1. science
2. chemistry

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14.1 Reaction Rates
Ch. 14: Chemical Kinetics
 Chemical Kinetics is the study of how fast
reactions take place.
 Reactions can be fast (reaction of sodium
metal with water) or very slow (rusting of iron).
 Why do you need to study kinetics?
 Drug delivery: time-release capsules
 Pollution control
 Food processing
Chemistry: Atoms First
Julia Burdge & Jason Overby
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Reaction Rates
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14.2 Collision Theory
 Reaction rate (or speed) = change in amount of
 Most reactions occur as a result of collisions
substance per unit of time.
between reacting molecules.
 Collision theory: rate of reaction is directly
proportional to the number of molecule
collisions per time.
 What factors can we change in a lab to alter
 See Figure 14.4 (pp. 554-555)
the rate of a reaction?
 Concentration
 Temperature
 Catalysts
 Surface area of solids
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Collision Theory
 Effective Collisions: molecules must collide
with sufficient kinetic energy and in the correct
orientation in order to make products.
Cl + NOCl → Cl2 + NO
Correct orientation to
facilitate reaction
An effective collision
results in reaction. 5
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Collision Theory
 Ineffective collisions: incorrect orientation of
molecules will NOT result in the formation of
products; molecules will simply bounce off of
each other.
Incorrect orientation does
not favor reaction
An ineffective collision
results in no reaction.6
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Activation Energy
Activation Energy
 Activation Energy (Ea): energy barrier
 Only a small fraction of molecules have enough
molecules have to overcome in order to react.
The high energy state is the transition state.
kinetic energy to overcome the activation
energy.
 Ea is different for each reaction.
 Reactions with low Ea are faster because more
molecules can overcome the Ea.
 At higher temperatures, a larger fraction of
molecules have enough kinetic energy to
overcome Ea.
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Reaction Energy Profile
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Reaction Energy Profile
 Is the reaction shown on the energy profile
 Enthalpy (H) is heat of reaction
 Endothermic:
 Energy absorbed
 H > 0
 Exothermic:
 Energy released
 H < 0
below endo- or exothermic?
 What is the value of Ea?
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14.3 Measuring Reaction Progress
and Expressing Reaction Rate
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Reaction Rates
 Reaction: A  B
 For the reaction: A  B, we can measure
rate of disappearance of reactants:
[A] decreases ( - sign)
and appearance of products:
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Reaction Rates
Reaction Rates
Br2(aq) + HCOOH(aq) → 2Br –(aq) + 2H+(aq) + CO2(g)
Br2(aq) + HCOOH(aq) → 2Br–(aq) + 2H+(aq) + CO2(g)
First 50
seconds:
First 100
seconds:
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Reaction Rates
N2O5(g) 
NO2(g) +
Reaction Rates
O2(g)
Rates are related by coefficients. O2 will appear ¼ as fast as
NO2. Can use coefficients as mol-mol ratio.
 2 N2O5(g)  4 NO2(g) + O2(g)
 This is called the “rate expression”. We can use
this expression (or equation) to determine the rate of
each substance and the rate of reaction over a given
time period.
 If N2O5 disappears at - 0.20 M/s, what is the rate of
appearance of NO2? Of O2?
 Two methods: Stoichiometry and rate expression
 What is the rate of reaction?
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Reaction Rates
Reaction Rates
 Write rate expressions for the following:
Write the rate expression for the following reaction:
 2HI(g)  H2(g) + I2(g)
CO2(g) + 2H2O(g) → CH4(g) + 2O2(g)
 5Br - (aq) + BrO3- (aq) + 6H+ (aq)  3Br2(aq) +
Solution:
Use the equation below to write the rate expression.
3H2O(l)
 C6H12O6(aq)  2C2H5OH(aq) + 2CO2(g)
 SiO3 + 4C  SiC + 3CO
 Worked examples: 14.1, 14.2
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Reaction Rates – Practice!
Measuring Reaction Rates
 5Br - (aq) + BrO3- (aq) + 6H+ (aq)  3Br2(aq) + 3H2O(l)
 Average Rate: rate of substance over a given time
interval
 Rate = (M2 – M1) / (t2 – t1)
 If the rate of appearance of Br2 is 1.2 x 10-3 M/s, what is the
rate of disappearance of Br -?
