Ms

49
11.2 (second half) Order of reaction with more
than one reactant

• Many reactions involve more than one reactant
• aA bB ? products
• 2 Reactants, A and B
• Rate kAm x Bn
• m n overall order

50
Solve using initial concentration

• aA bB ? products
• Initial concentrations of A differ, but that of B
  is held constant
• Rate1 kA1m kBn
• Rate2 kA2m kBn
• Divide the second equation by the first
• Rate2 / Rate1 kA2m kBn / kA1m kBn
  A2m / A1m (A2 /
  A1)m
• If you know the two rates and the ratio of the
  two concentrations, then you can find m just like
  before

51
example

• (CH3)3CBr OH- ? (CH3)3COH Br-
• (CH3)3CBr 0.50 1.0 1.5 1.0 1.0
• OH- 0.050 0.050 0.050 0.10 0.20
• Rate 0.0050 0.010 0.015 0.010 0.010
• Write the rate expression for the reaction

52

• Pick any two rates
• Rate 3/rate 1 ( (CH3)3CBr 3 / (CH3)3CBr 1)m
• 0.015 / 0.0050 (1.5/0.50)m
• 3 3m
• m 1
• Pick any two rates
• Rate 5/rate 2 ( OH- 5 / OH- 2)n
• 0.010 / 0.010 (0.20 / 0.050)n
• 1 4n
• n 0
• - the rate is independent of the concentration
  of OH-
• Rate expression is
• rate k (CH3)3CBr 1

53
Practice

• 2NO Cl2 ? 2NOCl
• Trial NO Cl2 Rate
• 1 0.10 0.10 0.18
• 2 0.10 0.20 0.36
• 3 0.20 0.20 1.45
• Your goal is the write the rate law expression

54
Answer

• Rate kNO2Cl21
• Lets go one step further
• What are the units for k is this rate law?
• Answer L2/mol2 s or M-2 s-1

55
Practice

• Rate kAmBn
• Trial A B rate
• 1 0.20 0.20 1 x 10-3
• 2 0.20 0.40 1 x 10-3
• 3 0.40 0.20 4 x 10-3
• Calculate the value for m and n.
• Tell me the overall reaction order

56
Practice
Calculate k and the rate when A 0.60 M and
B 0.40 M
57
Review write rate law
Trial A B C rate
58
(No Transcript)
59
11.3 reactant concentration and time

• It is more important to know the relation between
  concentration and time, rather than between
  concentration and rate
• Calculus was used to develop the integrated rate
  equations from the rate law

60
First-order reactions

• A ? products
• Rate kA
• Relationship between concentration and time is
• ln A0/A kt
• ln natural logarithm
• A0 initial concentration
• A concentration at time t
• K rate constant
• T time

61
Rearranging first order

• ln A0/A kt
• Can also be written as
• ln A0 ln A kt
• Or
• ln A ln A0 kt
• Or
• ln A /A0 - kt
• Or
• A A0e-kt

62
How it was derived First Order Reactions

• rate -kA
• rate dA/dt
• dA/dt -kA
• dA/A -k dt solving
• ln A -k t C
• At t 0 lnA0 k(0) C
• ln A/A0 -kt or A A0e-kt

63

• ln A ln A0 kt
• This equation resembles the general equation of a
  straight line
• Y b mx
• B y intercept
• m slope
• From a plot of ln A versus t
• Should be a straight line with a y-intercept of
  ln A0 and a slope of -k

64
Graphing First Order

• Graphing results

• Linear graph

y mx b
lnA -kt lnA0
Slope -k
65
(No Transcript)
66
First Order

• Suppose k 0.250/s and A0 1.00 M
• How much A at 10 s?
• ln A0/A kt
• ln 1.00 M / A 0.250 /s 10 s
• ln 1.00 M /A 2.5
• 1.00 M/A e2.5
• 1.00 M/A 12.18
• 1.00 M 12.18A
• 1.00 M/12.18 A
• A 0.082 M

67
(No Transcript)
68
(No Transcript)
69
Important feature of first order reactions

• The time required for one half of a reactant to
  decompose via a first order reaction has a fixed
  value, independent of concentration
• This value is called the half life

