Kinetics: Rates and Mechanisms of Reactions
Chemical Kinetics tells us:
…how fast a reaction will occur
…how molecules react (MECHANISM)
a mechanism is a sequence of steps that lead to the product
1
Factors Affecting the rate:
1. concentration:
2. temperature:
generally a 10oC increase will double the rate
3. nature of the reactant: i.e. surface area
4. catalyst: (two types: homogenous and heterogeneous)
5. mechanism: (orientation, shape, & order)
COLLISION THEORY = CAPPING A MARKER
2
Rate of RXN =
The increase in concentration of a product per unit time.
or
The decrease in concentration of a reactant per unit time.
3
Conc. is usually measured in M (Molarity= mol/L)
for solutions.
Rate =
M
time
=
mol
L•s
or mol•L-1•s-1 or M•s -1
Square brackets [ ] are often
used to express molarity (i.e.
[HCl] means Molarity of HCl)
Since many reactions involve gases, P is often used for
concentration.
n
n
from PV  nRT, P  RT and has units of ...
V
V
moles/L
4
Consider the reaction (net ionic eq.):
3ClO- (aq)  2Cl-(aq) + ClO3-(aq)
Rate could be defined in at least 3 ways: (3 coefficients and ions)

1. appearance of ClO3-
[ClO 3 ]
rate 
time
2. appearance of Cl-
[Cl - ]
rate 
time
3. disappearance of ClO-
Disappearance
 [ClO - ]
rate 
time
5
Consider the reaction (net ionic):
3ClO- (aq)  2Cl-(aq) + ClO3-(aq)

1. appearance of ClO3-
[ClO 3 ]
rate 
time
2. appearance of Cl-
[Cl - ]
rate 
time
3. disappearance of ClO-
Disappearance
 [ClO - ]
rate 
time
Question: Are these three rates equal?
6
Consider the reaction (net ionic):
3ClO- (aq)  2Cl-(aq) + ClO3-(aq)
Let’s make these three rates equal.
[ClO -3 ] [Cl - ]  [ClO - ]
rate 


time
2time
3time
Note the use of
coefficients and the sign
7
General Form:
aA + bB  cC + dD
rate =  [C]
ctime
= [D]
=
dtime
“PRODUCTS”
-
[A]
atime
=-
[B]
btime
“REACTANTS”
8
 conc
rate 
 time
slope
Average rate = slope (over time period)
0.30- 0.74
 0.0146
40 - 10
-
7 4 1 0. 0 
6 2. 0 -0 7. 0
ep o l s
01 - 04
negative sign
9
 conc
rate 
 time
instantaneous rate = tangent slope (changing)
WHY?
Collision Theory!
10
y
Slope
 tangentat pointof interest
x
11
IF we can now
somehow get a linear
plot in the form of:
y = mx + b.
The slope would be a
constant independent
of concentration!
12
We could call the slope the
rate constant and assign it the letter
k!
rate constant = k  instantaneous rate or rate
“call in the mathematicians”
13
rate constant: k
is conc. independent
Is still temperature dependent
and mechanism dependent!
General form of rate law :
for RXN: A  products
rate  k[A]
m
m = RXN order according to A.
Determined by experiment only!
conc. of A
rate constant
14
General form of rate law :
aA  bB D
cC
rate  k[A] [B] [D] [C]
w
x
y
z
RXN orders (w, x, y, and z)must be determined by exp. only!
Total (overall) order =  individual orders
15
General Equation Forms:
0 order:
1st order:
2nd order:
3rd order:
rate = k
rate = k[A]
rate = k[A]2
rate = k[A]3
or rate = k[A][B]
or ........
Simple experiments are done by doubling 1 conc. at a time
and looking at the effect.
16
General Equation Forms:
0 order:
rate = k
1st order:
rate = k[A]
2nd order: rate = k[A]2 or rate = k[A][B]
3rd order: rate = k[A]3 or ........
Simple experiments are done by doubling 1 conc. at a time
and looking at the effect.
order
0
1/2
1
2
doubling
[2]0 = 1
[2]1/2 = 1.41..
[2]1 = 2
[2]2 = 4
effect on rate
none
increase by 1.41..
doubles
quadruples
Question: suppose rate = k[A]2[B] what is the effect of
doubling both A and B?
17
Let’s look a some rate data for the RXN:
NH4+(aq) + NO2-(aq)  N2(g) + 2H2O(l)
18
NH4+(aq) + NO2-(aq)  N2(g) + 2H2O(l)
doubles
double
double
doubles
19
NH4+(aq) + NO2-(aq)  N2(g) + 2H2O(l)
2
2
2
+
[2]x
for NH4 :
=2
x = 1 (1st Order)
2
for NO2- : [2]y = 2
y = 1 (1st Order)
20
 1
4
 1
2
rate law : rate  k[NH ] [NO ]
first order with respect to each reactant, 2nd order overall.
orders usually have integer values, but can be fractional.
Can also be (-) (inhibitors).
Since rate has units, k must also have units.
M
mol
rate 

