Tro chapter 13 kinetics spring 2015 (1)

Definitions/Terminology
 Kinetics is the study of rates of chemical
reactions and the mechanisms (pathways) by
which they occur.
3
• The rate is the increase in concentration of a
product per unit time or decrease in concentration of
a reactant per unit time.
• Rates have units of +/-[conc/time] (e.g., Molar/sec).
• A mechanism is the series of molecular steps by
which a reaction occurs.

Reaction Rates
 Thermodynamics determines whether or not a
reaction can occur.
 Kinetics describes how quickly a reaction occurs.
 Some reactions are thermodynamically favorable (G0
< 0), but are kinetically so slow as to be imperceptible
at room temperature:
4
CH4(g) + O2(g)  CO2(g) + H2O(g) G0 = -800 kJ/mol
H2(g) + O2(g)  H2O(g) G0 = -230 kJ/mol
• These reactions do not occur at room temperature and
pressure without activation (e.g., a spark or a catalyst).

Reaction Rates
5
• Rates define the velocity at which reactants
disappear or products appear
• Rates have units of [concentration/time].

Reaction Rates
6
• in many reactions, the coefficients of reactants and products in
the balanced equation are not all the same, for example:
H2 (g) + I2 (g)  2 HI(g)
 For every 1 mole of H2 consumed, 1 mole of I2 will also be
consumed, and 2 moles of HI will be produced.
 The rate of disappearance of [H2] and [I2] will, therefore, be
½ the rate of appearance of [HI].
• The rate of the overall reaction is, thus, the change in the
concentration of each substance multiplied by
1/[coefficient in the balanced equation]:
Rate = -
D[H2 ]
Dt
= -
D[I2 ]
Dt
= +
1
2
æ
è
ç
ö
ø
÷
D[HI]
Dt

Reaction Rates
7
For [H2], the
instantaneous
rate at 50 s is:
For [HI], the
instantaneous
rate at 50 s is:
s
M0.0070Rate
s40
M28.0Rate


s
M0.0070Rate
s40
M56.0
2
1Rate







Reaction Rates
 Mathematically, the rate of a generic reaction:
aA + bB --> cC + dD can be written as:
 For X = A, B, C or D, [X] = [X]t - [X]0, where
[X]t is concentration at time, t after the start of the reaction, and [X]0 is
the initial concentration at the beginning of the reaction, t = 0.
Minus sign indicates that the reactants concentrations decrease with
time.
8
• For example, for a reaction 2A  B:
Rate = -(1/2)[A]/t = [B]/t
• Rates are often approximated as instantaneous
rates, and the notation of calculus is used:
Rate = -1/2d[A]/dt = d[B]/dt
The rate equation

Reaction Rates
 The rate of a simple one-reactant, one-step reaction
is directly proportional to the concentration of the
reacting substance:
 Rate has units of conc/unit time (e.g., Ms-1)
 [A] is the concentration of A, e.g., in molarity (M).
 The proportionality constant, k is called the rate constant.
 For this simple expression, k must have units of inverse
time (e.g., s-1).
 Rate constants are always positive numerical quantities. 9
A  B + C
rate  [A] or rate = k[A] the rate law expression
(different from the rate eqn!)

Reaction Rates
 For the simple expression, rate = k[A]:
 If the initial concentration of A is doubled, the
initial rate of the reaction is doubled.
 If the initial concentration of A is halved, the
initial rate of reaction is cut in half.
 The initial rate is, thus, directly proportional to
the initial [A].
10

Reaction Rates
 If more than one reactant molecule appears in
the chemical equation for a one-step reaction:
2A  B + C
The experimentally determined rate law is:
rate = k[A]2
11
• This relationship means that, If [A] is doubled,
the rate increases by a factor of 4 (= 22).
• If initial [A] is halved, the initial rate decreases
by a factor of 4 (=(1/2)2).

Reaction Rates
 Rate = k[X] or rate = k[X]2 are examples of a rate
law.
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 Rate laws can only be determined experimentally.
 The rate law cannot be determined from the balanced
chemical equation because:
 most chemical reactions are not one-step reactions.