70
Half life

• t½ ln 2 / k
• t ½ 0.693/ k
• For a first order reaction
• K rate constant

71
Nuclear decay

• Rate kX
• k rate constant
• X amount of a radioactive isotope present
• Ln X0/X kt

72
First Order Half-Life

• A A0/2
• A /A 0 1/2
• ln (1/2) -k(t1/2)
• t1/2 half life
• ln (2) kt1/2
• NOTE does not depend on A

73
example

• plutonium-240 has a half-life of 6.58 x 103 years
• Find the first-order rate constant for the decay
• t ½ 0.693/ k
• k 0.693 / t ½
• k 0.693 / 6.58 x 103 years
• k 1.05 x 10-4 years

74
(No Transcript)
75
Second-order reactions with a single reactant

• A ? products
• Rate kA2
• Relationship between concentration and time
• 1/A 1/Ao kt
• Or
• 1/A kt 1/A0

76
Second Order Reactions

• How it was derived from calculus
• rate -kA2 rate dA/dt
• dA/dt -kA2
• dA/A2 -k dt solving
• -1/ A -k t C
• At t 0 -1/A0 -k(0) C
• 1/A - 1A0 kt

77
Second order

• Suppose k 0.250/s and A0 1.00 M
• How much A at 10 s?
• 1/A 1/Ao kt

78
Second Order Reactions

• Graphing Results

• Linear graph

y mx b
1/A kt 1/A0
Slope k
79
(No Transcript)
80
Half life

• t½ 1 / kA0
• k rate constant
• NOTE does depend on A0
• How did we get this?
• A A0/2
• 2/A0 - 1/A0 kt1/2
• 1/A0 kt1/2
• t1/2 1/kA0

81
example

• What is the half life of a second order reaction
  with a k of 0.250/Ms and a concentration of 1.00
  M?
• t½ 1 / kA0
• t1/2 1/(0.250/Ms)(1.00 M) 4 s

82
Zero Order reactions

• A? products
• Rate kA0 k
• Any quantity raised to the zero power is eaul to
  1
• The rate of a zero-order reaction is constant,
  independent of concentration
• Relationship between concentration and time
• A A0 kt
• Or
• A - A0 - kt

83
Zero Order Reactions

• How did we get this?
• rate -kA0 rate dA/dt
• dA/dt -k
• dA -k dt solving
• A -k t C
• At t 0 A0 -k(0) C
• A - A0 -kt

84
Zero Order Reactions

• Graph Results

y mx b
A -kt A0
Slope -k
85
(No Transcript)
86
Zero Order Reactions

• Suppose k 0.250 M/s and A0 1.00 M
• How much A at 10 s?
• A - A0 - kt

87
Zero Order Half Life

• t1/2 A0/2k
• NOTE does depend on A0
• How did we get this?
• A A0/2
• A0/2 - A0 -kt1/2
• A0/2 kt1/2

88
example

• t1/2 A0/2k
• What is the half life of a substance that has a
  concentration of 1.00 M and a k of 0.250 M/s
• t1/2 (1.00 M) / 2(0.250 M/s) 2 s

89
(No Transcript)
90
Graphical Method of Determining Rate Law

• 1) Make 3 plots
• A vs time
• ln A vs. time
• 1/A vs. time.
• 2) The most linear plot gives the correct order
  for A the other 2 graphs should be curves.

91
(No Transcript)
92
Write rate law, what is the overall order, solve
for k
Trial A B rate
When A 0.553 M and B 0.300 M, what is the
rate?
Draw the appropriate straight line graph for the
overall order of reaction. Also, write the
equation for the straight line.
93
Review problems
Answer (3/8)(0.5) 0.1875
94
6. Rate kNO2O2 use Exp 1 3 to determine
order of O2 use 1 2 for order of NO
95
8. c rate (1)(1)2(1)0(1) 1 now double the
concentrations rate (1)(2)2(2)0(2) 8
96
(No Transcript)
97
(No Transcript)
98
(No Transcript)
99
b. Write the equation for this type of reaction.
100
answer
101
(No Transcript)
102
11.4 Models for reaction rate

