Chemistry Reference
In-Depth Information
confronting the resulting rate equation with the experimental data. This is
elaborated in detail in section 4.6.1, in the case where stage 1 is assumed to
be the RDS of mechanism 1. This elaboration will serve at the same time
as an illustration of the procedure, since this method was deliberately not
discussed in detail in Chapter 1 under section 1.7.2 addressing, in general
terms, formulated mechanisms (Equations 1.27-1.32).
4.6.1
Stage 1 as RDS
The rate equation of stage 1 is given by:
(
)
-
v
=
k
c
1
-
q
k
q
[4.15]
1
1
-
-
-
1
-
HO
HO
HO
2
2
2
By assuming a virtual balance for the other stages, it is possible to elimi-
nate the unknown, experimentally inaccessible parameters q and (1 - q) in
Equation 4.15 by equalising, in Equations 4.16 to 4.19, the two terms of the
right part (n=0).
(
)
(
)
(
)
=
k
q
1
-
q
1
-
q
k
q
q
1
-
q
[4.21]
2
-
-
•
-
2
-
•
-
O
OH
O
OH
HO
HO
2
2
(
)
=
(
)
k
q
c
1
-
q
k
q
1
-
q
[4.22]
3
•
-
-
-
3
-
•
OH
OH
O
O
OH
(
)
=
2
()
()
(
)
k
E
q
2
1
-
q
k
E
q
1
-
q
[4.23]
4
-
-
4
-
-
-
O
O
O
O
2
2
(
)
()
()
k
E
q
=
k
E c
2
1
-
q
[4.24]
5
-
-
5
O
-
O
O
2
2
By combining Equations 4.21 and 4.22, it is possible to deduce an expres-
sion for q
O
-
from which q
O
•
is eliminated. A second expression for q
O
-
results from the combination of Equations 4.23 and 4.24, in which the cov-
erage fraction of O
2
-
is eliminated. Equalising these two expressions for
q
O
-
results in an expression in which, besides the coverage of HO
2
-
and the
non-covered surface fraction (1 -q), only accessible concentrations occur:
q
() ()
() ()
1
k k
kk
EEc
EEc
-
HO
-
4
-
5
O
2
)
=
2
[4.25]
(
1
-
q
KK
23
4
5
-
-
OH
HO
2
where K (K
n
= k
n
/k
-
n
) is an equilibrium constant of a chemical reaction of
the corresponding substage. The (1 -q) term of HO
2
-
can be equalised to 1
if the degree of coverage of HO
2
-
is sufficiently small. Since stage 1 of the
reaction sequence is considered as the speed-determining one, all the other
stages of the sequence will be faster. This means that the HO
2
-
formed from
stage 1 can continue to react rapidly through these faster stages, and it can
be assumed that the degree of cover of HO
2
-
will be minimal. When com-
bining Equations 4.15 and 4.25 under this condition, the following is
obtained:
