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Chem 121
Kinetics
Reaction rate as change in concentration
What influences reaction rate?
Simple Collision Theory
Rate laws
Half-life
Determining a rate law experimentally
Activation energy, Activated Complex Theory
Why temperature affects rate
Arrhenius equation
Catalysts, enzymes
Reaction mechanism and the link to rate laws; ratedetermining step
Change of rate with time
A typical reaction: the decomposition of hydrogen peroxide: H2O2(aq)  H2O(ℓ) + ½ O2(g)
black line
0
2.32
200
2.01
400
1.72
600
1.49
1200
0.98
(Average rate)
[H2O2], mol/L
Time(s) [H2O2] M
blue line
Time, s
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black line
[H2O2], mol/L
blue line
Time, s
Reaction rates and stoichiometry
H2O2(aq)  H2O(ℓ) + ½ O2(g)
Example 1:
Rate 
[H 2O]
[H 2O2 ]

t
t
Rate 

Example 2:
Rate 
Example 3:

For a general reaction :
[H 2O] 2[O2 ]

t
t
aA+bBcC+dD
1 [A]
1 [B] 1 [C] 1 [D]



a t
b t
c t
d t
2 N2O5(g)  4 NO2(g) + O2(g)
Rate  
1 [N 2O5 ] 1 [NO2 ] [O2 ]


2 t
4 t
t

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Rate laws
Overall order of the reaction is
e.g. 2 HgCl2(aq) + C2O42(aq)  2 Cl(aq) + 2 CO2(g) + Hg2Cl2(s)
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Rate and concentration
The collision model explains why rate depends on concentration.
Using initial rates to determine rate laws
Example:
S2O82-(aq) + 3 I-(aq)  2 SO42-(aq) + I3-(aq)
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S2O82-(aq) + 3 I-(aq)  2 SO42-(aq) + I3-(aq)
Expt.
[S2O82-] (mol L-1)
[I-] (mol L-1)
Init. Rate (mol L-1s-1)
1
0.15
0.21
1.14
2
0.22
0.21
1.70
3
0.22
0.12
0.98
The change of concentration with time
First-order reactions
A  products
Rate  
[A]
 k[A]
t
For a small time interval, using calculus notation:

dA 
 k[A]
dt
d[A]


 k dt
[A]
integrate from initial concentration [A]o to
concentration at time t, [A]t , and time from 0
to
t.

[A] t

[A] 0
t
dA
 k  dt
[A]
ln[A]t  ln[A]0  k t  (k  0)
0

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Second order reactions
Rate  
A + A  product
[A]
 k[A]2
t
Integrate over the same limits as for first order:

[A]t

dA
[A]0 [A]
2
t
 k 
dt
0
Zero-order reactions
Integrate over the same limits:

Rate  
[A]
k
t
[A]t
t
[A]0
0
  d[A]  k  dt
Half-life
First order reaction
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Summary of kinetics from an experimental perspective:
finding the rate
Summary of kinetics from an experimental perspective:
finding the order
We determine order of a reaction from
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Activation energy
Molecules must collide with enough energy
to activate the key bonds in the reaction.
For a given T, molecules have a
specific distribution of energies.
Only a fraction of the molecules
have enough energy to react;
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Reaction mechanism
The sequence of steps that describes the pathway from reactants to products
Multistep mechanisms
overall reaction:
H2(g) + 2 ICl(g)  I2(g) + 2 HCl(g)
mechanism: (Step 1) H2(g) + ICl(g)  HI(g) + HCl(g)
(Step 2) HI(g) + ICl(g)  I2(g) + HCl(g) (fast)
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Rate-determining step for a multi-step mechanism
H2(g) + ICl(g)  HI(g) + HCl(g)
HI(g) + ICl(g)  I2(g) + HCl(g)
Overall rate = k[H2] [ICl]
mechanism:
potential energy
reactants
products
reaction progress
Mechanisms with a slow initial step
e.g. Reaction: 2N2O  2N2 + O2
N2O
N2O + O
experimental rate law: Rate = k [N2O]
N2 + O
N2 + O 2
Mechanisms with a fast initial step
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Catalysis
Homogeneous catalysis
2 H2O2(aq)  2 H2O(l) + O2(g)  use I(aq) as a catalyst
new reaction pathway:
H2O2(aq) + I(aq)  H2O(l) + IO(aq)
H2O2(aq) + IO(aq)  H2O(l) + O2(g) +I(aq)
Heterogeneous catalysis
Automobile catalytic converter
exhaust
to tail pipe
catalyst
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Enzymes
Enzymes catalyze biological reactions so that they can
occur at a reasonable rate under physiological conditions.
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