• So far we have approached reaction rate from an
  experimental point of view, describing what
  happens in the laboratory or the world around us
• Change emphasis and try to explain why certain
  reactions occur rapidly while others take place
  slowly

103
Collision theory

• Explains how these factors affect reaction rates
• Reactant molecules must collide
• Colliding particles need the right orientation to
  react
• The molecules must collide at just the right
  orientation and the right speed to react
• The kinetic energy of the reactants must have a
  minimum energy for an effective collision

104
Effective collision
105
Ineffective collision

• Molecules do not have proper orientation with
  respect to one another when they collide
• Not enough kinetic energy for reactant molecules
• Held together by strong chemical bonds
• Only if the colliding molecules are moving very
  rapidly will the kinetic energy be large enough
  to supply the energy required to break these
  bonds
• Molecules with small kinetic energies bounce off
  one another without reacting

106
Collision theory continued

• The chances for an effective collision increase
  when the concentration of the reactants increases
• Temperature affects collision theory
• Increase temperature, increase speed, increase
  effective collisions

107
Factors affecting reaction rate

• Reactions occur by collisions
• Not every collision leads to a reaction.
• Colliding molecules must have enough energy to
  break bonds and make an Activated Complex!
• Activation energy
• Remember the minimum amount of energy needed for
  a reaction to proceed
• A small hill means low energy is needed
• A tall hill lots of energy is needed
• As you increase the activation energy, you
  decrease the reaction rate

108
Activation Energy
Energy
109
(No Transcript)
110
Collision model of reaction rates

• K p Z f
• K rate constant
• P steric factor
• Takes into the fact that only certain
  orientations of colliding molecules are likely to
  lead to reaction
• Likely to be less than 1, sometimes much less
• Z collision frequency
• Gives the number of molecular collisions
  occurring in unit time at unit concentrations of
  reactants
• F fraction of collisions in which the energy of
  the colliding molecules is equal to or greater
  than Ea

111
F fraction of collisions

• f e-Ea/RT
• E base of natural logarithms
• R gas constant
• T absolute temperature in K
• Substitute this in for f
• k p Z e-Ea/RT
• This equations tells us that the larger the value
  of Ea, the smaller the rate constant

112
The larger the value of Ea, the smaller the rate
constant

• If Ea 0
• e-Ea/RT e0 1
• If Ea RT
• e-Ea/RT e-1 0.37
• If Ea 2RT
• e-Ea/RT e-2 0.14
• The larger the Ea, the smaller the fraction of
  molecules having enough energy to react on
  collision, so the slower the rate of reaction

113
(No Transcript)
114
11.5 Reaction rate and temperature

• The rates of most reactions increase as the
  temperature rises
• Rule increase of 10oC doubles the reaction rate

115
Recall from 11-4

• k p Z e-Ea/RT
• P steric factor is presumably
  temperature-independent
• Z collision number is relatively insensitive
  to temperature
• In regards to temperature dependence
• k Ae-Ea/RT
• A constant
• Take natural logarithm of both sides of the
  equation
• ln k ln A - Ea/RT
• Arrhenius equation

116
Rate Constant and Temp

• Compare it to the general formula for a straight
  line
• ln(k) -Ea/RT ln A
• R 8.31 J/K , Ea is in Joules
• Plot ln(k) vs 1/T, slope - Ea/R

117
Two Point Method relating k and T

• ln k ln A - Ea/RT
• Pick two different temperatures
• ln k2 ln A - Ea/RT2
• ln k1 ln A - Ea/RT1
• ln(k2 / k1) Ea /R(1/T1 - 1/T2)
• R 8.31 L/molK

118
(No Transcript)
119
example

• Suppose the rate doubles when T increases from 25
  C to 35 C. What is Ea?
• 53 kJ

120
Practice

• Write the following for the reaction
• 2 Na Cl2 ? 2 NaCl
• The rate expression for the reaction
• The order of the reaction in each of the reagents
• The overall order of the reaction
• Hint if you are not given experimental values,
  then the coefficients are your orders.

121
Practice

• The rate constant for the reaction HNO3 NH3 ?
  NH4NO3 is 12.5 L / molsec
• If the concentration of nitric acid is 0.020 M
  and the concentration of ammonia is 0.30 M, what
  will the rate of this reaction be?

