Chemical Kinetics

1
Chemical Kinetics

• How are chemical reaction rates determined and
  what factors affect these rates?

2
Objectives

• To understand rates of reaction and the
  conditions affecting rates.
• To derive the rate equation, rate constant, and
  reaction order from experimental data.
• To use integrated rate laws.
• To understand collision theory of reaction rates
  and the role of activation energy.
• To relate reaction mechanisms and rate laws.

3
Chemical Kinetics

• Ch 14 hmwk problems
• 11 15, 20, 21, 25, 29, 33, 38, 45, 47, 50, 55,
  58, 59, 62, 66, 69, 87, 98

4
Why are Kinetics Important?

• Medicinal How quickly or slowly will a
  medication work?
• Environmental Is the rate at which ozone
  depletion occurs equal to the rate at which it is
  formed?
• Industrial How long will it take to produce a
  chemical? Is there a way to change this?

5
What are Kinetics Anyway?

• Chemical Kinetics is the study of chemical
  reaction rates, or speeds.
• Can you think of an example of a chemical
  reaction with a fast rate?
• combustion of gasoline
• Can you think of an example of a chemical
  reaction with a slow rate?
• formation of a diamond

6
What Makes a Reactions Occur?

• If a reaction is to occur between two or more
  reactants, the reactants must come in contact
  with each other, and have enough energy to both
  beak existing bonds and form new bonds.
• The rate at which these reactants come in contact
  with each other affects the rate of the overall
  chemical reaction.

7
Factors Affecting Reaction Rates

• The Physical State of the Reactants Reactions
  are limited by the area of contact between two
  reactants. Therefore, the greater the surface
  area of a solid reactant, the faster a reaction
  can proceed.

8
Factors Affecting Reaction Rates

• The Concentration of Reactants As the
  concentration of one or more reactants increases,
  the more frequently the reactants can collide,
  leading to increased reaction rates.

9
Factors Affecting Reaction Rates

• The temperature at which the reaction occurs The
  rate of a chemical reaction occurs increases with
  increasing temperature. As the molecules move
  more rapidly, they collide more frequently and
  with greater energy, increasing the reaction rate.

10
Factors Affecting Reaction Rates

• The presence of a catalyst Catalysts are
  reagents that increase the speed of a reaction
  without being consumed in the reaction by
  affecting why the reactants collide together. We
  will discuss this in greater depth later.

11
Reaction Rates

• Since a rate is expressed as change over time, a
  reaction rate is expressed as the change in the
  concentration of reactants over time.
• For a simple reaction, A B, the rate would be
  can be described as the rate of appearance of B
  or the rate of disappearance of A, since one mole
  of A will produce to one mole of B.

12
Reaction Rates

• For a simple reaction, A B, the rate would be
  expressed as either ?B/ ?t or -?A/ ?t
• Rates are always expressed as positive
  quantities, therefore, a negative sign must be
  used because the A decreases as the reaction
  progresses

13
A reaction vessel initially contains 1.00 M of
reactant A. Twenty seconds later, the reaction
vessel contains only 0.54 M of reactant A and
0.46 M of product B. Calculate the average rate
of the disappearance of reactant A.

• Recall, A B, and rate -?A/ ?t .
• Rate -0.54M 1.00M/(20s 0s)
• Rate 0.46M/20s
• Rate 2.3 x 10-2 M/s

14
Reaction Rates and Stoichiometry

• What if the reaction stoichiometry is more
  complicated than our simple 11 ratio?
• Lets look at 2HI H2 I2
• Rate -½ ?HI/?t ?H2/ ?t
• I2/ ?t

15
Reaction Rates and Stoichiometry

• In general, for the reaction
• aA bB cC dD
• the rate is given by
• Rate
• -1/a?A/?t -1/b?B/?t
• 1/c?C/ ?t 1/d?D/ ?t

16
How is the rate at which ozone disappears related
to the rate at which oxygen appears in the
reaction 2O3 3O2?

• Rate -1/2?O3/?t 1/3?O2/?t
• If the rate at which O2 appears is 6.0 x 10-5 M/s
  at a particular instant, at what rate is O3
  disappearing at the same time?
• Rate -1/2?O3/?t 1/3?O2/?t
• Rate -1/2?O3/?t 1/3(6.0 x 10-5 M/s)
• Rate -?O3/?t 2/3(6.0 x 10-5 M/s)
• Rate -?O3/?t 4.0 x 10-5 M/s

17
You should now be able to complete 11 12
18
Rate Laws

• The rate law is an equation that relates the rate
  of a reaction to the concentration of reactants
  (and catalysts) raised to various powers
• The rate law for our last example,
• aA bB cC dD, would be expressed as
• Rate kAmBn, where k is the rate constant
  and m and n are typically small, whole numbers.

