Solving Differential Equations in Kinetics by Using Power Series Elvln Hughes, Jr. Southeastern Louisiana University, Hammond, LA 70402 This article descrihes an approach for solving eq 2, a differential equation that arises in the discussion of a set of consecutive chemical reactions that are outlined in eq 1. These coefficients form a series in k l and kz and can, in general, be represented by the following series:
Kl and K z are first-order rate constants. The differential equation representing the change in the concentration of species B with respect to time is the following:
[&I is the initial concentration of component A, and [B] is the concentration of species B. The solution to this equation is ordinarily materially inadequately presented in textbooks and provides very little to instruct students how to solve differential equations. This article descrihes a series solution for the equation and covers the following two objectives: a . 1)escribeageneral method for sohing differentia!equations, and
where Thus, the power series solution introduces another series that must he resolved in order t o calculate the coefficients. Equation 9 can he rewritten as S, = k,"
(10)
If one divides hoth sides of the equation by k l and multiplies by kz, one obtains
b. Provide background for solving drfierentia! equations, whrch can later be extended in quantum mechanics.
The solution t o eq. 2 can he expressed as a power series, in time, by the following equation:
+ k,"-'k, + k1"-2k,2 + . ..+ klk2"-'
S,k,lkl = kl"-'k,
+ k,"-'k,2 + . ..+ k,kZn-' + k,"
(11)
Now, subtract eq 11 from eq 10 to obtain S,
- S,k,lk,
= k,"
- k,"
(12)
Rearrangement of eq 12 leads to where a, is the coefficient of the time t, raised to the nth power. The initial boundary condition is [B] = 0 a t t = 0. Immediately, this implies that an is equal to zero. With this as set of conditions, differentiating eq 3 and expressing Lk1' a power series, eq 2 becomes Xna,tn-' = kl[A,,]X(-l)"k,"t"ln! - k,Xo,t"
By comparing the coefficients of tn-' on hoth sides of eq 4, one obtains the following recursion relationships for the coefficients:
- kza,_,
- k,a._,h
a, = (-l)""[A,,]k,(k," - k,")l[(k, - k,)n!]
(14)
a, = (-l)"[A,,lkl(kl"- k,")l[(k, - k,)n!]
(15)
or
[Bl = E(-l)"[Aolk,(k," - kp")tnl[k2 - k,)n!]
(16)
(5)
This series is now immediately recognizable as the expansion of (e-'I' - eLbat).With this in mind, we arrive at the solution, eq 17, frequently given in physical chemistry textbooksl.
(6)
[B] = [ ~ , ] k , ( e ~-"eCkz')l(kz ~ - k,)
or a . = k,"[A,](-1)"-'In!
(13)
where -1 was factored from the denominator in eq 14. Finally, eq 15 can he substituted into the original power series expansion in eq 3 and with this one obtains the following:
= X(-l)"kl"t"ln!
na, = kl"[A,,](-1)"-'l(n-1)
- kz")l(k,- k 2 )
Substitution of eq 13 into eq 8 leads us to the following
(4)
where e?'
S, = k,(k,"
(17)
Several of the coefficients, a,, are given by the following equations: a, = 0 a, = k@ol
46
Journal of Chemical Education
' Alberty. Robert A. Physical Chemistly, 7ih ed.; Wiley: New York.
1987; p 691.