Reaction Rates
 The order of a reaction expresses:
 the dependence on the concentrations of each reactant in the rate
law (= the exponent for each concentration).
 the sum of the orders for each reactant.
13
• For the reaction:
N2O5(g)  2NO2(g) + 1/2O2(g)
rate = k[N2O5]
• The reaction is said to be first order in N2O5 and first
order overall.

Reaction Rates
 First order reactions are dependent on the
concentration of only a single reactant.
 First order reactions are common for many
chemical reactions and all simple radioactive
decays.
14
 Two examples of first order reactions:
2N2O5(g)  2 N2O4(g) + O2(g)
238U  234Th + 4He
Rate = k[N2O5]
Rate = k[238U]

Reaction Rates
15
2NO(g) + O2(g)  2NO2(g)
• The experimentally determined rate law is:
rate = k[NO]2[O2]
• This reaction is second order in NO, first
order in O2 and third order overall.

Reaction Rates
16
(CH3)3CBr(aq) + OH-
(aq)  (CH3)3COH(aq) + Br -
(aq)
rate = k[(CH3)3CBr]
• This reaction is first order in (CH3)3CBr and
zero order in OH- and first order overall.

Concentrations of Reactants:
The Rate-Law Expression
 Example 1: The following rate data were obtained at 25oC for
the reaction: 2A(g) + B(g)  3C(g)
 What are the rate law and rate constant for this reaction?
Experiment
Number
Initial [A]
(M)
Initial [B]
(M)
Initial rate
(M/s)
1 0.10 0.10 2.0 x 10-4
2 0.20 0.10 4.0 x 10-4
3 0.10 0.20 2.0 x 10-4
22

23
• Compare experiments 1 and 3.
• When [A] is held constant and [B] is doubled, the
rate does not change.
• The reaction is zeroth order in [B].
• The rate law can be written generically as:
Rate = k[A]x[B]y
• The rate law requires experimental
determination of the exponents, x and y.
• Therefore, y = 0 and the rate law reduces to:
Rate = k[A]x

24
• Next compare experiments 1 and 2.
• If [A] is doubled, the rate increases by a
factor of 2.
• Therefore, (2)x = 2 and x = 1.
• The rate law reduces to:
rate = k[A]
• The reaction is first order in [A] and first
order overall.

25

 Example 2: The following data were obtained for the
reaction:
2 A(g) + B(g) + 2 C(g)  3 D(g) + 2 E(g)
Experiment
Initial [A]
(M)
Initial [B]
(M)
Initial [C]
(M)
Initial rate
(M/s)
1 0.20 0.10 0.10 2.0 x 10-4
2 0.20 0.30 0.20 6.0 x 10-4
3 0.20 0.10 0.30 2.0 x 10-4
4 0.60 0.30 0.40 1.8 x 10-3
26
From these initial rate data we can determine the rate-
law expression and the rate constant for this reaction.

27
• Comparing experiments 1 and 3:
• [A] and [B] are held constant, but [C] is
increased by 3-times.
• The rate does not change.
• Therefore, the exponent, z = 0, and the rate
law simplifies to:
Rate = k[A]x[B]y
Rate = k[A]x[B]y[C]z

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• [A] is held constant, but [B] is increased by 3-
times.
• The rate also increases by 3-times.
• Therefore, y = 1, and the rate law further
simplifies to:
Rate = k[A]x[B]

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• [B] is held constant, but [A] is increased by 3-
times.
• The rate also increases by 3-times.
• The reaction is, thus, first order in A, first
order in B and second order overall.
• Therefore, x = 1, and the rate law further
simplifies to:
Rate = k[A][B]

30

 Example 3: A reaction between compounds A
and B is determined to be first order in A, first
order in B, and second order overall. From the
information given below, fill in the blanks.
Experiment
Initial Rate
(M/s)
Initial [A]
(M)
Initial [B]
(M)
1 4.0 x 10-3 0.20 0.050
2 1.6 x 10-2 ? 0.050
3 3.2 x 10-2 0.40 ? 31
Rate = k[A][B]