19
Rate Laws

• k in the rate law is the rate constant defined as
  a proportionality constant in the relationship
  between rate and concentrations.
• The rate constant is fixed at a particular
  temperature (but will change as the temperature
  changes)

20
Rate Laws

• We can classify reactions by their orders
• The reaction order, with respect to a given
  reactant species, equals the exponent of the
  concentration of that species in the rate law, as
  determined experimentally
• The overall order of the reaction equals the sum
  of the orders of all reactant species in the rate
  law
• In the case Rate kAmBn, the overall order
  of this reaction is given by mn.

21
Rate Laws

• For the following reaction, the rate law can be
  written as is indicated below
• NH4 NO2- N2 2H2O
• Rate kNH4NO2-
• Because the exponent of NH4 is one, the rate
  in NH4 is first order. The rate in NO2- is
  also first order. The overall reaction order is
  second order.
• The exponents in a rate law indicate how the rate
  is affected by the concentration of each
  reactant.

22
For the following reaction, the rate law can be
written as is indicated below NH4 NO2-
N2 2H2O Rate kNH4NO2-,

• What is the effect on the rate if NO2- doubles?
• Because the exponent of NO2- in the rate law is
  one, doubling its concentration will double the
  rate.
• What is the effect on the rate if NH4 triples?
• Because the exponent of NH4 in the rate law is
  one, tripling its concentration will triple the
  rate.

23
For a given reaction, the rate law is given
by Rate kNO22

• What is the order of the reaction with respect to
  NO2?
• Because the exponent of NO2 in the equation is 2,
  the reaction is second order with respect to NO2.
  Because the concentration of only one reactant
  is expressed in the rate law, the overall order
  is also second.
• What is the effect on the rate if NO2 triples?
• If the concentration triples, then the rate
  increases by 9 since 32.

24
Rate Laws Additional Examples

• 2N2O5 4NO2 O2 Rate kN2O5
• CHCl3 Cl2 CCl4 HCl Rate
  kCHCl3Cl2½
• H2 I2 2HI Rate kH2I2
• Note While exponents in rate laws are often the
  same as the coefficients in a balanced equation,
  this is not a rule! Exponents in a rate law can
  only be determined experimentally!

25
Determining a Rate Law Experimentally Initial
Rate Method
Trial NO H2 R0 (M/s)
1 0.10 0.10 1.23 x 10-3
2 0.10 0.20 2.46 x 10-3
3 0.20 0.10 4.92 x 10-3
Using the data above, determine the rate law,
calculate the value of k and determine the rate
when NO 0.050M and H2 0.150 M.
26
Determining a Rate Law Experimentally Initial
Rate Method

• A. Determine the rate law
• Recall, the Rate kNOxH2y. We can determine
  the values of x and y by using the experimental
  data and the equation and solving for the unknown
  variables.

27
Determining a Rate Law Experimentally Initial
Rate Method

• Lets compare trial two to trial one
• Rate 2 NOxH2y 0.10x.20y
• Rate 1 NOxH2y 0.10x.10y
• We can simplify our expression and substitute the
  values for Rate 1 and Rate 2
• Rate 2 2.46 x 10-3 .20y
• Rate 1 1.23 x 10-3 .10y
• Notice, we are only left with one unknown
  variable to solve for.

28
Determining a Rate Law Experimentally Initial
Rate Method

• Rate 2 2.46 x 10-3 .20y
• Rate 1 1.23 x 10-3 .10y
• 2y 2
• y must equal 1.
• So far we have determined that
• Rate kNOxH21. We must repeat the process
  to solve for x by comparing trial 3 to trial 1.

