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5.68J/10.652J Feb 13, 2003
Numerical Solution to Kinetic Equations
Lecture 1
The Problem
We want to solve
l
H
1
dn
i
=
±
ν
r
l
(
T
n
)
−
Transport
V
of
out
i
of
Eq.(1)
,
'
'
V dt
i
Where n
i
is the number of moles of species ‘i’,
ν
is the matrix of stoichiometric
coefficients, and
r
l
=
Rateof Step
'
l
' (units: moles/volumesecond)
For a typical bimolecular elementary step reaction A+B = C+D:
T
A B
T C
r
l
=
k
forward
]
][
)[
(
−
k
][
)[
(
D
]
reverse
For a typical gasphase dissociation reaction A= C + D:
,
C
r
l
=
k
(
P
T
]
)[
−
k
(
P
T
][
)[
D
]
forward
,
A
reverse
For a typical gasphase isomerization reaction A = C:
,
A
,
C
r
l
=
k
forward
(
P
T
]
)[
−
k
(
P
T
]
)[
reverse
The rate constants also usually depend (weakly) on the chemical environment (e.g.
dielectric constant), so they strictly are not really constant but (very weak) functions of
all the n’s. Note that the forward and reverse rates are normally related by the equilibrium
constant, related to the free energy of reaction (thermochemistry).
It is important to appreciate that except in the simplest cases, there is no analytical
solution to this system of differential equations, so we MUST use the computer to solve
them numerically. Approximations like quasisteadystate extend the range of systems
where analytical (approximate) solutions are possible, but you don’t have to get very
complicated before even the steadystateapproximation solutions are intractable.
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²
³
´
µ
i
i
More About the Fundamental Equation
Chemistry textbooks usually give instead of Eq. (1) the equation:
i
d
l
H
,
]
[
=
±
ν
r
l
(
T
n
)
Eq.(2)
dt
i
But Eq.(2) is true only if dV/dt = 0 (“isochoric”) and there is no transport (i.e. perfectly
homogeneous, no convective flow). These textbook approximations are good for dilute
reactions in solution or for constant volume apparatus (e.g. bomb calorimeter, shock
tube). In many important cases the system is isobaric, i.e. dP/dt = 0, so if the temperature
increases or there is a change in the number of moles of gas V will change.
It is convenient to express the equation in terms of intensive quantities rather than the
extensive variables n and V. The intensive variables that work best are mass fractions y
ι
:
W
n
i
=
]
[
y
=
i
W
i
i
ρ
M
total
dy
W
i
1
dn
i
V dt
−
Transport
V
of
out
i
of
'
'
=
dt
You can plug in the same expression as above for the term in parentheses. Note that mole
fractions x
i
are not such a good choice for kinetics as mass fractions, since the number of
moles can change in chemical reactions, but the mass does not.
However one writes the chemical reaction equations, one must be explicit about the
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 Spring '03
 RogerD.Kamm
 Mole

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