32

33

34

Concentration vs. Time:
Integrated Rate Laws
 The integrated rate equation relates time and
concentration for chemical reactions.
 The integrated rate equation can be used to predict the
amount of product that is produced in a given amount of
time.
 Provides an alternative to the initial rate method for
obtaining rate constants.
 Initially we will look at the integrated rate equation
for first order reactions.
 These reactions are 1st order in one reactant and 1st order
overall.
 For a generic first order reaction with one reactant:
 A  products, rate = k[A]
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or -d[A]/dt = k[A]

where:
t = time elapsed since beginning of reaction.
[A]0= mol/L of A at t = 0 (sometimes called initial conc., [A]i)
[A] = mol/L of A at time t ([A] sometimes written as [A]t)
k = rate constant
 Take the integral of this rate law from the beginning (t =
0, [A]0) to [A] at any later reaction time, t.
 The integrated rate equation for a 1st order reaction:
36
ln([A]/[A]0) = -kt,
MEMORIZE and be able to
interconvert these equations!
or
[A] = [A]0e-kt
ln([A]0/[A]) = kt,
y = mx + bor ln[A] = -kt + ln[A]0
or ln[A]0 - ln[A] = kt,

2N2O5(g)  2 N2O4(g) + O2(g)
37
• The reaction is found experimentally to be first order in
N2O5 and first order overall.
• Rate = k[N2O5]
plots of the first order integrated rate equations
for this reaction…

38
[N2O5] = [N2O5]0e-kt
ln[N2O5] = -kt + ln[N2O5]0

 Define the half-time, t1/2, of a reaction as the time
required for half of the reactant to be consumed, i.e., the
time, t, at which [A]=1/2[A]0, then at t1/2, [A]0/[A] = 2.
39
ln 2 = 0.693 = kt1/2
• So, the half-time for any overall first order reaction is
given by:
t1/2 = 0.693/k MEMORIZE (or take ln 2)
• The half-time is independent of [A]0!
[A]0/[A] = 2 = ekt at t = t1/2

 Example: Cyclopropane decomposes to propene
according to the following equation:
The reaction is found to be first order in
cyclopropane with k = 9.2 s-1 at 1000 0C. What is
the half time for disappearance of cyclopropane
at 1000 0C?
40
t1/2 = 0.693/9.2 s-1 = 0.075 s

 Example: Refer to the previous example.
How much of a 3.0 g sample of cyclopropane
remains after 0.50 seconds?
 The integrated rate laws can be use for any unit
that’s proportional to moles or concentration.
 This example uses grams rather than mol/L.
41

42
ln [A]0 - ln [A] = kt
[A]0 = 3, t = 0.50 s, k = 9.2 s-1, solve for [A]:
ln 3 - ln [A] = 9.2 s-1(0.5 s)
1.1 - ln [A] = 4.6
ln [A] = -(4.6-1.1) = -3.5
[A] = e-3.5 = 0.03 g ~ 1% remains
Use the integrated first order rate equation:

 Example: The half-time for the first order reaction:
CS2(g)  CS(g) + S(g)
is 688 hours at 1000 0C. Calculate the rate constant,
k, at 1000 0C and the amount of a 3.0 g sample of
CS2 that remains after 48 hours.
43
k = 0.693/688 hr = 0.00101 hr-1.

Concentration vs. Time: The
44
ln [A]0 - ln [A] = kt
ln (3.0) - ln [A] = (0.00101 hr-1)(48 hr)
1.1 - ln [A] = 0.048
ln [A] = -(0.048 - 1.1) = 1.052
[A] = e1.052 = 2.86g ~ 2.9 g or 97% unreacted

 For reactions that are second order with respect to
a single reactant, A, and second order overall, with
rate = k[A]2 = (1/a)(d[A]/dt), the integrated rate law
is:
 Where a = stoichiometric coefficient of A in the balanced
overall equation.
 If A  P, then a = 1.
 If 2A  P, then a = 2
45
MEMORIZE!
• An alternative form of the 2nd order integrated
rate law:
[A] = [A]0/(1 + akt[A]0).