29
Determining a Rate Law Experimentally Initial
Rate Method

• Rate 3 NOxH2y 0.20x.10y
• Rate 1 NOxH2y 0.10x.10y
• Rate 3 4.92 x 10-3 .20x
• Rate 1 1.23 x 10-3 .10x
• 2x 4
• X must equal 2. Therefore, the experimentally
  determined rate law is Rate kNO2H21

30
Determining a Rate Law Experimentally Initial
Rate Method

• B. Calculate the value of k
• If Rate kNO2H21, then simple substitution
  from the data table will allow us to calculate k.
• Rate kNO2H21 (use trial 1)
• 1.23 x 10-3M/s k0.10M20.10M1
• 1.23 x 10-3M/s k(0.0010M3)
• 1.2 M-2s-1 k

31
Determining a Rate Law Experimentally Initial
Rate Method

• C. Determine the rate when NO 0.050M and H2
  0.150 M
• Since Rate 1.2M-2s-1NO2H21
• Rate 1.2M-2s-10.050M20.150M1
• Rate 4.5 x 10-4 M/s

32
You should now be able to complete 13, 14, 15,
20, 21, 25
33
Instantaneous Rates

• Calculated from a graph of experimental data that
  represents concentration vs. time.

34
Instantaneous Rates

• Recall, the slope of a graph is ?y/?x. In a
  concentration vs. time graph, the ?y
  ?reactant and ?x ?time.
• Therefore, the slope of the graph ?reactant/
  ?time and will yield the instantaneous rate of
  reaction progression at any given point in time.
• Average reaction rates can be determined from the
  average of several calculated instantaneous rates.

35
Rate Orders

• Recall, the overall order of a reaction is equal
  to the sum of the exponents in a rate law
• Rate kA0 or Rate k is zero order
• Rate kA1 is first order
• Rate kA2 or Rate kAB is second order
• Rate kA3 or Rate kA2B is third order

36
A closer look at Zero-order Reactions

• Rate kA0 or Rate k is zero order
• This is referred to as the differential rate law.
• If we integrate the above equation, we find a
  very useful equation
• A -kt A0
• This will allow us to solve for unknown
  concentration values of zero-order reactions

37
A closer look at Zero-order Reactions

• A zero-order reaction begins with X 0.50 M.
  How long does it take to react all but 0.10M if
  k0.0410 M/min? What is the half-life? What is
  the X after 5 min?

38
A closer look at Zero-order Reactions

• A. X 0.10 M, X0 0.50 M and k0.0410 M/min
• A -kt A0
• 0.10M -(0.0410M/min)t 0.50M
• -0.40M -(0.0410M/min)t
• 10 min t

39
A closer look at Zero-order Reactions

• B. What is the half-life?
• Recall, a half-life is the time it takes for half
  a reactant to be consumed.
• So, if X0 0.50 M, then X must equal 0.25 M
• If A -kt A0, then
• 0.25M -(0.0410M/min)t½ 0.50M
• -0.25M -(0.0410M/min)t½
• 6.1 min t ½

40
A closer look at Zero-order Reactions

• C. What is the X after 5 min?
• If A -kt A0, then
• X -(0.0410M/min)5min 0.50M
• X -.205M 0.50M
• X 0.3 M

41
A closer look at Zero-order Reactions

• Notice, A -kt A0 closely mimics
• y mx b
• Therefore, a graph of change in concentration vs.
  time will be linear for a zero-order reaction.

42
A closer look at First-order Reactions

• Rate kA1 is first order
• This is referred to as the differential rate law.
• If we integrate the above equation, we find a the
  integrated rate law for first order equations
• lnAt - lnA0 -kt
• Or
• lnAt -kt lnA0
• (y) (mx) b
• This will allow us to solve for unknown
  concentration values of first-order reactions

43
A closer look at Second-order Reactions

• Rate kA2 or is second order
• This is referred to as the differential rate law.
• If we integrate the above equation, we find a the
  integrated rate law for second order equations
• 1/At kt 1/A0
• (y) (mx) (b)
• This will allow us to solve for unknown
  concentration values of second-order reactions

44
Determining Order From Graphical Data
Time (min) X
0 0.0200
200 0.0160
400 0.0131
600 0.0106
800 0.0086
1000 0.0069
1200 0.0056
1600 0.0037
An experiment was conducted were a reaction was
allowed to proceed and the reactant concentration
was measured over a period of time. What is the
order of this reaction?
45
Determining Order From Graphical Data

• Recall, the integrated rate law of zero, first
  and second order reactions can be rearranged in
  the form of ymx b
• Zero-order A -kt A0
• First-order lnAt -kt lnA0
• Second-order 1/At kt 1/A0
• If we create three plots of our data, we can
  determine order X vs. t lnX vs. t 1/X
  vs. t. The plot that yield a linear graph will
  determine its order.

