Ch 12 Chemical Kinetics
Sec 12.1 Reaction Rates
Sec 12.2 Rate Laws: An introduction
Sec 12.3 Determining the form of the Rate law
Big Idea 4:
 Rates of chemical reactions are
determined by details of the molecular
collisions.
Enduring understanding 4.A:
 Reaction rates that depend on
temperature and other environmental
factors are determined by measuring
changes in concentrations of reactants
or products over time.
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Essential knowledge 4.A.1:
 The rate of a reaction is influenced by
the concentration or pressure of
reactants, the phase of the reactants
and products, and environmental factors
such as temperature and solvent.
 a. The rate of a reaction is measured by
the amount of reactants converted to
products per unit of time.
Kinetics
• Def
•
• The study of reaction rates; how
fast chemical reactions will occur
Spontaneous reactions (Thermo)
• reactions that will happen - but we
can’t tell how fast.
• Example: Diamond will spontaneously turn to
graphite – eventually…takes a very, very long time
A
B
time
rate = -
[A]
t
rate =
[B]
t
13.1
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Review- Collision Theory
• Particles have to collide to react.
•
• Have to hit hard enough, possess enough
energy
• Right orientation
Things that increase this increase rate
• High temp – faster reaction
• High concentration – faster reaction
• Small particles = greater surface area
means faster reaction
Reaction Rate
 Rate = Conc. of A at t2 -Conc. of A at t1
 Rate = [A]
t 2- t 1
t
Change in concentration per unit time
 As the reaction progresses the
concentration H2 decreases
C
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N2 + 3H2 → 2NH3
[H2]
Time
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 As the reaction progresses the concentration
N2 decreases
C
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N2 + 3H2 → 2NH3
[N2]
[H2]
Time
 As the reaction progresses the concentration H2
decreases 3 times as fast as N2 (Bal Eqn Coeff)
 This is called RELATIVE RATES
C
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c
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t
r
a
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N2 + 3H2 → 2NH3
[N2]
[H2]
Time
 Or, it can also be said
As the reaction progresses the concentration N2
decreases 1/3 as fast as H2
C
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r
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N2 + 3H2 → 2NH3
[N2]
[H2]
Time
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Relative rates of reaction
 For this reaction
N2 + 3H2
2NH3
 As the reaction progresses the
concentration NH3 increases
C
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t
r
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N2 + 3H2 → 2NH3
[N2]
[H2]
[NH3]
Time
Relative rates of reaction
 For this reaction
N2 + 3H2
2NH3
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Relative rates of reaction
 For ANY reaction
aA + bB → cC + dD
 Relative Rates of Reaction:
∆
∆
=
∆
∆
=
∆
∆
=
∆
∆
Calculating Rates
 Average rates are taken over long
intervals
 Instantaneous rates are determined by
finding the slope of a line tangent to the
curve at any given point because the
rate can change over time
 Average slope method
C
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[H2]
t
Time
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 Instantaneous slope method.
C
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r
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[H2]
d[H
t
dt
Time
Rate Laws: Initial rate method
• Reactions are reversible.
• As products accumulate they can begin to
turn back into reactants.
• Early on the rate will depend on only the
amount of reactants present.
• We want to measure the reactants as soon
as they are mixed.
• We will assume that the reverse reaction is
negligible at this point in time.
Rate Laws: Initial rate method
• Two key points
• The concentration of the products do not
appear in the rate law because this is an
initial rate…reactants only
• The order of the reaction and the rate
constant must be determined
experimentally, can’t be obtained from the
equation
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How to write a Rate Law Expression
 For one reactant
2 NO2
2 NO + O2
 Rate = k [NO2]n
• k ≡ rate law constant
• n ≡ order of reaction with respect to NO2
• [NO2] ≡ concentration of NO2
How to write a Rate Law Expression
 For two reactants
2 NO(g) + Cl2
2 NOCl
 Rate = k [NO]n [Cl2]m
•
•
•
•
•
k ≡ rate law constant
n ≡ order of reaction with respect to NO
[NO2] ≡ concentration of NO2
m ≡ order of reaction with respect to Cl2
[Cl2] ≡ concentration of Cl2
How to write a Rate Law Expression
 For three reactants
BrO3- + 5Br - + 6H +
 Rate = k
•
•
•
•
•
•
•
[BrO3- ]n
3Br2 + 3 H2O
[Br - ]m
[H + ]p
k ≡ rate law constant
n ≡ order of reaction with respect to BrO3[BrO3- ] ≡ concentration of BrO3m ≡ order of reaction with respect to Br [Br - ] ≡ concentration of Br –
p ≡ order of reaction with respect to H +
[H + ] ≡ concentration of H +
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2 NO2
2 NO + O2
 You will find that the rate will only
depend on the concentration of the
reactants. (Initially)
 Rate = k[NO2]n
 This is called a rate law expression.