 Half-time, t1/2, for second order reactions:
 Use the second order integrated rate-law as a starting
point.
 At the half-time, t1/2 is [A] = 1/2[A]0.
46

Integrated Rate Equations
 If we solve for t1/2:
 So the half-time of a second order reaction depends
on [A]0.
 Less [A]0, longer half-time
 Half-time increases as the reaction proceeds (unlike an
overall first order reaction, where t remains constant).
47

 Example: Acetaldehyde, CH3CHO, undergoes gas
phase thermal decomposition to methane and carbon
monoxide.
The rate law was found to be:
rate = k[CH3CHO]2, and k = 2.0 x 10-2 M-1hr-1 at 527 oC.
(a) What is the half-time (t1/2) for disappearance of
CH3CHO if 0.10 mole is injected into a 1.0 L vessel at
527 oC? 48

49

 (b) How many moles of CH3CHO remain after 200
hours in the 1 L vessel?
50

 (c) What percent of the CH3CHO remains
after 200 hours in the 1 L vessel?
51

52
(a) For the same reaction as the previous example, what is
the half-time (t1/2) for disappearance of CH3CHO if 0.10
mole is injected into a 10.0 L vessel at 527 oC?
● So, if [A]0 decreases by 10-times, t1/2 must increase by 10-
times,
● 10 x 5.0 x 102 hr = 5.0 x 103 hr
● Note that the vessel size is increased by a factor of 10
compared to the previous example, which decreases the
initial concentration by a factor of 10.
● Recall the equation for half time of this 2nd order
reaction (in which the coefficient, a = 1) is:
t1/2 = 1/k[A]0)

 For an overall zeroth order reaction the rate
expression is:
57
 Which gives the zeroth order integrated rate
equation:

Using Integrated Rate Laws to
Determine Reaction Order
58

59
y = mx + b
ln [A] = -kt + ln [A]0
 Plots of linear form of the integrated rate
equations can determine the reaction order and
rate constant.
1st order:
zeroth order: [A] = -kt + [A]0
2nd order:

 Example: For the thermal decomposition of ethyl bromide:
C2H5Br(g)  C2H4(g) + HBr(g)
60
Time (min) 0 1 2 3 4 5
[C2H5Br] 1.00 0.82 0.67 0.55 0.45 0.37
ln [C2H5Br] 0.00 -0.20 -0.40 -0.60 -0.80 -0.99
1/[C2H5Br] 1.0 1.2 1.5 1.8 2.2 2.7


Using Rate Integrated Laws to
 Make an x,y plot for each of the three sets of
values in the table:
1. [C2H5Br] (y-axis) vs. time (x-axis)
 If the plot is linear, then the reaction is zeroth order
with respect to [C2H5Br].
61
2. ln [C2H5Br] (y-axis) vs. time (x-axis)
 If the plot is linear, then the reaction is first order
with respect to [C2H5Br].
3. 1/ [C2H5Br] (y-axis) vs. time (x-axis)
 If the plot is linear, then the reaction is second
order with respect to [C2H5Br].

 Plot of [C2H5Br] versus time.
62Is it linear?
[C2H5Br]

Using Rate Laws to Determine
Reaction Order
 Plot of ln [C2H5Br] versus time.
Is it linear?
ln [C2H5Br]

Using Integrated Rate Laws
to Determine Reaction Order
 Plot of 1/[C2H5Br] versus time.
64Is it linear?
1/[C2H5Br]

Using Integrated Rate Laws
to Determine Reaction Order
 ln[C2H5Br] vs. time is the only linear plot.
 The reaction is, therefore, first order with respect to
[C2H5Br] and first order overall.
 From the integrated rate law for a first order reaction:
ln[A] = -kt + ln[A]0, slope = -k.
66
● Slope = y2 - y1/x2 - x1.
• So k = 0.2 min-1

