Evaluating Rate Constants in Systems of Differential Equations

Evaluating Rate Constants in Systems of Differential Equations. William Ames. Ind. Eng. Chem. , 1960, 52 (6), pp 517–518. DOI: 10.1021/ie50606a032...
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I

WILLIAM

F. AMES

Mechanical Engineering Department, University of Delaware, Newark, Del.

Evaluating Rate Constants in Systems of Differential Equations This technique is more useful than graphical methods for developing proper chemical mechanisms, reducing parameters needed to solve reaction equations, and solving rate constant ratios more accurately

A natural processes can be described by differential equations which NUMBER OF

are analogous to those which arise in chemical reactions. These processes may be lumped under the general title of “birth and death” processes. U1timately their mathematical formulation rests on some observed or postulated natural law such as the law of mass action describing the probabilistic statement of how the process occurs. I n such systems constants invariably arise, the estimation of which is difficult. In certain simple cases explicit expressions are obtainable for the constants. For more complex reactions the system of differential equations describing them, even in simpler cases, is nonlinear and the actual estimate of rate constants, using observed data, is usually very difficult. Mostly, the technique used is to allow one or more of the reactants to be in excess. Certain derivatives can then be assumed zero and approximate values for the constants obtained. However, the question always arises whether or not these constants are completely applicable where the reactants are not in excess but the process is operating as intended. The technique developed here is generally an implicit method-i.e., desired quantities are obtained in the form of implicit algebraic functions which must be solved by a numerical procedure. Usually ratios of the rate constants and not the constants themselves can be computed with this procedure. The differential equations are revised by eliminating the time variable and the resulting differential equations are solved by using the general theory of homogeneous differential equations ( 4 ) . A function, f ( x , y , z ) , is said to be homogeneous in its independent variables and of degree n if f(ux,uy,uz) = unf( x J , ~ ) . Thus y expb/x) y 2 / x is homogeneous of degree 1. The differential equation, P(x,y)dx Q(x,y)dy, is said to be homogeneous if P and Q are a

+

+

homogeneous functions of x and y of the same degree. Homogeneous differential equations and equations reducible to this form are important in solving certain nonlinear differential equations. This method applies to reactions of the general type A1

+ A2 3-A3

A1

+

or setting a = k z / k ~ Equation , 4 equals (ax8

- x2)dxz - xz d x a

(5)

dx 9 + P ( x ) y = Q ( x ) as dxz -2 -

dx

=

(;).XI

kz ’43-

= 0

which is a first order differential equation of homogeneous type. Equation 5 is also a simple first order linear differential equation expressible in the standard form

-1

A4

as well as to variations on this system and other reaction systems. Equations of this type are often the detailed mechanism for seemingly more complicated reactions. The technique is illustrated by several general examples where rate constant ratios appear in implicit form:

Example 1. Consider the reaction

However this approach is not used here because the homogeneous theory is more general in that it will handle certain equations which are nonlinear and not expressible in the first order linear form. T o solve Equation 5, yet x3 = uxz. Then dx3 = vdxz x2du and upon substitution, Equation 5 becomes

+

(avxz

- x2)dx2 - xz(udxz + xzdv) = 0 or dxe/xz = d o / [ u ( a - 1) - 11 (6)

Upon integration of Equation 6 and evaluating the integration constant there results which is characteristic of the first two steps in polymerization reactions as well as other systems. The differential equations for Equations l are (where x i = concentration of At per unit volume)

- xi(kixz + kzxs)

dxi/dt = dx:/dt d ~ s / d t=

a

- kixixz - kz~s)

xi(kiX2

dxa/dt =

k2~1~3

x1-x2+x4=a-p

+ +

x8

=

xa

(Xl

x4

=

(3) X2)

If in Equations 2, dxs/dt is divided by dxz/dt, then

(a

-

1)

2 xz

- 11

(sa)

must be solved for its positive root(s); and (a 1) X Q / X Z 1 < 0 where the implicit function

-

-

In [l

P

- a) f 2(P -

-

(2)

in which redundancies occur. From the mathematics or from material balances with xl(0) = a, xz(0) = p, X s ( 0 ) = X d ( 0 ) = 0, xz

Equation 7 is an implicit expression for a. Two possibilities occur in Equation 7: (a 1) x3/x2 1 > 0, where the implicit function

- (u - 1) 2](8b)

must be solved for its positive root(s). Choice of Equation 8a or 8b depends on knowledge of the relative speeds for the reaction. Example 2. The same approach to a more complicated system generates expressions for ratios of the rate constants. VOL. 52, NO. 6

JUNE 1960

517

Consider the kinetics problem governed by the chemical equations A~

I

that the iteration converges to p, take as E where E the initial guess 6, = bM is some positive number. Consider the following example. Let xs(0) = 1 , x j = 0.5, xg = 0.3. In general

\

+

+ A~ LA~ (9)