time L  time
so units of k must work with [ ] to match units.
21
Determine the units of k in each of the following:
Since rate has units, k must also have units.
M
rate 
time
(so units of k must work with [ ] to match units.)
1. rate = k[A]
M

time
M
M

t
M2
M M
1

kunits must = time so that :
t
?
2. rate = k[A]2
t
M M
1
1

kunits must = time M so that :
t
?
t
3. rate  k[NH4 ]1[NO2 ]1
M
k MM
t
so k must have units of M-1t-1
22
4. rate = k[A][B]2[C]
kunits=M-3time-1
5. rate = k[A]0
M
k units 
time
23
Ebbing 4th ed. P 490
14.38 In a kinetic study of the reaction:
2NO(g) + O2(g)  2NO2(g)
the following data were obtained for the initial rates disappearance of NO:
INITIAL
CONC.
NO
Exp. 1
Exp. 2
Exp. 3
0.0125 M
0.0250 M
0.0125 M
INITIAL Initial rate
CONC. of RXN of
O2
NO
0.0253 M
0.0253 M
0.0506 M
0.0281 M/s
0.112 M/s
0.0561 M/s
Obtain the rate law. What is the value of the rate constant?
.
24
Ebbing 4th ed. P 490
14.38 In a kinetic study of the reaction:
2NO(g) + O2(g)  2NO2(g)
the following data were obtained for the initial rates disappearance of NO:
INITIAL
CONC.
NO
Exp. 1
Exp. 2
Exp. 3
0.0125 M
0.0250 M
0.0125 M
INITIAL Initial rate
CONC. of RXN of
O2
NO
0.0253 M
0.0253 M
0.0506 M
0.0281 M/s
0.112 M/s
0.0561 M/s
Obtain the rate law. What is the value of the rate constant?
Write overall rate equation:
rate = k[NO]x[O2]y
For NO: select a pair of experiments in which the conc. of
[NO] is changed, but other concentrations are unchanged.
Let’s use Exp. 1 and 2
25
Ebbing 4th ed. P 490
14.38 In a kinetic study of the reaction:
2NO(g) + O2(g)  2NO(g)
INITIAL
CONC.
NO
Exp. 1
Exp. 2
Exp. 3
0.0125 M
0.0250 M
0.0125 M
INITIAL Initial rate
CONC. of RXN of
O2
NO
0.0253 M
0.0253 M
0.0506 M
0.0281 M/s
0.112 M/s
0.0561 M/s
Obtain the rate law. What is the value of the rate constant?
overall rate equation: rate = k[NO]x[O2]y
“Let’s divide Exp.1 by Exp.2 to allow us to cancel terms.
x
1
x
2
y
2 1
y
2 2
Exp. 1 rate 1 k[NO] [O ]