Using Rate Equations to
 Example: Concentration-versus-time data for the reaction:
2NO2(g) --> 2NO(g) + O2(g),
are given in the table below. Plot each of these time
functions to determine the rate of the reaction and the value
of the rate constant.
67
Time(min) 0 1 2 3 4 5
[NO2] 1.0 0.53 0.36 0.27 0.22 0.18
ln [NO2] 0.0 -0.63 -1.0 -1.3 -1.5 -1.7
1/[NO2] 1.0 1.9 2.8 3.7 4.6 5.5

 Plot of [NO2] versus time.
69
Is it linear?
[NO2]

 Plot of ln [NO2] versus time.
70
Is it linear?
ln [NO2]

 Plot of 1/[NO2] versus time.
71Is it linear?
1/[NO2]

 1/[NO2] vs. time is the only linear plot.
 The reaction is, therefore, second order in
[NO2] and second order overall.
72
● Determine the value of the rate constant from
the slope of the line.

 From the equation for a second order reaction
the slope = a k
 In this reaction a = 2.
73
Slope = 0.90 = 2 k
k = 0.45 M-1min-1

Collision Theory of
Reaction Rates
 Three basic events must occur for a
bimolecular reaction to occur.
 The atoms, molecules or ions must:
1. Collide.
2. Collide with enough energy to break and form
bonds.
3. Collide with the proper orientation for a reaction
to occur.
79

Collision Theory of
Reaction Rates
 For a simple second order reaction between two different
gaseous atoms, A and B:
80
A B
A B
A B
B
A B
A B
A B
A B
4 different possible A-B
collisions
collisions
collisions
A(g) + B(g)  products, where rate = k[A][B]
Increasing the number of atoms per unit volume (equivalent
to increasing their concentrations) increases the probability
of A/B collisions per unit time and, thereby increases the
reaction rate.

Collision Theory of
Reaction Rates
 An example of an effective collision is:
81
X Y
X Y
X Y
X Y
X Y
+
X Y
 Colliding molecules (i.e., individual reactants that
contain more than one atom) must be oriented
properly for reaction.
X2(g) + Y2(g) 2 XY(g)

Collision Theory of
Reaction Rates
82
X
X
Y Y
Y
Y
X X X X Y Y
 Some possible ineffective collisions are :

Collision Theory of
Reaction Rates
 Effective and ineffective molecular collisions for:
NO + N2O --> NO2 + N2
83
Effective
reactants collision
products
Ineffective (wrong orientation)
reactants collision
reactants
(blue spheres = N, red spheres = O)

Reaction Mechanisms and the
Rate-Law Expression
 Many chemical reactions occur by more than one step.
 The sequence of individual chemical steps by which the
reaction occurs is called the reaction mechanism.
 Each step consists of either a unimolecular or bimolecular
chemical reaction.
 The sum of the individual steps must equal that of the overall
balanced reaction.
 Rate laws can be used to propose reaction mechanisms.
84

Rate-Law Expression
 The slowest step (the “bottleneck”) in a reaction mechanism
is the rate-determining step (r.d.s.) (sometimes called the
rate-limiting step).
85
 Rules for a mechanism to be consistent with the
experimentally determined rate law.
 Reaction orders indicate the number of molecules
involved in:
 The r.d.s. only
or
 The r.d.s. and any steps preceding the r.d.s.
 Any step after the r.d.s. cannot be part of the rate law.
 Sum of the steps in the mechanism must equal the
balanced equation for the overall reaction.
 Intermediates in a mechanism cannot appear in the rate
law.

Rate-Law Expression
 Use the experimental rate law to postulate a
mechanism.
 Consider the iodide ion-catalyzed
decomposition of hydrogen peroxide to water
and oxygen.
86
2H2O2(aq) ---> 2H2O(l) + O2(g)
I-

Rate-Law Expression
 This reaction has been experimentally determined to
be first order in H2O2, first order in I- , and second
order overall, i.e.
 Rate = k[H2O2][I-]
87
Step 1, slow: H2O2 + I- --> IO- +H2O
Step 2, fast: IO- + H2O2 --> H2O + O2 + I-
Overall reaction: 2H2O2 --> 2H2O + O2
 A mechanism consistent with this rate law is:

Rate-Law Expression
 According to this mechanism:
 One hydrogen peroxide molecule and one iodide
ion react in the slow step (the r.d.s.).
88
 The iodide ion functions as a catalyst: it is
consumed in step 1 and regenerated in step 2.
 Hypoiodite (IO-) should not accumulate
appreciably because it is produced in the r.d.s.,
then rapidly consumed in the subsequent fast
reaction with hydrogen peroxide.
 (IO- has been detected in very small amounts as a
short-lived reaction intermediate).

