Chemistry 132, Winter 2008
Solutions to Homework No. 9
Problem 1.
In a chain reaction, the chain carriers are reactive intermedi
ates that occur in chain propagation steps (Levine p. 565). A
step that generates chain carriers from relatively unreactive
species is an initiation step.
A termination step consumes
chain carriers without generating new ones.
A branching
step is one that generates more chain carriers than it con
sumes (Levine p. 566). Although not clearly stated in Levine
p. 565, it can be inferred that a propagation step generates
as many chain carriers as it consumes.
The chain carriers
in the given mechanism are O, H, and OH, as mentioned in
the problem statement. Based on the above definitions, the
steps of the given mechanism can be identified as follows:
step
type
reason
1
initiation
generates H
2
initiation
generates O
3
initiation
generates H
4
branching
consumes H, generates OH and O
5
branching
consumes O, generates OH and H
6
propagation
consumes OH, generates H
7
termination
consumes H
8
termination
consumes O
9
termination
consumes H
Problem 4.
The Michaelis constant
K
M
is a parameter of the Michaelis
Menten equation (Levine eq. 17.125), which gives the initial
rate
r
0
(or
v
0
) of an enzymatic reaction as a function of
initial substrate concentration
S
0
. Taking the reciprocal of
both sides of the MichaelisMenten equation gives (Levine
eq. 17.126):
1
r
0
=
K
M
k
2
E
0
1
S
0
+
1
k
2
E
0
,
where
k
2
is the rate constant for the second step of the as
sumed mechanism (Levine eq.
17.120) and
E
0
is the ini
tial concentration of the enzyme. Then, a plot of
y
= 1
/r
0
against
x
= 1
/S
0
should yield a straight line with slope
a
=
K
M
/k
2
E
0
and intercept
b
= 1
/k
2
E
0
, so that
K
M
=
a/b
.
Figure 1 shows that performing linear regression on the given
data points and taking the ratio of slope to intercept gives
K
M
= 2
.
2
×
10

3
M.
Problem 7.
See Levine problem 22.8. Solve this problem for argon gas
(
M
= 39
.
948
×
10

3
kg
/
mol) in a 20cm
3
box at 300 T. The
number
n
of available translational states with energy less
Figure 1: Excel spreadsheet for problem 4
1
2
3
4
5
6
7
8
9
10
11
12
13
A
B
C
D
E
F
y
=1/
r
0
x
=1/
S
0
/
(10
3
M
−1
)
2.8
0.7
3.2
0.9
4.2
1.3
6.2
2.2
9
3.3
a
1.08125
b
/ (10
3
M)
2.3802083
b/a
/ (10
3
M)
2.2013487
cell
formula
F7
=INTERCEPT(E2:E6,F2:F6)
F8
=SLOPE(E2:E6,F2:F6)
F9
=F8/F7
y = 2.3802x + 1.0813
0
2
4
6
8
10
0
1
2
3
4
x
y
than 3
kT
is given by (Levine p. 836)
n
≈
60
mkT
h
2
3
/
2
V
= 60
MkT
N
A
h
2
3
/
2
V
= 60
(39
.
948
×
10

3
kg
/
mol)(1
.
38065
×
10

23
J
/
K)
(6
.
022142
×
10
23
mol

1
)
×
(300 K)
(6
.
62607
×
10

34
J s)
2
3
/
2
(20
×
10

6
m
3
)
= 1
.
879
×
10
28
.
Problem 10.
See Levine problem 22.15. Solve this problem for O
2
in a 10
cm
3
cubic box. The kinetic energy associated with motion
in the
x
direction is given by (Levine eq. 18.42)
tr,x
(
n
x
) =
n
2
x
h
2
8
ma
2
,
where
n
x
= 1
,
2
, ...
is a quantum number and
a
=
V
1
/
3
is the
length of the side of the cubic box. The difference between
adjacent
x
translational energies is
Δ
x
=
tr,x
(
n
x
+ 1)

tr,x
(
n
x
)
=
h
2
8
ma
2
(
n
x
+ 1)
2

n
2
x
=
h
2
8
ma
2
(2
n
x
+ 1)
.
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 Winter '08
 Lindenberg
 Reaction, Energy, Entropy, Chemical reaction, Levine 22.32.

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