 k is called the rate constant.
 n is the order of the reactant -usually a
positive integer.
2 NO2
2 NO + O2
 The rate of appearance of O2 can be
said to be.
 Rate' = [O2] = k'[NO2]
t
 Because there are 2 NO2 for each O2
 Rate = 2 x Rate'
 So k[NO2]n = 2 x k'[NO2]n
 So k = 2 x k'
The Rate Law
The rate law expresses the relationship of the rate of a
reaction to the rate constant and the concentrations of the
reactants raised to some powers.
aA + bB
cC + dD
Rate = k [A]x[B]y
reaction is xth order in A
reaction is yth order in B
reaction is (x +y)th order overall
13.2
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Rate Laws
•
Rate laws are always determined experimentally.
•
Reaction order is always defined in terms of reactant
(not product) concentrations.
•
The order of a reactant is not related to the
stoichiometric coefficient of the reactant in the balanced
chemical equation.
F2 (g) + 2ClO2 (g)
2FClO2 (g)
rate = k [F2][ClO2] 1
13.2
Types of Rate Laws
 Differential Rate law - describes how rate
depends on concentration.
 Integrated Rate Law - Describes how
concentration depends on time.
 For each type of differential rate law
there is an integrated rate law and vice
versa.
 Rate laws can help us better understand
reaction mechanisms.
Determining Rate Laws
 The first step is to determine the form of
the rate law (especially its order).
 Must be determined from experimental
data.
 For this reaction
2 N2O5 (aq)
4NO2 (aq) + O2(g)
 The reverse reaction won’t play a role
because the gas leaves
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[N2O5] (mol/L)
1.00
0.88
0.78
0.69
Now graph
0.61
0.54
0.48
0.43
0.38
0.34
0.30
Time (s)
0
200
400
600
the data
800
1000
1200
1400
1600
1800
2000
 To find rate we have to find the slope
at two points
 We will use the tangent method.
At .80 M the rate is (.88 - .68) = 0.20 =- 5.0x 10 -4
(200-600) -400
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At .40 M the rate is (.52 - .32) = 0.20 =- 2.5 x 10 -4
(1000-1800) -800
 Since the rate at twice as fast when the
concentration is twice as big the rate law
must be..
 First power
 Rate = -[N2O5] = k[N2O5]1 = k[N2O5]
t
 We say this reaction is first order in N2O5
 The only way to determine order is to run
the experiment.
The method of Initial Rates
 This method requires that a reaction be
run several times.
 The initial concentrations of the
reactants are varied.
 The reaction rate is measured just after
the reactants are mixed.
 Eliminates the effect of the reverse
reaction.
