Page 177 - Essentials of physical chemistry

Basic Chemical Kinetics 139

RADIUM

There are many isotopes of Radium from 201 Ra to 234 Ra but the one with the longest half-life is
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226 Ra with a half-life of 1599 years and decays by emitting an alpha particle (He ) with an energy
2þ
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of 4.870 million electron volts (MeV).
Example 1
In their original fractional crystallization separation method, the Curies eventually obtained about
0.1 g of radium chloride. How much of a 0.1 g sample of 226 RaCl 2 isolated in 1911 would remain
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4
today (2011)? Thinking like a chemist, if you assume that radium decays to Rn by emitting He 2þ to
2
form 222 Rn, the Rn would escape as a gas. Then the Cl 2 originally associated with the radium cation
86
might leave as a gas along with the radon but what matters is that the radium is gone.
0:693
a(t) ¼ a(0)e ð 1599 years Þ(2011 1911) years ¼ (100 mg)(0:958) ¼ 95:8 mg:
So, we see as a laboratory source of alpha particles the supply would be pretty constant over a long
period of time. Another consideration is that radium is in the same column of the periodic chart as
Ca and so biologically it might have similar chemistry to Ca and become trapped in bone tissue
where it would be radioactive for a long time. Thus, this interlude regarding the fact that first-order
decay is a useful model for nuclear processes has provided an opportunity to discuss some aspects of
nuclear chemistry. Considering the crossover of physics and chemistry in the work of the Curies
(Marie, Pierre, and Irene) and information in the popular domain regarding nuclear chemistry, we
think this brief discussion is justified as an essential part of physical chemistry.

SECOND-ORDER RATE PROCESSES: [A] ¼ [B]
In some texts, the [A] ¼ [B] case is the only one shown for second-order rate processes because it is
easy to derive, but actually it is not very general and applies only to cases of reagents which dimerize
or to cases where the concentrations have been carefully prepared so that in fact [A] ¼ [B] by initial
preparation. We will see below that the general case of [A] 6¼ [B] is not very difficult and much more
general. Nevertheless, let us derive this case and see how we could treat the data graphically.
A þ A ! B with [B] ¼ x and initially we have [A] ¼ a and [B] ¼ 0at t ¼ 0 so, we can write
dx 2
¼ k 2 (a x)(a x) ¼ k 2 (a x) . We now see why the rate is second order because the sum of
dt
þ
the exponents of the concentration terms is two. Once again we separate the variables and integrate
the two sides of the equation separately but with a combined integration constant.
ð ð
dx 1 1
¼ k 2 dt so, we find ¼ k 2 t þ C.At t ¼ 0, ¼ 0 þ C so we can write
(a x) 2 (a x) a
1 1 a (a x) x
¼ k 2 t or ¼ k 2 t which is the same as ¼ k 2 t. Note that for the
(a x) a a(a x) a(a x)

d[x] 2
second-order case we see that here that ¼ k 2 [a x] so k 2 has (1=time concentration) units
dt
and if the concentrations are in (mol=L) the units of k 2 ¼ (L=mol time). The value of the rate constant

1 1
can be obtained by plotting y ¼ mx þ b using ¼ k 2 t þ , the slope ¼ k 2 .
a x a
SECOND-ORDER RATE PROCESSES: [A] 6¼ [B]
This is the more general case of a second-order rate process where the concentrations of the two
species are not the same. In some texts, this is avoided because of the problem with integrating the
rate expression but we show here a special theorem [4] that permits solution of this problem