Chemistry Reference
In-Depth Information
where L is a constant and ∆ H ad is the adsorption enthalpy. In their turn, the rate con-
stants k ef and k 2 have the Arrhenius temperature dependence. Therefore, Eq. 4.82
can be written as:
k A E
RT
H
RT P mm
(
)
2
ad
=
exp
L
exp
.
(4.85)
ef
2
0
Normally, the pressure of C g is maintained constant in kinetic experiments that
means that the P m term is constant and independent of temperature. However, the
equilibrium pressure is temperature dependent in accord with Eq. 4.75. Considering
this dependence, the effective activation energy of the process can be determined
as follows:
m
ER k
T
d
d
ln
EHmH P
PP
=−
=+ +
(4.86)
ef
0
.
ef
2
ad
r
1
mm
0
Equation 4.86 is the result obtained by Pawlyutschenko and Prodan [ 106 ]. It should
be noted that if one replaces the Freundlich isotherm with the Langmuir isotherm
[ 1 ] at lower pressure that would be equivalent to using m = 1 in Eq. 4.83. Then
Eq. 4.86 for E ef would take the following form:
EEHH P
PP
0
=+ +
,
(4.87)
ef
2
ad
r
0
which allows one to arrive at the same conclusions as Eq. 4.86 but without the need
for guessing the value of m .
At any rate, the third term in the right-hand side of Eq. 4.86 (and 4.87) deter-
mines the dependence of the effective activation energy of reversible decomposi-
tion on pressure. It is clear that when P is maintained slightly below the equilibrium
value P 0 the third term tends to infinity and so does the value of E ef . A physically
meaningful value of E ef can only be obtained when the gaseous pressure is main-
tained markedly below the equilibrium pressure, i.e., when the reaction is run in
vacuum. Then P can be neglected relative to P 0 so that the pressure term P 0 /(  P 0 P )
becomes approximately equal to 1 and the pressure dependence of the effective ac-
tivation energy vanishes. Figure 4.37 demonstrates the dynamics of decreasing the
pressure term as a function of the gaseous product pressure. The dependencies sug-
gest that the pressure dependence of E ef should practically vanish when the gaseous
product pressure is at least ten times lower than the equilibrium pressure.
Another important conclusion that can be drawn from Eqs. 4.87 and 4.86 is that
one may not need to use vacuum to eliminate the pressure dependence of the ef-
fective activation energy. Alternatively, one can run a reaction at ambient pressure
but at temperatures significantly above the equilibrium temperature. Such condi-
tions can be relatively easy to realize when thermal decomposition is studied under
nonisothermal conditions because they allow one to stretch the temperature range
of a study to significantly higher temperatures than those covered by isothermal
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