Exp. 2 rate 2 k[NO] [O ]
26
Ebbing 4th ed. P 490
14.38 In a kinetic study of the reaction:
2NO(g) + O2(g)  2NO(g)
INITIAL
CONC.
NO
Exp. 1
Exp. 2
Exp. 3
0.0125 M
0.0250 M
0.0125 M
INITIAL Initial rate
CONC. of RXN of
O2
NO
0.0253 M
0.0253 M
0.0506 M
0.0281 M/s
0.112 M/s
0.0561 M/s
Obtain the rate law. What is the value of the rate constant?
overall rate equation: rate = k[NO]x[O2]y
x
1
x
2
y
2 1
y
2 2
Exp. 1 rate 1 k[NO] [O ]


Exp. 2 rate 2 k[NO] [O ]
0.0281 [0.0125]x

0.112 [0.0250]x
27
Ebbing 4th ed. P 490
14.38 In a kinetic study of the reaction:
2NO(g) + O2(g)  2NO(g)
overall rate equation: rate = k[NO]x[O2]y
Exp. 1 rate 1 k[NO]1x [O 2 ]1y


x
y
Exp. 2 rate 2 k[NO]2 [O 2 ]2
0.0281 [0.0125]x

0.112 [0.0250]x
x
0.0281  0.0125
How do we solve for x?


0.112  0.0250
Use logarithms
 0.0281
 0.0125
ln
 x ln


0.112
0.0250




ln0.250   x ln0.5
28
Ebbing 4th ed. P 490
14.38 In a kinetic study of the reaction:
2NO(g) + O2(g)  2NO(g)
overall rate equation: rate = k[NO]x[O2]y
x
0.0281  0.0125
How do we solve for x?


0.112  0.0250
Use logarithms
 0.0281
 0.0125
ln
 x ln


 0.112 
 0.0250
ln0.250   x ln0.5
ln[0.250]
x
2
ln[0.5]
Therefore the rate equation becomes: rate = k[NO]2[O2]y
29
Now let’s determine the reaction order with respect to [O2]
30
Ebbing 4th ed. P 490
14.38 In a kinetic study of the reaction:
2NO(g) + O2(g)  2NO2(g)
the following data were obtained for the initial rates disappearance of NO:
INITIAL
CONC.
NO
Exp. 1
Exp. 2
Exp. 3
0.0125 M
0.0250 M
0.0125 M
INITIAL Initial rate
CONC. of RXN of
O2
NO
0.0253 M
0.0253 M
0.0506 M
0.0281 M/s
0.112 M/s
0.0561 M/s
Write overall rate equation: rate = k[NO]2[O2]y
For O2: select a pair of experiments in which the conc. of
[O2] is changed, but all other concentrations are unchanged.
Let’s use Exp. 1 and 3
31
Ebbing 4th ed. P 490
14.38 In a kinetic study of the reaction:
2NO(g) + O2(g)  2NO2(g)
the following data were obtained for the initial rates disappearance of NO:
INITIAL
CONC.
NO
Exp. 1
Exp. 2
Exp. 3
0.0125 M
0.0250 M
0.0125 M
INITIAL Initial rate
CONC. of RXN of
O2
NO
0.0253 M
0.0253 M
0.0506 M
0.0281 M/s
0.112 M/s
0.0561 M/s
Write overall rate equation: rate = k[NO]2[O2]y
Let’s use Exp. 1 and 3
k[NO x3] [O2 ]y3
Exp. 3 rate 3


Exp. 1 rate1 k[NO x] [O ]y
1 21
32
Ebbing 4th ed. P 490
14.38 In a kinetic study of the reaction:
2NO(g) + O2(g)  2NO2(g)
INITIAL
CONC.
NO
Exp. 1
Exp. 2
Exp. 3
0.0125 M
0.0250 M
0.0125 M
INITIAL Initial rate
CONC. of RXN of
O2
NO
0.0253 M
0.0253 M
0.0506 M
0.0281 M/s
0.112 M/s
0.0561 M/s
Write overall rate equation: rate = k[NO]2[O2]y
k[NO x3] [O2 ]y3
Exp. 3 rate 3