46
Determining Order From Graphical Data
Notice, only one graph yields a straight line
The plot of lnX vs. t. Therefore, the reaction
must be first order.
47
You should now be able to complete 29, 33, 38
48
A Microscopic View of Reaction Rates

• Consider the reaction
• NO O3 NO2 O2
• where Rate kNOO3
• Imagine these reactants in rapid, random motion
  inside a flask. They strike the walls of the
  reaction vessel and collide with other molecules.
• Will all collisions result in a chemical
  reaction?
• What conditions must be met for these reactants
  to react together?

49
How a Chemical Reaction Takes Place

• Only a tiny fraction of reactant collisions leads
  to reaction. Why?
• Reactant molecules must collide in the proper
  orientation.
• They must collide with the minimum energy
  required to initiate a chemical reaction. This
  is called activation energy (Ea) and the value
  varies from reaction to reaction.

50
Activation Energy
The diagrams shows the change in potential energy
of the molecules during the reaction. Energy
must be supplied to the reactants to stretch and
break reactant bonds into an intermediate form
(represented at the peak of the graph) before
proceeding to the final product. The energy
difference between the starting molecule and the
highest energy along the path represent the
activation energy. The difference in the
potential energies of the reactants and products
indicate whether the reaction was endo- or
exothermic.
51
The Arrhenius Equation

• Most reaction rate data is based on three factors
• The fraction of molecules possessing the minimum
  Ea or greater
• The number of collisions occurring per second
• The fraction of collisions that have the
  appropriate orientation
• Arrhenius related these factors mathematically
• k Ae-Ea/RT
• The frequency factor, A, is considered constant.

52
The Arrhenius Equation

• The Arrhenius can be written in several forms
• k Ae-Ea/RT
• lnk -Ea/RT lnA
• ln(k1/k2) Ea/R (1/T2 1/T1)
• (notice this last formula closely mimics ymx b)

53
Using the Arrhenius Equation

• Use the data to determine the activation energy
  for the zero-order decomposition of HI on a
  platinum surface.

Temp (K) Rate Constant (M/s)
573 2.91 x 10-6
673 8.38 x 10-4
773 7.65 x 10-2
We can determine the activation energy for a
reaction from a plot of the lnk vs. 1/T.
According to ln(k1/k2) Ea/R (1/T2 1/T1), the
slope of this graph will equal Ea/R.
54
Using the Arrhenius Equation
Temp (K) Rate Constant (M/s) lnk 1/T (K-1)
573 2.91 x 10-6 -12.75 0.00175
673 8.38 x 10-4 -7.08 0.00149
773 7.65 x 10-2 -2.57 0.00129
55
Using the Arrhenius Equation

• The graph yields a straight line with a slope of
  -2.2x104K. According to the equation, ln(k1/k2)
  Ea/R (1/T2 1/T1), this equals Ea/R.
• So, -2.2x104 K Ea/R
• -2.2x104 K Ea/8.314 J/Kmol
• Ea 1.8 x 102 kJ/mol

56
You should now be able to complete 45, 47, 50
57
Reaction Mechanisms

• We know that reactants must change form (often
  forming an intermediate state) in order to
  proceed to becoming products.
• This process can be described in much greater
  detail if we look at a chemical reaction as
  occurring over a series of steps.
• This is referred to as a reaction mechanism.

58
Reaction Mechanisms

• Reaction mechanisms can be very simple and
  involve only one step, or very complicated,
  involving many steps.
• Reaction mechanisms are often speculated. It is
  very difficult to scientifically prove that a
  chemical reaction happens in a specific way.

59
Reaction Mechanisms

• The steps of a reaction mechanism can be
  described by the number of molecules that
  participate as reactants in the step. This is
  called the molecularity of the reaction.
• H3CNC gt H3CCN unimolecular
• NO O2 gt NO2 O2 bimolecular
• And so on

60
Reaction Mechanisms

• Reactions involving only one step are referred to
  as elementary reactions
• Reaction involving several steps are referred to
  as multistep reactions. This often involves the
  formation and consumption of an intermediate.

