Concise Physical Chemistry

c10 JWBS043-Rogers September 13, 2010 11:26 Printer Name: Yet to Come

FIRST-ORDER KINETIC RATE LAWS 145

decay per second. You have observed arateof decay−dX/dtin the units (number

of atoms)/time, usually given as a frequency (s−^1 ). Decay causes a decrease in the

number of atomsX, which accounts for the minus sign in front of the rate. If you

take 200 billion atoms, you find twice as many radioactive events. That only stands

to reason. Radioactive decay is a random phenomenon. Observing twice as many

atoms, one expects to see twice as many events in a specified time interval. Takingk

times as many atoms givesktimes as many radioactive events: rate=kX. Now you

have arate,arate law, and arate equation. The rate of disintegration is

−

dX

dt

=kX

The rate of product production is equal but opposite in sign to the rate of decay.

Provided that the daughter is not radioactive, starting with a number of atomsX 0 at

a timet=t 0 =0 (an arbitrary time that you choose to start watching your counter),

the average count rate gradually decreases with time. If you watch long enough,^1 the

count rate will be reduced to half of what it was att 0. This is called thehalf-timeof

the radioactive element.

The rate equation can be rearranged to give

dX

X

=−kdt

The ratiodX/Xbeing unitless,−kdtmust be unitless as well, so the unit ofkis s−^1

becausetis in seconds. Integration betweent 0 andtis straightforward:

∫t

t 0

1

X

dX=−k

∫t

t 0

dt

gives

ln

X

X 0

=−k(t−t 0 )

In exponential form

X

X 0

=e−k(t−t^0 )

X 0 is the initial concentration (a constant) andt 0 is usually set to zero so that

X 0 =X(0) and

X=X 0 e−kt

(^1) About 1600 years in the case of radium.