122
(No Transcript)
123
review
124
(No Transcript)
125
11.6 catalyst

• Catalyst
• Definition a chemical that is added to a
  reaction that speeds up the reaction rate, but is
  not consumed during the reaction
• Provide a shortcut to the reaction pathway
• Adding a catalyst increases the reaction rate
• Help the molecules line up in the specific way so
  reactions can occur
• Increases reaction rate while not being consumed
• Provides alternative path or activated complex
  with lower activation energy
• Changes the reaction path to one with a lower
  activation energy
• Frequently the path consists of two or more steps

126
(No Transcript)
127
(No Transcript)
128
Catalyst

• In homogeneous catalysis, the catalyst exists in
  the same phase as the reaction system (e.g.
  solution)
• In heterogeneous catalysis, the catalyst is a
  different phase than the reaction system. (eg.
  Solid catalyst, gas phase reaction
• Most commonly, the catalyst is a solid that
  increases the rate of a gas-phase or liquid-phase
  reaction

129
Heterogeneous catalysis

• Catalytic converters (cars)
• Heterogonous catalysis
• Pt or other metal catalyst
• Purpose reduce pollution
• m. cat.
• 2 CO (g) O2 (g) ? 2CO2 (g)
• CO byproduct of combustion of gasoline
• The reaction occurs before the gas can leave the
  car
• Problems with these catalysts
• The solid catalyst is easily poisoned
• Foreign materials deposit on the catalyst surface
  during the reaction reduce or even destroy its
  effectiveness

130
Catalysis Example

• Homogeneous example
• 2H2O2 ? 2H2O O2
• Direct reaction requires the correct collision of
  two molecules
• Adding I- is two steps with lower Ea
• H2O2 I- ? H2O IO-
• H2O2 IO- ? H2O O2 I-
• Adding 2H2O2 ? 2H2O O2
• Many reactions that take place slowly under
  ordinary conditions occur readily in living
  organisms in the presence of catalysts called
  enzymes
• Protein molecules of high molar mass

131
(No Transcript)
132
11.7 - Reaction Mechanism

• Reaction mechanism a description of a path, or
  a sequence of steps, by which a reaction occurs
  at the molecular level
• Most reactions are a series of steps.
• Eliminate unstable intermediates

133
Example

• CO (g) NO2 (g) ? NO (g) CO2 (g)
• At lower temperatures
• NO2 (g) NO2 ? NO3 (g) NO (g)
• CO (g) NO3 (g) ? CO2 (g) NO2 (g)
• --------------------------------------------------
  ----
• CO (g) NO2 (g) ? NO (g) CO2 (g)
• The overall reaction, obtained by summing the
  individual steps, is identical with that for the
  one-step process

134
The rate expressions are quite different though

• High temperatures
• Rate kCONO2
• Low temperatures
• Rate kNO22
• The nature of the rate expression and hence the
  reaction order depend on the mechanism by which
  the reaction takes place

135
Elementary steps

• Elementary steps
• The individual steps that constitute a reaction
  mechanism
• Unimolecular
• A ? B C
• Rate kA
• Bimolecular
• A B ? C D
• Rate kAB
• Termolecular
• A B C ? D E
• Rate kABC
• The rate of an elementary step is equal to a rate
  constant k multiplied by the concentration of
  each reactant molecule

136
Example

• The 2 step mechanism for the overall reaction
• Br2 2NO ? 2BrNO is
• Step 1 Br2 NO ? Br2NO (Bimolecular step)
• Step 2 Br2NO NO ? 2BrNO (Bimolecular step)
• Intermediates are short lived species that are
  formed during the reaction, then are subsequently
  consumed. Intermediates do not appear in the
  overall balanced equation. e.g. Br2NO for the
  example above
• For an elementary step, the rate law can be
  written using the stoichiometric coefficients of
  the reactants (molecularity order).