61
Multistep Mechanisms

• Lets consider the reaction
• NO2 CO gt NO CO2
• Studies of the reaction have shown that it most
  likely occurs in two steps,
• NO2 NO2 gt NO3 NO
• NO3 CO gt NO2 CO2
• Thus, we say the reaction occurs by a two-step
  mechanism.
• Each step is represented by an elementary
  reaction and the rate law can be written directly
  from its molecularity.

62
Multistep Mechanisms

• The chemical equations for the elementary
  reactions in a multistep mechanism must always
  add to give the chemical equation of the overall
  process.
• NO2 NO2 gt NO3 NO
• NO3 CO gt NO2 CO2
• 2 NO2 NO3 CO gt NO2 NO3 NO CO2
• This equation can be simplified by eliminating
  substances that appear on both sides to yield
• NO2 CO gt NO CO2
• Notice, NO3 appears in the reaction mechanism,
  but it neither a reactant nor a product. It is
  formed in one elementary step and consumed in the
  next. NO3 is an intermediate.

63
Multistep Mechanisms

• Each step of a mechanism has its own rate law.
• Each step is represented by an elementary
  reaction and the rate law can be written directly
  from its molecularity.
• Step 1 NO2 NO2 gt NO3 NO Rate1 k1NO22
• Step 2 NO3 CO gt NO2 CO2 Rate2 k2NO3CO
• It is important to remember that the overall rate
  law of a chemical reaction does not necessarily
  follow the molecularity of the balanced chemical
  equation! It must be determined experimentally.

64
Multistep Mechanisms

• Most chemical reactions occur by mechanisms
  involving two or more steps.
• These steps often occur at different rates.
• The overall rate of a chemical reaction is
  determined by the slowest step of its mechanism.
  This is referred to as the rate-determining step.

65
Multistep Mechanisms Slow Initial Step

• Lets take another look at our example.
• Step 1 NO2 NO2 gt NO3 NO
• Step 2 NO3 CO gt NO2 CO2
• Overall NO2 CO gt NO CO2
• If we know that step 1 is much, much slower than
  step two, step 1 must be rate-determining.
  Therefore, the overall rate law for this reaction
  must be Rate kNO22

66
Multistep Mechanisms Fast Initial Step

• If the initial step is fast, and therefore not
  rate-determining, it is likely that the
  rate-determining step involves an intermediate as
  a reactant.
• Lets look at another example
• Step 1 NO Br2 ltgt NOBr2 (fast)
• Step 2 NOBr2 NO gt 2NOBr (slow)
• Overall 2NO Br2 gt 2NOBr
• How would you write a rate law for the overall
  reaction?

67
Multistep Mechanisms Fast Initial Step

• You may have been tempted to say
• Rate kNOBr2NO
• However, notice that an intermediate appears in
  the rate law. The concentrations of
  intermediates are usually unknown. Thus, our
  rate law depends on an unknown concentration. We
  must rewrite our rate law to include only known
  quantities.
• But how???

68
Multistep Mechanisms Fast Initial Step

• Lets go back to the mechanism
• Step 1 NO Br2 ltgt NOBr2 (fast)
• Step 2 NOBr2 NO gt 2NOBr (slow)
• Overall 2NO Br2 gt 2NOBr
• Write a rate law the first step
• Step 1 NO Br2 ltgt NOBr2 (fast)
• Rate1 k1NOBr2
• But notice the equilibrium? The rate above is for
  the forward reaction, what about the reverse?
• Rate-1 k-1NOBr2

69
Multistep Mechanisms Fast Initial Step

• Write a rate law the second step
• Step 2 NOBr2 NO gt 2NOBr (slow)
• Rate2 k2NOBr2NO
• Recall, the original overall rate law was written
  as Rate kNOBr2NO. We must rewrite the rate
  law without NOBr2.
• The three elementary rate laws written from the
  mechanism should give us plenty of substitution
  to choose from.

