Outline:
 Seminar

Report - in front
Extra Seminar – Wednesday @ 4pm
 Pick

2/5/07
up Quiz #3 – from me
Exam 1 – one week from Friday…
 Outline
Chapter 15 - Kinetics (cont’d):
- integrated rate law calcs
Quiz #3

Average = 7.9
Quiz #3
25
20
15
10
5
0
0 to 2
2 to 4
4 to 6
6 to 8
8 to 10
Summary:
Can build a rate law from observed data:
(initial rates)
Rate = k [A]m [B]n
 m, n depend only on the chemical reaction
under consideration….
 Can use integrated rate laws to predict rates,
concentrations at various times, etc.

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
Note something weird about k:
The definition of rate constant varies
from reaction to reaction….
e.g. Rate = k [A] [B] or Rate = k [C]
Therefore the units of k vary from
reaction to reaction….
e.g. k is in M-1 sec-1 or k is in sec-1
or k is in atm-1 sec-1
The integrated forms of the rate laws are
important:
Can predict [concentration] as a
function of time!
Can predict rate as a function
of time too!
e.g. CAPA-7 Question 10
“Popcorn kernels pop unimolecularly….”
= first-order reaction
ln(A) - ln(A0) = -kt
29 kernels pop in 10 seconds when 400 total
kernels are present. After 100 have popped,
how many will pop over the next 10 seconds?
A0 = 400 (initial number of kernels)
t = 10 sec
A = 371 (number of kernels at time t)
 k = 0.00753 sec-1
1st ORDER:
ln(A) - ln(A0) = -kt
and k = 0.00753 sec-1
After 100 have popped, how many will
pop over the next 10 seconds?
A0 = 300 (initial number of kernels)
t = 10 sec
k = 0.00753 sec-1
A = 278.25
DA = 21.75
Try Worksheet #5….
Work with somebody nearby and
complete “Version 1”…
On Worksheet #5….answer (a):
25%
1.
25%
2.
25%
3.
25%
4.
1
2
3
Rate = k pSO2Cl2
Rate = k pSO2Cl22
Rate = k pSO2Cl23
Rate = k
4
5
Try Worksheet #5….
First Order
2.00
y = -0.1675x + 1.6013
1.50
R2 = 0.9999
ln (p)
1.00
0.50
0.00
-0.50 0
5
10
-1.00
-1.50
time (h)
15
20
Try Worksheet #5….
2nd Order
3.50
3.00
2.50
y = 0.1695x - 0.0713
2
R = 0.9131
1/P
2.00
1.50
1.00
0.50
0.00
-0.50 0
5
10
Time (h)
15
20
On Worksheet #5….answer (b):
25%
1.
25%
2.
25%
3.
25%
4.
5.
1
2
3
k = 5.95 /hr
k = -5.95 atm/hr
k = 0.168 /hr
k = -0.168 atm/hr
None of the above
4
5
Worksheet #5….
ln P(SO2Cl2)
Version #1
2
1.5
1
0.5
0
-0.5 0
-1
-1.5
Rate = k (pSO2Cl2)
k = 1/5.95 =
0.168 hr-1
5
10
time (hrs)
SO2Cl2  SO2 + Cl2
15
20
Practice!
CAPA problems (sets 7-8)
textbook examples (15.25-15.34)
 How
do you tell which integrated rate
equation to use if you aren’t told?
 How
do we actually use “kinetics”
for anything useful?
Like determining mechanisms?
A + B  C + D
Rate = k [A]n[B]m
“Isolation experiment”
 Keep
one reagent constant (in excess),
vary the other reagent….
O3 + isoprene  prod
Rate = k [O3]n[iso]m
Exp # 1: Let [O3]>>[iso]
[O3] = 5.410- 4 M [iso] = 2.5 10- 6 M
time
0
0.215
0.495
0.955
O3
(E-4 M)
5.4
5.4
5.4
5.4
[isoprene]
(E-6 M)
2.5
1.11
0.305
0.037
the rate becomes:
Rate = k’ [iso]m
where k’ = k[O3]n
Plot the data…. ln[iso] = ln[iso]o - k t
1/[iso] = 1/[iso]o + k t
ln([iso])
1st Order
2
1
0
-1
-2
-3
-4
R2 = 0.9988
0
2nd Order
0.5
Time (s)
1/[iso]
30
20
R2 = 0.8335
10
0
-10
0
0.5
Time (s)
1
Only one will be
truly linear….
Rate = k’ [iso]1
m=1, k’ = 4.4 s-1
1
“Isolation experiment”
 To
find n, you change [O3] but keep
[O3]>>[iso]
Exp # 2: Let [O3]>>[iso]
[O3] = 2.710- 4 M [iso] = 2.510- 6 M
[O3]II = 1/2 [O3]I
Perform the exp. again and graph
data as 1st order to find k”
k” = 2.2 s-1
“Isolation experiment”
n
k ' k[O ]  k[O3 ]I   5.4 
n
  

 
  (2.0)
k ' ' k[O ]  k[O3 ]II   2.7 
n
3 I
n
3 II
k'
n
 (2.0)
k''
 k' 
ln  
k'' 

n
ln (2.0)
n
 k' 
ln    ln (2.0) n
 k'' 
 k' 
ln    n ln (2.0)
 k'' 
 4.4 s -1 

ln 
-1 
2.2 s  ln( 2.0)



1
ln (2.0)
ln( 2.0)
“Isolation experiment”
 That
is all there is to it!
Rate = k [O3]1[iso]1
k’ = k[O3]
so k = k’/[O3] = 8.110+3 M-1s-1
You’re about to do this….
Chem 114: CV lab!
Variation on a theme: Half-life



Radioactive decay = 1st order decay
Integrated rate law: ln(At/Ao) = - kt
In the case that At = 0.5 Ao
(half is left)
t = t1/2
therefore: ln(0.5) = - kt1/2
half-life is just another way to
define the rate constant k.