Page 80 - Basic physical chemistry for the atmospheric sciences

66 Basic physical chemistry
3 . 22. Show that for a reaction that is second-order in just one
1
reactant A, the half-life of A is / k''[A]0• (Note: In this case,
the half-life depends on the initial concentration of A for
every half-life interval.)
3 . 23. A wooden carving, found on an archaeological site, is sub­
jected to radiocarbon dating 7 The carbon-1 4 activity is 1 2
.
counts per minute per gram of carbon, compared to 1 5
counts per minute per gram of carbon fo r a living tree.
What is the maximum age of the carving? (Hint: Radioactive
decay, as measured by the activity in counts pe£ minute per
gram of sample, is directly proportional to the number of
radioactive atoms present in the sample. Therefore, it is
described by the relations for a first-order chemical reaction
given in Sections 3 1 and 3 . 7. The half-life of carbon-1 4 is
.
5 . 7 x 1 0 3 yr.)
)
3.24. Ammonia (NH3 , nitrous oxide (N20), and methane (CH4)
4
comprise 1 .0 x 1 0 - 7 , 3 . 0 x 1 0 - 5 , and 1 . 7 x 1 0 - % by mass of
the Earth's atmosphere, respectively. If the efftuxes of these
1
chemicals from the atmosphere are . 0 x 1 0 1 1 , 1 . 0 x 1 0 1 1 , and
7 .5 x 0 1 1 kg y r - 1 , respectively, what are the residence
1
times (with respect to their efftuxes) of NH3, N 2 0 , and CH4
'
in the Earth s atmosphere? (Mass of the Earth s atmo­
'
1
=
sphere 5 . 1 4 x 1 0 8 kg. )
1
a
3 . 2 5 . A lake has n area of 0 hectares and is 6 m deep. The lake
initially contains 5% by volume of a liquid chemical X. If
more of the chemical X is put into the lake at a rate of 500
3
m per minute, what is the initial (instantaneous) residence
time of chemical X in the lake with respect to (a) the influx
of X and (b) the efflux of X? Assume that, due to seepage,
the level of the lake does not change when liquid is put into
3
it at a rate of 500 m per minute.
Notes
"M" in the equation for a chemical reaction indicates any molecule that can take up
excess energy that needs to be removed in order for the reaction to proceed. Thus, in
the example given, the molecule M can collide with the OH and S02 molecules and
reduce their energies so that they can combine to form HOS02.
2 In practice, the principle of detailed balancing works moderately well as an
approximation; it is better for slow reactions than fast.
3 A strict mathematical treatment of this collision problem, in which the motion of all
the molecules is taken into account, gives a collision rate per molecule of 11Tp2cn.