 1.2 x 10-3 M/s * (5 mol Br -/3 mol Br2) = - 2.0 x 10-3 M/s
 Instantaneous rate: rate at a specific time
 Determined by measuring slope of tangent at a point
 C6H12O6(aq)  2C2H5OH(aq) + 2CO2(g)
 If rate of appearance of ethanol is 4.2 x 10-2 M/s, what is rate
 Initial rate: rate at time = 0
 Determined by measuring slope of tangent from t = 0
of disappearance of glucose? - 2.1 x 10-2 M/s
 SiO3 + 4C  SiC + 3CO
 These rates will all have different values. Rates are not
 If the rate of appearance of CO is +0.78 M/s, what is the rate
constant; they change over time. If they were constant,
graphs would be straight lines.
with respect to C? +1.0 M/s
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14.4 Dependence of Reaction Rate on
Reactant Concentration – Rate Laws
Measuring Reaction Rates
 Rate law relationship between concentration of each
 Initial
reactant and the reaction rate.
 Instantaneous






(pink)
 Average
(orange and
blue)
aA + bB  cC + dD
rate = k[A]x[B]y
Rate is the rate of reaction
k is the rate constant
[A] and [B] are reactant concentrations
x and y are orders of reactants (NOT necessarily the
same as coefficients).
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Rate Laws
Rate Laws
 5Br
 Rate Laws have to be determined from
experimental data (concentrations and
reaction rates), using initial rates (at time = 0)
 Br2(aq) + HCOOH(aq)  2Br - (aq) + 2H+ (aq) +
CO2(g)
 rate =
 OR rate= k[Br2]
k[Br2]1[HCOOH]0
- (aq)
+ BrO3- (aq) + 6H+ (aq)  3Br2(aq) + 3H2O(l)
Experiment
[Br-](M)
[BrO3-](M)
[H+](M)
Initial Rate (M/s)
1
0.10
0.010
0.15
1.2 x 10–3
2
0.10
0.020
0.15
4.8 x 10–3
3
0.30
0.010
0.15
3.6 x 10–3
4
0.10
0.020
0.30
4.8 x 10–3
 Need two trials where only one reactant concentration
changes (all others are constant).
 Compare change in rates to change in concentration to
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determine the order of the reactant (the exponent in the
rate law).
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Rate Laws
5Br
- (aq)
+
BrO3- (aq)
+
6H+ (aq)
Rate Laws
 3Br2(aq) + 3H2O(l)
Experiment
[Br-](M)
[BrO3-](M)
[H+](M)
Initial Rate (M/s)
1
0.10
0.010
0.15
1.2 x 10–3
2
triples
0.10
0.020
0.15
3
0.30
0.010
[H+] 0.15
[BrO3-]
constant
constant 0.020
0.10
0.30
[Br-]
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[Br-]
triples, rate triples:
Br-
is
1st
Rate
triples
 Shortcut for determining orders:
 If the concentration doubles and the rate doesn’t
change, then the concentration has no effect on rate
and the reactant is 0 order
 If the concentration doubles and the rate doubles,
then the reactant is 1st order.
 What would happen to rate if the concentration of
a 1st order reactant tripled?
 If the concentration doubles and rate quadruples,
then the reactant is 2nd order.
 What would happen to rate if the concentration of
a 2nd order reactant tripled?
4.8 x 10–3
3.6 x 10–3
4.8 x 10–3
order (x = 1)
Which two trials should be used for BrO3-?
Which two for H+?
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Rate Laws
Rate Laws
Experiment
[Br-](M)
[BrO3-](M)
[H+](M)
Initial Rate (M/s)
1
0.10
0.010
0.15
1.2 x 10–3
2
0.10
0.020
0.15
4.8 x 10–3
3
0.30
0.010
0.15
3.6 x 10–3
4
0.10
0.020
0.30
4.8 x 10–3
1) A rate law is an equation that is always defined in
terms of reactant concentrations, never products.
2) The exponents (orders of reactants) in a rate law
must be determined from a table of experimental
data.
rate = k[Br-][BrO3-]2
3) Comparing changes in individual reactant
concentrations to changes in rate allows you to
find the order of the reactant.
Now we can solve for the rate constant. How? What
information do we need to solve for k?
Rate = 120 M-2s-1 [Br-][BrO3-]2
Three important things to remember about rate laws:
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Exp’l Determination of Rate Laws
Determine the rate law and calculate the rate
constant (with units):
2NO + Cl2  2NOCl
Experiment
[NO]
[Cl2]
Rate (M/s)
1
2
3
4
5
0.10 M
0.10 M
0.10 M
0.20 M
0.30 M
0.10 M
0.20 M
0.30 M
0.30 M
0.30 M
1.7x10-4
6.7x10-4
1.5x10-3
3.0x10-3
4.5x10-3
Rate = 0.17 M-2s-1 [NO] [Cl2]2
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Rate Laws – Group Practice
 S2O82–(aq) + 3I–(aq) → 2SO42–(aq) + I3–(aq)
 Determine the rate law and calculate the rate
constant, including its units.