Exp. 1 rate1 k[NO x] [O ]y
1 21
0.0561  0.0506

0.0281  0.0253
y
y
2  [2]
33
Ebbing 4th ed. P 490
14.38 In a kinetic study of the reaction:
2NO(g) + O2(g)  2NO2(g)
Write overall rate equation: rate = k[NO]2[O2]y
k[NO x3] [O2 ]y3
Exp. 3 rate 3


Exp. 1 rate1 k[NO x] [O ]y
1 21
0.0561  0.0506

0.0281  0.0253
y
y
2  [2]
y=1
Therefore:
rate = k[NO]2[O2]1
Now solve for k.
34
Ebbing 4th ed. P 490
14.38 In a kinetic study of the reaction:
2NO(g) + O2(g)  2NO2(g)
INITIAL
CONC.
NO
Exp. 1
Exp. 2
Exp. 3
0.0125 M
0.0250 M
0.0125 M
Overall rate equation:
INITIAL Initial rate
CONC. of RXN of
O2
NO
0.0253 M
0.0253 M
0.0506 M
0.0281 M/s
0.112 M/s
0.0561 M/s
rate = k[NO]2[O2]1
Choose any Exp. and substitute in experimental to obtain k.
i.e. Exp1. : rate = k[NO]2[O2]1
so :
0.0281 = k[0.0125]2[0.0253]
k = 7100 M-2s-1
35
Ebbing 4th ed. P 490
14.40 Iodine ion is oxidized to hypoiodite ion, IO-, by hypochlorite
ion, ClO-, in basic solution.
I - (aq)  ClO- (aq) OH - IO - (aq)  Cl- (aq)
the following initial rate experiments were run:
INITIAL
CONC.
IExp. 1
Exp. 2
Exp. 3
Exp. 4
0.010 M
0.020 M
0.010 M
0.010 M
INITIAL
CONC.
ClO0.020 M
0.010 M
0.010 M
0.010 M
INITIAL Initial rate
CONC. of RXN of
OHNO (M/s)
0.010 M
0.010 M
0.010 M
0.020 M
12.2 x 10-2
12.2 x 10-2
6.1 x 10-2
3.0 x 10-2
Obtain the rate law. What is the value of the rate constant?
.
36

-
k[I ][ClO ]
rate 
[OH ]

