Page 172 - Essentials of physical chemistry

134 Essentials of Physical Chemistry

The method we want to teach here is to identify the extent of reaction ‘‘x’’ relative to the initial
concentrations. An increase of B corresponds to the appearance of ‘‘x.’’ Let ‘‘a’’ be [A] at time t ¼ 0
and for simplicity, let [B] ¼ 0at t ¼ 0. Then, k 1 has units of (1=t) and where ‘‘x’’ is now the
concentration of [B] for times greater than zero (zero is whenever you start your clock!).

þdx
¼ k 1 (a x) and x ¼ 0 at t ¼ 0:
dt
Perhaps it is a good idea to write the variables under the chemical reaction as follows:

A ! B
(a x) x; t > 0

Here, we use the key concept that if there is a ‘‘proportionality’’ such as C / D we can immediately
write C ¼ kD using the basic idea that a proportionality symbol ‘‘/’’ can be replaced by ‘‘¼ k,’’
where k is called the proportionality constant. In kinetics, the proportionality constant is called the
rate constant. Here we add a subscript to the rate constant to indicate the order of the reaction as k 1
for a first-order reaction. We will try to do this for higher orders but eventually in complicated cases
we may abandon this simpler convention. Next, we can rearrange the kinetic equation to separate x
and t variables.

dx
¼ k 1 dt,
(a x)

so we can integrate this to

ð ð
dx
¼ k 1 dt:
(a x)
ð
d(cabin)
If we recall our joke in the math review chapter ¼ ln (cabin) þ C, we can integrate the
cabin
rate equation and apply the boundary conditions to the indefinite expression. If we use cabin ¼
(a x), then d(cabin) ¼ dx so we need a minus sign in the answer. We find
ln (a x) ¼ k 1 t þ C,

then at t ¼ 0, x ¼ 0, so we have ln (a) ¼ C.
Note that both sides of the indefinite equation would have a constant of integration but since we
do not know either of them we combine the constants into one value, C. Then we have
a

ln (a x) ¼ k 1 t ln (a) which can be rearranged to ln ¼ kt and then we take the anti-
a x
a k 1 t

ln of the whole equation to find the final result in what should be a familiar form ¼ e .
a x
A first-order reaction is usually treated as a decay of the original concentration (invert the
equation), so, finally, we have
(a x) ¼ a e k 1 t or a(t) ¼ a 0 e k 1 t :

Although that is the solution, in nuclear chemistry it is common to refer to the ‘‘half-life’’ of
an original amount. Let us see what the equation looks like when (a x) ¼ a=2, which occurs at a