Experiment
[S2O82–](M)
[I–](M)
Initial Rate (M/s)
1
0.080
0.034
2.2 x 10–4
2
0.080
0.017
1.1 x 10–4
3
0.16
0.017
2.2 x 10–4
 rate = 0.081 M-1s-1 [S2O82–] [I–]
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14.5 Dependence of Reactant Conc. on
Time – Integrated Rate Laws
Rate Laws – Practice
 Determine the rate law and calculate the rate
constant for: 2O3(g)  3O2(g)

[O3] (M)
0.00600
0.00300
0.00150
 In a 1st order ( A  B ),
Rate (M/s)
5.03 x 10-7
1.28 x 10-7
3.08 x 10-8
 Rate also can be written, rate = k[A]
 Setting them equal:
 What is the rate if [O3] = 0.00500 M?
 Hint: What other information is needed to solve
 Integrate to get 1st order integrated rate law:
this?
 Rate = 0.0140 M-1s-1 [O3]2; rate = 3.49 x 10-7 M/s
 Worked Example 14.3
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Integrated Rate Laws
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First Order IRL
First order integrated rate law:
•
ln is the natural logarithm
In linear format:
ln[A]t = –kt + ln[A]0
y
= mx + b
Math note: ln (5/7) = ln 5 - ln 7
ln (x) = 0.23; What is x?
x = 1.3
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IRL – Half Life
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Second Order IRL
 Two options for rates of second order reactions:
 Rate = k[A]2 OR Rate = k[A][B]
Half-life: the time
required for the
amount of a
reactant to
decrease by one
half (for 1st order).
 2nd option is too complex, we’ll only work with the
first example.
 Rate = k[A]2 = -[A]/t
 Integrate: 1/[A]t = kt + 1/[A]o (linear eqn)
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Second Order IRL
IRL – Summary Table
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Integrated Rate Laws –
Graphing
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IRL – Practice
 The half life for the reaction 2A → B is 8.06
 Make three plots:
 [A] vs. time (zero order)
 ln [A] vs time (first order)
 1/[A] vs. time (second order)
minutes at 110oC. The reaction is first order
in A. How long will it take for [A] to decrease
from 1.25 M to 0.71 M?
 The one that is most linear determines the
 k = 8.6x10-2 min-1; t = 6.58 minutes
order of the reactant (the other graphs should
be curves).
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IRL – Practice
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IRL – Practice
 If the half life is 18.2 min for a first order reactant
 A second order reaction has a half life of 699
with initial concentration of 2.4 M, what is the
concentration of A after 7.5 minutes?
 k = 0.0381 min-1, [A]t = 1.8 M
seconds. If the initial reaction concentration is
0.0355 M, what will the reactant concentration
be after 855 seconds?
 k = 0.040239 M-1s-1, [A]t = 0.0160 M
 If a first order reaction has a rate constant of 4.73 x
102 s-1, how long would it take for 75% to react?
 t = 2.93 x 10-3 s
 Worked Example: 14.7
 Worked Examples 14.4, 14.5, 14.6
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14.6 Dependence of Reaction Rate on
Temperature
Arrhenius Equation
 The Arrhenius Equation can be used to relate the
 The initial equation can be modified to give a
temperature of a reaction to the rate constant.
linear variation.
Ea is the activation energy (in kJ/mol).
R is the gas constant (8.314 J/mol K).
T is the absolute temperature (in K).
A represents the collision frequency and is called
the frequency factor (a constant for each
reaction).
 This variation works well for sets of 4 or more
data points.
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Worked Example 14.8
Arrhenius Equation
• Worked Example 14.8: Rate constants for the
reaction CO(g) + NO2(g) → CO2(g) + NO(g)
were measured at four different temperatures.
The data are shown in the table. Determine
the activation energy for the reaction.
T (K)
288
298
308
318
k (M-1·s-1)
0.0521
0.101
0.184
0.332
1/T (1/K)
0.00347
0.00336
0.00325
0.00314
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Solution
0
-0.5
-1
-1.5
-2
ln k
-2.95
-2.29
-1.69
-1.10
-2.5
y = -5590.9x + 16.47
-3
-3.5
0.0031
m = -Ea / R 
0.00315
0.0032
0.00325
0.0033
0.00335
0.0034
0.00345
0.0035
Ea = m * -R = -5590.0 K * -8.314 J / K·mol
Ea = 46,482 J/mol = 46.5 kJ/mol
Worked Example 14.9, 14.10
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Worked Example 14.10
Arrhenius Equation
A certain first-order reaction has an activation energy of 83 kJ/mol. If the rate
constant for this reaction is 2.1×10-2 s-1 at 150oC, what is the rate constant at
300oC?
