Page 178 - Essentials of physical chemistry

140 Essentials of Physical Chemistry

‘‘by inspection.’’ Here the reaction is A þ B ! C where [C] ¼ x with [A] ¼ (a x) and
[B] ¼ (b x) for t > 0 but at t ¼ 0 we initially have [A] ¼ a, [B] ¼ b and [C] ¼ 0. We can set
this up as before for the first-order case to keep track of the concentrations as

A þ B ! C
(a x) (b x) x
ð ð
dx dx
¼ k 2 (a x)(b x) which leads to ¼ k 2 dt.At first glance this looks
Then þ
dt (a x)(b x)
like a difficult integral on the left side and maybe some of you will dig out your calculus text to look
up ‘‘the method of partial fractions’’ which is a tedious way to split the integral into two simpler
integrals, but there is an easy way!

Theorem:

Let p(x) and q(x) denote polynomials in the variable ‘‘x’’ with no factor common (cancel all such
beforehand) and let the degree of p(x) be less than q(x). Consider the case in which q(x) has a linear
factor (x a), not repeated. Then, provided p(x), q(x), f(x), and g(x) are all continuous, we can
determine the partial fraction coefficient c ¼ c(a) for the term c=(x a).

p(x) g(x) c
Proof: Let g(x) ¼ (x a)f (x) so that we have f (x) ¼ ¼ ¼ þ h(x). Here h(x)
q(x) (x a) (x a)
c
is the left over part of f (x) after we separate the part and c is the number we seek. Then we
(x a)
g(x) c
can rearrange ¼ þ h(x) to g(x) ¼ c þ (x a)h(x). Now take the limit as
(x a) (x a)
x ! a: lim [(x a)f (x)] ¼ lim g(x) ¼ lim [c þ (x a)h(x)] ¼ g(a) ¼ c, Q.E.D.
x!a x!a x!a

We hope this little proof has not made the process more mysterious than it is in an actual
c
application. The point is that mentally we can separate a factor of from a function with
(x a)
(x a) in the denominator and determine the coefficient ‘‘c’’ simply by letting x ¼ a everywhere in
the remainder of the function. We need a simple example. Consider

1 c 1 c 2 1 1 (x b) (x a)
:
þ ¼ þ ¼
¼
(x a)(x b) (x a) (x b) (x a)(a b) (x b)(b a) (x a)(x b)(a b)
1 1
So, using the theorem above we have shown that ¼ . You will see that
(x a)(x b) (x a)(x b)
when we use this theorem what we do mentally is factor out the (x a) part of the denominator and
just substitute x ¼ a in the remainder of the expression and then move on to the next factor of
(x b) in the denominator and substitute x ¼ b in the rest of the function, including the (x a) part.
With practice on a few examples this can be performed mentally by inspection and can even be used
in some other forms of kinetic reactions [4] including the rarely used third-order case. Thus, we can
proceed to solve the A þ B ! C case posed above.

dx dx dx
ð ð ð ð
¼ k 2 dt, which is much easier to integrate:
(a x)(b x) (a x)(b a) (b x)(a b)
þ
¼