 1
rate  k[I ][ClO ][OH ]
-
k = 6.1 s-1
37
This data isn’t linear! What can we do?
Integrated Rate Laws
38
IF we can now
somehow get a linear
plot in the form of:
y = mx + b.
The slope would be a
constant, independent
of concentration!
39
We could call the slope the
rate constant and assign it the letter
k!
rate constant = k  rate
“call in the mathematicians”
40
Key Equations:
Order
in [A]
0
1
2
Rate Law*
Integrated Rate Law
(in y = mx + b form)
r at e = k
rate = k[A]
rate = k[A]2
[A]t = -kt + [A]0
ln[A]t = -kt + ln[A]0
1/[A]t = kt + 1/[A]0
*Since
Linear
Graph
? vs t
[A]t
ln[A]t
1/[A]t
Slope of
Line
Equals
-k
-k
k
Half life
Equations
t1/2 = [A]0/2k
t1/2 = 0.693/k
t1/2 = 1/k[A]0
the units of rate are concentration/time, the
units of k (the rate constant) must dimensionally
agree. So for each order, k will have different units and
those units can tell one which equation to use. [ ]
means the concentration of the enclosed species in
Molarity (M).
41
The data below was collected for the reaction:
NOCl(g)  NO(g) + 1/2Cl2(g)
Time (s)
0
30
60
100
200
300
400
[NOCl] (M)
0.100
0.064
0.047
0.035
0.021
0.015
0.012
Prepare THREE graphs to determine if the RXN is ZERO,
1st, or 2nd order. Then determine the value and units of
the rate constant k.
42
Zero Order Plot
[A]t vs. time
rate = k
[A]t = -kt + [A]0
y = mx + b
43
First Order Plot
ln[A]t vs. time
rate =k[A]
ln[A]t = -kt + ln[A]0
y
= mx + b
44
2nd Order Plot
1/[A]t vs. time
rate = k[A]2
1/[A]t = kt + 1/[A]0
y
= mx + b
Plot is linear so 2nd Order
k = slope = 0.185 M-1s-1
45
Integrated rate laws:
Zero Order:
rate = k[A]0 = k
rate = k
integrated gives:
[A]t = -kt + [A]0
y = mx + b
slope = -k
If a RXN is zero order, a plot
of [A] vs. time should be linear
and the slope = -k.
46
Integrated rate laws:
1st order rate laws:
rate = k[A] integrated gives:
[ A ]t
ln
 - kt
[ A ]0
rearranged to : y = mx + b gives:
ln[A]t = -kt + ln[A]0
slope = -k
If reaction data is 1st order, a plot
of ln[A] vs. time should be linear.
47
2nd Order Integrated Rate Equations:
rate =
k[A]2
integrated gives:
1
1
 kt
[ A ]t
[A]0
y = mx + b
slope = k
If a RXN is 2nd order, a plot of
1/[A] vs. time should be linear
and the slope = k.
48
RXN is first order with respect to CH3NC
Zero Order Plot
1st Order Plot
slope = -k
ln[CH3NC]t = -kt + ln[CH3NC]0
49
RXN is 2nd order with respect to [NO2]
First order plot
2nd order Plot
Slope = k
50
General form 1st order: ln[A]t = -kt + ln[A]0
Note: This is a formula that can be used to solve (1st order)
problems. (If all but one of the variables are given)
1. Given the RXN: C3H6  CH2=CHCH3
Where k = 6.0 x 10-4 s-1 @500oC.
Looking at the units of k, determine the order.
Problem: if [C3H6]0 = 0.0226 M, find [C3H6] @ 955 s.
ln[A]t = -kt + ln[A]0
- 6.0 x 10-4 (955s)
ln[A]t 
 ln[0.0226]
s
ln[A]t = -4.362
[A]t = 0.0127 M
51
The other equations can
be used in a similar fashion.
52
Half-life: the time it takes to decrease the concentration to 1/2
its initial value.
“fold paper to view subsequent half-lives”
53
Half-lives:
Formulas:
1st order:
t1/2 = 0.693/k
Derivation: ln [ A]t  kt
[ A]0
and:
so:
ln2  kt1/
@t1/2 :
1
ln   kt1/
2
2
2
ln 2 0.693
t1/ 

2
k
k
54
Half-life formulas:
Zero Order
[ A]0
t1/ 
2
2k
First Order
ln 2 0.693
t1/ 

2
k
k
2nd Order
1
t1/ 
2
k[ A]0
These 2 are conc. dependent (and not very useful).
55
Energy Diagrams:
All chemical and physical changes are accompanied by energy
changes.
E = Activation energy
energy 
a
reactants
E Exothermic
products
time 
Question: What keeps the reactants from rolling down the hill?
56
57
58
59
60
61
rate constant (k) varies with temperature.
but not with concentration
62
The Arrhenius Equation:
Ea /RT
k  Ae
Temp in K
8.31 J/mol•K
activation energy
base e (natural ln)
frequency factor (1/time), fraction of
collisions with correct geometry.
rate constant
-Ea/RT is always <1 and refers to the fraction of molecules
having minimum energy for a RXN.
63
Ea /RT
k  Ae
How can we make this a linear equation in the form of
y = mx + b?
Take the ln of each side.
ln k = ln A - Ea/RT
or:
ln k  ln A
-
Ea 1
R T
y = b - mx
A plot of ln k vs. 1/T gives a straight line with the
slope = -Ea/R (Ea = -8.31 x slope)
64
At two diff. temps. we get:
k 2 Ea  1