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An example
 For the reaction
BrO3- + 5 Br- + 6H+
3Br2 + 3 H2O
 Rate Law is
Rate = k[BrO3-]n[Br-]m[H+]p
 We use experimental data to determine
the values of n,m,and p
Initial concentrations (M)
BrO30.10
0.20
0.20
0.10
Br0.10
0.10
0.20
0.10
H+
0.10
0.10
0.10
0.20
Rate (mol/L·s)
8.0 x 10-4
1.6 x 10-3
3.2 x 10-3
3.2 x 10-3
EXPERIMENTAL DATA
FINDING n, the order of BrO3-
Rxn#1
Rxn#2
Rxn#3
Rxn#4
BrO3-
Br-
H+
0.10
0.20
0.20
0.10
0.10
0.10
0.20
0.10
0.10
0.10
0.10
0.20
Rate (mol/L·s)
8.0 x 10-4
1.6 x 10-3
3.2 x 10-3
3.2 x 10-3
Look for the rxns where BrO3- changes
but Br- and H+ do not
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FINDING n, the order of BrO3-
Rxn#1
Rxn#2
Rxn#3
Rxn#4
BrO3-
Br-
H+
0.10
0.20
0.20
0.10
0.10
0.10
0.20
0.10
0.10
0.10
0.10
0.20
Rate (mol/L·s)
8.0 x 10-4
1.6 x 10-3
3.2 x 10-3
3.2 x 10-3
Look for the rxns where BrO3- changes
but Br- and H+ do not
FINDING n, the order of BrO3-
Rxn#1
Rxn#2
Rxn#3
Rxn#4
BrO3-
Br-
H+
0.10
0.20
0.20
0.10
0.10
0.10
0.20
0.10
0.10
0.10
0.10
0.20
Rate (mol/L·s)
8.0 x 10-4
1.6 x 10-3
3.2 x 10-3
3.2 x 10-3
To find the value of n, set up the rate law ratio
of these two reactions
FINDING n, the order of BrO3-
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FINDING m, the order of Br-
Rxn#1
Rxn#2
Rxn#3
Rxn#4
BrO3-
Br-
H+
0.10
0.20
0.20
0.10
0.10
0.10
0.20
0.10
0.10
0.10
0.10
0.20
Rate (mol/L·s)
8.0 x 10-4
1.6 x 10-3
3.2 x 10-3
3.2 x 10-3
To find the value of m, set up the rate law
ratio of these two reactions
FINDING m, the order of Br-
FINDING p, the order of H+
Rxn#1
Rxn#2
Rxn#3
Rxn#4
BrO3-
Br-
H+
0.10
0.20
0.20
0.10
0.10
0.10
0.20
0.10
0.10
0.10
0.10
0.20
Rate (mol/L·s)
8.0 x 10-4
1.6 x 10-3
3.2 x 10-3
3.2 x 10-3
To find the value of n, set up the rate law ratio
of these two reactions
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FINDING p, the order of H+
An example
 For the reaction
BrO3- + 5 Br- + 6H+
3Br2 + 3 H2O
 Rate Law is
We found n=1, m=1, p=1
Therefore,
• Rate = k[BrO3-]1[Br-]1[H+]1
An example
 For the reaction
BrO3- + 5 Br- + 6H+
3Br2 + 3 H2O
 Now, find the rate constant, k
Pick one of the experiments
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FINDING k, the rate constant
BrO3-
Br-
H+
0.10
0.20
0.20
0.10
0.10
0.10
0.20
0.10
0.10
0.10
0.10
0.20
Rxn#1
Rxn#2
Rxn#3
Rxn#4
Rate (mol/L·s)
8.0 x 10-4
1.6 x 10-3
3.2 x 10-3
3.2 x 10-3
Pick one of the experimental data sets and
plug in the values
Rate = k[BrO3-]1[Br-]1[H+]1
8.0 x 10-4 = k [0.10][0.10][0.10]
K=0.8
Determine the rate law and calculate the rate constant
for the following reaction from the following data:
2SO42- (aq) + I3- (aq)
S2O82- (aq) + 3I- (aq)
Experiment
[S2O82-]
[I-]
Initial Rate
(M/s)
1
0.08
0.034
2.2 x 10-4
2
0.08
0.017
1.1 x 10-4
3
0.16
0.017
2.2 x 10-4
13.2
Determine the rate law and calculate the rate constant for
the following reaction from the following data:
2SO42- (aq) + I3- (aq)
S2O82- (aq) + 3I- (aq)
Initial Rate
(M/s) rate = k [S2O82-]x[I-]y
Experiment
[S2O82-]
[I-]
1
0.08
0.034
2.2 x 10-4 y = 1
2
0.08
0.017
1.1 x 10-4 x = 1
3
0.16
0.017
2.2 x 10-4
rate = k [S2O82-][I-]
Double [I-], rate doubles (experiment 1 & 2)
Double [S2O82-], rate doubles (experiment 2 & 3)
k=
2.2 x 10-4 M/s
rate
=
= 0.08/M•s
[S2O82-][I-] (0.08 M)(0.034 M)
13.2
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