Biomedical Engineering Reference
In-Depth Information
Therefore, the apparent order of reaction can be first order when the concentration is high,
while second order when the concentration is low.
Case 2. U
A. The collisional activation is achieved
by collision of A with any molecule in the system, whereas the deactivation is achieved by
collision with another A molecule only. In this case, C
U
¼
¼
any molecule in the system and W
¼
C
T
and
Eqn (6.54)
is reduced to
k
k
C
T
C
1
3
A
r
¼
(6.57)
k
C
A
þ
k
2
3
One can infer that
8
<
:
C
2
A
k
k
1
3
z
; w n
C
A
z
C
T
k
C
A
þ
k
2
3
k
k
C
T
C
k
1
3
A
3
r
¼
(6.58)
ð
k
C
T
Þ
C
A
;
when
C
T
¼
constant and
C
A
1
k
C
A
þ
k
k
2
3
2
k
k
C
T
when
k
1
3
3
;
C
A
and
C
T
¼
constant
k
k
2
2
When A is of small fraction in an inert mixture, the total concentration C
T
does not change
noticeably with the change of C
A
. If there
n
P
¼
0, C
T
¼
constant. The last two scenarios indi-
cate that when C
T
z
constant, the apparent order of reaction can be either first order (at low
concentrations) or zero
th
order (at high concentrations).
6.5. FREE RADICALS
We next look into the active complexes as appeared in the unimolecular kinetic discus-
sions. As we noted earlier, the active complex may or may not be a stable molecule, which
is the reason it is active. To start with this discussion, we look into the thermal decomposition
of acetaldehyde
(6.59)
occurs in gas phase at about 500
C. The rate of this reaction is empirically found to be 3/2
order in acetaldehyde in a wide range of concentrations. That is
CH
CHO
/
CH
4
þ
CO
3
3=2
r
¼
k
½
CH
CHO
(6.60)
3
This reaction is of the formA
C, but it is clearly not an elementary reaction as the order
of reaction is not first order in the only reactant. It does not fit directly to the simplistic colli-
sion theory based unimolecular mechanism as described in the preceding section.
This reaction is actually a multiple-reaction system that involves the major steps
CH
/
B
þ
CHO
/$
CH
3
þ $
CHO
r
(6.61)
3
1
$
CH
3
þ
CH
CHO
/
CH
4
þ
CH
CO
$
r
(6.62)
3
3
2
CH
CO
$/$
CH
3
þ
CO
r
(6.63)
3
3
$
CH
3
þ $
CH
3
/
C
H
r
(6.64)
2
6
4
$
þ $
/
þ
CO
r
(6.65)
CHO
CHO
HCHO
5
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