70
Multistep Mechanisms Fast Initial Step

• We must rewrite Rate kNOBr2NO using the
  elementary rate laws shown below.
• Rate1 k1NOBr2
• Rate-1 k-1NOBr2
• Rate2 k2NOBr2NO
• Recall that because of equilibrium, Rate1
  Rate-1. So, k1NOBr2 k-1NOBr2
• Notice, we can now solve for NOBr2.
• (k1/k-1) NOBr2 NOBr2

71
Multistep Mechanisms Fast Initial Step

• Lets make the substitution.
• Rate kNOBr2NO
• Rate k(k1/k-1) NOBr2NO
• Since ks are constant, lets rename k(k1/k-1) as
  K.
• Rate K NOBr2NO
• And now simplify
• Rate K NO2Br2 where K k(k1/k-1).
• This is a viable rate law since the
  concentrations of all species in the rate law are
  known.

72
You should now be able to complete 55, 58, 59,
62
73
Catalysis

• A catalyst is a substance that changes the speed
  of a chemical reaction without undergoing a
  permanent chemical change.
• A catalyst can be homogeneous (in the same phase
  as the reacting molecules) or heterogeneous (in a
  different phase as the reacting molecules)
• Enzymes in the human body are examples of organic
  catalysts.

74
Homogeneous Catalysis

• H2O2(aq) gt H2O(l) O2(g)
• The decomposition of hydrogen peroxide can be
  catalyzed by Br- as shown.
• Step 1 2Br-(aq) 2H(aq) H2O2(aq) gt Br2(aq)
  O2(g)
• Step 2 Br2(aq) H2O2(aq) gt 2Br-(aq) 2H(aq)
  O2(g)
• Notice, adding steps 1 2 yields
• H2O2(aq) gt H2O(l) O2(g)
• How does this work?

75
Homogeneous Catalysis

• As a general rule, a catalyst lowers the overall
  activation energy for a chemical reaction

76
Heterogeneous Catalysis

• The initial set usually begins with adsorption of
  reactants onto the catalyst surface.
• Atoms/Ions on the surface of a solid are very
  reactive because of the unused bonding capacity.
• Be sure to read this section of your text book
  carefully so you understand how heterogeneous
  catalysis works!!!

77
The decomposition reaction is determined to be
first order. A graph of the partial pressure of
HCOOH versus time for decomposition at 838 K is
shown as the red curve in Figure 14.28. When a
small amount of solid ZnO is added to the
reaction chamber, the partial pressure of acid
versus time varies as shown by the blue curve in
Figure 14.28.
78
SAMPLE INTEGRATIVE EXERCISE continued

• (a) Estimate the half-life and first-order rate
  constant for formic acid decomposition.
• (b) What can you conclude from the effect of
  added ZnO on the decomposition of formic acid?

• (c) The progress of the reaction was followed by
  measuring the partial pressure of formic acid
  vapor at selected times. Suppose that, instead,
  we had plotted the concentration of formic acid
  in units of mol/L. What effect would this have
  had on the calculated value of k?
• (d) The pressure of formic acid vapor at the
  start of the reaction is 3.00 ? 10 2 torr.
  Assuming constant temperature and ideal-gas
  behavior, what is the pressure in the system at
  the end of the reaction? If the volume of the
  reaction chamber is 436 cm3, how many moles of
  gas occupy the reaction chamber at the end of the
  reaction?
• (e) The standard heat of formation of formic acid
  vapor is ?Hf -378.6 kJ/mol.
• Calculate ?Hº for the overall reaction. Assuming
  that the activation energy (Ea) for the reaction
  is 184 kJ/mol, sketch an approximate energy
  profile for the reaction, and label Ea, ?Hº, and
  the transition state.

79
You should now be able to complete 66, 87, 88,
98
80
You Should Now Be Able To

• Identify factors which affect reaction rates.
• Calculate the rate of production of a product or
  consumption of a reactant using mole ratios and
  the given rate.
• Determine the rate law for a reaction from given
  data, overall order and value of the rate
  constant (including units!)
• Determine the instantaneous rate of a reaction.

81
You Should Now Be Able To

• Use integrated rate laws to determine
  concentrations at a certain time, t, and create
  graphs to determine the order of a reaction.
  Also, determine the half-life reaction.
• Determine the activation energy for the reaction
  using the Arrhenius equation.
• Graphically determine the activation energy using
  the Arrhenius equation.
• Write the rate law from a given mechanism and
  identify catalysts and intermediates present.