Numerical Solution of Kinetic Equations

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Numerical Solution of Kinetic Equations Very
quickly the solution of the differential
equations that describe even simple kinetic
processes become intractable and numerical
methods are required for their solution. As an
illustration of the approach to solving
differential equations numerically we will use
the simple 1st order irreversible process that we
have already analytically solved as an
example k1 A
------gt products The differential equation
governing the disappearence of A is d A / dt
- k1 A Since the derivative is just the
slope of a tangent to the curve of a plot of A
versus t, we can approximate that slope with a
chord
The slope of the chord approaches the slope of
the tangent and hence the derivative in the limit
that D t approaches zero d A /
d t lim ( A t D t - A t ) / ( (t D t)
- t ) D t --gt 0 ( A t D t - A t )
/ D t
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This approach which represents the derivative as
a finite difference between the variables is
known as the method of finite differences. Substit
uing this finite difference appoximation for the
derivative into the 1st order differential
equation d A / d t ( A t D t - A t )
/ D t - k1 A t gives a recursion relation
which will allow the concentration of A at the
advanced time step, A t D t, to be calculated
from the concentration of A at the current time
step, At A t D t A t - k1 A t
D t In developing this numerical model a
compromise needs to be reached in choosing the
size of the the time step, Dt. Dt must be kept
small in order to satisfy the limit condition
that allowed us to approximate the derivative
d A / d t lim ( A t D t
- A t ) / D t) D t --gt 0 However, if
Dt is too small, then the recursion relation for
A A t D t A t - k1 A t D
t will be used so many times that the cumulative
error in A that is inherently present in this
calculation will become too large at large times.
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The time evolution of A can then be modeled on a
spreadsheet
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As can be seen in the plot below, the fit of the
numerical solution and the analytically exact
solution is pretty good!
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Numerical Solution of Kinetic Equations Use the
finite difference method on a spreadsheet to
calculate the concentrations of A, B, and C
as functions of time in the kinetic system
k1 k2 A ltgt B
--------gt C k-1 Initially no B or
C are present and the concentration of A is to be
taken as 1.2 in some arbitrary units. Use a time
step which is less than the inverse of the
largest rate constant. The concentration of C as
a function of time should be calculated from a
mole conservation relation. You should hand
in a. A plot for k1 1.000 sec-1
k-1 0.100 sec-1 k2 0.200 sec-1 b. A
plot illustrating a situation where A and B reach
a pseudo equilibrium (why can A and B never be
truly at equilibrium?). The values of the rate
constants for this case should indicated on the
plot. c. A plot illustrating a situation where B
has reached steady state. In this situation the
concentration of B never becomes very great and
changes only slowly with time. The values of
the rate constants for this case should be
indicated on the plot. In each graph asked for
the above all three concentrations, A, B, and
C, should be plotted as a function of time.
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Take Aways Even for simple chemical reactions,
numerical solution of the differential equations
describing the dynamics of these reactions may
the only or perhaps the most straight forward
approach to solving these equations. In the
method of finite differences derivatives in the
differential equations are represented by the
finite difference expressions, e.g. d A / d t
( A t Dt - A t ) / Dt which yields a
recursion relation in A, allowing a later value
of the concentration, Ai1, to be calculated
from an earlier value, Ai. In using this
numerical method a balance, typically, has to be
struck between keeping Dt small enough that the
numerical approximation is not violated while at
the same time keeping the number of iterations
small enough that the errors inherent in the
method do not accumulate beyond acceptable
levels.