(22)

where ratios of the rate constants are to be determined from experimental data. I n this process reactant A1 has an effective constant concentration C because it is in excess and goes into solution as reaction goes on so that C moles per unit volume are always present. The method applies equally well without this assumption. The differential equations for chemical Reactions 9 are dxzldt =

- xz(kiC + k 2 ~ 3+ kaxs + kaxc)

dxg/dt = kiCxz

- k2~2~3

dxc/dt = kpXzX3 d ~ s / d t= - ksxnxs dx,/dt = kgx~x5- k4xpxB

(loa) (lob) (1 OC)

(10d)

= xS(0)

x2

=

XI(())

f(b) has a negative minimum for a value of b, greater than one, and it becomes positive as b increases indefinitely and substitution for x6 in Equation 17 an equation reducible to homogeneous form results: x j d ~ 7f b ( X g ( 0 ) - x,j - X ? ) dxg = 0 (18) where b = k4/k3. To reduce Equation 18 to homogeneous form, introduce the new variables

(IOe)

d+/dt = k4xzxs (10f) where x i is concentration of A , in moles per unit volume, and ka is rate constant in liters/g. mole sec. Certain redundancies in System 10 result: x6

and for these data

- 2 6 - X, - xg - 2x4 - x6 - 2x7

x5

u =

b[X6(0) - xs

- x7]

- [$ dv + d u ] .

(u

- u)du - ;du

= 0

(19)

which is homogeneous. Solving it as before (or in the first order linear form) and substituting for u and u, the solution of Equation 18 is

and letting a = kiC/kz, a solution is found by separating variables to be

(12) To solve for the second ratio (k3/kl) note that Equation 10d is equivalent to -1 = x z dt or by integrating k3

*‘ x5

Equations 12 and 20 are implicit equations for kl/kz and k4/k3, respectively, while 16 is an explicit expression for ka/kl. The solution of Equations 12 or 20 for the ratios is accomplished most easily by a numerical procedure wherein the experimental data for the reactants are used. A second order procedure such as the Newton-Raphson Method is recommended ( 5 ) . For Equation 20 write

Also by adding Equations 10b and 1Oc d(xa

+ ~ 4 ) l d =t kiCxz

(14) x6

or

Equating Equations 13 and 15

TQ obtain the third and final ratio kr/ko, note that from Equation 10f and 1Od

-

From Equation log, xg = xg(0) x5 that upon the elimination of dt

x7 S O

5 18

fI(b) = 1

+ ([0.5]blog, [0.5])/0.3 =

1

(23)

1 - 2.3105 [0.5]* (24) When b = 1.208 (approximately) fi(b) = 0 and this value of b gives a minimum of f ( b ) which is negative. To start the iteration take bo = 2. The iteration will now converge in a few steps to the proper root p and as much accuracy as desired can be obtained by continuing the procedure. A few steps are tabulated :

Step 0 1 2

b, 2 1.605 1.6125

f@n)

f’@J

0.1667 -0.002 -0.0005

0.4224 0.2655

...

Equation 18 then becomes

(log)

Dividing Equation 10b by 1Oc

-

Iteration Table

u =

so that dx5 = du, dx7 =

+ ([0.5]b - 0.5)/0.3

f(b) = b

(21) and determine that numerical value of b which makes Equation 21 equal to zero. Preliminary analysis shows that b = 1 is always a solution regardless of the values of X g ( o ) , x5, and x6. This is not the required solution. To assure that Equation 21 has another solution fl and that the iterative procedure converges to the solution (and not to b = 1) one can easily demonstrate that f ( b ) has a negative minimum at (say) b M for some bM > 1 andf(6) becomes positive as b increases indefinitely. The graph o f f ( 6 ) has the approximate shape shown in the illustration. To assure

INDUSTRIAL AND ENGINEERING CHEMISTRY

I n some applications these ratios of rate constants may be the only information required. The use of these ratios and the statistics associatcd with theme.g., the mean and standard deviation of the experimental data and the ratioscan aid in developing the proper chemical mechanism. If the course of the reaction with time is important, then for computer solution, the knowledge of the ratios reduces the parameters of the system from n to one (four to one in the second example). That is, adjusting one rate constant automatically fixes values of the remainder. References

(1) Alfrey, T., Coldfinger, G., J . Chem. Phys. 12, 205 (1944). (2) Flory, P. J., “Principles of Polymer Chemistry,” chap. V, Cornel1 University Press, 1953. (3) Mayo, F., Lewis, F., J . Am. Chem. SOC. 66, 1594 (1944). (4) Morris, M., Brown, 0.E., “Differential Equations,” Prentice Hall, New York, 1932. (5) Scarborough, J. B., “Numerical Mathematical Analysis,” 3rd ed., Johns Hopkins Press, Baltimore, Md., 1955. RECEIVED for review October 20, 1959 ACCEPTED March 2, 1960 When this work was done, the author was not aware of previous efforts in kinetics and other rate processes where elimination of the time variable played an important role. He is indebted to the referee for pointing out this omission and for supplying several references ( 7-3).