Rate-Law Expression
 Ozone, O3, reacts very rapidly with nitric oxide, NO,
in a reaction that is first order in each reactant and
second order overall.
89

Rate-Law Expression
 One possible mechanism that is consistent with the
rate law is:
90
k1
k2
Rate = k1[O3][NO]
• k2 >> k1, so the slow step is the “bottleneck” which
limits the overall reaction rate
• k1 = k, i.e., the rate constant for the overall reaction

Rate-Law Expression
 A mechanism that is inconsistent with the
rate-law is:
91
• Experimentally determined rate laws can, thus,
DISPROVE, but cannot, by themselves, PROVE a
particular mechanism

Rate-Law Expression
92
• Consider the reaction:
2NO(g) + Br2(g)  2NOBr(g)
• The experimentally determined rate law is:
Rate = k[NO]2[Br2]
• One possible mechanism consistent with the rate law is a
simultaneous collision and reaction of two NO molecules
and one Br2 molecule. However, simultaneous three-body
collisions are highly improbable.
• More likely mechanisms would consist of a sequence
of bimolecular steps.

Rate-Law Expression
93
NO + Br2 NOBr2 fast pre-equilibrium
NOBr2 + NO ---> 2NOBr
2NO + Br2 --> 2NOBr overall reaction
• According to this mechanism, the rate law for the slow step is:
Rate = k2[NOBr2][NO]
How to reconcile with the experimental rate law, rate = k[NO]2[Br2]?
• For the fast pre-equilibrium, the rates of the forward and
reverse reactions must be equal: k1[NO][Br2] = k-1[NOBr2].
• Rearrange to [NOBr2] = k1/k-1[NO][Br2], substitute into the rate
law for the r.d.s.:
• Rate = k2(k1/k-1[NO][Br2])[NO] = k[NO]2[Br2], where k = k2k1/k-1
k2
k-1
k1
slow

Transition State Theory
 Transition state theory postulates that the
reactants form a high energy intermediate,
the transition state, which then rearranges
into the lower energy products.
 For a reaction to occur, the reactants must
acquire sufficient energy to form the transition
state.
 This energy is called the activation energy, Ea.
94

95
Energy changes during a chemical reaction.
Ea = activation energy
transition state
ΔE ~ ΔH
2H2(g) + O2(g)  2H2O(g)

97
NO2(g) + CO(g)  NO(g) + CO2(g)
experimental rate law: Rate = k[NO2]2
• Step 1) is the r.d.s (k1 << k2).
• Step 1) is slower than step 2)
because Ea1 > Ea2.
• The rate law of step 1) is the
same as the rate law of the
overall reaction.
A mechanism consistent with the rate law:
step 1: NO2(g) + NO2(g)  NO3(g) + NO(g) Rate1 = k1[NO2]2 slow
step 2: NO3(g) + CO(g)  NO2(g) + CO2(g) Rate2 = k2[NO3][CO] fast

 The relationship between the activation energy
for forward and reverse reactions is:
 Forward reaction activation energy = Ea
 Reverse reaction activation energy = Ea + E
 difference = E (~Hrxn, a thermodynamic quantity)
 Higher Ea, slower reaction, lower rate constant, k.
 In order for a reaction to reach the transition state and
convert to products, the combined kinetic energy of
the colliding molecules must be ≥ Ea.
 One way to increase kinetic energy of molecules is to
raise the temperature.
 Therefore, the rates of most chemical reaction
increase with increasing temperature. 98

Temperature Effects:
 Molecular kinetic energies increase as the
temperature is increased.
99

