# Lecture 10

```Outline:
1/31/07
 Turn
in Research Symposium
Seminar reports – to me

Exam 1 – two weeks from Friday…
 Today:
Start Chapter 15: Kinetics
Kinetics &amp; Reaction mechanisms
Chapters 6 and 14
introduced Thermodynamics:
heat, work, energy, 1st , 2nd laws, state
vs. path variables, spontaneity, etc. as
related to chemical reactions….
Chapter 15 introduces:
 the rate of reactions (kinetics)
 the mechanisms of reactions
These two concepts are closely
related on a molecular level!
Is the rate of a reaction important?
e.g. airbags….
Is the exact mechanism important?
e.g. Ozone destruction
(i) O3 + Cl  O2 + ClO
(ii) O + ClO  O2 + Cl
O3 + O  2 O2
Both rate of reaction and mechanism
are vital to understanding this problem!
CFC + ultraviolet light  free Cl atoms
hn
Reaction Rates
• Speed of a reaction is measured by the
change in concentration with time.
• For a reaction A  B
change in number of moles of B
Average rate 
change in time
moles of B 

t
moles of A 
Average rate with respect to A  
t
Reaction Rates
Consider:
C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)
Most useful units for this rate =
molarity/time.
(Since volume is constant, molarity and
moles are directly proportional.)
The average rate decreases with time?
Reaction Rates
How do we get a useful number?
Plot [C4H9Cl] versus time:
The rate at any instant in time
(instantaneous rate) is the slope of
the tangent to the curve.
Instantaneous rate is different from
average rate.
All reaction “rates” slow down
when viewed this way…
 Go
What
back
is the
to a“rate”
molecular
of a reaction?
picture…
= the number of reactions/unit time
Why does the reaction slow down?
Answer: Per molecule it doesn’t !!!
(For some reactions…)
6/12 = 50%
3/6 = 50%
Reaction Rates
In general:
rates decrease as concentrations decrease...
there are stoichiometric factors…
For a reaction (in general):
aA + bB  cC + dD
1 A
1 B 1 C 1 D
Rate  



a t
b t
c t
d t
(if rate is not specified for a particular substance!)
Concentration and Rate
For example:
NH4+(aq) + NO2(aq)  N2(g) + 2H2O(l)
Concentration and Rate
• For the reaction
NH4+(aq) + NO2-(aq)  N2(g) + 2H2O(l)
– as [NH4+] doubles, the rate doubles...
– as [NO2-] doubles, the rate doubles...
– rate  [NH4+][NO2-].
• Rate law:


Rate  k[ NH 4 ][ NO2 ]
• The constant k is the rate constant.
Rate Laws
Rate = k [A]x[B]y
where: k is the rate constant
[A],[B] are the concentrations of A,B
x,y exponents are the reaction “order”
For a general reaction with rate law
Rate  k[reactant 1]m[reactant 2]n
the reaction is mth order in reactant 1
and nth order in reactant 2.
• The overall order of reaction is m + n
• A reaction can be zeroth order if m, n,
are zero.
• Note the values of the exponents
(orders) have to be determined
experimentally. They are not simply
related to stoichiometry.
• A reaction is zero order in a reactant if
the change in concentration of that
reactant produces no effect.
• A reaction is first order if doubling the
concentration causes the rate to double.
• A reaction is nth order if doubling the
concentration causes an 2n increase in
rate.
• Note that the rate constant (k) does not
depend on concentration.
Examples of reaction order:
 1st
order: x = 1 or y = 1
e.g. Rate = k [A]

2nd order: x = 2 or y = 2 or (x = 1, y = 1)
e.g. Rate = k[A]2 or Rate = k[A][B]

3rd order: x = 3 or (x = 2 , y = 1) etc.
e.g. Rate = k[A]3 or Rate = k[A]2[B]
Determining the Reaction Rate:
Two proposed mechanisms for
2 NO2  2 NO + O2
A) step 1: NO2  NO + O
(slow)
step 2: NO2 + O  NO + O2 (fast)
B) step 1: 2 NO2  NO3 + NO
step 2: NO3  NO + O2
Which is correct???
(slow)
(fast)
Determining the Reaction Rate:
Two proposed mechanisms for
2 NO2  2 NO + O2
A) step 1: NO2  NO + O
(slow)
step 2: NO2 + O  NO + O2 (fast)
Unimolecular, so Rate = k[NO2]
B) step 1: 2 NO2  NO3 + NO (slow)
step 2: NO3  NO + O2
(fast)
Bimolecular, so Rate = k[NO2]2
Rate = k [NO2]x
0.20 M/s = k [4.1]x
0.08 M/s = k [2.5]x 
2.5 = [1.6]x x = 2
0.20 0.12 0.08
M/s
Kinetics tells us about the mechanism!
Determining the Reaction Rate:
Find the rate limiting step and use the
reactant(s) and coefficient(s) in the rate
law.
B) step 1: 2 NO2  NO3 + NO
step 2: NO3  NO + O2
Bimolecular, so Rate = k[NO2]2
(slow)
(fast)
Rate = k [NO2]2
Rate-determining step
The rate equation cannot be predicted,
it can only be measured empirically.
Bottom line:
Rate Law is related to the mechanism
of the rate-determining step!
The rate equation cannot be predicted,
it can only be measured empirically.
Calculate k from initial rates
 Use the integrated form of the rate eqn.
to solve for concentration (Section 15.4)

There are two forms to know:
15-3
First order:
15-5
Second order:
ln[A] = ln[A]o  k t
1/[A] = 1/[A]o  k t
Can use data to find
k and reaction order
How do you find
the reaction order?
Plot both….
ln[A] = ln[A]o  k t
1st Order Plot
-3.5
slope
1/[A] = 1/[A]o  k t
-4.5
-5.0
-5.5
R2 = 0.8389
2nd Order Plot
0
400.0
200
-6.0
-6.5
300
Time (s)
R2 = 0.9949
300.0
1/[NO2]
100
Ln[NO2]
-4.0
Series2
200.0
Linear
(Series2)
100.0
0.0
0
100
200
Time (s)
300
Only one will be
truly linear….
Rate = k [NO2]2
Series2
Linear
(Series2)
```