I'Hospital's Rule Applied to Kinetic Equations

I'Hospital's Rule Applied to. Kinetic Equations and when a is not equal to b the solution becomes. Students are often confused because substitution of...
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Seymour Lowell College of Long lslond University Brookville, L.I., N e w York

C.W. Post

I'Hospital's Rule Applied to Kinetic Equations

The differential equation describing a second order reaction is

Where x is the concentration of products, a and b are the initial reactant concentrations, k is the rate constant and t is time. The solution to equation (1) when a equals b is

and when a is not equal to b the solution becomes

Students are often confused because substitution of a for b in equation (3) does not yield equation (2). Actually the value obt,ained is indeterminate since this substitution gives kt = 0/0. The form 0/0 suggests using 1'Hospital's rule' in order to determine the limit of equation (3) as b a. Therefore,?

-

Upon taking the derivative df(b)ldb -ds(b)ldb

~

1

L'I'Haspital'srule" is:

.+.

_

lim f(2) - - l.m f'b) 7 Y (2) g(z) See for example, STEPHENSON, G., "Msthematical Methods for Science Students," John Wiley & Sons, Inc., New Yark, 1961,

.+,

- - -..

nn 9697 FL-.

In cases where a is near, but not equal to b, the method of WIDEQ~EST, S., Arkiv Kemi,Z6A (1948), can be employed.

552

/

Journol o f Chemical Education

and substituting a for b one obtains equation (2). Another application of 1'Hospital's rule can be made to a second order reaction accompanied by a first order reaction producing the same products. An example of this is the nucleophilic displacement of an alkyl halide accompanied by solvolytic displacement. The rate equation for this type of reaction is

where k, is the bimolecular rate constant and kz is the solvolytic constant, The solution to equation (5) by the method of partial fractions give

The term (a - x) can be substituted for (b - x) in equation (6) only under the following two conditions: (1) kz = O a n d a t t = O , a = b . Then(a-x)isequalto(b-x) a t all times; (2) k2 # 0 and a t t = 0, a > b. Then a t time t', (a - x) will equal (b - $)but beforeand after t', (a - x) cannot equal (b - x). Since equation (6) applies a t all times, condition (1) is the only circumstance that permits substitution of (a - a) for (b - x). Therefore equation (6) becomes

since ka is zero, substitution gives the indeterminate form, t = 0/0 and again applying I'Hospital's rule one obtaim t = lim k*+O

d k*

(8)

After taking the derivative, simplifying,and substituting kt = 0, the expression for a second order reaction with initially equal is obtained