Post on 22-Feb-2016
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CHAPTER 12: KINETICS
Dr. Aimée TomlinsonChem1212
Reaction Rates
Section 12.1
Reaction RateThe rates at which products are formed and reactants are consumed are connected
They are always represented as concentration change / time change
For the general case aA + bB cC + dD the relationships are:
where t = tf - ti, [A] is the concentration of A (moles/L) and [A] = [A]f - [A]i
we loose reactants as products are formed which is why the rates of A & B are negative
1 [ ] 1 [ ] 1 [ ] 1 [ ]A B C DRatea t b t c t d t
Example Application of Definition1. What is the rate relationship between the production of O2 and O3?
2O3(g) 3O2(g)
the rate of O2 production is 1.5 times faster than the ate of consumption of O3the rate of O3 consumption is 2/3 times the production of O2
3 3 32 2 2[ ] [ ] [ ][ ] [ ] [ ]1 1 3 2 or 2 3 2 3
O O OO O Ot t t t t t
2. The decomposition of N2O5 proceeds according to the equation: 2N2O5(g) 4NO2(g) + O2(g)
If the rate of decomposition of N2O5 at a particular instant is 4.2 x 10-7 M/s, what is the rate of production of NO2 and O2?
2 5 2 2
7 72 52
72 52
[ ] [ ] [ ]1 12 4
[ ][ ] 4 2 4.2 10 8.4 102
[ ][ ] 1 2.1 102
N O NO ORatet t tN ONO M M
s st tN OO M
st t
Example: 2 5( ) 2( ) 2( )2 4g g gN O NO O
52 52 5
622
22
0.0101 0.0120[ ]rate of decomposition of N O 1.9 10400 300
0.00049 0.0040[ ]rate of formation of O 9 10400 300
0.0197 0.0160[ ]rate of formation of NO 3.7400 300
MN O Mst s
MO Mst s
MNOt s
510 M
s
Different Types of Rates
Average RateConcentration change over a time interval (colored region in plot)
Instantaneous RateSlope of the tangent line at a given time (purple line in plot)
Rate Law & Reaction Order
Section 12.2
Rate Law
Relates the rate directly to reactant concentrations
For the General case aA + bB cC + dD, the rate law is:
Rate = k[A]m[B]n
where m and n are determined experimentallyk is called the rate constant
CAUTION!!! Rate law is NOT related to stoichiometry!2 5( ) 2( ) 2( ) 2 5
12
3( ) 2( ) 4( ) ( ) 3 2
2 4 [ ]
[ ][ ]
g g g
g g g g
N O NO O Rate k N O
CHCl Cl CCl HCl Rate k CHCl Cl
Reaction OrderThe power to which each reactant is raised
For the General case aA + bB cC + dD, with
Rate = k[A]m[B]n
“m” is the order of reactant A and “n” is the order of reactant BOverall reaction order is the sum of all or m+n in this caseName the reactant orders and overall reaction order for
It is 1st order in CHCl3 and ½ order for Cl2 with 1 ½
order overall
12
3 2[ ][ ]Rate k CHCl Cl
Experimental Determination of Rate
Law
Section 12.3
Illustrative Example
Experiment [A]0 [B]0 Initial Rate M/s
1 0.100
0.100 4.0 x 10-5
2 0.100
0.200 4.0 x 10-5
3 0.200
0.100 16.0 x 10-5
Determine the rate law, the rate constant and the reaction orders for each reactant and the overall reaction order using the data given below.
Rate constant k & Overall OrderIt is an indicator of the overall rate order of the equation
?
1
?
1?
n
M Ms
n
n n
rate k A
M Ms
Ms M s M
zeroth 0 /first 1 1/second 2 1/
order m k unitsM s
sM s
Final Thoughts
What to do when finding ‘m’ isn’t obvious
1 1 1 1ln ln4 2 4 2
1ln4 21ln2
m
m
m
First-Order Integrated Rate Law & its Half-Life
Section 12.4 & 12.5
First-Order Integrated Rate Law• Final equation is the integrated rate law• Change in reactant concentration over time
may be plotted to get rate law• We plot ln[A]t versus t:
0
0
[ ]
0
[ ]
[ ][ ] 0
0
[ ][ ]
[ ][ ]
ln[
[ ][ ]
] | |
ln[
ln[
]
] ln[ ]
ln[ ]
tA t
A o
A t t
t
A
t
Arate k A
A k tA
d A kdtA
A kdt
A
A t
t
t
k
A
A
k
0ln[ ] ln[ ]tm xy b
A k t A
Half-Life for First-Order Reaction
0
102
120
12
12
12
[ ]ln
[ ][ ]
ln[ ]1ln2
0.693
0.693
tAkt
AA
ktA
kt
kt
t k
Second-Order Reactions
Section 12.6
Second-Order Integrated Rate Law• Final equation is the integrated rate law• This time we plot 1/[A]t versus t to get
linearity
0
0
2
[ ]
2[ ]
[
2
0
][ ] 0
0
[
1 1
][ ]
[ ][ ]
1 |
][ ]
|
1 1
[
tA t
A
t
o
A t tA
t
ktA A
A k tA
d A kdtA
kdtA
ktA A
Arate k At
1 1[ ] [ ]t m x
by
k tA A
Zeroth-Order Reactions
Section 12.7
Zeroth-Order Integrated Rate Law
0
0
[ ][ ][ ] 0 0
[ ]
0
[ ]
[
[ ] [
] [ ] | | [ ]
[
]
]
]
[tA t
A t t
o
t
A tA
A k t
d A kdt A
A
kdt A A
A k
rate kt
t
kt
A
Summary of Integrated Equations
Reaction Mechanisms
Section 12.8
Definitions
Reaction Mechanisms: how electrons move during a reaction
IntermediateSpecies that is produced then consumedNever partakes or appears in the rate law
Elementary StepA step that occurs in a reactionMost reactions have multiple elementary stepsThe stoichiometry of these steps CAN be used to get the reaction orderThe sum of these steps leads to the overall reaction
MolecularityNumber of reacting particles in an elementary stepUni- (1 species), Bi- (2 species), Ter- (3 species)
Example ApplicationGiven the mechanism below state the overall reaction, rate law and molecularity of each step as well as the intermediate.