 The Arrhenius Equation can also be used to
determine the rate constant after a change in
temperature (i.e., 2 data points) or to find Ea
given two k’s and T’s:
Strategy Rearrange and solve for k2 using the following
k1
k2 =
E
1
1
a
R
e
T2 – T2
Ea = 8.3×104 J/mol, T1 = 423 K, T2 = 573 K, R = 8.314 J/K·mol, and
k1 = 2.1×10-2 s-1.
Solution
2.1×10-2 s-1
k2 =
e
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8.3×104 J/mol
8.314 J/K·mol
1
1
–
573 K
423 K
The rate constant of 300oC is 1.0×101 s-1.
= 1.0×101 s-1
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14.7 Reaction Mechanisms
Reaction Mechanisms
A balanced chemical equation does not indicate
how a reaction actually takes place.
Given elementary steps in a mechanism, we can:
A chemical reaction takes place through a series
of simpler steps called elementary steps
(typically 2 or more). These elementary steps
determine the step-wise reaction mechanism for
the overall reaction.
 Determine the molecularity of each step
 Write the overall (or net) equation
 Use relative rates (slow or fast) to identify the ratedetermining step
 Identify the intermediate (if it exists)
 Is produced and then consumed in a later step
 Identify the catalyst (if it exists)
 Is consumed and then produced in a later step
(recycled)
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 Write the rate law of each step
Reaction Mechanisms
Multistep Mechanisms
Each step in a mechanism is defined by its molecularity:
the number of reactant molecules.
 One step in a multistep mechanism will be slower
unimolecular (one reactant molecule)
 Step 1 (slow): Br2 + NO  Br2NO
A → products
rate = k[A]
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than the others. The slow step is the ratedetermining step (RDS).
1st order
 Step 2 (fast): Br2NO + NO  2BrNO
bimolecular (two reactant molecules)
A + B → products
rate = k[A][B]
2nd order
 1) Write the overall equation.
A + A → products
rate = k[A]2
2nd order
 2) What is the molecularity of each step?
termolecular (three reactant molecules) – rare!
 3) Which step is the rate-determining step?
 4) What is the intermediate? The catalyst?
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Rate Laws: Elementary Steps
Rate Laws: Mechanisms
 For elementary steps, the rate law can be written
 Group Work:
 S2O82- + I-  2SO42- + I+
from the coefficients in the equation. The
molecularity of the step = the overall order of the
step.
I+ + I-  I2
 Step 1 (slow): 2NO2 (g)  NO3(g) + NO(g)


 Step 2 (fast): NO3(g) + CO(g) NO2(g) + CO2(g)

 Determine: overall equation, molecularity of each
step, RDS, intermediate, and catalyst.

slow
fast
Which step is rate determining?
What is the intermediate?
What is the rate law for each step? What is the
overall rate law?
Worked Example 14.11, 14.12
 Write the rate law for each step.
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Rate Laws: Mechanisms
Rate Laws: Mechanisms
 If the 1st step is slow, the rate law of the overall reaction
Step 1:
is the same as the rate law of the first step. If the 2nd
step is slow, rate laws are more complicated to
determine.
(slow)
Step 2:
H2O2 + I–
H2O2 +
Overall reaction:
 Step 1: Tl3+ + Fe2+  Tl2+ + Fe3+
2H2O2
 Step 2: Tl2+ + Fe2+  Tl+ + Fe3+
2H2O + O2
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14.8 Catalysis
Catalysis
A catalyst is a substance that increases the rate of a
chemical reaction without itself being consumed.
H2O2 + I–
H2O2 + IO–
Overall reaction:
H 2O + O 2 + I –
Rate = k[H2O2][I–]
 Which step is rate-determining?
 Second
Step 2:
H2O + IO–
Which step is slow?
 Overall rate law:
Step 1: (slow)
IO–
2H2O2
 Energy profile without a catalyst (left) and with
a catalyst (right). Figure 14.17.
H2O + IO–
H2O + O2 + I–
2H2O + O2
A catalyst speeds up a reaction by providing a set of
elementary steps with more favorable kinetics than
those that exist in its absence; it usually does this by
lowering the activation energy.
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Catalysis
Catalysis
 Catalysts can be classified as
 Catalysts can also be classified as
 Heterogeneous: the catalyst and reactants are in
different phases (e.g., catalytic converter)
 Homogeneous : the catalyst and reactants are
in the same phase.
 The advantages of homogeneous catalysts:
 Reactions can be carried out under atmospheric
conditions
 Can be designed to function selectively
 Are generally cheaper
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