 One method to increase the energy necessary to
break and reform bonds is to heat the molecules.
 Recall the reaction of methane with molecular oxygen
is exothermic (H0 < 0) but slow at 25 0C:
102
 The reaction can be started with a spark or match.
 This provides the initial activation energy necessary to break
the first few bonds.
 Afterwards the reaction is self-sustaining, i.e., heat
generated by reaction of the first few molecules provides the
activation energy to break bonds in other molecules.
CH4(g) + O2(g)  CO2(g) + H2O(g) H0 = -880 kJ/mol

The Arrhenius Equation
103
MEMORIZE and be able to use the Arrhenius equation!
• k is the rate constant
• T is the absolute temperature
• Ea is the activation energy
• R is the gas constant (8.314 J/mol∙K)
• A is the “pre-exponential factor” or “frequency factor”
• The number of approaches to the activation barrier per unit time.
• e(-Ea/RT) is the “exponential factor”
• the fraction of molecules that convert to product after approaching
the activation barrier.
ork = Ae
-Ea
RT
ln k =
-Ea
R
1
T
æ
è
ç
ö
ø
÷ + ln A

104
Temp, K k, M-1∙s-1 Temp, K k, M-1∙s-1
600 3.37 x 103 1300 7.83 x 107
700 4.83 x 104 1400 1.45 x 108
800 3.58 x 105 1500 2.46 x 108
900 1.70 x 106 1600 3.93 x 108
1000 5.90 x 106 1700 5.93 x 108
1100 1.63 x 107 1800 8.55 x 108
1200 3.81 x 107 1900 1.19 x 109
Example: Determine the activation energy and frequency factor
for the reaction O3(g)  O2(g) + O(g) given the following data:

105
Plot these data as lnk vs I/T:

106
Ea = m(-R)
solve for Ea
Ea
= 1.12 x104
K( ) 8.314 J
mol•K( )= 9.31x104 J
mol
Ea
= 93.1 kJ
mol
A = ey-intercept
solve for A 11-11
118.26
sM1036.4
1036.4



A
eA
y = mx + b, where m = -(Ea/R) and b = ln(A)
ln k =
-Ea
R
1
T
æ
è
ç
ö
ø
÷ + ln A

 If the Arrhenius equation is written for two
temperatures, T2 and T1 with T2 > T1.
107
2
a
2
1
a
1
-Alnln
and
-Alnln
RT
E
k
RT
E
k



Subtract one equation from the other:
108
ln k2
- ln k1
= ln A - ln A -
Ea
RT2
- -
Ea
RT1
æ
è
ç
ö
ø
÷
simplify to: ln k2
- ln k1
=
Ea
RT1
-
Ea
RT2
rearrange to: ln
k2
k1
æ
è
ç
ö
ø
÷ =
Ea
R
1
T1
-
1
T2
æ
è
ç
ö
ø
÷

 Consider the rate of a reaction for which Ea = 50
kJ/mol, at 20 oC (293 K) and at 30 oC (303 K).
 How much do the two rates differ?
109

 Many reactions have Ea  50 kJ/mol.
 For these reactions the rate approximately doubles for
every 10 0C rise in temperature (near room temperature).
 This is a good “rule of thumb” for dependence of reaction
rates (strictly speaking, rate constants) on temperature.
110

Catalysts
 Catalysts increase reaction rates by providing an
alternative reaction pathway with a lower activation
energy.
 There is no net consumption of the catalyst during
the reaction it catalyzes.
112

Catalysts
 Homogeneous catalysts exist in same phase as the
reactants.
 Heterogeneous catalysts exist in different phases
than the reactants.
 Solid catalysts often catalyze solution reactions.
113
2H2O2 --> O2(g) + 2H2O
MnO2

Catalysts
 Examples of commercial catalyst systems include:
114

Catalysts
 This movie shows catalytic converter chemistry on the
molecular scale
115