137
Slow Steps

• Often, one step in a mechanism is much slower
  than any other
• Focus on slowest step rate-determining step
• The rate of the overall reaction can be taken to
  be that of the slow step
• Example
• Step 1 A ? B (fast)
• Step 2 B ? C (slow)
• Step 3 C ? D (fast)
• -------------------------
• A ? D

The rate at which A is converted to D (the
overall reaction) is approximately equal to the
rate of conversion of B to C (the slow step)
138
Reaction plot for a reaction with a two step
mechanism
139

• The overall rate of the reaction cannot exceed
  that of the slowest step
• If that step is by far the slowest, its rate will
  be approximately equal to that of the overall
  reaction
• The slowest step in a mechanism is ordinarily the
  one with the highest activation energy

140
Deducing a rate expression from a proposed
mechanism

• Find the slowest step and equate the rate of the
  overall reaction to the rate of that step.
• Find the rate expression for the slowest step.

141

• NO2 (g) NO2 ? NO3 (g) NO (g) (slow)
• CO (g) NO3 (g) ? CO2 (g) NO2 (g) (fast)
• --------------------------------------------------
  ----
• CO (g) NO2 (g) ? NO (g) CO2 (g)
• Rate NO2NO2
• Or
• Rate NO22

142
Sample Reaction

• X2(g) ? 2X(g) fast
• X(g) A2(g) ? AX(g) A(g) slow
• A(g) X2(g) ? AX(g) X(g) fast
• rate k2 X A2

143
Elimination of intermediates

• Sometimes the rate expression obtained involves a
  reactive intermediate
• Occurs with a reaction that has a fast first step
  that is a reversible reaction
• Example
• Step 1 NO (g) Cl2(g) NOCl2 (g)
  (fast)
• Step 2 NOCl2 (g) NO (g) ? 2 NOCl (g) (slow)
• --------------------------------------------------
  -------------------
• 2 NO (g) Cl2(g) ? 2 NOCl (g)

144
Rate of reaction

• From second step
• Rate k2NOCl2NO
• Cannot include the intermediate
• Reverse reaction means the rates of the forward
  and reverse are equal
• K1 (forward) NOCl2 k--1 (reverse) NOCl2
• We need to get rid of the intermediate in our
  reaction rate, so solve for it in the reversible
  rate equation
• NOCl2 k1 NOCl2 / k -1
• Plug into rate for intermediate
• Rate k2NOCl2NO k2 k1 NO2Cl2 / k -1
• Book might simplify to be rate k NO2Cl2

145
Consider this reaction mechanism

• A 2C ? AC2 fast
• AC2 A ? 2AC slow
• What is the net reaction?
• What is the rate expression?
• It should not have any intermediates!

146
(No Transcript)
147
(No Transcript)
148
(No Transcript)
149
practice

• N2(g) H2(g) ? NH3 (g)
• Write the reaction rate expression for this
  reaction
• Write the rate law for the expression using the
  following information
• N2 0.50 1.0 1.5 1.0 1.0
• H2 0.050 0.050 0.050 0.10 0.20
• Rate 0.0050 0.010 0.015 0.010 0.010
• What is the order of each reactant? What is the
  order of the overall reaction?
• Solve for k. Solve for rate if N2 is 2.0 M
  and H2 is 3.0 M.

150

• plutonium-240 has a half-life of 6.58 x 103 years
• Find the first-order rate constant for the decay
• t ½ 0.693/ k
• k 0.693 / t ½
• k 0.693 / 6.58 x 103 years
• k 1.05 x 10-4 years

151
example

• t1/2 A0/2k
• What is the order of a reaction with this
  half-life? What is the half life of a substance
  that has a concentration of 1.00 M and a k of
  0.250 M/s
• t1/2 (1.00 M) / 2(0.250 M/s) 2 s

152

• What is the half life of a second order reaction
  with a k of 0.250/Ms and a concentration of 1.00
  M?
• t½ 1 / kA0
• t1/2 1/(0.250/Ms)(1.00 M) 4 s

153

• A 2C ? AC2 fast
• AC2 A ? 2AC slow
• What is the net reaction?
• What is the rate expression?
• It should not have any intermediates!

154

• Calculate Keq for the synthesis of ammonia at
  4000C if the following concentrations are present
  at equilibrium.
• N2(g) H2(g) ? NH3(g)
• N2 4.5 mol/L
• H2 1.80 mol/L
• NH3 3.28 mol/L