ln

k1
R  T1
1 
- 
T2 
65
Reaction Mechanisms: Reaction is broken into steps with
intermediates being formed.
“some RXNS occur in one step, but most occur in
in multiple steps.”
Each Step is called an elementary step, and the
number of molecules involved in each step defines
the molecularity of the step.
uni-molecular: = 1 i.e. O3*  O2 + O
bi-molecular: = 2 (these are the most common)
i.e. HI + HI  activated complex  H2 + I2
ter-molecular: = 3 (rare, due to probability of orientation
and energy both being correct.)
i.e. O(g) + O2(g) + N2(g)  O3(g) + “energetic” N2(g)66
The Raschig process for the preparation of hydrazine (N2H4)
Overall RXN: 2NH3(g) + NaOCl(aq)  N2H4(aq) + NaCl(aq) + H2O(l)
Proposed Mechanism: (Only from experiment)
Step 1: NH3(aq) + OCl-(aq)  NH2Cl(aq) ‡ + OH-‡ (aq)
Step 2: NH Cl(aq) ‡ + NH (aq)  N H +‡ + Cl-(aq)
2
3
2
5
Step 3: N2H5+(aq) ‡ + OH-‡ (aq)  N2H4(aq) + H2O(l)
“Cancel intermediates and “add steps” to give overall RXN.”
2NH3(g) + OCl-(aq)  N2H4(aq) + Cl-(aq) + H2O(l)
The overall rate law, mechanism, and the total order can’t be
predicted from the stoichiometry, only by experiment.
67
The following is only true for individual steps:
The rate law of an elementary step is given by the product
of a rate constant and the conc. of the reactants in the step.
Step
A  Product(s)
Molecularity
uni
rate law
rate = k[A]
A + B  Product(s)
bi
rate = k[A][B]
A + A  Product(s)
bi
rate = k[A]2
ter
rate = k[A]2[B]
2A + B  Product(s)
The overall mechanism must match the observed rate law.
Usually one STEP is assumed to be the rate determining step.
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Example:
Overall RXN:
2NO2(g) + F2(g)  2NO2F(g)
Observed Experimental rate law:
rate = k[NO2][F2]
Question: Why does this rule out a single step RXN?
Answer:
rate law for single step process would be: rate = k[NO2]2[F2]
“Let’s try to work out a Mechanism that matches the
observed rate law.”
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Example:
Overall RXN:
2NO2(g) + F2(g)  2NO2F(g)
Observed Experimental rate law:
rate = k[NO2][F2]
k1
slow: NO2(g) + F2(g)  NO2F(g) + F(g) ‡ rate = k[NO2][F2]
k2
fast: NO2(g) + F(g) ‡  NO2F(g)
2NO2(g) + F2(g)  2NO2F(g)
The rate law is dependent upon the slow step.
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The rate law is dependent upon the slow step.
Let’s try making the 2nd RXN slow and the first fast
Overall RXN:
2NO2(g) + F2(g)  2NO2F(g)
Observed Experimental rate law:
rate = k[NO2][F2]
k1
fast: NO2(g) + F2(g)  NO2F(g) + F(g) ‡
k2
rate = k[NO2][F ‡]
slow: NO2(g) + F(g) ‡  NO2F(g)
2NO2(g) + F2(g)  2NO2F(g)
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1. At low temperatures, the rate law for
the reaction:
CO(g) + NO2(g)  CO2(g) + NO(g)
2
is:rate
=
k[NO
]
2
Which mechanism
is consistent with the
law?
a. COrate
+ NO
2  CO2 + NO
rate = k[CO][NO2]
b. 2NO2  N2O4‡
N2O4 ‡ + 2CO  2CO2 + 2NO
(fast)
(slow)
c. 2NO2  NO3 ‡ + NO
NO3 ‡ + CO  NO2 + CO2
(slow)
(fast)
d. 2NO2  2NO + O2 ‡
2CO + O2 ‡  2CO2
(slow)
(fast)
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For the following mechanism:
2NO  N2O2
(fast)
N2O2 + H2  H2O + N20
(slow)
N2O + H2  N2 + H2O
(fast)
a. Determine the overall reaction
b. Does the mechanism agree with the rate law:
rate = k[NO]2[H2]
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