Catalysts
116
 Catalytic converters also catalyze oxidation of CO
to CO2
 Overall reaction
2 CO(g)+ O2(g) 2CO2(g)
 Adsorption
CO(g) CO(surface)
O2(g) O2(surface)
 Activation
O2(surface)  O(surface)
 Reaction
CO(surface) +O(surface)  CO2(surface)
 Desorption
CO2(surface)  CO2(g)

Catalysts
 The Haber process is used industrially for “fixing” nitrogen
by producing ammonia, NH3, from N2 and H2:
117
N2(g) + 3H2(g) --------> 2NH3(g)
Fe,Fe2O3
• A good illustration of the interplay between kinetics and
thermodynamics.
• This reaction is actually an equilibrium that lies far towards the right
at 25 0C and is also exothermic.
N2(g) + 3H2(g) 2NH3(g) Kc = 108 at 25 0C, H = -92 kJ/mol
• However, the rate of this reaction is negligible at 25 0C.
• The reaction is carried out at 500 0C with catalyst to increase the rate.
• But high temperature favors the reverse reaction (because it’s
exothermic), so high pressure (200-1000 atm) is used to shift the
equilibrium to the right (fewer moles of gaseous product than
reactants).

The Haber Process: Balancing
Factors that Affect Equilibria
118

Enzymes
119
Catalase: 2H2O2  O2(g) + 2H2O
 Enzymes are biological catalysts (usually proteins)
that:
 increase reaction rates
 provide specificity, optimized to bind only particular
substrates and convert them to single sets of products.

After studying Tro Chapter 13 text and lecture
and working the problems, you should be able
to:
123
1. Determine reaction orders, rate laws, and values of rate constants
given a set of initial rates and initial concentrations.
2. Identify overall orders of reactions and order in each reactant given
a rate law.
3. Know units for zeroth, 1st and 2nd order rate constants.
4. Write the rate laws for simple overall zeroth, 1st and 2nd order
reactions.
5. Write and use the integrated rate equations for zeroth, 1st and 2nd
order rate laws with one reactant.
6. Describe how concentrations change with time or how the rate will
change for a given concentration change for each of the rate laws in
#3.
7. Use these rate laws to determine: half-times, concentrations
remaining after a given reaction time or from a given starting
concentration and rate constants.
8. Identify or draw and describe equations for linear plots of zeroth, 1st
and 2nd order reactions.
9. Determine order of reaction given plots in #8.

After studying Tro Chapter 13 text and lecture
and working the problems, you should be able
to:
10. Write reasonable mechanisms given rate laws and vice versa.
11. Define the rate-determining step of a reaction mechanism.
12. Effects of collisions and molecular orientations on reactions.
13. Diffusion-controlled reactions are limited only by collision rates.
14. Define activation energy draw reaction progress diagrams and
label activation energy, transition state and Hrxn.
15. Qualitatively describe temperature dependence of reactions in
terms of molecular kinetic energies.
16. Qualitatively draw and/or interpret Maxwell-Boltzmann distributions
of molecular kinetic energies and label areas equal to and
exceeding activation energy.
17. Memorize and use Arrhenius equation(s) to calculate temperature
dependence of reaction rates and activation energies.
18. Define catalyst and its effect on activation energy and reaction rate.
19. Not responsible for enzyme catalysis or chain reactions.
124

Mid-term Exam 4
125
Chapter 18 (Electrochemistry) including:
1. Use the Nernst equation (memorize) to calculate Ecell under non-
standard conditions given half-reaction E0 values, initial concentrations
and the Nernst equation.
2. Calculate values of K and Grxn from E0
cell using the Nernst equation
and G = -nFEcell (memorize).
3. Describe and calculate Eo for a concentration cell.
4. Identify components in an alkaline dry cell given the half reactions and
a diagram of the cell. Show direction of current flow, and explain why
voltage stays fairly constant during discharge.
5. Draw a diagram of a lead acid car battery given the half reactions, and
describe the discharge and recharge cycles.
6. Describe and draw a diagram of a H2/O2 fuel cell.
7. Given E0’ values (pH 7), describe and calculate how energy can be
derived from biological redox reactions.
All of Chapter 13 (kinetics)

Tro chapter 13 kinetics spring 2015 (1)

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