2 2 3
3 2 2
2 2
1:2 :
step NO NO NO NO elementarystep NO CO NO CO elementary
NO CO NO CO overall
Rate law for step 1: rate1 = k1[NO2]2
Rate law for step 2: rate2 = k2[NO3][CO]Both steps have 2 interacting species so bimolecularIntermediate is NO3 which is produced then consumed
Rate Laws & Reaction Mechanisms
Section 12.9
Examples of Molecularity
Molecularity Elementary step Rate LawUnimolecular A products Rate = k[A]Bimolecular A + A products Rate = k[A]2
Bimolecular A + B products Rate = [A][B]Termolecular A + A + A products Rate = k[A]3
Termolecular A + A + B products Rate = k[A]2[B]Termolecular A + B + C products Rate = k[A][B][C]
Example ProblemWhat is the molecularity & rate law for each of the following?
2 2H NO N H O
2N NO N O
2 2H O H O
2[ ][ ]rate k H NObimolecular
bimolecular
bimolecular
[ ][ ]rate k N NO
2[ ][ ]rate k H O
molecularity Rate law
Rate Laws for Overall Reactions
Section 12.10
Rate Determining StepThe slowest step in a mechanism
Just like the slowest person in a relay race – the slowest step in the mechanism will ruin the speed/time of the reaction
If the slow step is first it is very easy to get the rate law:1
2
2 3
3 2
1: 2 ( )
2 : ( )
k
k
step NO NO NO slow
step NO NO O fast
21 2the rate law would then be: [ ]rate k NO
Example for Fast Step FirstWhat is the rate law for the mechanism below?
1
2
2 3
3 2
1: ( )
2 : 2 ( )
k
k
step NO O NO fast
step NO NO NO slow
Fast Equilibrium Applied ExampleThe rate laws for the thermal and photochemical decomposition of NO2 are different. Which of the following mechanisms are possible for thermal and photochemical rates given the information below?
Thermal rate = k[NO2]2
Photochemical rate = k[NO2] 2
2 2
.)
slow
fast
a NO NO O
O NO NO O
2 2 2 4
2 4 3
3 2
.)
fast
slow
fast
b NO NO N O
N O NO NO
NO NO O
2 2 3
3 2
.)
slow
fast
c NO NO NO NO
NO NO O
Reaction Rates & Temperatures –
Arrhenius Equation
Section 12.11
Arrhenius EquationRelates rate to energy and temperature
aERTk Ae
Frequency Factor AIndicative of the number of successful collisions
Energy of Activation, Ea
Energy that must be overcome to form products
Using the Arrhenius Equation
Section 12.12
Finding the Ea through plotting
Mechanisms & Energy Profiles
14_19.jpg
2 2 3
3 2
slow
fast
NO NO NO NO
NO NO O
The slow step has the larger Ea since it takes longer to generate more energy
Transition StateDefn: state at which reactant bonds are broken and product bonds begin to form
Located at the top of the hump for each elementary step in the reaction profile
Catalysts
Section 12.13 & 12.14
CatalystA species that lowers the activation energy of a chemical reaction and does not undergo any permanent chemical change
it is not present in the overall reaction expression
it is not present in the rate law
it must be consumed and produced in the elementary steps
Two types of catalysts:Homogeneous: in the same phase as the reactantsHeterogeneous: a different phase from reactants
Example of Catalysis
14_21.jpg
Uncatalyzed mechanism - blue line in the figure
Cl Catalyzed mechanism - red line
3 2
3 2
3 2
1:
2 : 2
2 317.7 /
hv
a
step O O O
step O O O
O OE kJ mol
3 2
3 2
3 2
1:
2 : 2
2 32.2 /a
step Cl O O ClO
step ClO O Cl O
O OE kJ mol